Recently, nonlinear optics has become one of the important fields of science that have wide range of physical and engineering applications. The significance of this field has been enhanced since the appearance of optical fiber as a common type of optical waveguide that transmits light and signals over longe distances [1, 2]. Further to this, the continuous theoretical and experimental research works confirm that optical fiber has potential influences on developing photonic and optoelectronic devices [3–6]. One of the diagnostic tools to examine the physical properties of optical fiber is the optical pulses. The controllable interaction of dispersion and nonlinearity of the pulse propagation leads to the formation of stable and undistorted pulses known as soliton. There are several mathematical models that study the dynamic of soliton in optical fibers. One of these models that is accounted as a generalized form of the nonlinear Schrödinger equation is the Fokas–Lenells equation (FLE). In literature, FLE is dealt with by many authors to obtain exact solutions by utilizing various powerful techniques. The employed integration schemes in the previous studies are Sine-Gordon expansion method, Riccati equation method, mapping method, trial equation method, Kudryashov’s method, semi-inverse variational principle, modified simple equation method, Laplace-Adomian decomposition method, auxiliary equation method and many others. For more details, readers are referred to references [7–18].

Soliton propagation along an optical fiber can be subject to the low count of chromatic dispersion (CD) which severely affects the transmission process. To overcome this effect, a variety of novel techniques have recently been proposed. One of the most popular technologies employed in the research studies is based on adding another form of dispersion such as Bragg gratings dispersion, pure–cubic dispersion, pure–quartic dispersion, cubic–quartic dispersion and many others. For example, the combination of fourth-order dispersion (4OD) and third-order dispersion (3OD) terms can completely compensate for low CD and gives rise to creation of the so-called cubic–quartic (CQ) solitons, see the references [19–24]. Later, the model of FLE is developed to include 4OD and 3OD terms and that means CQ solitons can be constructed in polarization preserving fibers [25–27]. The current study mainly discusses CQ-FLE with perturbation terms of Hamiltonian type. The proposed model takes the form(1)$$i{\mathrm{\Psi }}_t+\mathrm{ia}{\mathrm{\Psi }}_{\mathrm{xxx}}+b{\mathrm{\Psi }}_{\mathrm{xxxx}}+\left|\mathrm{\Psi }{|}^2\right(c\mathrm{\Psi }+\mathrm{id}{\mathrm{\Psi }}_x)=i\{\alpha {\mathrm{\Psi }}_x+\lambda \left(\right|\mathrm{\Psi }{|}^{2n}\mathrm{\Psi }{)}_x+\mu \left(\right|\mathrm{\Psi }{|}^{2n}{)}_x\mathrm{\Psi }\},$$ where Ψ(x, t) is is a complex-valued function representing optical soliton profile. The independent variables x and t denote the distance along the fiber and the elapsed time, respectively. The first term indicates the time evolution while the terms with a and b account for the third- and fourth-order dispersions. The nonlinear influence has the form of Kerr law and is given by the coefficient of c. The term with d is the coefficient of nonlinear dispersion. On the right-hand side of equation (1), the perturbation terms with α, λ and μ are defined as inter-modal dispersion, self-steepening effect and higher-order dispersion, respectively. The parameter n represents the full nonlinearity effect and$i=\sqrt{-1}$.

The model (1) is investigated with the help of the improved projective Riccati equations method [28, 29] to derive distinct exact solutions. The rest of this paper is organized as follows. In Section 2, we describe the suggested scheme. Section 3 demonstrates how the FLE is reduced to a simple form using the traveling wave transformation. In Section 4, various solution expressions illustrating different wave structures are extracted. In Section 5, the modulation instability by means of standard linear stability analysis is examined. Section 6 displays the remarks and discussion of the obtained results. Finally, our conclusion is given in Section 7.

Herein, we present the process of applying the improved projective Riccati equations method as follows. Consider a nonlinear evolution equation (NLEE) in the form(2)$$P(u,u_t,u_x,u_{\mathrm{xx}},u_{\mathrm{xt}},u_{\mathrm{tt}},u_{\mathrm{xxx}}\dots )=0,$$ where u = u(x, t) is an unknown function and P is a polynomial in u and its various partial derivatives.

Based on the traveling wave transformation given by(3)$$u(x,t)=U\left(\xi \right),\hspace{1em}\xi =x-\mathrm{ct},$$ the NLEE (2) reduces to a nonlinear ordinary differential equation (NLODE) of the form(4)$$Q(U,U\mathrm{\prime },U\mathrm{\text{'}\text{'}},U\mathrm{\text{'}\text{'}\text{'}},\dots )=0,$$ where prime denotes the derivative with respect to ξ.

We assume that equation (4) has a solution in the form of a finite series as(5)$$U\left(\xi \right)=a_0+b_{0}f\left(\xi \right) g\left(\xi \right)+\sum _{j=1}^m\mathrm{}\left[a_j f^j\left(\xi \right)+b_j g^j\left(\xi \right)\right],$$ where a_j, b_j, (j = 0, 1, 2, …, m) are constants to be determined. The parameter m is a positive integer which can be identified by balancing the highest order derivative term with the nonlinear term in equation (4).

The variables f(ξ) and g(ξ) satisfy the the following improved projective Riccati equations(6)$$\begin{array}{c}f\mathrm{\prime }\left(\xi \right)=\delta A{g}^2\left(\xi \right),\hspace{1em} g\mathrm{\prime }\left(\xi \right)=-\mathrm{Af}\left(\xi \right) g\left(\xi \right)-\frac{B}{A} g\left(\xi \right) (R-\mathrm{Bf}(\xi \left)\right),\hspace{1em}\\ g^2\left(\xi \right)=\delta \left[\frac{1}{A^2} (R-\mathrm{Bf}(\xi ){)}^2-f^2(\xi )\right],\end{array}$$$$\begin{array}{c}f\prime (\xi )=\delta A{g}^2(\xi ),g\prime (\xi )=-Af(\xi )g(\xi )-\frac{B}{A}g(\xi )(R-Bf(\xi )),\\ g^2(\xi )=\delta [\frac{1}{A^2}(R-Bf(\xi ){)}^2-f^2(\xi )],\end{array}$$ where A, B and R are arbitrary constants and δ = ±1. The third equation in the system (6), which gives the relation between the functions f(ξ) and g(ξ), represents the first integral of the couple ODEs in this system.

The set of equations (6) is found to possess solutions in the form(7)$$f_1\left(\xi \right)=\frac{R\tanh \left(\mathrm{R\xi }\right)}{A+B\tanh \left(\mathrm{R\xi }\right)},\hspace{1em} g_1\left(\xi \right)=\frac{R\mathrm{sech}\left(\mathrm{R\xi }\right)}{A+B\tanh \left(\mathrm{R\xi }\right)},$$ which implies δ = 1, and(8)$$f_2\left(\xi \right)=\frac{R\coth \left(\mathrm{R\xi }\right)}{A+B\coth \left(\mathrm{R\xi }\right)},\hspace{1em} g_2\left(\xi \right)=\frac{R\mathrm{csch}\left(\mathrm{R\xi }\right)}{A+B\coth \left(\mathrm{R\xi }\right)},$$ provided that δ = −1.

Substituting (5) along with (6) into equation (4) gives a polynomial in f^j (ξ) and f^j (ξ) g(ξ). Then, we equate each coefficient of f^j (ξ) and f^j (ξ)g(ξ) in this polynomial to zero to get a set of algebraic equations for a_j, b_j. Finally, solving this system of equations, we obtain various exact solutions of equation (2) according to (7) and (8).

Now, we aim to reduce the complex form of the model (1) to an NLODE with a view to deriving the optical soliton solutions. Therefore, we assume the traveling wave transformation of the form(9)$$\mathrm{\Psi }(x,t)=\psi \left(\xi \right)e^{\mathrm{i\varphi }(x,t)},$$ where ψ(ξ) accounts for the amplitude of the soliton while ϕ(x, y, t) denotes the phase component. The wave variable ξ is given by(10)$$\xi =x-\nu t,$$ and the function ϕ(x, y, t) is introduced as(11)$$\varphi (x,y,t)=-\kappa x+\omega t+\theta ,$$ where the parameters ν, κ, ω and θ represent the soliton velocity, frequency, wave number and phase constant, respectively.

Substituting (8) into equation (1) leads to a couple of equations for real and imaginary parts given, respectively, as(12)$$b{\psi }^{\mathrm{iv}}+(3\mathrm{a\kappa }-6b{\kappa }^2)\psi \mathrm{\text{'}\text{'}}-(\omega +\alpha \kappa +a{\kappa }^3-b{\kappa }^4)\psi +(c+\mathrm{d\kappa }){\psi }^3-\lambda \kappa {\psi }^{2n+1}=0,$$(13)$$(a-4\mathrm{a\kappa })\psi \mathrm{\text{'}\text{'}\text{'}}-(\alpha +\nu +3a{\kappa }^2-4b{\kappa }^3)\psi \mathrm{\prime }+d{\psi }^2\psi \mathrm{\prime }-(2\mathrm{n\mu }+(2n+1\left)\lambda \right){\psi }^{2n}\psi \mathrm{\prime }=0,$$ where the prime denotes the derivative with respect to ξ. The system of equations (12) and (13) is reduced to(14)$$b{\psi }^{\mathrm{iv}}+6b{\kappa }^2\psi \mathrm{\text{'}\text{'}}-(\omega +\alpha \kappa +3b{\kappa }^4)\psi +(c+\mathrm{d\kappa }-\lambda \kappa ){\psi }^3=0,$$ with the expression for the velocity of the soliton presented as(15)$$\nu =-\alpha -8b{\kappa }^3,$$ under the constraints(16)$$n=1,$$(17)$$a=4\mathrm{b\kappa },$$(18)$$d-2\mu -3\lambda =0.$$

Now, we embark on deriving the solutions of the perturped CQ-FLE through implementing the improved projective Riccati equations method stated in Section 2. The proposed technique is basically used to handle equation (14) and then its obtained solutions are plugged into the transformation (9) so as to extract the optical solitons of the governing model.

According to the series formula given in (5) and the balance between the terms ψ^iv and ψ³ in equation (14), this leads to m = 2. Hence, the general solution form of equation (14) reads(19)$$\psi \left(\xi \right)=a_0+b_{0}f\left(\xi \right) g\left(\xi \right)+\sum _{j=1}^2\mathrm{}\left[a_j f^j\left(\xi \right)+b_j g^j\left(\xi \right)\right].$$

Substituting (19) together with equations (6) into equation (14) gives rise to an equation having different powers of f^lg^s. Collecting all the terms with the same power of f^lg^s together and equating each coefficient to zero, yields a set of algebraic equations. Solving these equations simultaneously leads to the following results.

Set I. If δ = 1, then the following cases of solutions in the hyperbolic secant and tangent functions are retrieved.

Case 1.

(20)$$a_0=a_1=a_2=b_2=0, \hspace{1em}{b}_0=\pm 2\sqrt{\frac{30b(A^2-B^2{)}^3}{A^2(c+\mathrm{d\kappa }-\lambda \kappa )}},\hspace{1em} b_1=\frac{b_0\mathrm{RB}}{A^2-B^2},\hspace{1em} \omega =-\alpha \kappa +\frac{24}{25}b{\kappa }^4,\hspace{1em} R=\pm \sqrt{\frac{3}{5}}\kappa .$$

Inserting (20) in accompany with (7) into (19) yields(21)$$\mathrm{\Psi }(x,t)=\pm 2R^2\sqrt{\frac{30b(A^2-B^2)}{c+\mathrm{d\kappa }-\lambda \kappa }} \frac{A\tanh \left(\mathrm{R\xi }\right)\mathrm{sech}\left(\mathrm{R\xi }\right)+B\mathrm{sech}\left(\mathrm{R\xi }\right)}{{\left(A+B\tanh \left(\mathrm{R\xi }\right)\right)}^2} e^{\mathrm{i\varphi }(x,t)},$$ where b(A² − B²)(c + dκ − λκ) > 0, ξ = x + (α + 8bκ³)t and $\varphi (x,t)=-\kappa x-\left(\alpha \kappa -\frac{24}{25}b{\kappa }^4\right)t+\theta$.

Case 2.

(22)$$b_0=b_1=b_2=0,\hspace{1em} a_0=\frac{3a_2{\kappa }^2}{10(A^2-B^2)},\hspace{1em} a_1=\frac{2a_2\mathrm{RB}}{A^2-B^2}, {\hspace{1em}a}_2=\pm \frac{2(A^2-B^2{)}^2}{A^2}\sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }}, \hspace{1em}\omega =-\alpha \kappa -\frac{219}{25}b{\kappa }^4,\hspace{1em} R=\pm \sqrt{-\frac{3}{10}}\kappa .$$

Plugging (22) along with (7) into (19) brings about(23)$$\mathrm{\Psi }(x,t)=\pm 2(A^2-B^2)R^2\sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }} \times \frac{\sec h^2\left(\mathrm{R\xi }\right)}{{\left(A+B\tanh \left(\mathrm{R\xi }\right)\right)}^2} e^{\mathrm{i\varphi }(x,t)},$$ where b(c + dκ − λκ) < 0, ξ = x + (α + 8bκ³)t and $\varphi (x,t)=-\kappa x-\left(\alpha \kappa +\frac{219}{25}b{\kappa }^4\right)t+\theta$.

Case 3.(24)$$a_1=a_2=b_0=b_1=0,\hspace{1em} a_0=-\frac{b_2(10R^2+3{\kappa }^2)}{30(A^2-B^2)}, \hspace{1em}{b}_2=\pm 2\left(A^2-B^2\right)\sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }},\hspace{1em}\omega =-\alpha \kappa -18b{R}^2{\kappa }^2-\frac{6}{5}b{\kappa }^4, 40R^4-30R^2{\kappa }^2+9{\kappa }^4=0.$$

Substituting (24) together with (7) into (19) generates(25)$$\mathrm{\Psi }(x,t)=\pm \sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }} \times \left\{\frac{10R^2+3{\kappa }^2}{15}-\frac{2(A^2-B^2)R^2\sec {\mathrm{h}}^2\left(\mathrm{R\xi }\right)}{{\left(A+B\tanh \left(\mathrm{R\xi }\right)\right)}^2}\right\}{e}^{\mathrm{i\varphi }(x,t)},$$ where b(c + dκ − λκ) < 0, ξ = x + (α + 8bκ³)t and $\varphi \left(x,t\right)=-\kappa x-\left(\alpha \kappa +18b{R}^2{\kappa }^2+\frac{6}{5}b{\kappa }^4\right)t+\theta$.

Case 4.(26)$$b_0=b_1=b_2=0, a_0=\frac{3a_2{\kappa }^2}{10(A^2-B^2)}, a_1=\frac{2a_2\mathrm{RB}}{A^2-B^2}, a_2=\pm \frac{2(A^2-B^2{)}^2}{A^2}\sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }}, \omega =-\alpha \kappa -\frac{219}{25}b{\kappa }^4, R=\pm \sqrt{-\frac{3}{10}}\kappa .$$

Putting (26) in addition to (7) into (19) gives us the soliton solution (23).

Case 5.(27)$$a_0=a_1=a_2=0, \hspace{1em}{b}_0=\pm \frac{b_2\sqrt{B^2-A^2}}{A}, \hspace{1em}{b}_1=\pm \frac{b_2\mathrm{RB}\sqrt{B^2-A^2}}{A(A^2-B^2)},\hspace{1em} b_2=\pm \left(A^2-B^2\right)\sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }},\hspace{1em}\omega =-\alpha \kappa -\frac{219}{25}b{\kappa }^4,\hspace{1em} R=\pm \sqrt{-\frac{6}{5}}\kappa .$$

Plugging (27) and (7) into (19) leads to(28)$$\mathrm{\Psi }(x,t)=\pm R^2\sqrt{\frac{30b(A^2-B^2)}{c+\mathrm{d\kappa }-\lambda \kappa }} \times \frac{\mathrm{sech}\left(\mathrm{R\xi }\right)\left[B+A\tanh \left(\mathrm{R\xi }\right)\pm \sqrt{B^2-A^2}\mathrm{sech}\left(\mathrm{R\xi }\right)\right]}{{\left(A+B\tanh \left(\mathrm{R\xi }\right)\right)}^2}{e}^{\mathrm{i\varphi }(x,t)},$$ where b(c + dκ − λκ) < 0, A² < B², ξ = x + (α + 8bκ³)t and $\varphi (x,t)=-\kappa x-\left(\alpha \kappa -\frac{219}{25}b{\kappa }^4\right)t+\theta$.

Case 6.(29)$$b_2=0, {\hspace{1em}a}_0=\frac{6b_0{\kappa }^2}{5A\sqrt{B^2-A^2}}, \hspace{1em}{a}_1=\pm \frac{2b_0\mathrm{RB}\sqrt{B^2-A^2}}{A(A^2-B^2)}, {\hspace{1em}a}_2=\pm \frac{b_0\sqrt{B^2-A^2}}{A}, {\hspace{1em}b}_1=\frac{b_0\mathrm{RB}}{A^2-B^2},{\hspace{1em}b}_0=\pm \sqrt{\frac{30b(A^2-B^2{)}^3}{A^2(c+\mathrm{d\kappa }-\lambda \kappa )}},\hspace{1em} \omega =-\alpha \kappa -\frac{219}{25}b{\kappa }^4,\hspace{1em} R=\pm \sqrt{-\frac{6}{5}}\kappa .$$

Substituting (29) in accompany with (7) into (19) produces the soliton solution (28).

Case 7.(30)$$b_0=b_1=b_2=0,\hspace{1em}{a}_0=-\frac{a_2(20R^2{A}^2-30R^2{B}^2-3A^2{\kappa }^2)}{30(A^2-B^2{)}^2}, {\hspace{1em}a}_1=\frac{2a_2\mathrm{RB}}{A^2-B^2}, {\hspace{1em}a}_2=\pm \frac{2(A^2-B^2{)}^2}{A^2}\sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }}, \hspace{1em}\omega =-\alpha \kappa -18b{R}^2{\kappa }^2-\frac{6}{5}b{\kappa }^4, 40R^4-30R^2{\kappa }^2+9{\kappa }^4=0.$$

Inserting (30) along with (7) into (19), we secure(31)$$\begin{array}{cc}\mathrm{\Psi }\left(x,t\right)=& \pm \sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }}\left\{\frac{20A^2{R}^2-3A^2{\kappa }^2-30B^2{R}^2}{15A^2}\right.\\ & \left.-\frac{2(A^2-B^2)R^2\tanh \left(\mathrm{R\xi }\right)\left\{(A^2+B^2)\tanh \left(\mathrm{R\xi }\right)+2\mathrm{AB}\right\}}{A^2{\left(A+B\tanh \left(\mathrm{R\xi }\right)\right)}^2}\right\}{e}^{\mathrm{i\varphi }(x,t)},\end{array}$$ where b(c + dκ − λκ) < 0, ξ = x + (α + 8bκ³)t and $\varphi (x,t)=-\kappa x-\left(\alpha \kappa +18b{R}^2{\kappa }^2+\frac{6}{5}b{\kappa }^4\right)t+\theta$.

Case 8.(32)$$\begin{array}{c}{a}_1=a_2=0, \hspace{1em}{a}_0=-\frac{b_2(5R^2+6{\kappa }^2)}{30(A^2-B^2)},\hspace{1em} b_0=\pm \frac{b_2\sqrt{B^2-A^2}}{A}, \hspace{1em}{b}_1=\pm \frac{b_2\mathrm{RB}\sqrt{B^2-A^2}}{A(A^2-B^2)},\hspace{1em}\\ b_2=\pm (A^2-B^2)\sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }},\hspace{1em} \omega =-\alpha \kappa -\frac{9}{2}b{R}^2{\kappa }^2-\frac{6}{5}b{\kappa }^4,\hspace{1em} 5R^4-15R^2{\kappa }^2+18{\kappa }^4=0.\end{array}$$

Substituting (32) as well as (7) into (19) provides the soliton solution(33)$$\begin{array}{c}\mathrm{\Psi }(x,t)=\pm \sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }}\left\{\frac{5R^2+6{\kappa }^2}{30}\pm \frac{R^2\sqrt{B^2-A^2}\mathrm{sech}\left(\mathrm{R\xi }\right)\left[B+A\tanh \left(\mathrm{R\xi }\right)\right]}{{\left(A+B\tanh \left(\mathrm{R\xi }\right)\right)}^2}\right.\\ \left.-\frac{R^2(A^2-B^2)\sec {\mathrm{h}}^2\left(\mathrm{R\xi }\right)}{{\left(A+B\tanh \left(\mathrm{R\xi }\right)\right)}^2}\right\}{e}^{\mathrm{i\varphi }(x,t)},\end{array}$$ where$b(c+\mathrm{d\kappa }-\lambda \kappa )<0,A^2<B^2,\xi =x+(\alpha +8b{\kappa }^3)t$ and$\varphi (x,t)=-\kappa x-\left(\alpha \kappa +\frac{9}{2}b{R}^2{\kappa }^2+\frac{6}{5}b{\kappa }^4\right)t+\theta$.

Case 9.

(34)$$\begin{array}{c}{b}_2=0, \hspace{1em}{a}_0=\frac{b_0(25R^2{A}^2-30R^2{B}^2-6A^2{\kappa }^2)}{30A(A^2-B^2)\sqrt{B^2-A^2}}, \hspace{1em}{a}_1=\pm \frac{2b_0\mathrm{RB}\sqrt{B^2-A^2}}{A(A^2-B^2)}, \hspace{1em}{a}_2=\pm \frac{b_0\sqrt{B^2-A^2}}{A},\\ b_1=\frac{b_0\mathrm{RB}}{A^2-B^2},\hspace{1em} b_0=\pm \sqrt{\frac{30b(A^2-B^2{)}^3}{A^2(c+\mathrm{d\kappa }-\lambda \kappa )}},\hspace{1em} \omega =-\alpha \kappa -\frac{9}{2}b{R}^2{\kappa }^2-\frac{6}{5}b{\kappa }^4,\hspace{1em} 5R^4-15R^2{\kappa }^2+18{\kappa }^4=0.\end{array}$$

Substituting (34) together with (7) into (19) creates(35)$$\mathrm{\Psi }\left(x,t\right)=\pm \sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }}\left\{\frac{25A^2{R}^2-6A^2{\kappa }^2-30B^2{R}^2}{30A^2}\right.\pm \frac{R^2{A}^2\sqrt{B^2-A^2}\mathrm{sech}\left(\mathrm{R\xi }\right)\left[B+A\tanh \left(\mathrm{R\xi }\right)\right]}{A^2{\left(A+B\tanh \left(\mathrm{R\xi }\right)\right)}^2}\left.-\frac{R^2(A^2-B^2)\tanh \left(\mathrm{R\xi }\right)\left[2\mathrm{AB}+(A^2+B^2)\tanh \left(\mathrm{R\xi }\right)\right]}{A^2{\left(A+B\tanh \left(\mathrm{R\xi }\right)\right)}^2}\right\}{e}^{\mathrm{i\varphi }(x,t)},$$ where b(c + dκ − λκ) < 0, A² < B², ξ = x + (α + 8bκ³)t and $\varphi (x,t)=-\kappa x-\left(\alpha \kappa +\frac{9}{2}b{R}^2{\kappa }^2+\frac{6}{5}b{\kappa }^4\right)t+\theta$.

Set II. If δ = −1, then the following cases of solutions in the hyperbolic cosecant and cotangent functions are retrieved.

Case 1.(36)$$a_0=a_1=a_2=b_2=0,\hspace{1em} b_0=\pm 2\sqrt{-\frac{30b(A^2-B^2{)}^3}{A^2(c+\mathrm{d\kappa }-\lambda \kappa )}}, \hspace{1em}{b}_1=\frac{b_0\mathrm{RB}}{A^2-B^2}, \hspace{1em}\omega =-\alpha \kappa +\frac{24}{25}b{\kappa }^4,\hspace{1em} R=\pm \sqrt{\frac{3}{5}}\kappa .$$

Inserting (36) along with (7) into (19) yields(37)$$\mathrm{\Psi }(x,t)=\pm 2R^2\sqrt{-\frac{30b(A^2-B^2)}{c+\mathrm{d\kappa }-\lambda \kappa }} \times \frac{A\coth \left(\mathrm{R\xi }\right)\mathrm{csch}\left(\mathrm{R\xi }\right)+B\mathrm{csch}\left(\mathrm{R\xi }\right)}{{\left(A+B\coth \left(\mathrm{R\xi }\right)\right)}^2} e^{\mathrm{i\varphi }(x,t)},$$ where b(A² − B²)(c + dκ − λκ) < 0, ξ = x + (α + 8bκ³)t and $\varphi (x,t)=-\kappa x-\left(\alpha \kappa -\frac{24}{25}b{\kappa }^4\right)t+\theta$.

Case 2.

(38)$$b_0=b_1=b_2=0,\hspace{1em} a_0=\frac{3a_2{\kappa }^2}{10(A^2-B^2)},\hspace{1em} a_1=\frac{2a_2\mathrm{RB}}{A^2-B^2}, {\hspace{1em}a}_2=\pm \frac{2(A^2-B^2{)}^2}{A^2}\sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }}, \hspace{1em}\omega =-\alpha \kappa -\frac{219}{25}b{\kappa }^4,\hspace{1em} R=\pm \sqrt{-\frac{3}{10}}\kappa .$$

Plugging (38) in addition to (7) into (19) brings about(39)$$\mathrm{\Psi }(x,t)=\pm 2(A^2-B^2)R^2\sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }} \frac{\csc {\mathrm{h}}^2\left(\mathrm{R\xi }\right)}{{\left(A+B\coth \left(\mathrm{R\xi }\right)\right)}^2} e^{\mathrm{i\varphi }(x,t)},$$ where$b(c+\mathrm{d\kappa }-\lambda \kappa )<0,\xi =x+(\alpha +8b{\kappa }^3)t$ and$\varphi (x,t)=-\kappa x-\left(\alpha \kappa +\frac{219}{25}b{\kappa }^4\right)t+\theta$.

Case 3.(40)$$a_1=a_2=b_0=b_1=0, \hspace{1em}{a}_0=\frac{b_2(10R^2+3{\kappa }^2)}{30(A^2-B^2)}, \hspace{1em}{b}_2=\pm 2\left(A^2-B^2\right)\sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }},\hspace{1em}\omega =-\alpha \kappa -18b{R}^2{\kappa }^2-\frac{6}{5}b{\kappa }^4,\hspace{1em} 40R^4-30R^2{\kappa }^2+9{\kappa }^4=0.$$

Substituting (40) and (7) into (19) generates(41)$$\mathrm{\Psi }(x,t)=\pm \sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }}\left\{\frac{10R^2+3{\kappa }^2}{15}+\frac{2(A^2-B^2)R^2\csc h^2\left(\mathrm{R\xi }\right)}{{\left(A+B\coth \left(\mathrm{R\xi }\right)\right)}^2}\right\}{e}^{\mathrm{i\varphi }(x,t)},$$ where b(c + dκ − λκ) < 0, ξ = x + (α + 8bκ³)t and $\varphi (x,t)=-\kappa x-\left(\alpha \kappa +18b{R}^2{\kappa }^2+\frac{6}{5}b{\kappa }^4\right)t+\theta$.

Case 4.

(42)$$b_0=b_1=b_2=0, \hspace{1em}{a}_0=\frac{3a_2{\kappa }^2}{10(A^2-B^2)},\hspace{1em} a_1=\frac{2a_2\mathrm{RB}}{A^2-B^2},\hspace{1em} a_2=\pm \frac{2(A^2-B^2{)}^2}{A^2}\sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }}, \hspace{1em}\omega =-\alpha \kappa -\frac{219}{25}b{\kappa }^4, \hspace{1em}R=\pm \sqrt{-\frac{3}{10}}\kappa .$$

Putting (42) as well as (7) into (19) gives us the soliton solution (23).

Case 5.

(43)$$a_0=a_1=a_2=0,\hspace{1em} b_0=\pm \frac{b_2\sqrt{A^2-B^2}}{A}, \hspace{1em}{b}_1=\pm \frac{b_2\mathrm{RB}\sqrt{A^2-B^2}}{A(A^2-B^2)}, {\hspace{1em}b}_2=\pm \left(A^2-B^2\right)\sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }},\hspace{1em} \omega =-\alpha \kappa -\frac{219}{25}b{\kappa }^4,\hspace{1em} R=\pm \sqrt{-\frac{6}{5}}\kappa .$$

Plugging (43) along with (7) into (19) results in(44)$$\mathrm{\Psi }(x,t)=\pm R^2\sqrt{-\frac{30b(A^2-B^2)}{c+\mathrm{d\kappa }-\lambda \kappa }} \times \frac{\mathrm{csch}\left(\mathrm{R\xi }\right)\left[B+A\coth \left(\mathrm{R\xi }\right)\pm \sqrt{A^2-B^2}\mathrm{csch}\left(\mathrm{R\xi }\right)\right]}{{\left(A+B\coth \left(\mathrm{R\xi }\right)\right)}^2}{e}^{\mathrm{i\varphi }(x,t)},$$ where$b(c+\mathrm{d\kappa }-\lambda \kappa )<0,A^2>B^2,\xi =x+(\alpha +8b{\kappa }^3)t$ and$\varphi (x,t)=-\kappa x-(\alpha \kappa -\frac{219}{25}b{\kappa }^4)t+\theta$.

Case 6.

(45)$$b_2=0, a_0=\frac{6b_0{\kappa }^2}{5A\sqrt{A^2-B^2}}, \hspace{1em}{a}_1=\pm \frac{2b_0\mathrm{RB}\sqrt{A^2-B^2}}{A(A^2-B^2)}, \hspace{1em}{a}_2=\pm \frac{b_0\sqrt{A^2-B^2}}{A}, \hspace{1em}{b}_1=\frac{b_0\mathrm{RB}}{A^2-B^2},\hspace{1em} b_0=\pm \sqrt{-\frac{30b(A^2-B^2{)}^3}{A^2(c+\mathrm{d\kappa }-\lambda \kappa )}},\hspace{1em} \omega =-\alpha \kappa -\frac{219}{25}b{\kappa }^4,\hspace{1em} R=\pm \sqrt{-\frac{6}{5}}\kappa .$$

Substituting (45) in accompany with (7) into (19) produces the soliton solution (44).

Case 7.(46)$$b_0=b_1=b_2=0, \hspace{1em}{a}_0=-\frac{a_2(20R^2{A}^2-30R^2{B}^2-3A^2{\kappa }^2)}{30(A^2-B^2{)}^2}, \hspace{1em}{a}_1=\frac{2a_2\mathrm{RB}}{A^2-B^2}, \hspace{1em}{a}_2=\pm \frac{2(A^2-B^2{)}^2}{A^2}\sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }}, \hspace{1em}\omega =-\alpha \kappa -18b{R}^2{\kappa }^2-\frac{6}{5}b{\kappa }^4, \hspace{1em}40R^4-30R^2{\kappa }^2+9{\kappa }^4=0.$$

Inserting (46) and (7) into (19), we come by(47)$$\mathrm{\Psi }(x,t)=\pm \sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }} \times \left\{\frac{20A^2{R}^2-3A^2{\kappa }^2-30B^2{R}^2}{15A^2}\right.\left.-\frac{2(A^2-B^2)R^2\coth \left(\mathrm{R\xi }\right)\left\{(A^2+B^2)\coth \left(\mathrm{R\xi }\right)+2\mathrm{AB}\right\}}{A^2{\left(A+B\coth \left(\mathrm{R\xi }\right)\right)}^2}\right\}{e}^{\mathrm{i\varphi }(x,t)},$$ where$b(c+\mathrm{d\kappa }-\lambda \kappa )<0,\xi =x+(\alpha +8b{\kappa }^3)t$ and$\varphi (x,t)=-\kappa x-\left(\alpha \kappa +18b{R}^2{\kappa }^2+\frac{6}{5}b{\kappa }^4\right)t+\theta$.

Case 8.

(48)$$a_1=a_2=0, \hspace{1em}{a}_0=\frac{b_2(5R^2+6{\kappa }^2)}{30(A^2-B^2)}, \hspace{1em}{b}_0=\pm \frac{b_2\sqrt{A^2-B^2}}{A}, {\hspace{1em}b}_1=\pm \frac{b_2\mathrm{RB}\sqrt{A^2-B^2}}{A(A^2-B^2)}, \hspace{1em}{b}_2=\pm (A^2-B^2)\sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }},\hspace{1em} \omega =-\alpha \kappa -\frac{9}{2}b{R}^2{\kappa }^2-\frac{6}{5}b{\kappa }^4,\hspace{1em} 5R^4-15R^2{\kappa }^2+18{\kappa }^4=0.$$

Substituting (48) in accompany with (7) into (19) provides the soliton solution(49)$$\mathrm{\Psi }(x,t)=\pm \sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }} \times \left\{\frac{5R^2+6{\kappa }^2}{30}\pm \frac{R^2\sqrt{A^2-B^2}\mathrm{csch}\left(\mathrm{R\xi }\right)\left[B+A\coth \left(\mathrm{R\xi }\right)\right]}{{\left(A+B\coth \left(\mathrm{R\xi }\right)\right)}^2}\right.\left.+\frac{R^2(A^2-B^2)\csc {\mathrm{h}}^2\left(\mathrm{R\xi }\right)}{{\left(A+B\coth \left(\mathrm{R\xi }\right)\right)}^2}\right\}{e}^{\mathrm{i\varphi }(x,t)},$$ where$b(c+\mathrm{d\kappa }-\lambda \kappa )<0,A^2>B^2,\xi =x+(\alpha +8b{\kappa }^3)t$ and$\varphi (x,t)=-\kappa x-\left(\alpha \kappa +\frac{9}{2}b{R}^2{\kappa }^2+\frac{6}{5}b{\kappa }^4\right)t+\theta$.

Case 9.

(50)$$b_2=0, \hspace{1em}{a}_0=-\frac{b_0(25R^2{A}^2-30R^2{B}^2-6A^2{\kappa }^2)}{30A(A^2-B^2)\sqrt{A^2-B^2}}, \hspace{1em}{a}_1=\pm \frac{2b_0\mathrm{RB}\sqrt{A^2-B^2}}{A(A^2-B^2)}, {\hspace{1em}a}_2=\pm \frac{b_0\sqrt{A^2-B^2}}{A},\hspace{1em} b_1=\frac{b_0\mathrm{RB}}{A^2-B^2},\hspace{1em} b_0=\pm \sqrt{\frac{-30b(A^2-B^2{)}^3}{A^2(c+\mathrm{d\kappa }-\lambda \kappa )}},\hspace{1em} \omega =-\alpha \kappa -\frac{9}{2}b{R}^2{\kappa }^2-\frac{6}{5}b{\kappa }^4,\hspace{1em} 5R^4-15R^2{\kappa }^2+18{\kappa }^4=0.$$

Substituting (50) together with (7) into (19) gives rise to(51)$$\begin{array}{cc}\mathrm{\Psi }\left(x,t\right)=& \pm \sqrt{-\frac{30b}{c+\mathrm{d\kappa }-\lambda \kappa }}\left\{\frac{25A^2{R}^2-6A^2{\kappa }^2-30B^2{R}^2}{30A^2}\right.\\ & \pm \frac{R^2{A}^2\sqrt{A^2-B^2}\csc \mathrm{h}\left(\mathrm{R\xi }\right)\left[B+A\coth \left(\mathrm{R\xi }\right)\right]}{A^2{\left(A+B\coth \left(\mathrm{R\xi }\right)\right)}^2}\\ & \left.-\frac{R^2(A^2-B^2)\coth \left(\mathrm{R\xi }\right)\left[2\mathrm{AB}+(A^2+B^2)\coth \left(\mathrm{R\xi }\right)\right]}{A^2{\left(A+B\coth \left(\mathrm{R\xi }\right)\right)}^2}\right\}{e}^{i\varphi (x,t)},\end{array}$$ where$b(c+\mathrm{d\kappa }-\lambda \kappa )<0,A^2>B^2,\xi =x+(\alpha +8b{\kappa }^3)t$ and$\varphi (x,t)=-\kappa x-\left(\alpha \kappa +\frac{9}{2}b{R}^2{\kappa }^2+\frac{6}{5}b{\kappa }^4\right)t+\theta$.