Integrand size = 33, antiderivative size = 175 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {64 a^3 (15 A+13 B) \tan (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (15 A+13 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {2 a (15 A+13 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac {2 (9 A-2 B) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 B (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d} \]
2/105*a*(15*A+13*B)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/63*(9*A-2*B)*(a+ a*sec(d*x+c))^(5/2)*tan(d*x+c)/d+2/9*B*(a+a*sec(d*x+c))^(7/2)*tan(d*x+c)/a /d+64/315*a^3*(15*A+13*B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+16/315*a^2*( 15*A+13*B)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d
Time = 0.46 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.55 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 a^3 \left (690 A+584 B+(345 A+292 B) \sec (c+d x)+3 (60 A+73 B) \sec ^2(c+d x)+5 (9 A+26 B) \sec ^3(c+d x)+35 B \sec ^4(c+d x)\right ) \tan (c+d x)}{315 d \sqrt {a (1+\sec (c+d x))}} \]
(2*a^3*(690*A + 584*B + (345*A + 292*B)*Sec[c + d*x] + 3*(60*A + 73*B)*Sec [c + d*x]^2 + 5*(9*A + 26*B)*Sec[c + d*x]^3 + 35*B*Sec[c + d*x]^4)*Tan[c + d*x])/(315*d*Sqrt[a*(1 + Sec[c + d*x])])
Time = 0.98 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4498, 27, 3042, 4489, 3042, 4280, 3042, 4280, 3042, 4279}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \sec ^2(c+d x) (a \sec (c+d x)+a)^{5/2} (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4498 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (\sec (c+d x) a+a)^{5/2} (7 a B+a (9 A-2 B) \sec (c+d x))dx}{9 a}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec (c+d x) (\sec (c+d x) a+a)^{5/2} (7 a B+a (9 A-2 B) \sec (c+d x))dx}{9 a}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (7 a B+a (9 A-2 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{9 a}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {\frac {3}{7} a (15 A+13 B) \int \sec (c+d x) (\sec (c+d x) a+a)^{5/2}dx+\frac {2 a (9 A-2 B) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{7} a (15 A+13 B) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}dx+\frac {2 a (9 A-2 B) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\) |
| \(\Big \downarrow \) 4280 |
| \(\displaystyle \frac {\frac {3}{7} a (15 A+13 B) \left (\frac {8}{5} a \int \sec (c+d x) (\sec (c+d x) a+a)^{3/2}dx+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a (9 A-2 B) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{7} a (15 A+13 B) \left (\frac {8}{5} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}dx+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a (9 A-2 B) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\) |
| \(\Big \downarrow \) 4280 |
| \(\displaystyle \frac {\frac {3}{7} a (15 A+13 B) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a (9 A-2 B) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{7} a (15 A+13 B) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a (9 A-2 B) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {\frac {3}{7} a (15 A+13 B) \left (\frac {8}{5} a \left (\frac {8 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a (9 A-2 B) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\) |
(2*B*(a + a*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*a*d) + ((2*a*(9*A - 2*B)* (a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d) + (3*a*(15*A + 13*B)*((2*a* (a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (8*a*((8*a^2*Tan[c + d*x] )/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d *x])/(3*d)))/5))/7)/(9*a)
3.2.36.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*b*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]])), x] /; Free Q[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ Symbol] :> Simp[(-b)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m - 1)/(f*m)), x] + Simp[a*((2*m - 1)/m) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && Intege rQ[2*m]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(b*(m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B , e, f, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b *(m + 1), 0] && !LtQ[m, -2^(-1)]
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]* ((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int [Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B) *Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a *B, 0] && !LtQ[m, -1]
Time = 34.91 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.76
| method | result | size |
| default | \(\frac {2 a^{2} \left (690 A \cos \left (d x +c \right )^{4}+584 B \cos \left (d x +c \right )^{4}+345 A \cos \left (d x +c \right )^{3}+292 B \cos \left (d x +c \right )^{3}+180 A \cos \left (d x +c \right )^{2}+219 B \cos \left (d x +c \right )^{2}+45 A \cos \left (d x +c \right )+130 B \cos \left (d x +c \right )+35 B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}\) | \(133\) |
| parts | \(\frac {2 A \,a^{2} \left (46 \cos \left (d x +c \right )^{3}+23 \cos \left (d x +c \right )^{2}+12 \cos \left (d x +c \right )+3\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{21 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 B \,a^{2} \left (584 \cos \left (d x +c \right )^{4}+292 \cos \left (d x +c \right )^{3}+219 \cos \left (d x +c \right )^{2}+130 \cos \left (d x +c \right )+35\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}\) | \(162\) |
2/315*a^2/d*(690*A*cos(d*x+c)^4+584*B*cos(d*x+c)^4+345*A*cos(d*x+c)^3+292* B*cos(d*x+c)^3+180*A*cos(d*x+c)^2+219*B*cos(d*x+c)^2+45*A*cos(d*x+c)+130*B *cos(d*x+c)+35*B)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*tan(d*x+c)*sec(d *x+c)^3
Time = 0.27 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.78 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 \, {\left (2 \, {\left (345 \, A + 292 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + {\left (345 \, A + 292 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (60 \, A + 73 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 5 \, {\left (9 \, A + 26 \, B\right )} a^{2} \cos \left (d x + c\right ) + 35 \, B a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]
2/315*(2*(345*A + 292*B)*a^2*cos(d*x + c)^4 + (345*A + 292*B)*a^2*cos(d*x + c)^3 + 3*(60*A + 73*B)*a^2*cos(d*x + c)^2 + 5*(9*A + 26*B)*a^2*cos(d*x + c) + 35*B*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*co s(d*x + c)^5 + d*cos(d*x + c)^4)
\[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}} \left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \]
\[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \]
4/315*(315*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*((A*a^2*d*cos(2*d*x + 2*c)^4 + A*a^2*d*sin(2*d*x + 2*c)^4 + 4*A* a^2*d*cos(2*d*x + 2*c)^3 + 6*A*a^2*d*cos(2*d*x + 2*c)^2 + 4*A*a^2*d*cos(2* d*x + 2*c) + A*a^2*d + 2*(A*a^2*d*cos(2*d*x + 2*c)^2 + 2*A*a^2*d*cos(2*d*x + 2*c) + A*a^2*d)*sin(2*d*x + 2*c)^2)*integrate((((cos(8*d*x + 8*c)*cos(2 *d*x + 2*c) + 3*cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 3*cos(4*d*x + 4*c)*cos (2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(8*d*x + 8*c)*sin(2*d*x + 2*c) + 3 *sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 3*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(9/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2*d*x + 2*c)*sin(8*d*x + 8*c) + 3*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 3*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(8*d*x + 8*c)*sin(2*d*x + 2*c) - 3*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 3*cos(4*d*x + 4*c)*sin(2*d*x + 2* c))*sin(9/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(5/2*arctan2( sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - ((cos(2*d*x + 2*c)*sin(8*d*x + 8*c) + 3*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 3*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(8*d*x + 8*c)*sin(2*d*x + 2*c) - 3*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 3*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*cos(9/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (cos(8*d*x + 8*c)*cos(2*d*x + 2*c) + 3*cos(6*d* x + 6*c)*cos(2*d*x + 2*c) + 3*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d* x + 2*c)^2 + sin(8*d*x + 8*c)*sin(2*d*x + 2*c) + 3*sin(6*d*x + 6*c)*sin...
\[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{2} \,d x } \]
Time = 22.64 (sec) , antiderivative size = 723, normalized size of antiderivative = 4.13 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,a^2\,4{}\mathrm {i}}{3\,d}-\frac {a^2\,\left (60\,A+73\,B\right )\,8{}\mathrm {i}}{315\,d}\right )+\frac {a^2\,\left (5\,A+2\,B\right )\,4{}\mathrm {i}}{3\,d}\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (-\frac {A\,a^2\,4{}\mathrm {i}}{5\,d}+\frac {a^2\,\left (3\,A+4\,B\right )\,16{}\mathrm {i}}{105\,d}+\frac {a^2\,\left (9\,A+10\,B\right )\,4{}\mathrm {i}}{5\,d}\right )-\frac {a^2\,\left (5\,A+2\,B\right )\,4{}\mathrm {i}}{5\,d}+\frac {a^2\,\left (5\,A+16\,B\right )\,4{}\mathrm {i}}{5\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,a^2\,4{}\mathrm {i}}{7\,d}+\frac {a^2\,\left (A+2\,B\right )\,20{}\mathrm {i}}{7\,d}+\frac {B\,a^2\,32{}\mathrm {i}}{63\,d}-\frac {a^2\,\left (A+B\right )\,40{}\mathrm {i}}{7\,d}\right )+\frac {a^2\,\left (A-8\,B\right )\,4{}\mathrm {i}}{7\,d}+\frac {a^2\,\left (5\,A+2\,B\right )\,4{}\mathrm {i}}{7\,d}-\frac {a^2\,\left (5\,A+9\,B\right )\,8{}\mathrm {i}}{7\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,a^2\,4{}\mathrm {i}}{9\,d}+\frac {a^2\,\left (3\,A+4\,B\right )\,20{}\mathrm {i}}{9\,d}-\frac {a^2\,\left (5\,A+2\,B\right )\,4{}\mathrm {i}}{9\,d}-\frac {a^2\,\left (11\,A+10\,B\right )\,4{}\mathrm {i}}{9\,d}\right )-\frac {A\,a^2\,4{}\mathrm {i}}{9\,d}-\frac {a^2\,\left (3\,A+4\,B\right )\,20{}\mathrm {i}}{9\,d}+\frac {a^2\,\left (5\,A+2\,B\right )\,4{}\mathrm {i}}{9\,d}+\frac {a^2\,\left (11\,A+10\,B\right )\,4{}\mathrm {i}}{9\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}-\frac {a^2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (345\,A+292\,B\right )\,4{}\mathrm {i}}{315\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )} \]
((exp(c*1i + d*x*1i)*((A*a^2*4i)/(3*d) - (a^2*(60*A + 73*B)*8i)/(315*d)) + (a^2*(5*A + 2*B)*4i)/(3*d))*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d *x*1i)/2))^(1/2))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)) + (( a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d *x*1i)*((a^2*(3*A + 4*B)*16i)/(105*d) - (A*a^2*4i)/(5*d) + (a^2*(9*A + 10* B)*4i)/(5*d)) - (a^2*(5*A + 2*B)*4i)/(5*d) + (a^2*(5*A + 16*B)*4i)/(5*d))) /((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) + ((a + a/(exp(- c* 1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*a^2* 4i)/(7*d) + (a^2*(A + 2*B)*20i)/(7*d) + (B*a^2*32i)/(63*d) - (a^2*(A + B)* 40i)/(7*d)) + (a^2*(A - 8*B)*4i)/(7*d) + (a^2*(5*A + 2*B)*4i)/(7*d) - (a^2 *(5*A + 9*B)*8i)/(7*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1 )^3) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp (c*1i + d*x*1i)*((A*a^2*4i)/(9*d) + (a^2*(3*A + 4*B)*20i)/(9*d) - (a^2*(5* A + 2*B)*4i)/(9*d) - (a^2*(11*A + 10*B)*4i)/(9*d)) - (A*a^2*4i)/(9*d) - (a ^2*(3*A + 4*B)*20i)/(9*d) + (a^2*(5*A + 2*B)*4i)/(9*d) + (a^2*(11*A + 10*B )*4i)/(9*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) - (a^2 *exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2) )^(1/2)*(345*A + 292*B)*4i)/(315*d*(exp(c*1i + d*x*1i) + 1))