Integrand size = 43, antiderivative size = 187 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {8 a^2 (63 A+57 B+47 C) \tan (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (63 A+57 B+47 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {2 (63 A-18 B+22 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{315 d}+\frac {2 C \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{9 d}+\frac {2 (3 B+C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{21 a d} \]
2/315*(63*A-18*B+22*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/9*C*sec(d*x+c )^2*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/21*(3*B+C)*(a+a*sec(d*x+c))^(5/2 )*tan(d*x+c)/a/d+8/315*a^2*(63*A+57*B+47*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^ (1/2)+2/315*a*(63*A+57*B+47*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d
Time = 2.38 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.80 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 (693 A+702 B+752 C+(567 A+648 B+748 C) \cos (c+d x)+(882 A+858 B+748 C) \cos (2 (c+d x))+189 A \cos (3 (c+d x))+156 B \cos (3 (c+d x))+136 C \cos (3 (c+d x))+189 A \cos (4 (c+d x))+156 B \cos (4 (c+d x))+136 C \cos (4 (c+d x))) \sec ^4(c+d x) \tan (c+d x)}{630 d \sqrt {a (1+\sec (c+d x))}} \]
Integrate[Sec[c + d*x]^2*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
(a^2*(693*A + 702*B + 752*C + (567*A + 648*B + 748*C)*Cos[c + d*x] + (882* A + 858*B + 748*C)*Cos[2*(c + d*x)] + 189*A*Cos[3*(c + d*x)] + 156*B*Cos[3 *(c + d*x)] + 136*C*Cos[3*(c + d*x)] + 189*A*Cos[4*(c + d*x)] + 156*B*Cos[ 4*(c + d*x)] + 136*C*Cos[4*(c + d*x)])*Sec[c + d*x]^4*Tan[c + d*x])/(630*d *Sqrt[a*(1 + Sec[c + d*x])])
Time = 1.15 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.279, Rules used = {3042, 4576, 27, 3042, 4498, 27, 3042, 4489, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a \sec (c+d x)+a)^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4576 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sec ^2(c+d x) (\sec (c+d x) a+a)^{3/2} (a (9 A+4 C)+3 a (3 B+C) \sec (c+d x))dx}{9 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec ^2(c+d x) (\sec (c+d x) a+a)^{3/2} (a (9 A+4 C)+3 a (3 B+C) \sec (c+d x))dx}{9 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (9 A+4 C)+3 a (3 B+C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{9 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d}\) |
| \(\Big \downarrow \) 4498 |
| \(\displaystyle \frac {\frac {2 \int \frac {1}{2} \sec (c+d x) (\sec (c+d x) a+a)^{3/2} \left (15 (3 B+C) a^2+(63 A-18 B+22 C) \sec (c+d x) a^2\right )dx}{7 a}+\frac {6 (3 B+C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \sec (c+d x) (\sec (c+d x) a+a)^{3/2} \left (15 (3 B+C) a^2+(63 A-18 B+22 C) \sec (c+d x) a^2\right )dx}{7 a}+\frac {6 (3 B+C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (15 (3 B+C) a^2+(63 A-18 B+22 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx}{7 a}+\frac {6 (3 B+C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {\frac {\frac {3}{5} a^2 (63 A+57 B+47 C) \int \sec (c+d x) (\sec (c+d x) a+a)^{3/2}dx+\frac {2 a^2 (63 A-18 B+22 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {6 (3 B+C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3}{5} a^2 (63 A+57 B+47 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}dx+\frac {2 a^2 (63 A-18 B+22 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {6 (3 B+C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d}\) |
| \(\Big \downarrow \) 4280 |
| \(\displaystyle \frac {\frac {\frac {3}{5} a^2 (63 A+57 B+47 C) \left (\frac {4}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a^2 (63 A-18 B+22 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {6 (3 B+C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3}{5} a^2 (63 A+57 B+47 C) \left (\frac {4}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a^2 (63 A-18 B+22 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {6 (3 B+C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (63 A-18 B+22 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}+\frac {3}{5} a^2 (63 A+57 B+47 C) \left (\frac {8 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )}{7 a}+\frac {6 (3 B+C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d}\) |
(2*C*Sec[c + d*x]^2*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(9*d) + ((6*( 3*B + C)*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d) + ((2*a^2*(63*A - 18*B + 22*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (3*a^2*(63*A + 57*B + 47*C)*((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*a))/(9*a)
3.5.92.3.1 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]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs c[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Cs c[e + f*x])^m*(d*Csc[e + f*x])^n*Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b* B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m , n}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
Time = 0.99 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.88
| method | result | size |
| default | \(\frac {2 a \left (378 A \cos \left (d x +c \right )^{4}+312 B \cos \left (d x +c \right )^{4}+272 C \cos \left (d x +c \right )^{4}+189 A \cos \left (d x +c \right )^{3}+156 B \cos \left (d x +c \right )^{3}+136 C \cos \left (d x +c \right )^{3}+63 A \cos \left (d x +c \right )^{2}+117 B \cos \left (d x +c \right )^{2}+102 C \cos \left (d x +c \right )^{2}+45 B \cos \left (d x +c \right )+85 C \cos \left (d x +c \right )+35 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}\) | \(164\) |
| parts | \(\frac {2 A a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (6 \sin \left (d x +c \right )+3 \tan \left (d x +c \right )+\sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{5 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 B a \left (104 \cos \left (d x +c \right )^{3}+52 \cos \left (d x +c \right )^{2}+39 \cos \left (d x +c \right )+15\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{105 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C a \left (272 \cos \left (d x +c \right )^{4}+136 \cos \left (d x +c \right )^{3}+102 \cos \left (d x +c \right )^{2}+85 \cos \left (d x +c \right )+35\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}\) | \(217\) |
int(sec(d*x+c)^2*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, method=_RETURNVERBOSE)
2/315*a/d*(378*A*cos(d*x+c)^4+312*B*cos(d*x+c)^4+272*C*cos(d*x+c)^4+189*A* cos(d*x+c)^3+156*B*cos(d*x+c)^3+136*C*cos(d*x+c)^3+63*A*cos(d*x+c)^2+117*B *cos(d*x+c)^2+102*C*cos(d*x+c)^2+45*B*cos(d*x+c)+85*C*cos(d*x+c)+35*C)*(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 = 135, normalized size of antiderivative = 0.72 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (2 \, {\left (189 \, A + 156 \, B + 136 \, C\right )} a \cos \left (d x + c\right )^{4} + {\left (189 \, A + 156 \, B + 136 \, C\right )} a \cos \left (d x + c\right )^{3} + 3 \, {\left (21 \, A + 39 \, B + 34 \, C\right )} a \cos \left (d x + c\right )^{2} + 5 \, {\left (9 \, B + 17 \, C\right )} a \cos \left (d x + c\right ) + 35 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]
integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="fricas")
2/315*(2*(189*A + 156*B + 136*C)*a*cos(d*x + c)^4 + (189*A + 156*B + 136*C )*a*cos(d*x + c)^3 + 3*(21*A + 39*B + 34*C)*a*cos(d*x + c)^2 + 5*(9*B + 17 *C)*a*cos(d*x + c) + 35*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d *x + c)/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4)
\[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \]
Integral((a*(sec(c + d*x) + 1))**(3/2)*(A + B*sec(c + d*x) + C*sec(c + d*x )**2)*sec(c + d*x)**2, x)
Timed out. \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="maxima")
\[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{2} \,d x } \]
integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="giac")
Time = 26.67 (sec) , antiderivative size = 709, normalized size of antiderivative = 3.79 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\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\,4{}\mathrm {i}}{5\,d}+\frac {a\,\left (3\,A+6\,B+4\,C\right )\,4{}\mathrm {i}}{5\,d}+\frac {a\,\left (3\,B+C\right )\,16{}\mathrm {i}}{105\,d}\right )-\frac {a\,\left (3\,A+2\,B\right )\,4{}\mathrm {i}}{5\,d}+\frac {a\,\left (A+4\,B+12\,C\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\,\left (3\,A+2\,B\right )\,4{}\mathrm {i}}{9\,d}-\frac {A\,a\,4{}\mathrm {i}}{9\,d}+\frac {a\,\left (5\,A+6\,B+4\,C\right )\,4{}\mathrm {i}}{9\,d}-\frac {a\,\left (7\,A+8\,B+12\,C\right )\,4{}\mathrm {i}}{9\,d}\right )-\frac {a\,\left (3\,A+2\,B\right )\,4{}\mathrm {i}}{9\,d}+\frac {A\,a\,4{}\mathrm {i}}{9\,d}-\frac {a\,\left (5\,A+6\,B+4\,C\right )\,4{}\mathrm {i}}{9\,d}+\frac {a\,\left (7\,A+8\,B+12\,C\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 {\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\,4{}\mathrm {i}}{3\,d}-\frac {a\,\left (21\,A+39\,B+34\,C\right )\,8{}\mathrm {i}}{315\,d}\right )+\frac {a\,\left (3\,A+2\,B\right )\,4{}\mathrm {i}}{3\,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 )}+\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\,4{}\mathrm {i}}{7\,d}+\frac {C\,a\,32{}\mathrm {i}}{63\,d}-\frac {a\,\left (2\,A+3\,B+2\,C\right )\,8{}\mathrm {i}}{7\,d}+\frac {a\,\left (3\,A+2\,B+8\,C\right )\,4{}\mathrm {i}}{7\,d}\right )+\frac {a\,\left (3\,A+2\,B\right )\,4{}\mathrm {i}}{7\,d}+\frac {a\,\left (A-8\,C\right )\,4{}\mathrm {i}}{7\,d}-\frac {a\,\left (2\,A+3\,B+6\,C\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 {a\,{\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 (189\,A+156\,B+136\,C\right )\,4{}\mathrm {i}}{315\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )} \]
((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*(3*A + 6*B + 4*C)*4i)/(5*d) - (A*a*4i)/(5*d) + (a*(3*B + C)*1 6i)/(105*d)) - (a*(3*A + 2*B)*4i)/(5*d) + (a*(A + 4*B + 12*C)*4i)/(5*d)))/ ((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) + ((a + a/(exp(- c*1 i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((a*(3*A + 2*B)*4i)/(9*d) - (A*a*4i)/(9*d) + (a*(5*A + 6*B + 4*C)*4i)/(9*d) - (a*(7 *A + 8*B + 12*C)*4i)/(9*d)) - (a*(3*A + 2*B)*4i)/(9*d) + (A*a*4i)/(9*d) - (a*(5*A + 6*B + 4*C)*4i)/(9*d) + (a*(7*A + 8*B + 12*C)*4i)/(9*d)))/((exp(c *1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) + ((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*4i)/(3*d) - (a*(21*A + 39*B + 34*C)*8i)/(315*d)) + (a*(3*A + 2*B)*4i)/(3*d)))/((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*a*4i)/(7*d) + (C*a*32i)/(63*d) - (a*(2*A + 3*B + 2*C)*8i)/(7*d) + (a*(3*A + 2*B + 8*C)* 4i)/(7*d)) + (a*(3*A + 2*B)*4i)/(7*d) + (a*(A - 8*C)*4i)/(7*d) - (a*(2*A + 3*B + 6*C)*8i)/(7*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1) ^3) - (a*exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d* x*1i)/2))^(1/2)*(189*A + 156*B + 136*C)*4i)/(315*d*(exp(c*1i + d*x*1i) + 1 ))