Integrand size = 41, antiderivative size = 144 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {8 a^2 (35 A+21 B+19 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (35 A+21 B+19 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 (7 B-2 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 a d} \]
2/35*(7*B-2*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/7*C*(a+a*sec(d*x+c))^ (5/2)*tan(d*x+c)/a/d+8/105*a^2*(35*A+21*B+19*C)*tan(d*x+c)/d/(a+a*sec(d*x+ c))^(1/2)+2/105*a*(35*A+21*B+19*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d
Time = 2.06 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.82 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 (70 A+126 B+164 C+(525 A+462 B+468 C) \cos (c+d x)+2 (35 A+63 B+52 C) \cos (2 (c+d x))+175 A \cos (3 (c+d x))+126 B \cos (3 (c+d x))+104 C \cos (3 (c+d x))) \sec ^3(c+d x) \tan (c+d x)}{210 d \sqrt {a (1+\sec (c+d x))}} \]
(a^2*(70*A + 126*B + 164*C + (525*A + 462*B + 468*C)*Cos[c + d*x] + 2*(35* A + 63*B + 52*C)*Cos[2*(c + d*x)] + 175*A*Cos[3*(c + d*x)] + 126*B*Cos[3*( c + d*x)] + 104*C*Cos[3*(c + d*x)])*Sec[c + d*x]^3*Tan[c + d*x])/(210*d*Sq rt[a*(1 + Sec[c + d*x])])
Time = 0.76 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.220, Rules used = {3042, 4570, 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.
| \(\displaystyle \int \sec (c+d x) (a \sec (c+d x)+a)^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (\sec (c+d x) a+a)^{3/2} (a (7 A+5 C)+a (7 B-2 C) \sec (c+d x))dx}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec (c+d x) (\sec (c+d x) a+a)^{3/2} (a (7 A+5 C)+a (7 B-2 C) \sec (c+d x))dx}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (7 A+5 C)+a (7 B-2 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {\frac {1}{5} a (35 A+21 B+19 C) \int \sec (c+d x) (\sec (c+d x) a+a)^{3/2}dx+\frac {2 a (7 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} a (35 A+21 B+19 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}dx+\frac {2 a (7 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 4280 |
| \(\displaystyle \frac {\frac {1}{5} a (35 A+21 B+19 C) \left (\frac {4}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a (7 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} a (35 A+21 B+19 C) \left (\frac {4}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a (7 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {\frac {1}{5} a (35 A+21 B+19 C) \left (\frac {8 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a (7 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\) |
(2*C*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*a*d) + ((2*a*(7*B - 2*C)* (a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (a*(35*A + 21*B + 19*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)
3.5.93.3.1 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_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e _.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S ymbol] :> Simp[(-C)*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*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Time = 0.94 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.91
| method | result | size |
| default | \(\frac {2 a \left (175 A \cos \left (d x +c \right )^{3}+126 B \cos \left (d x +c \right )^{3}+104 C \cos \left (d x +c \right )^{3}+35 A \cos \left (d x +c \right )^{2}+63 B \cos \left (d x +c \right )^{2}+52 C \cos \left (d x +c \right )^{2}+21 B \cos \left (d x +c \right )+39 C \cos \left (d x +c \right )+15 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{105 d \left (\cos \left (d x +c \right )+1\right )}\) | \(131\) |
| parts | \(\frac {2 A a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (5 \sin \left (d x +c \right )+\tan \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 B a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (6 \sin \left (d x +c \right )+3 \tan \left (d x +c \right )+\sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{5 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C a \left (104 \cos \left (d x +c \right )^{3}+52 \cos \left (d x +c \right )^{2}+39 \cos \left (d x +c \right )+15\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{105 d \left (\cos \left (d x +c \right )+1\right )}\) | \(178\) |
int(sec(d*x+c)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,me thod=_RETURNVERBOSE)
2/105*a/d*(175*A*cos(d*x+c)^3+126*B*cos(d*x+c)^3+104*C*cos(d*x+c)^3+35*A*c os(d*x+c)^2+63*B*cos(d*x+c)^2+52*C*cos(d*x+c)^2+21*B*cos(d*x+c)+39*C*cos(d *x+c)+15*C)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*tan(d*x+c)*sec(d*x+c)^ 2
Time = 0.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.78 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left ({\left (175 \, A + 126 \, B + 104 \, C\right )} a \cos \left (d x + c\right )^{3} + {\left (35 \, A + 63 \, B + 52 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, B + 13 \, C\right )} a \cos \left (d x + c\right ) + 15 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \]
integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="fricas")
2/105*((175*A + 126*B + 104*C)*a*cos(d*x + c)^3 + (35*A + 63*B + 52*C)*a*c os(d*x + c)^2 + 3*(7*B + 13*C)*a*cos(d*x + c) + 15*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)
\[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \]
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), x)
\[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right ) \,d x } \]
integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="maxima")
2/105*(105*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*(((3*A + 2*B)*a*d*cos(2*d*x + 2*c)^2 + (3*A + 2*B)*a*d*sin(2*d*x + 2*c)^2 + 2*(3*A + 2*B)*a*d*cos(2*d*x + 2*c) + (3*A + 2*B)*a*d)*integrat e((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4) *(((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(7/2*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(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(3/2*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*c os(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(8*d*x + 8*c)*sin(2*d*x + 2*c) - 3*c os(6*d*x + 6*c)*sin(2*d*x + 2*c) - 3*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*co s(7/2*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)*c os(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...
\[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right ) \,d x } \]
integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="giac")
Time = 22.45 (sec) , antiderivative size = 585, normalized size of antiderivative = 4.06 \[ \int \sec (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 (\frac {A\,a\,2{}\mathrm {i}}{d}-\frac {a\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (175\,A+126\,B+104\,C\right )\,2{}\mathrm {i}}{105\,d}\right )}{{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1}+\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 )\,2{}\mathrm {i}}{7\,d}-\frac {A\,a\,2{}\mathrm {i}}{7\,d}+\frac {a\,\left (5\,A+6\,B+4\,C\right )\,2{}\mathrm {i}}{7\,d}-\frac {a\,\left (7\,A+8\,B+12\,C\right )\,2{}\mathrm {i}}{7\,d}\right )+\frac {a\,\left (3\,A+2\,B\right )\,2{}\mathrm {i}}{7\,d}-\frac {A\,a\,2{}\mathrm {i}}{7\,d}+\frac {a\,\left (5\,A+6\,B+4\,C\right )\,2{}\mathrm {i}}{7\,d}-\frac {a\,\left (7\,A+8\,B+12\,C\right )\,2{}\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\,\left (3\,A+2\,B\right )\,2{}\mathrm {i}}{5\,d}+\frac {a\,\left (7\,A-8\,C\right )\,2{}\mathrm {i}}{35\,d}-\frac {a\,\left (2\,A+3\,B+6\,C\right )\,4{}\mathrm {i}}{5\,d}\right )-\frac {A\,a\,2{}\mathrm {i}}{5\,d}+\frac {a\,\left (2\,A+3\,B+2\,C\right )\,4{}\mathrm {i}}{5\,d}-\frac {a\,\left (3\,A+2\,B+8\,C\right )\,2{}\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 )\,2{}\mathrm {i}}{3\,d}-\frac {a\,\left (35\,A+28\,B+52\,C\right )\,2{}\mathrm {i}}{105\,d}\right )-\frac {A\,a\,2{}\mathrm {i}}{3\,d}+\frac {a\,\left (3\,A+6\,B+4\,C\right )\,2{}\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 )} \]
((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((A*a*2i)/d - (a*exp(c*1i + d*x*1i)*(175*A + 126*B + 104*C)*2i)/(105*d)))/(exp(c*1i + d*x*1i) + 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*(3*A + 2*B)*2i)/(7*d) - (A*a*2i)/(7*d) + (a*( 5*A + 6*B + 4*C)*2i)/(7*d) - (a*(7*A + 8*B + 12*C)*2i)/(7*d)) + (a*(3*A + 2*B)*2i)/(7*d) - (A*a*2i)/(7*d) + (a*(5*A + 6*B + 4*C)*2i)/(7*d) - (a*(7*A + 8*B + 12*C)*2i)/(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)*(e xp(c*1i + d*x*1i)*((a*(3*A + 2*B)*2i)/(5*d) + (a*(7*A - 8*C)*2i)/(35*d) - (a*(2*A + 3*B + 6*C)*4i)/(5*d)) - (A*a*2i)/(5*d) + (a*(2*A + 3*B + 2*C)*4i )/(5*d) - (a*(3*A + 2*B + 8*C)*2i)/(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*(3*A + 2*B)*2i)/(3*d) - (a*(35*A + 2 8*B + 52*C)*2i)/(105*d)) - (A*a*2i)/(3*d) + (a*(3*A + 6*B + 4*C)*2i)/(3*d) ))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1))