3.9 \(\int \tan ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=141 \[ -\frac {a^2 (4 A-5 i B) \tan ^3(c+d x)}{12 d}+\frac {a^2 (B+i A) \tan ^2(c+d x)}{d}+\frac {2 a^2 (A-i B) \tan (c+d x)}{d}+\frac {2 a^2 (B+i A) \log (\cos (c+d x))}{d}-2 a^2 x (A-i B)+\frac {i B \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d} \]

[Out]

-2*a^2*(A-I*B)*x+2*a^2*(I*A+B)*ln(cos(d*x+c))/d+2*a^2*(A-I*B)*tan(d*x+c)/d+a^2*(I*A+B)*tan(d*x+c)^2/d-1/12*a^2
*(4*A-5*I*B)*tan(d*x+c)^3/d+1/4*I*B*tan(d*x+c)^3*(a^2+I*a^2*tan(d*x+c))/d

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Rubi [A]  time = 0.25, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3594, 3592, 3528, 3525, 3475} \[ -\frac {a^2 (4 A-5 i B) \tan ^3(c+d x)}{12 d}+\frac {a^2 (B+i A) \tan ^2(c+d x)}{d}+\frac {2 a^2 (A-i B) \tan (c+d x)}{d}+\frac {2 a^2 (B+i A) \log (\cos (c+d x))}{d}-2 a^2 x (A-i B)+\frac {i B \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

-2*a^2*(A - I*B)*x + (2*a^2*(I*A + B)*Log[Cos[c + d*x]])/d + (2*a^2*(A - I*B)*Tan[c + d*x])/d + (a^2*(I*A + B)
*Tan[c + d*x]^2)/d - (a^2*(4*A - (5*I)*B)*Tan[c + d*x]^3)/(12*d) + ((I/4)*B*Tan[c + d*x]^3*(a^2 + I*a^2*Tan[c
+ d*x]))/d

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3525

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[(b*d*Tan[e + f*x])/f, x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3592

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(B*d*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e
+ f*x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] &&  !LeQ[m, -1]

Rule 3594

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*f
*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] &&  !LtQ[n, -1]

Rubi steps

\begin {align*} \int \tan ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx &=\frac {i B \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac {1}{4} \int \tan ^2(c+d x) (a+i a \tan (c+d x)) (a (4 A-3 i B)+a (4 i A+5 B) \tan (c+d x)) \, dx\\ &=-\frac {a^2 (4 A-5 i B) \tan ^3(c+d x)}{12 d}+\frac {i B \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac {1}{4} \int \tan ^2(c+d x) \left (8 a^2 (A-i B)+8 a^2 (i A+B) \tan (c+d x)\right ) \, dx\\ &=\frac {a^2 (i A+B) \tan ^2(c+d x)}{d}-\frac {a^2 (4 A-5 i B) \tan ^3(c+d x)}{12 d}+\frac {i B \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac {1}{4} \int \tan (c+d x) \left (-8 a^2 (i A+B)+8 a^2 (A-i B) \tan (c+d x)\right ) \, dx\\ &=-2 a^2 (A-i B) x+\frac {2 a^2 (A-i B) \tan (c+d x)}{d}+\frac {a^2 (i A+B) \tan ^2(c+d x)}{d}-\frac {a^2 (4 A-5 i B) \tan ^3(c+d x)}{12 d}+\frac {i B \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}-\left (2 a^2 (i A+B)\right ) \int \tan (c+d x) \, dx\\ &=-2 a^2 (A-i B) x+\frac {2 a^2 (i A+B) \log (\cos (c+d x))}{d}+\frac {2 a^2 (A-i B) \tan (c+d x)}{d}+\frac {a^2 (i A+B) \tan ^2(c+d x)}{d}-\frac {a^2 (4 A-5 i B) \tan ^3(c+d x)}{12 d}+\frac {i B \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}\\ \end {align*}

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Mathematica [B]  time = 6.49, size = 305, normalized size = 2.16 \[ \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \left (-4 d x (A-i B) (\cos (2 c)-i \sin (2 c)) \cos ^3(c+d x)+2 (A-i B) (\cos (2 c)-i \sin (2 c)) \cos ^3(c+d x) \tan ^{-1}(\tan (3 c+d x))+\frac {1}{3} (7 A-8 i B) \sec (c) (\cos (2 c)-i \sin (2 c)) \sin (d x) \cos ^2(c+d x)+(B+i A) (\cos (2 c)-i \sin (2 c)) \cos ^3(c+d x) \log \left (\cos ^2(c+d x)\right )-\frac {1}{6} (\cos (2 c)-i \sin (2 c)) (2 (A-2 i B) \tan (c)-6 i A-9 B) \cos (c+d x)+\frac {1}{3} (A-2 i B) \cos (c) (\tan (c)+i)^2 \sin (d x)-\frac {1}{4} B (\cos (2 c)-i \sin (2 c)) \sec (c+d x)\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

((-4*(A - I*B)*d*x*Cos[c + d*x]^3*(Cos[2*c] - I*Sin[2*c]) + 2*(A - I*B)*ArcTan[Tan[3*c + d*x]]*Cos[c + d*x]^3*
(Cos[2*c] - I*Sin[2*c]) + (I*A + B)*Cos[c + d*x]^3*Log[Cos[c + d*x]^2]*(Cos[2*c] - I*Sin[2*c]) - (B*Sec[c + d*
x]*(Cos[2*c] - I*Sin[2*c]))/4 + ((7*A - (8*I)*B)*Cos[c + d*x]^2*Sec[c]*(Cos[2*c] - I*Sin[2*c])*Sin[d*x])/3 + (
(A - (2*I)*B)*Cos[c]*Sin[d*x]*(I + Tan[c])^2)/3 - (Cos[c + d*x]*(Cos[2*c] - I*Sin[2*c])*((-6*I)*A - 9*B + 2*(A
- (2*I)*B)*Tan[c]))/6)*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c +
d*x] + B*Sin[c + d*x]))

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fricas [A]  time = 0.45, size = 230, normalized size = 1.63 \[ \frac {{\left (30 i \, A + 42 \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (66 i \, A + 72 \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (50 i \, A + 58 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (14 i \, A + 16 \, B\right )} a^{2} + {\left ({\left (6 i \, A + 6 \, B\right )} a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (24 i \, A + 24 \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (36 i \, A + 36 \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (24 i \, A + 24 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (6 i \, A + 6 \, B\right )} a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/3*((30*I*A + 42*B)*a^2*e^(6*I*d*x + 6*I*c) + (66*I*A + 72*B)*a^2*e^(4*I*d*x + 4*I*c) + (50*I*A + 58*B)*a^2*e
^(2*I*d*x + 2*I*c) + (14*I*A + 16*B)*a^2 + ((6*I*A + 6*B)*a^2*e^(8*I*d*x + 8*I*c) + (24*I*A + 24*B)*a^2*e^(6*I
*d*x + 6*I*c) + (36*I*A + 36*B)*a^2*e^(4*I*d*x + 4*I*c) + (24*I*A + 24*B)*a^2*e^(2*I*d*x + 2*I*c) + (6*I*A + 6
*B)*a^2)*log(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I
*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)

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giac [B]  time = 0.84, size = 408, normalized size = 2.89 \[ \frac {6 i \, A a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 6 \, B a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 24 i \, A a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 24 \, B a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 36 i \, A a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 36 \, B a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 24 i \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 24 \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 30 i \, A a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 42 \, B a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 66 i \, A a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 72 \, B a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 50 i \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 58 \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i \, A a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 6 \, B a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 14 i \, A a^{2} + 16 \, B a^{2}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/3*(6*I*A*a^2*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 6*B*a^2*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x +
2*I*c) + 1) + 24*I*A*a^2*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 24*B*a^2*e^(6*I*d*x + 6*I*c)*log(
e^(2*I*d*x + 2*I*c) + 1) + 36*I*A*a^2*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 36*B*a^2*e^(4*I*d*x +
4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 24*I*A*a^2*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 24*B*a^2*
e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 30*I*A*a^2*e^(6*I*d*x + 6*I*c) + 42*B*a^2*e^(6*I*d*x + 6*I*
c) + 66*I*A*a^2*e^(4*I*d*x + 4*I*c) + 72*B*a^2*e^(4*I*d*x + 4*I*c) + 50*I*A*a^2*e^(2*I*d*x + 2*I*c) + 58*B*a^2
*e^(2*I*d*x + 2*I*c) + 6*I*A*a^2*log(e^(2*I*d*x + 2*I*c) + 1) + 6*B*a^2*log(e^(2*I*d*x + 2*I*c) + 1) + 14*I*A*
a^2 + 16*B*a^2)/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x +
2*I*c) + d)

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maple [A]  time = 0.03, size = 193, normalized size = 1.37 \[ \frac {2 i a^{2} B \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {a^{2} B \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}+\frac {i a^{2} A \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {a^{2} A \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {2 i a^{2} B \tan \left (d x +c \right )}{d}+\frac {a^{2} B \left (\tan ^{2}\left (d x +c \right )\right )}{d}+\frac {2 a^{2} A \tan \left (d x +c \right )}{d}-\frac {i a^{2} A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {a^{2} B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}+\frac {2 i a^{2} B \arctan \left (\tan \left (d x +c \right )\right )}{d}-\frac {2 a^{2} A \arctan \left (\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x)

[Out]

2/3*I/d*a^2*B*tan(d*x+c)^3-1/4/d*a^2*B*tan(d*x+c)^4+I/d*a^2*A*tan(d*x+c)^2-1/3/d*a^2*A*tan(d*x+c)^3-2*I/d*a^2*
B*tan(d*x+c)+1/d*a^2*B*tan(d*x+c)^2+2*a^2*A*tan(d*x+c)/d-I/d*a^2*A*ln(1+tan(d*x+c)^2)-1/d*a^2*B*ln(1+tan(d*x+c
)^2)+2*I/d*a^2*B*arctan(tan(d*x+c))-2/d*a^2*A*arctan(tan(d*x+c))

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maxima [A]  time = 0.57, size = 112, normalized size = 0.79 \[ -\frac {3 \, B a^{2} \tan \left (d x + c\right )^{4} + 4 \, {\left (A - 2 i \, B\right )} a^{2} \tan \left (d x + c\right )^{3} - {\left (12 i \, A + 12 \, B\right )} a^{2} \tan \left (d x + c\right )^{2} + 24 \, {\left (d x + c\right )} {\left (A - i \, B\right )} a^{2} - 12 \, {\left (-i \, A - B\right )} a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 24 \, {\left (A - i \, B\right )} a^{2} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*(3*B*a^2*tan(d*x + c)^4 + 4*(A - 2*I*B)*a^2*tan(d*x + c)^3 - (12*I*A + 12*B)*a^2*tan(d*x + c)^2 + 24*(d*
x + c)*(A - I*B)*a^2 - 12*(-I*A - B)*a^2*log(tan(d*x + c)^2 + 1) - 24*(A - I*B)*a^2*tan(d*x + c))/d

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mupad [B]  time = 6.10, size = 153, normalized size = 1.09 \[ \frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {a^2\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{3}+\frac {B\,a^2\,1{}\mathrm {i}}{3}\right )}{d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (-A\,a^2+a^2\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}+B\,a^2\,1{}\mathrm {i}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {a^2\,\left (B+A\,1{}\mathrm {i}\right )}{2}+\frac {B\,a^2}{2}+\frac {A\,a^2\,1{}\mathrm {i}}{2}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (2\,B\,a^2+A\,a^2\,2{}\mathrm {i}\right )}{d}-\frac {B\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^2*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^2,x)

[Out]

(tan(c + d*x)^3*((a^2*(A*1i + B)*1i)/3 + (B*a^2*1i)/3))/d - (tan(c + d*x)*(a^2*(A*1i + B)*1i - A*a^2 + B*a^2*1
i))/d + (tan(c + d*x)^2*((A*a^2*1i)/2 + (a^2*(A*1i + B))/2 + (B*a^2)/2))/d - (log(tan(c + d*x) + 1i)*(A*a^2*2i
+ 2*B*a^2))/d - (B*a^2*tan(c + d*x)^4)/(4*d)

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sympy [A]  time = 1.26, size = 243, normalized size = 1.72 \[ \frac {2 i a^{2} \left (A - i B\right ) \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {- 14 i A a^{2} - 16 B a^{2} + \left (- 50 i A a^{2} e^{2 i c} - 58 B a^{2} e^{2 i c}\right ) e^{2 i d x} + \left (- 66 i A a^{2} e^{4 i c} - 72 B a^{2} e^{4 i c}\right ) e^{4 i d x} + \left (- 30 i A a^{2} e^{6 i c} - 42 B a^{2} e^{6 i c}\right ) e^{6 i d x}}{- 3 d e^{8 i c} e^{8 i d x} - 12 d e^{6 i c} e^{6 i d x} - 18 d e^{4 i c} e^{4 i d x} - 12 d e^{2 i c} e^{2 i d x} - 3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**2*(a+I*a*tan(d*x+c))**2*(A+B*tan(d*x+c)),x)

[Out]

2*I*a**2*(A - I*B)*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-14*I*A*a**2 - 16*B*a**2 + (-50*I*A*a**2*exp(2*I*c) -
58*B*a**2*exp(2*I*c))*exp(2*I*d*x) + (-66*I*A*a**2*exp(4*I*c) - 72*B*a**2*exp(4*I*c))*exp(4*I*d*x) + (-30*I*A*
a**2*exp(6*I*c) - 42*B*a**2*exp(6*I*c))*exp(6*I*d*x))/(-3*d*exp(8*I*c)*exp(8*I*d*x) - 12*d*exp(6*I*c)*exp(6*I*
d*x) - 18*d*exp(4*I*c)*exp(4*I*d*x) - 12*d*exp(2*I*c)*exp(2*I*d*x) - 3*d)

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