3.71 \(\int \cos (c+d x) (a+i a \tan (c+d x))^5 \, dx\)

Optimal. Leaf size=130 \[ -\frac {35 i a^5 \sec (c+d x)}{2 d}-\frac {35 a^5 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {35 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{6 d}-\frac {7 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^2}{3 d}-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^4}{d} \]

[Out]

-35/2*a^5*arctanh(sin(d*x+c))/d-35/2*I*a^5*sec(d*x+c)/d-7/3*I*a^3*sec(d*x+c)*(a+I*a*tan(d*x+c))^2/d-2*I*a*cos(
d*x+c)*(a+I*a*tan(d*x+c))^4/d-35/6*I*sec(d*x+c)*(a^5+I*a^5*tan(d*x+c))/d

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Rubi [A]  time = 0.10, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3496, 3498, 3486, 3770} \[ -\frac {35 i a^5 \sec (c+d x)}{2 d}-\frac {35 a^5 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {7 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^2}{3 d}-\frac {35 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{6 d}-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^4}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*(a + I*a*Tan[c + d*x])^5,x]

[Out]

(-35*a^5*ArcTanh[Sin[c + d*x]])/(2*d) - (((35*I)/2)*a^5*Sec[c + d*x])/d - (((7*I)/3)*a^3*Sec[c + d*x]*(a + I*a
*Tan[c + d*x])^2)/d - ((2*I)*a*Cos[c + d*x]*(a + I*a*Tan[c + d*x])^4)/d - (((35*I)/6)*Sec[c + d*x]*(a^5 + I*a^
5*Tan[c + d*x]))/d

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3496

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(m + 2*n - 2))/(d^2*m), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rule 3498

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

Rule 3770

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

Rubi steps

\begin {align*} \int \cos (c+d x) (a+i a \tan (c+d x))^5 \, dx &=-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^4}{d}-\left (7 a^2\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^3 \, dx\\ &=-\frac {7 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^2}{3 d}-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^4}{d}-\frac {1}{3} \left (35 a^3\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^2 \, dx\\ &=-\frac {7 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^2}{3 d}-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^4}{d}-\frac {35 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{6 d}-\frac {1}{2} \left (35 a^4\right ) \int \sec (c+d x) (a+i a \tan (c+d x)) \, dx\\ &=-\frac {35 i a^5 \sec (c+d x)}{2 d}-\frac {7 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^2}{3 d}-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^4}{d}-\frac {35 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{6 d}-\frac {1}{2} \left (35 a^5\right ) \int \sec (c+d x) \, dx\\ &=-\frac {35 a^5 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {35 i a^5 \sec (c+d x)}{2 d}-\frac {7 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^2}{3 d}-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^4}{d}-\frac {35 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{6 d}\\ \end {align*}

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Mathematica [A]  time = 1.69, size = 151, normalized size = 1.16 \[ \frac {a^5 \cos ^2(c+d x) (\tan (c+d x)-i)^5 \left ((\cos (4 c-d x)-i \sin (4 c-d x)) (-i (49 \sin (c+d x)+57 \sin (3 (c+d x)))+511 \cos (c+d x)+153 \cos (3 (c+d x)))-840 i (\cos (5 c)-i \sin (5 c)) \cos ^3(c+d x) \tanh ^{-1}\left (\cos (c) \tan \left (\frac {d x}{2}\right )+\sin (c)\right )\right )}{24 d (\cos (d x)+i \sin (d x))^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*(a + I*a*Tan[c + d*x])^5,x]

[Out]

(a^5*Cos[c + d*x]^2*((-840*I)*ArcTanh[Sin[c] + Cos[c]*Tan[(d*x)/2]]*Cos[c + d*x]^3*(Cos[5*c] - I*Sin[5*c]) + (
Cos[4*c - d*x] - I*Sin[4*c - d*x])*(511*Cos[c + d*x] + 153*Cos[3*(c + d*x)] - I*(49*Sin[c + d*x] + 57*Sin[3*(c
+ d*x)])))*(-I + Tan[c + d*x])^5)/(24*d*(Cos[d*x] + I*Sin[d*x])^5)

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fricas [A]  time = 0.63, size = 216, normalized size = 1.66 \[ \frac {-96 i \, a^{5} e^{\left (7 i \, d x + 7 i \, c\right )} - 462 i \, a^{5} e^{\left (5 i \, d x + 5 i \, c\right )} - 560 i \, a^{5} e^{\left (3 i \, d x + 3 i \, c\right )} - 210 i \, a^{5} e^{\left (i \, d x + i \, c\right )} - 105 \, {\left (a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{5}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) + 105 \, {\left (a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{5}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{6 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+I*a*tan(d*x+c))^5,x, algorithm="fricas")

[Out]

1/6*(-96*I*a^5*e^(7*I*d*x + 7*I*c) - 462*I*a^5*e^(5*I*d*x + 5*I*c) - 560*I*a^5*e^(3*I*d*x + 3*I*c) - 210*I*a^5
*e^(I*d*x + I*c) - 105*(a^5*e^(6*I*d*x + 6*I*c) + 3*a^5*e^(4*I*d*x + 4*I*c) + 3*a^5*e^(2*I*d*x + 2*I*c) + a^5)
*log(e^(I*d*x + I*c) + I) + 105*(a^5*e^(6*I*d*x + 6*I*c) + 3*a^5*e^(4*I*d*x + 4*I*c) + 3*a^5*e^(2*I*d*x + 2*I*
c) + a^5)*log(e^(I*d*x + I*c) - I))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c)
+ d)

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giac [B]  time = 3.49, size = 510, normalized size = 3.92 \[ \frac {8295 \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 24885 \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 24885 \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 18585 \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 55755 \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 55755 \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 8295 \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 24885 \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 24885 \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 18585 \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) + 55755 \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) + 55755 \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 24576 i \, a^{5} e^{\left (7 i \, d x + 7 i \, c\right )} - 118272 i \, a^{5} e^{\left (5 i \, d x + 5 i \, c\right )} - 143360 i \, a^{5} e^{\left (3 i \, d x + 3 i \, c\right )} - 53760 i \, a^{5} e^{\left (i \, d x + i \, c\right )} + 8295 \, a^{5} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 18585 \, a^{5} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 8295 \, a^{5} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 18585 \, a^{5} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right )}{1536 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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(cos(d*x+c)*(a+I*a*tan(d*x+c))^5,x, algorithm="giac")

[Out]

1/1536*(8295*a^5*e^(6*I*d*x + 6*I*c)*log(I*e^(I*d*x + I*c) + 1) + 24885*a^5*e^(4*I*d*x + 4*I*c)*log(I*e^(I*d*x
+ I*c) + 1) + 24885*a^5*e^(2*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c) + 1) - 18585*a^5*e^(6*I*d*x + 6*I*c)*log(I*
e^(I*d*x + I*c) - 1) - 55755*a^5*e^(4*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) - 1) - 55755*a^5*e^(2*I*d*x + 2*I*c
)*log(I*e^(I*d*x + I*c) - 1) - 8295*a^5*e^(6*I*d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 24885*a^5*e^(4*I*d*x
+ 4*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 24885*a^5*e^(2*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) + 18585*a^5*
e^(6*I*d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) - 1) + 55755*a^5*e^(4*I*d*x + 4*I*c)*log(-I*e^(I*d*x + I*c) - 1) +
55755*a^5*e^(2*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 24576*I*a^5*e^(7*I*d*x + 7*I*c) - 118272*I*a^5*e^(
5*I*d*x + 5*I*c) - 143360*I*a^5*e^(3*I*d*x + 3*I*c) - 53760*I*a^5*e^(I*d*x + I*c) + 8295*a^5*log(I*e^(I*d*x +
I*c) + 1) - 18585*a^5*log(I*e^(I*d*x + I*c) - 1) - 8295*a^5*log(-I*e^(I*d*x + I*c) + 1) + 18585*a^5*log(-I*e^(
I*d*x + I*c) - 1))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)

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maple [A]  time = 0.48, size = 214, normalized size = 1.65 \[ \frac {i a^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{3}}-\frac {34 i a^{5} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{3 d}-\frac {i a^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}-\frac {i a^{5} \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{d}-\frac {83 i a^{5} \cos \left (d x +c \right )}{3 d}+\frac {5 a^{5} \left (\sin ^{5}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {5 a^{5} \left (\sin ^{3}\left (d x +c \right )\right )}{2 d}+\frac {37 a^{5} \sin \left (d x +c \right )}{2 d}-\frac {35 a^{5} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}-\frac {10 i a^{5} \left (\sin ^{4}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a+I*a*tan(d*x+c))^5,x)

[Out]

1/3*I/d*a^5*sin(d*x+c)^6/cos(d*x+c)^3-34/3*I/d*a^5*cos(d*x+c)*sin(d*x+c)^2-I/d*a^5*sin(d*x+c)^6/cos(d*x+c)-I/d
*a^5*cos(d*x+c)*sin(d*x+c)^4-83/3*I/d*a^5*cos(d*x+c)+5/2/d*a^5*sin(d*x+c)^5/cos(d*x+c)^2+5/2*a^5*sin(d*x+c)^3/
d+37/2*a^5*sin(d*x+c)/d-35/2/d*a^5*ln(sec(d*x+c)+tan(d*x+c))-10*I/d*a^5*sin(d*x+c)^4/cos(d*x+c)

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maxima [A]  time = 0.64, size = 173, normalized size = 1.33 \[ -\frac {15 \, a^{5} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) - 4 \, \sin \left (d x + c\right )\right )} + 120 i \, a^{5} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} + 4 i \, a^{5} {\left (\frac {6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )} + 60 \, a^{5} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )} + 60 i \, a^{5} \cos \left (d x + c\right ) - 12 \, a^{5} \sin \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+I*a*tan(d*x+c))^5,x, algorithm="maxima")

[Out]

-1/12*(15*a^5*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + 3*log(sin(d*x + c) + 1) - 3*log(sin(d*x + c) - 1) - 4*sin
(d*x + c)) + 120*I*a^5*(1/cos(d*x + c) + cos(d*x + c)) + 4*I*a^5*((6*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 + 3*co
s(d*x + c)) + 60*a^5*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1) - 2*sin(d*x + c)) + 60*I*a^5*cos(d*x + c)
- 12*a^5*sin(d*x + c))/d

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mupad [B]  time = 7.14, size = 222, normalized size = 1.71 \[ -\frac {35\,a^5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {37\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,27{}\mathrm {i}-118\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,48{}\mathrm {i}+139\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,55{}\mathrm {i}}{3}-\frac {166\,a^5}{3}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,1{}\mathrm {i}+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,3{}\mathrm {i}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,3{}\mathrm {i}+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)*(a + a*tan(c + d*x)*1i)^5,x)

[Out]

- (35*a^5*atanh(tan(c/2 + (d*x)/2)))/d - (139*a^5*tan(c/2 + (d*x)/2)^2 - a^5*tan(c/2 + (d*x)/2)^3*48i - 118*a^
5*tan(c/2 + (d*x)/2)^4 + a^5*tan(c/2 + (d*x)/2)^5*27i + 37*a^5*tan(c/2 + (d*x)/2)^6 - (166*a^5)/3 + (a^5*tan(c
/2 + (d*x)/2)*55i)/3)/(d*(tan(c/2 + (d*x)/2) - tan(c/2 + (d*x)/2)^2*3i - 3*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d
*x)/2)^4*3i + 3*tan(c/2 + (d*x)/2)^5 - tan(c/2 + (d*x)/2)^6*1i - tan(c/2 + (d*x)/2)^7 + 1i))

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sympy [A]  time = 0.54, size = 201, normalized size = 1.55 \[ \frac {35 a^{5} \left (\frac {\log {\left (e^{i d x} - i e^{- i c} \right )}}{2} - \frac {\log {\left (e^{i d x} + i e^{- i c} \right )}}{2}\right )}{d} + \frac {- 87 a^{5} e^{5 i c} e^{5 i d x} - 136 a^{5} e^{3 i c} e^{3 i d x} - 57 a^{5} e^{i c} e^{i d x}}{- 3 i d e^{6 i c} e^{6 i d x} - 9 i d e^{4 i c} e^{4 i d x} - 9 i d e^{2 i c} e^{2 i d x} - 3 i d} + \begin {cases} - \frac {16 i a^{5} e^{i c} e^{i d x}}{d} & \text {for}\: d \neq 0 \\16 a^{5} x e^{i c} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+I*a*tan(d*x+c))**5,x)

[Out]

35*a**5*(log(exp(I*d*x) - I*exp(-I*c))/2 - log(exp(I*d*x) + I*exp(-I*c))/2)/d + (-87*a**5*exp(5*I*c)*exp(5*I*d
*x) - 136*a**5*exp(3*I*c)*exp(3*I*d*x) - 57*a**5*exp(I*c)*exp(I*d*x))/(-3*I*d*exp(6*I*c)*exp(6*I*d*x) - 9*I*d*
exp(4*I*c)*exp(4*I*d*x) - 9*I*d*exp(2*I*c)*exp(2*I*d*x) - 3*I*d) + Piecewise((-16*I*a**5*exp(I*c)*exp(I*d*x)/d
, Ne(d, 0)), (16*a**5*x*exp(I*c), True))

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