\(\int \cos ^3(c+d x) (a+i a \tan (c+d x))^5 \, dx\) [72]

Optimal result

Integrand size = 24, antiderivative size = 98 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {5 a^5 \text {arctanh}(\sin (c+d x))}{d}+\frac {5 i a^5 \sec (c+d x)}{d}+\frac {10 i a^3 \cos (c+d x) (a+i a \tan (c+d x))^2}{3 d}-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d} \]

[Out]

5*a^5*arctanh(sin(d*x+c))/d+5*I*a^5*sec(d*x+c)/d+10/3*I*a^3*cos(d*x+c)*(a+I*a*tan(d*x+c))^2/d-2/3*I*a*cos(d*x+
c)^3*(a+I*a*tan(d*x+c))^4/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3577, 3567, 3855} \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {5 a^5 \text {arctanh}(\sin (c+d x))}{d}+\frac {5 i a^5 \sec (c+d x)}{d}+\frac {10 i a^3 \cos (c+d x) (a+i a \tan (c+d x))^2}{3 d}-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d} \]

[In]

Int[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^5,x]

[Out]

(5*a^5*ArcTanh[Sin[c + d*x]])/d + ((5*I)*a^5*Sec[c + d*x])/d + (((10*I)/3)*a^3*Cos[c + d*x]*(a + I*a*Tan[c + d
*x])^2)/d - (((2*I)/3)*a*Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^4)/d

Rule 3567

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 3577

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 3855

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d}-\frac {1}{3} \left (5 a^2\right ) \int \cos (c+d x) (a+i a \tan (c+d x))^3 \, dx \\ & = \frac {10 i a^3 \cos (c+d x) (a+i a \tan (c+d x))^2}{3 d}-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d}+\left (5 a^4\right ) \int \sec (c+d x) (a+i a \tan (c+d x)) \, dx \\ & = \frac {5 i a^5 \sec (c+d x)}{d}+\frac {10 i a^3 \cos (c+d x) (a+i a \tan (c+d x))^2}{3 d}-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d}+\left (5 a^5\right ) \int \sec (c+d x) \, dx \\ & = \frac {5 a^5 \text {arctanh}(\sin (c+d x))}{d}+\frac {5 i a^5 \sec (c+d x)}{d}+\frac {10 i a^3 \cos (c+d x) (a+i a \tan (c+d x))^2}{3 d}-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.97 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.33 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {a^5 \cos ^4(c+d x) \left (30 \text {arctanh}\left (\sin (c)+\cos (c) \tan \left (\frac {d x}{2}\right )\right ) \cos (c+d x) (i \cos (5 c)+\sin (5 c))-(\cos (3 c-2 d x)-i \sin (3 c-2 d x)) (10+13 \cos (2 (c+d x))-17 i \sin (2 (c+d x)))\right ) (-i+\tan (c+d x))^5}{3 d (\cos (d x)+i \sin (d x))^5} \]

[In]

Integrate[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^5,x]

[Out]

(a^5*Cos[c + d*x]^4*(30*ArcTanh[Sin[c] + Cos[c]*Tan[(d*x)/2]]*Cos[c + d*x]*(I*Cos[5*c] + Sin[5*c]) - (Cos[3*c
- 2*d*x] - I*Sin[3*c - 2*d*x])*(10 + 13*Cos[2*(c + d*x)] - (17*I)*Sin[2*(c + d*x)]))*(-I + Tan[c + d*x])^5)/(3
*d*(Cos[d*x] + I*Sin[d*x])^5)

Maple [A] (verified)

Time = 26.35 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.13

method	result	size
risch	\(-\frac {4 i a^{5} {\mathrm e}^{3 i \left (d x +c \right )}}{3 d}+\frac {8 i a^{5} {\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {2 i a^{5} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {5 a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {5 a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\)	\(111\)
derivativedivides	\(\frac {i a^{5} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+5 a^{5} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+\frac {10 i a^{5} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}-\frac {10 a^{5} \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {5 i a^{5} \left (\cos ^{3}\left (d x +c \right )\right )}{3}+\frac {a^{5} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(165\)
default	\(\frac {i a^{5} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+5 a^{5} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+\frac {10 i a^{5} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}-\frac {10 a^{5} \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {5 i a^{5} \left (\cos ^{3}\left (d x +c \right )\right )}{3}+\frac {a^{5} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(165\)

[In]

int(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^5,x,method=_RETURNVERBOSE)

[Out]

-4/3*I/d*a^5*exp(3*I*(d*x+c))+8*I/d*a^5*exp(I*(d*x+c))+2*I*a^5*exp(I*(d*x+c))/d/(exp(2*I*(d*x+c))+1)+5/d*a^5*l
n(exp(I*(d*x+c))+I)-5/d*a^5*ln(exp(I*(d*x+c))-I)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.24 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {-4 i \, a^{5} e^{\left (5 i \, d x + 5 i \, c\right )} + 20 i \, a^{5} e^{\left (3 i \, d x + 3 i \, c\right )} + 30 i \, a^{5} e^{\left (i \, d x + i \, c\right )} + 15 \, {\left (a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{5}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 15 \, {\left (a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{5}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^5,x, algorithm="fricas")

[Out]

1/3*(-4*I*a^5*e^(5*I*d*x + 5*I*c) + 20*I*a^5*e^(3*I*d*x + 3*I*c) + 30*I*a^5*e^(I*d*x + I*c) + 15*(a^5*e^(2*I*d
*x + 2*I*c) + a^5)*log(e^(I*d*x + I*c) + I) - 15*(a^5*e^(2*I*d*x + 2*I*c) + a^5)*log(e^(I*d*x + I*c) - I))/(d*
e^(2*I*d*x + 2*I*c) + d)

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.51 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {2 i a^{5} e^{i c} e^{i d x}}{d e^{2 i c} e^{2 i d x} + d} + \frac {5 a^{5} \left (- \log {\left (e^{i d x} - i e^{- i c} \right )} + \log {\left (e^{i d x} + i e^{- i c} \right )}\right )}{d} + \begin {cases} \frac {- 4 i a^{5} d e^{3 i c} e^{3 i d x} + 24 i a^{5} d e^{i c} e^{i d x}}{3 d^{2}} & \text {for}\: d^{2} \neq 0 \\x \left (4 a^{5} e^{3 i c} - 8 a^{5} e^{i c}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**3*(a+I*a*tan(d*x+c))**5,x)

[Out]

2*I*a**5*exp(I*c)*exp(I*d*x)/(d*exp(2*I*c)*exp(2*I*d*x) + d) + 5*a**5*(-log(exp(I*d*x) - I*exp(-I*c)) + log(ex
p(I*d*x) + I*exp(-I*c)))/d + Piecewise(((-4*I*a**5*d*exp(3*I*c)*exp(3*I*d*x) + 24*I*a**5*d*exp(I*c)*exp(I*d*x)
)/(3*d**2), Ne(d**2, 0)), (x*(4*a**5*exp(3*I*c) - 8*a**5*exp(I*c)), True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.57 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {10 i \, a^{5} \cos \left (d x + c\right )^{3} + 20 \, a^{5} \sin \left (d x + c\right )^{3} + 2 i \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a^{5} + 20 i \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{5} + 5 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a^{5} + 2 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{5}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^5,x, algorithm="maxima")

[Out]

-1/6*(10*I*a^5*cos(d*x + c)^3 + 20*a^5*sin(d*x + c)^3 + 2*I*(cos(d*x + c)^3 - 3/cos(d*x + c) - 6*cos(d*x + c))
*a^5 + 20*I*(cos(d*x + c)^3 - 3*cos(d*x + c))*a^5 + 5*(2*sin(d*x + c)^3 - 3*log(sin(d*x + c) + 1) + 3*log(sin(
d*x + c) - 1) + 6*sin(d*x + c))*a^5 + 2*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^5)/d

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1683 vs. \(2 (84) = 168\).

Time = 1.18 (sec) , antiderivative size = 1683, normalized size of antiderivative = 17.17 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^5 \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^5,x, algorithm="giac")

[Out]

-1/6144*(39225*a^5*e^(16*I*d*x + 8*I*c)*log(I*e^(I*d*x + I*c) + 1) + 313800*a^5*e^(14*I*d*x + 6*I*c)*log(I*e^(
I*d*x + I*c) + 1) + 1098300*a^5*e^(12*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 2196600*a^5*e^(10*I*d*x + 2*
I*c)*log(I*e^(I*d*x + I*c) + 1) + 2196600*a^5*e^(6*I*d*x - 2*I*c)*log(I*e^(I*d*x + I*c) + 1) + 1098300*a^5*e^(
4*I*d*x - 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 313800*a^5*e^(2*I*d*x - 6*I*c)*log(I*e^(I*d*x + I*c) + 1) + 2745
750*a^5*e^(8*I*d*x)*log(I*e^(I*d*x + I*c) + 1) + 39225*a^5*e^(-8*I*c)*log(I*e^(I*d*x + I*c) + 1) + 8520*a^5*e^
(16*I*d*x + 8*I*c)*log(I*e^(I*d*x + I*c) - 1) + 68160*a^5*e^(14*I*d*x + 6*I*c)*log(I*e^(I*d*x + I*c) - 1) + 23
8560*a^5*e^(12*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) - 1) + 477120*a^5*e^(10*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*
c) - 1) + 477120*a^5*e^(6*I*d*x - 2*I*c)*log(I*e^(I*d*x + I*c) - 1) + 238560*a^5*e^(4*I*d*x - 4*I*c)*log(I*e^(
I*d*x + I*c) - 1) + 68160*a^5*e^(2*I*d*x - 6*I*c)*log(I*e^(I*d*x + I*c) - 1) + 596400*a^5*e^(8*I*d*x)*log(I*e^
(I*d*x + I*c) - 1) + 8520*a^5*e^(-8*I*c)*log(I*e^(I*d*x + I*c) - 1) - 39225*a^5*e^(16*I*d*x + 8*I*c)*log(-I*e^
(I*d*x + I*c) + 1) - 313800*a^5*e^(14*I*d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 1098300*a^5*e^(12*I*d*x + 4
*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 2196600*a^5*e^(10*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 2196600*a^5
*e^(6*I*d*x - 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 1098300*a^5*e^(4*I*d*x - 4*I*c)*log(-I*e^(I*d*x + I*c) + 1)
- 313800*a^5*e^(2*I*d*x - 6*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 2745750*a^5*e^(8*I*d*x)*log(-I*e^(I*d*x + I*c)
+ 1) - 39225*a^5*e^(-8*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 8520*a^5*e^(16*I*d*x + 8*I*c)*log(-I*e^(I*d*x + I*c
) - 1) - 68160*a^5*e^(14*I*d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 238560*a^5*e^(12*I*d*x + 4*I*c)*log(-I*e
^(I*d*x + I*c) - 1) - 477120*a^5*e^(10*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 477120*a^5*e^(6*I*d*x - 2*
I*c)*log(-I*e^(I*d*x + I*c) - 1) - 238560*a^5*e^(4*I*d*x - 4*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 68160*a^5*e^(2
*I*d*x - 6*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 596400*a^5*e^(8*I*d*x)*log(-I*e^(I*d*x + I*c) - 1) - 8520*a^5*e^
(-8*I*c)*log(-I*e^(I*d*x + I*c) - 1) + 15*a^5*e^(16*I*d*x + 8*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 120*a^5*e^(14
*I*d*x + 6*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 420*a^5*e^(12*I*d*x + 4*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 840*a
^5*e^(10*I*d*x + 2*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 840*a^5*e^(6*I*d*x - 2*I*c)*log(I*e^(I*d*x) + e^(-I*c))
+ 420*a^5*e^(4*I*d*x - 4*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 120*a^5*e^(2*I*d*x - 6*I*c)*log(I*e^(I*d*x) + e^(-
I*c)) + 1050*a^5*e^(8*I*d*x)*log(I*e^(I*d*x) + e^(-I*c)) + 15*a^5*e^(-8*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 15*
a^5*e^(16*I*d*x + 8*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 120*a^5*e^(14*I*d*x + 6*I*c)*log(-I*e^(I*d*x) + e^(-I*
c)) - 420*a^5*e^(12*I*d*x + 4*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 840*a^5*e^(10*I*d*x + 2*I*c)*log(-I*e^(I*d*x
) + e^(-I*c)) - 840*a^5*e^(6*I*d*x - 2*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 420*a^5*e^(4*I*d*x - 4*I*c)*log(-I*
e^(I*d*x) + e^(-I*c)) - 120*a^5*e^(2*I*d*x - 6*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 1050*a^5*e^(8*I*d*x)*log(-I
*e^(I*d*x) + e^(-I*c)) - 15*a^5*e^(-8*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 8192*I*a^5*e^(19*I*d*x + 11*I*c) + 1
6384*I*a^5*e^(17*I*d*x + 9*I*c) - 176128*I*a^5*e^(15*I*d*x + 7*I*c) - 1003520*I*a^5*e^(13*I*d*x + 5*I*c) - 243
7120*I*a^5*e^(11*I*d*x + 3*I*c) - 3411968*I*a^5*e^(9*I*d*x + I*c) - 2953216*I*a^5*e^(7*I*d*x - I*c) - 1568768*
I*a^5*e^(5*I*d*x - 3*I*c) - 471040*I*a^5*e^(3*I*d*x - 5*I*c) - 61440*I*a^5*e^(I*d*x - 7*I*c))/(d*e^(16*I*d*x +
8*I*c) + 8*d*e^(14*I*d*x + 6*I*c) + 28*d*e^(12*I*d*x + 4*I*c) + 56*d*e^(10*I*d*x + 2*I*c) + 56*d*e^(6*I*d*x -
2*I*c) + 28*d*e^(4*I*d*x - 4*I*c) + 8*d*e^(2*I*d*x - 6*I*c) + 70*d*e^(8*I*d*x) + d*e^(-8*I*c))

Mupad [B] (verification not implemented)

Time = 6.14 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.65 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {10\,a^5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {8\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,34{}\mathrm {i}-\frac {82\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,38{}\mathrm {i}+\frac {46\,a^5}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,3{}\mathrm {i}-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,4{}\mathrm {i}+3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )} \]

[In]

int(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^5,x)

[Out]

(10*a^5*atanh(tan(c/2 + (d*x)/2)))/d - (a^5*tan(c/2 + (d*x)/2)^3*34i - (82*a^5*tan(c/2 + (d*x)/2)^2)/3 + 8*a^5
*tan(c/2 + (d*x)/2)^4 + (46*a^5)/3 - a^5*tan(c/2 + (d*x)/2)*38i)/(d*(3*tan(c/2 + (d*x)/2) - tan(c/2 + (d*x)/2)
^2*4i - 4*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d*x)/2)^4*3i + tan(c/2 + (d*x)/2)^5 + 1i))