\(\int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx\) [51]

Optimal result

Integrand size = 24, antiderivative size = 124 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {5 i a^3 \cos ^7(c+d x)}{63 d}+\frac {5 a^3 \sin (c+d x)}{9 d}-\frac {5 a^3 \sin ^3(c+d x)}{9 d}+\frac {a^3 \sin ^5(c+d x)}{3 d}-\frac {5 a^3 \sin ^7(c+d x)}{63 d}-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^2}{9 d} \]

[Out]

-5/63*I*a^3*cos(d*x+c)^7/d+5/9*a^3*sin(d*x+c)/d-5/9*a^3*sin(d*x+c)^3/d+1/3*a^3*sin(d*x+c)^5/d-5/63*a^3*sin(d*x
+c)^7/d-2/9*I*a*cos(d*x+c)^9*(a+I*a*tan(d*x+c))^2/d

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 124, 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, 2713} \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {5 a^3 \sin ^7(c+d x)}{63 d}+\frac {a^3 \sin ^5(c+d x)}{3 d}-\frac {5 a^3 \sin ^3(c+d x)}{9 d}+\frac {5 a^3 \sin (c+d x)}{9 d}-\frac {5 i a^3 \cos ^7(c+d x)}{63 d}-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^2}{9 d} \]

[In]

Int[Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(((-5*I)/63)*a^3*Cos[c + d*x]^7)/d + (5*a^3*Sin[c + d*x])/(9*d) - (5*a^3*Sin[c + d*x]^3)/(9*d) + (a^3*Sin[c +
d*x]^5)/(3*d) - (5*a^3*Sin[c + d*x]^7)/(63*d) - (((2*I)/9)*a*Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^2)/d

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^2}{9 d}+\frac {1}{9} \left (5 a^2\right ) \int \cos ^7(c+d x) (a+i a \tan (c+d x)) \, dx \\ & = -\frac {5 i a^3 \cos ^7(c+d x)}{63 d}-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^2}{9 d}+\frac {1}{9} \left (5 a^3\right ) \int \cos ^7(c+d x) \, dx \\ & = -\frac {5 i a^3 \cos ^7(c+d x)}{63 d}-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^2}{9 d}-\frac {\left (5 a^3\right ) \text {Subst}\left (\int \left (1-3 x^2+3 x^4-x^6\right ) \, dx,x,-\sin (c+d x)\right )}{9 d} \\ & = -\frac {5 i a^3 \cos ^7(c+d x)}{63 d}+\frac {5 a^3 \sin (c+d x)}{9 d}-\frac {5 a^3 \sin ^3(c+d x)}{9 d}+\frac {a^3 \sin ^5(c+d x)}{3 d}-\frac {5 a^3 \sin ^7(c+d x)}{63 d}-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^2}{9 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.91 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.82 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^3 (-i \cos (3 (c+d x))+\sin (3 (c+d x))) \left (210 \sqrt {\cos ^2(c+d x)}+\left (32+567 \sqrt {\cos ^2(c+d x)}\right ) \cos (2 (c+d x))+\left (32-162 \sqrt {\cos ^2(c+d x)}\right ) \cos (4 (c+d x))-7 \sqrt {\cos ^2(c+d x)} \cos (6 (c+d x))-32 i \sin (2 (c+d x))-378 i \sqrt {\cos ^2(c+d x)} \sin (2 (c+d x))-32 i \sin (4 (c+d x))+216 i \sqrt {\cos ^2(c+d x)} \sin (4 (c+d x))+14 i \sqrt {\cos ^2(c+d x)} \sin (6 (c+d x))\right )}{2016 d \sqrt {\cos ^2(c+d x)}} \]

[In]

Integrate[Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(a^3*((-I)*Cos[3*(c + d*x)] + Sin[3*(c + d*x)])*(210*Sqrt[Cos[c + d*x]^2] + (32 + 567*Sqrt[Cos[c + d*x]^2])*Co
s[2*(c + d*x)] + (32 - 162*Sqrt[Cos[c + d*x]^2])*Cos[4*(c + d*x)] - 7*Sqrt[Cos[c + d*x]^2]*Cos[6*(c + d*x)] -
(32*I)*Sin[2*(c + d*x)] - (378*I)*Sqrt[Cos[c + d*x]^2]*Sin[2*(c + d*x)] - (32*I)*Sin[4*(c + d*x)] + (216*I)*Sq
rt[Cos[c + d*x]^2]*Sin[4*(c + d*x)] + (14*I)*Sqrt[Cos[c + d*x]^2]*Sin[6*(c + d*x)]))/(2016*d*Sqrt[Cos[c + d*x]
^2])

Maple [A] (verified)

Time = 215.94 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.97

method	result	size
risch	\(-\frac {i a^{3} {\mathrm e}^{9 i \left (d x +c \right )}}{576 d}-\frac {3 i a^{3} {\mathrm e}^{7 i \left (d x +c \right )}}{224 d}-\frac {3 i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}}{64 d}-\frac {9 i a^{3} \cos \left (d x +c \right )}{64 d}+\frac {21 a^{3} \sin \left (d x +c \right )}{64 d}-\frac {19 i a^{3} \cos \left (3 d x +3 c \right )}{192 d}+\frac {7 a^{3} \sin \left (3 d x +3 c \right )}{64 d}\)	\(120\)
derivativedivides	\(\frac {-i a^{3} \left (-\frac {\left (\cos ^{7}\left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{9}-\frac {2 \left (\cos ^{7}\left (d x +c \right )\right )}{63}\right )-3 a^{3} \left (-\frac {\left (\cos ^{8}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{9}+\frac {\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{63}\right )-\frac {i a^{3} \left (\cos ^{9}\left (d x +c \right )\right )}{3}+\frac {a^{3} \left (\frac {128}{35}+\cos ^{8}\left (d x +c \right )+\frac {8 \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {48 \left (\cos ^{4}\left (d x +c \right )\right )}{35}+\frac {64 \left (\cos ^{2}\left (d x +c \right )\right )}{35}\right ) \sin \left (d x +c \right )}{9}}{d}\)	\(166\)
default	\(\frac {-i a^{3} \left (-\frac {\left (\cos ^{7}\left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{9}-\frac {2 \left (\cos ^{7}\left (d x +c \right )\right )}{63}\right )-3 a^{3} \left (-\frac {\left (\cos ^{8}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{9}+\frac {\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{63}\right )-\frac {i a^{3} \left (\cos ^{9}\left (d x +c \right )\right )}{3}+\frac {a^{3} \left (\frac {128}{35}+\cos ^{8}\left (d x +c \right )+\frac {8 \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {48 \left (\cos ^{4}\left (d x +c \right )\right )}{35}+\frac {64 \left (\cos ^{2}\left (d x +c \right )\right )}{35}\right ) \sin \left (d x +c \right )}{9}}{d}\)	\(166\)

[In]

int(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

-1/576*I/d*a^3*exp(9*I*(d*x+c))-3/224*I/d*a^3*exp(7*I*(d*x+c))-3/64*I/d*a^3*exp(5*I*(d*x+c))-9/64*I/d*a^3*cos(
d*x+c)+21/64*a^3*sin(d*x+c)/d-19/192*I/d*a^3*cos(3*d*x+3*c)+7/64/d*a^3*sin(3*d*x+3*c)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.84 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {{\left (-7 i \, a^{3} e^{\left (12 i \, d x + 12 i \, c\right )} - 54 i \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} - 189 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 420 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 945 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 378 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 21 i \, a^{3}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{4032 \, d} \]

[In]

integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4032*(-7*I*a^3*e^(12*I*d*x + 12*I*c) - 54*I*a^3*e^(10*I*d*x + 10*I*c) - 189*I*a^3*e^(8*I*d*x + 8*I*c) - 420*
I*a^3*e^(6*I*d*x + 6*I*c) - 945*I*a^3*e^(4*I*d*x + 4*I*c) + 378*I*a^3*e^(2*I*d*x + 2*I*c) + 21*I*a^3)*e^(-3*I*
d*x - 3*I*c)/d

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 275 vs. \(2 (112) = 224\).

Time = 0.42 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.22 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx=\begin {cases} \frac {\left (- 270582939648 i a^{3} d^{6} e^{13 i c} e^{9 i d x} - 2087354105856 i a^{3} d^{6} e^{11 i c} e^{7 i d x} - 7305739370496 i a^{3} d^{6} e^{9 i c} e^{5 i d x} - 16234976378880 i a^{3} d^{6} e^{7 i c} e^{3 i d x} - 36528696852480 i a^{3} d^{6} e^{5 i c} e^{i d x} + 14611478740992 i a^{3} d^{6} e^{3 i c} e^{- i d x} + 811748818944 i a^{3} d^{6} e^{i c} e^{- 3 i d x}\right ) e^{- 4 i c}}{155855773237248 d^{7}} & \text {for}\: d^{7} e^{4 i c} \neq 0 \\\frac {x \left (a^{3} e^{12 i c} + 6 a^{3} e^{10 i c} + 15 a^{3} e^{8 i c} + 20 a^{3} e^{6 i c} + 15 a^{3} e^{4 i c} + 6 a^{3} e^{2 i c} + a^{3}\right ) e^{- 3 i c}}{64} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**9*(a+I*a*tan(d*x+c))**3,x)

[Out]

Piecewise(((-270582939648*I*a**3*d**6*exp(13*I*c)*exp(9*I*d*x) - 2087354105856*I*a**3*d**6*exp(11*I*c)*exp(7*I
*d*x) - 7305739370496*I*a**3*d**6*exp(9*I*c)*exp(5*I*d*x) - 16234976378880*I*a**3*d**6*exp(7*I*c)*exp(3*I*d*x)
- 36528696852480*I*a**3*d**6*exp(5*I*c)*exp(I*d*x) + 14611478740992*I*a**3*d**6*exp(3*I*c)*exp(-I*d*x) + 8117
48818944*I*a**3*d**6*exp(I*c)*exp(-3*I*d*x))*exp(-4*I*c)/(155855773237248*d**7), Ne(d**7*exp(4*I*c), 0)), (x*(
a**3*exp(12*I*c) + 6*a**3*exp(10*I*c) + 15*a**3*exp(8*I*c) + 20*a**3*exp(6*I*c) + 15*a**3*exp(4*I*c) + 6*a**3*
exp(2*I*c) + a**3)*exp(-3*I*c)/64, True))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.17 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {105 i \, a^{3} \cos \left (d x + c\right )^{9} + 5 i \, {\left (7 \, \cos \left (d x + c\right )^{9} - 9 \, \cos \left (d x + c\right )^{7}\right )} a^{3} - 3 \, {\left (35 \, \sin \left (d x + c\right )^{9} - 135 \, \sin \left (d x + c\right )^{7} + 189 \, \sin \left (d x + c\right )^{5} - 105 \, \sin \left (d x + c\right )^{3}\right )} a^{3} - {\left (35 \, \sin \left (d x + c\right )^{9} - 180 \, \sin \left (d x + c\right )^{7} + 378 \, \sin \left (d x + c\right )^{5} - 420 \, \sin \left (d x + c\right )^{3} + 315 \, \sin \left (d x + c\right )\right )} a^{3}}{315 \, d} \]

[In]

integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/315*(105*I*a^3*cos(d*x + c)^9 + 5*I*(7*cos(d*x + c)^9 - 9*cos(d*x + c)^7)*a^3 - 3*(35*sin(d*x + c)^9 - 135*
sin(d*x + c)^7 + 189*sin(d*x + c)^5 - 105*sin(d*x + c)^3)*a^3 - (35*sin(d*x + c)^9 - 180*sin(d*x + c)^7 + 378*
sin(d*x + c)^5 - 420*sin(d*x + c)^3 + 315*sin(d*x + c))*a^3)/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. 1039 vs. \(2 (106) = 212\).

Time = 0.78 (sec) , antiderivative size = 1039, normalized size of antiderivative = 8.38 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/516096*(119511*a^3*e^(11*I*d*x + 5*I*c)*log(I*e^(I*d*x + I*c) + 1) + 478044*a^3*e^(9*I*d*x + 3*I*c)*log(I*e^
(I*d*x + I*c) + 1) + 717066*a^3*e^(7*I*d*x + I*c)*log(I*e^(I*d*x + I*c) + 1) + 478044*a^3*e^(5*I*d*x - I*c)*lo
g(I*e^(I*d*x + I*c) + 1) + 119511*a^3*e^(3*I*d*x - 3*I*c)*log(I*e^(I*d*x + I*c) + 1) + 128898*a^3*e^(11*I*d*x
+ 5*I*c)*log(I*e^(I*d*x + I*c) - 1) + 515592*a^3*e^(9*I*d*x + 3*I*c)*log(I*e^(I*d*x + I*c) - 1) + 773388*a^3*e
^(7*I*d*x + I*c)*log(I*e^(I*d*x + I*c) - 1) + 515592*a^3*e^(5*I*d*x - I*c)*log(I*e^(I*d*x + I*c) - 1) + 128898
*a^3*e^(3*I*d*x - 3*I*c)*log(I*e^(I*d*x + I*c) - 1) - 119511*a^3*e^(11*I*d*x + 5*I*c)*log(-I*e^(I*d*x + I*c) +
1) - 478044*a^3*e^(9*I*d*x + 3*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 717066*a^3*e^(7*I*d*x + I*c)*log(-I*e^(I*d*
x + I*c) + 1) - 478044*a^3*e^(5*I*d*x - I*c)*log(-I*e^(I*d*x + I*c) + 1) - 119511*a^3*e^(3*I*d*x - 3*I*c)*log(
-I*e^(I*d*x + I*c) + 1) - 128898*a^3*e^(11*I*d*x + 5*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 515592*a^3*e^(9*I*d*x
+ 3*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 773388*a^3*e^(7*I*d*x + I*c)*log(-I*e^(I*d*x + I*c) - 1) - 515592*a^3*e
^(5*I*d*x - I*c)*log(-I*e^(I*d*x + I*c) - 1) - 128898*a^3*e^(3*I*d*x - 3*I*c)*log(-I*e^(I*d*x + I*c) - 1) + 93
87*a^3*e^(11*I*d*x + 5*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 37548*a^3*e^(9*I*d*x + 3*I*c)*log(I*e^(I*d*x) + e^(-
I*c)) + 56322*a^3*e^(7*I*d*x + I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 37548*a^3*e^(5*I*d*x - I*c)*log(I*e^(I*d*x)
+ e^(-I*c)) + 9387*a^3*e^(3*I*d*x - 3*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 9387*a^3*e^(11*I*d*x + 5*I*c)*log(-I*
e^(I*d*x) + e^(-I*c)) - 37548*a^3*e^(9*I*d*x + 3*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 56322*a^3*e^(7*I*d*x + I*
c)*log(-I*e^(I*d*x) + e^(-I*c)) - 37548*a^3*e^(5*I*d*x - I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 9387*a^3*e^(3*I*d
*x - 3*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 896*I*a^3*e^(20*I*d*x + 14*I*c) - 10496*I*a^3*e^(18*I*d*x + 12*I*c)
- 57216*I*a^3*e^(16*I*d*x + 10*I*c) - 195584*I*a^3*e^(14*I*d*x + 8*I*c) - 509696*I*a^3*e^(12*I*d*x + 6*I*c) -
861696*I*a^3*e^(10*I*d*x + 4*I*c) - 768768*I*a^3*e^(8*I*d*x + 2*I*c) + 88704*I*a^3*e^(4*I*d*x - 2*I*c) + 5913
6*I*a^3*e^(2*I*d*x - 4*I*c) - 236544*I*a^3*e^(6*I*d*x) + 2688*I*a^3*e^(-6*I*c))/(d*e^(11*I*d*x + 5*I*c) + 4*d*
e^(9*I*d*x + 3*I*c) + 6*d*e^(7*I*d*x + I*c) + 4*d*e^(5*I*d*x - I*c) + d*e^(3*I*d*x - 3*I*c))

Mupad [B] (verification not implemented)

Time = 5.73 (sec) , antiderivative size = 330, normalized size of antiderivative = 2.66 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {2\,a^3\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-3{}\mathrm {i}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {2048\,a^3\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\mathrm {i}\right )}{9\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^9}-\frac {1024\,a^3\,\left (8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-9{}\mathrm {i}\right )}{9\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^8}-\frac {4\,a^3\,\left (14\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-39{}\mathrm {i}\right )}{3\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2}+\frac {8\,a^3\,\left (43\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-97{}\mathrm {i}\right )}{3\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^3}-\frac {16\,a^3\,\left (188\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-357{}\mathrm {i}\right )}{7\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4}+\frac {128\,a^3\,\left (263\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-333{}\mathrm {i}\right )}{21\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7}-\frac {64\,a^3\,\left (1598\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-2289{}\mathrm {i}\right )}{63\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6}+\frac {32\,a^3\,\left (2041\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-3339{}\mathrm {i}\right )}{63\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]

[In]

int(cos(c + d*x)^9*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

(2*a^3*(tan(c/2 + (d*x)/2) - 3i))/(d*(tan(c/2 + (d*x)/2)^2 + 1)) + (2048*a^3*(tan(c/2 + (d*x)/2) - 1i))/(9*d*(
tan(c/2 + (d*x)/2)^2 + 1)^9) - (1024*a^3*(8*tan(c/2 + (d*x)/2) - 9i))/(9*d*(tan(c/2 + (d*x)/2)^2 + 1)^8) - (4*
a^3*(14*tan(c/2 + (d*x)/2) - 39i))/(3*d*(tan(c/2 + (d*x)/2)^2 + 1)^2) + (8*a^3*(43*tan(c/2 + (d*x)/2) - 97i))/
(3*d*(tan(c/2 + (d*x)/2)^2 + 1)^3) - (16*a^3*(188*tan(c/2 + (d*x)/2) - 357i))/(7*d*(tan(c/2 + (d*x)/2)^2 + 1)^
4) + (128*a^3*(263*tan(c/2 + (d*x)/2) - 333i))/(21*d*(tan(c/2 + (d*x)/2)^2 + 1)^7) - (64*a^3*(1598*tan(c/2 + (
d*x)/2) - 2289i))/(63*d*(tan(c/2 + (d*x)/2)^2 + 1)^6) + (32*a^3*(2041*tan(c/2 + (d*x)/2) - 3339i))/(63*d*(tan(
c/2 + (d*x)/2)^2 + 1)^5)