Integrand size = 24, antiderivative size = 124 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {5 i a^3 \cos ^7(c+d x)}{63 d}+\frac {5 a^3 \sin (c+d x)}{9 d}-\frac {5 a^3 \sin ^3(c+d x)}{9 d}+\frac {a^3 \sin ^5(c+d x)}{3 d}-\frac {5 a^3 \sin ^7(c+d x)}{63 d}-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^2}{9 d} \] Output:
-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*ta n(d*x+c))^2/d
Time = 0.85 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.82 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^3 (-i \cos (3 (c+d x))+\sin (3 (c+d x))) \left (210 \sqrt {\cos ^2(c+d x)}+\left (32+567 \sqrt {\cos ^2(c+d x)}\right ) \cos (2 (c+d x))+\left (32-162 \sqrt {\cos ^2(c+d x)}\right ) \cos (4 (c+d x))-7 \sqrt {\cos ^2(c+d x)} \cos (6 (c+d x))-32 i \sin (2 (c+d x))-378 i \sqrt {\cos ^2(c+d x)} \sin (2 (c+d x))-32 i \sin (4 (c+d x))+216 i \sqrt {\cos ^2(c+d x)} \sin (4 (c+d x))+14 i \sqrt {\cos ^2(c+d x)} \sin (6 (c+d x))\right )}{2016 d \sqrt {\cos ^2(c+d x)}} \] Input:
Integrate[Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^3,x]
Output:
(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])*Cos[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)] - (3 2*I)*Sin[2*(c + d*x)] - (378*I)*Sqrt[Cos[c + d*x]^2]*Sin[2*(c + d*x)] - (3 2*I)*Sin[4*(c + d*x)] + (216*I)*Sqrt[Cos[c + d*x]^2]*Sin[4*(c + d*x)] + (1 4*I)*Sqrt[Cos[c + d*x]^2]*Sin[6*(c + d*x)]))/(2016*d*Sqrt[Cos[c + d*x]^2])
Time = 0.43 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.86, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3042, 3977, 3042, 3967, 3042, 3113, 2009}
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.
| \(\displaystyle \int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^3}{\sec (c+d x)^9}dx\) |
| \(\Big \downarrow \) 3977 |
| \(\displaystyle \frac {5}{9} a^2 \int \cos ^7(c+d x) (i \tan (c+d x) a+a)dx-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^2}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5}{9} a^2 \int \frac {i \tan (c+d x) a+a}{\sec (c+d x)^7}dx-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^2}{9 d}\) |
| \(\Big \downarrow \) 3967 |
| \(\displaystyle \frac {5}{9} a^2 \left (a \int \cos ^7(c+d x)dx-\frac {i a \cos ^7(c+d x)}{7 d}\right )-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^2}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5}{9} a^2 \left (a \int \sin \left (c+d x+\frac {\pi }{2}\right )^7dx-\frac {i a \cos ^7(c+d x)}{7 d}\right )-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^2}{9 d}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {5}{9} a^2 \left (-\frac {a \int \left (-\sin ^6(c+d x)+3 \sin ^4(c+d x)-3 \sin ^2(c+d x)+1\right )d(-\sin (c+d x))}{d}-\frac {i a \cos ^7(c+d x)}{7 d}\right )-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^2}{9 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {5}{9} a^2 \left (-\frac {a \left (\frac {1}{7} \sin ^7(c+d x)-\frac {3}{5} \sin ^5(c+d x)+\sin ^3(c+d x)-\sin (c+d x)\right )}{d}-\frac {i a \cos ^7(c+d x)}{7 d}\right )-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^2}{9 d}\) |
Input:
Int[Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^3,x]
Output:
(5*a^2*(((-1/7*I)*a*Cos[c + d*x]^7)/d - (a*(-Sin[c + d*x] + Sin[c + d*x]^3 - (3*Sin[c + d*x]^5)/5 + Sin[c + d*x]^7/7))/d))/9 - (((2*I)/9)*a*Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^2)/d
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
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] + Simp[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])
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] - Simp[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]) || (ILtQ[m, 0] & & LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]
Time = 198.50 (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 | \(\frac {-i a^{3} \left (-\frac {\cos \left (d x +c \right )^{7} \sin \left (d x +c \right )^{2}}{9}-\frac {2 \cos \left (d x +c \right )^{7}}{63}\right )-3 a^{3} \left (-\frac {\cos \left (d x +c \right )^{8} \sin \left (d x +c \right )}{9}+\frac {\left (\frac {16}{5}+\cos \left (d x +c \right )^{6}+\frac {6 \cos \left (d x +c \right )^{4}}{5}+\frac {8 \cos \left (d x +c \right )^{2}}{5}\right ) \sin \left (d x +c \right )}{63}\right )-\frac {i a^{3} \cos \left (d x +c \right )^{9}}{3}+\frac {a^{3} \left (\frac {128}{35}+\cos \left (d x +c \right )^{8}+\frac {8 \cos \left (d x +c \right )^{6}}{7}+\frac {48 \cos \left (d x +c \right )^{4}}{35}+\frac {64 \cos \left (d x +c \right )^{2}}{35}\right ) \sin \left (d x +c \right )}{9}}{d}\) | \(166\) |
| default | \(\frac {-i a^{3} \left (-\frac {\cos \left (d x +c \right )^{7} \sin \left (d x +c \right )^{2}}{9}-\frac {2 \cos \left (d x +c \right )^{7}}{63}\right )-3 a^{3} \left (-\frac {\cos \left (d x +c \right )^{8} \sin \left (d x +c \right )}{9}+\frac {\left (\frac {16}{5}+\cos \left (d x +c \right )^{6}+\frac {6 \cos \left (d x +c \right )^{4}}{5}+\frac {8 \cos \left (d x +c \right )^{2}}{5}\right ) \sin \left (d x +c \right )}{63}\right )-\frac {i a^{3} \cos \left (d x +c \right )^{9}}{3}+\frac {a^{3} \left (\frac {128}{35}+\cos \left (d x +c \right )^{8}+\frac {8 \cos \left (d x +c \right )^{6}}{7}+\frac {48 \cos \left (d x +c \right )^{4}}{35}+\frac {64 \cos \left (d x +c \right )^{2}}{35}\right ) \sin \left (d x +c \right )}{9}}{d}\) | \(166\) |
Input:
int(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
-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)
Time = 0.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.84 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {{\left (-7 i \, a^{3} e^{\left (12 i \, d x + 12 i \, c\right )} - 54 i \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} - 189 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 420 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 945 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 378 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 21 i \, a^{3}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{4032 \, d} \] Input:
integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
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
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 = 0.41 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.22 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx=\begin {cases} \frac {\left (- 270582939648 i a^{3} d^{6} e^{13 i c} e^{9 i d x} - 2087354105856 i a^{3} d^{6} e^{11 i c} e^{7 i d x} - 7305739370496 i a^{3} d^{6} e^{9 i c} e^{5 i d x} - 16234976378880 i a^{3} d^{6} e^{7 i c} e^{3 i d x} - 36528696852480 i a^{3} d^{6} e^{5 i c} e^{i d x} + 14611478740992 i a^{3} d^{6} e^{3 i c} e^{- i d x} + 811748818944 i a^{3} d^{6} e^{i c} e^{- 3 i d x}\right ) e^{- 4 i c}}{155855773237248 d^{7}} & \text {for}\: d^{7} e^{4 i c} \neq 0 \\\frac {x \left (a^{3} e^{12 i c} + 6 a^{3} e^{10 i c} + 15 a^{3} e^{8 i c} + 20 a^{3} e^{6 i c} + 15 a^{3} e^{4 i c} + 6 a^{3} e^{2 i c} + a^{3}\right ) e^{- 3 i c}}{64} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**9*(a+I*a*tan(d*x+c))**3,x)
Output:
Piecewise(((-270582939648*I*a**3*d**6*exp(13*I*c)*exp(9*I*d*x) - 208735410 5856*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) + 811748818944*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))
Time = 0.07 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.17 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {105 i \, a^{3} \cos \left (d x + c\right )^{9} + 5 i \, {\left (7 \, \cos \left (d x + c\right )^{9} - 9 \, \cos \left (d x + c\right )^{7}\right )} a^{3} - 3 \, {\left (35 \, \sin \left (d x + c\right )^{9} - 135 \, \sin \left (d x + c\right )^{7} + 189 \, \sin \left (d x + c\right )^{5} - 105 \, \sin \left (d x + c\right )^{3}\right )} a^{3} - {\left (35 \, \sin \left (d x + c\right )^{9} - 180 \, \sin \left (d x + c\right )^{7} + 378 \, \sin \left (d x + c\right )^{5} - 420 \, \sin \left (d x + c\right )^{3} + 315 \, \sin \left (d x + c\right )\right )} a^{3}}{315 \, d} \] Input:
integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
Output:
-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*si n(d*x + c)^5 - 420*sin(d*x + c)^3 + 315*sin(d*x + c))*a^3)/d
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 = 0.61 (sec) , antiderivative size = 1039, normalized size of antiderivative = 8.38 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
1/516096*(119511*a^3*e^(11*I*d*x + 5*I*c)*log(I*e^(I*d*x + I*c) + 1) + 478 044*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) + 51 5592*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) + 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)) + 5632 2*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(...
Time = 1.80 (sec) , antiderivative size = 330, normalized size of antiderivative = 2.66 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {2\,a^3\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-3{}\mathrm {i}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {2048\,a^3\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\mathrm {i}\right )}{9\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^9}-\frac {1024\,a^3\,\left (8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-9{}\mathrm {i}\right )}{9\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^8}-\frac {4\,a^3\,\left (14\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-39{}\mathrm {i}\right )}{3\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2}+\frac {8\,a^3\,\left (43\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-97{}\mathrm {i}\right )}{3\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^3}-\frac {16\,a^3\,\left (188\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-357{}\mathrm {i}\right )}{7\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4}+\frac {128\,a^3\,\left (263\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-333{}\mathrm {i}\right )}{21\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7}-\frac {64\,a^3\,\left (1598\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-2289{}\mathrm {i}\right )}{63\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6}+\frac {32\,a^3\,\left (2041\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-3339{}\mathrm {i}\right )}{63\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \] Input:
int(cos(c + d*x)^9*(a + a*tan(c + d*x)*1i)^3,x)
Output:
(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)
Time = 0.17 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.10 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^{3} \left (-28 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8} i +103 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} i -141 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} i +85 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} i -19 \cos \left (d x +c \right ) i +28 \sin \left (d x +c \right )^{9}-117 \sin \left (d x +c \right )^{7}+189 \sin \left (d x +c \right )^{5}-147 \sin \left (d x +c \right )^{3}+63 \sin \left (d x +c \right )+19 i \right )}{63 d} \] Input:
int(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^3,x)
Output:
(a**3*( - 28*cos(c + d*x)*sin(c + d*x)**8*i + 103*cos(c + d*x)*sin(c + d*x )**6*i - 141*cos(c + d*x)*sin(c + d*x)**4*i + 85*cos(c + d*x)*sin(c + d*x) **2*i - 19*cos(c + d*x)*i + 28*sin(c + d*x)**9 - 117*sin(c + d*x)**7 + 189 *sin(c + d*x)**5 - 147*sin(c + d*x)**3 + 63*sin(c + d*x) + 19*i))/(63*d)