\(\int \frac {\tan ^{\frac {8}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx\) [234]

Optimal result

Integrand size = 26, antiderivative size = 319 \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=-\frac {5 \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {5 \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {2 i \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{\sqrt {3} a d}+\frac {5 \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {2 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}+\frac {5 \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {5 \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{3 a d}-\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{a d}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))} \] Output:

5/12*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))/a/d+5/12*arctan(3^(1/2)+2*tan(d*x 
+c)^(1/3))/a/d-2/3*I*arctan(1/3*(1-2*tan(d*x+c)^(2/3))*3^(1/2))*3^(1/2)/a/ 
d+5/6*arctan(tan(d*x+c)^(1/3))/a/d+2/3*I*ln(1+tan(d*x+c)^(2/3))/a/d+5/24*l 
n(1-3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*3^(1/2)/a/d-5/24*ln(1+3^(1/ 
2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*3^(1/2)/a/d-1/3*I*ln(1-tan(d*x+c)^(2 
/3)+tan(d*x+c)^(4/3))/a/d-2*I*tan(d*x+c)^(2/3)/a/d-1/2*tan(d*x+c)^(5/3)/d/ 
(a+I*a*tan(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 3.44 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.06 \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {\frac {6 \tan ^{\frac {11}{3}}(c+d x)}{a+i a \tan (c+d x)}+\frac {2 i \left (4 \sqrt {3} \arctan \left (\frac {-1+2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )+4 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )-2 \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )-12 \tan ^{\frac {2}{3}}(c+d x)+3 \tan ^{\frac {8}{3}}(c+d x)\right )}{a}+\frac {\tan ^{\frac {5}{3}}(c+d x) \left (5 i \log \left (1-i \sqrt [6]{\tan ^2(c+d x)}\right )-5 i \log \left (1+i \sqrt [6]{\tan ^2(c+d x)}\right )+5 \sqrt [6]{-1} \log \left (1-\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )-5 \sqrt [6]{-1} \log \left (1+\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )+5 (-1)^{5/6} \log \left (1-(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )-5 (-1)^{5/6} \log \left (1+(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )-6 \tan ^2(c+d x)^{5/6}\right )}{a \tan ^2(c+d x)^{5/6}}}{12 d} \] Input:

Integrate[Tan[c + d*x]^(8/3)/(a + I*a*Tan[c + d*x]),x]
 

Output:

((6*Tan[c + d*x]^(11/3))/(a + I*a*Tan[c + d*x]) + ((2*I)*(4*Sqrt[3]*ArcTan 
[(-1 + 2*Tan[c + d*x]^(2/3))/Sqrt[3]] + 4*Log[1 + Tan[c + d*x]^(2/3)] - 2* 
Log[1 - Tan[c + d*x]^(2/3) + Tan[c + d*x]^(4/3)] - 12*Tan[c + d*x]^(2/3) + 
 3*Tan[c + d*x]^(8/3)))/a + (Tan[c + d*x]^(5/3)*((5*I)*Log[1 - I*(Tan[c + 
d*x]^2)^(1/6)] - (5*I)*Log[1 + I*(Tan[c + d*x]^2)^(1/6)] + 5*(-1)^(1/6)*Lo 
g[1 - (-1)^(1/6)*(Tan[c + d*x]^2)^(1/6)] - 5*(-1)^(1/6)*Log[1 + (-1)^(1/6) 
*(Tan[c + d*x]^2)^(1/6)] + 5*(-1)^(5/6)*Log[1 - (-1)^(5/6)*(Tan[c + d*x]^2 
)^(1/6)] - 5*(-1)^(5/6)*Log[1 + (-1)^(5/6)*(Tan[c + d*x]^2)^(1/6)] - 6*(Ta 
n[c + d*x]^2)^(5/6)))/(a*(Tan[c + d*x]^2)^(5/6)))/(12*d)
 

Rubi [A] (warning: unable to verify)

Time = 0.84 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.82, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.808, Rules used = {3042, 4033, 27, 3042, 4011, 3042, 4021, 3042, 3957, 266, 807, 750, 16, 824, 27, 216, 1142, 25, 1083, 217, 1103}

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.

\(\displaystyle \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^{8/3}}{a+i a \tan (c+d x)}dx\)

\(\Big \downarrow \) 4033

\(\displaystyle \frac {\int \frac {1}{3} \tan ^{\frac {2}{3}}(c+d x) (5 a-8 i a \tan (c+d x))dx}{2 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \tan ^{\frac {2}{3}}(c+d x) (5 a-8 i a \tan (c+d x))dx}{6 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \tan (c+d x)^{2/3} (5 a-8 i a \tan (c+d x))dx}{6 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {\int \frac {5 \tan (c+d x) a+8 i a}{\sqrt [3]{\tan (c+d x)}}dx-\frac {12 i a \tan ^{\frac {2}{3}}(c+d x)}{d}}{6 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {5 \tan (c+d x) a+8 i a}{\sqrt [3]{\tan (c+d x)}}dx-\frac {12 i a \tan ^{\frac {2}{3}}(c+d x)}{d}}{6 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 4021

\(\displaystyle \frac {5 a \int \tan ^{\frac {2}{3}}(c+d x)dx+8 i a \int \frac {1}{\sqrt [3]{\tan (c+d x)}}dx-\frac {12 i a \tan ^{\frac {2}{3}}(c+d x)}{d}}{6 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {8 i a \int \frac {1}{\sqrt [3]{\tan (c+d x)}}dx+5 a \int \tan (c+d x)^{2/3}dx-\frac {12 i a \tan ^{\frac {2}{3}}(c+d x)}{d}}{6 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {\frac {8 i a \int \frac {1}{\sqrt [3]{\tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}+\frac {5 a \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}-\frac {12 i a \tan ^{\frac {2}{3}}(c+d x)}{d}}{6 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {\frac {24 i a \int \frac {\sqrt [3]{\tan (c+d x)}}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {15 a \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {12 i a \tan ^{\frac {2}{3}}(c+d x)}{d}}{6 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {\frac {12 i a \int \frac {1}{\tan (c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)}{d}+\frac {15 a \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {12 i a \tan ^{\frac {2}{3}}(c+d x)}{d}}{6 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 750

\(\displaystyle \frac {\frac {12 i a \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)\right )}{d}+\frac {15 a \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {12 i a \tan ^{\frac {2}{3}}(c+d x)}{d}}{6 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {\frac {15 a \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {12 i a \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {12 i a \tan ^{\frac {2}{3}}(c+d x)}{d}}{6 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 824

\(\displaystyle \frac {\frac {15 a \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int -\frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{2 \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int -\frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{2 \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}\right )}{d}+\frac {12 i a \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {12 i a \tan ^{\frac {2}{3}}(c+d x)}{d}}{6 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {15 a \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )}{d}+\frac {12 i a \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {12 i a \tan ^{\frac {2}{3}}(c+d x)}{d}}{6 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {15 a \left (-\frac {1}{6} \int \frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {12 i a \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {12 i a \tan ^{\frac {2}{3}}(c+d x)}{d}}{6 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {\frac {15 a \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int -\frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {12 i a \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{2} \int \left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )d\tan ^{\frac {2}{3}}(c+d x)\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {12 i a \tan ^{\frac {2}{3}}(c+d x)}{d}}{6 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {15 a \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {12 i a \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {12 i a \tan ^{\frac {2}{3}}(c+d x)}{d}}{6 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {15 a \left (\frac {1}{6} \left (-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}-\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {12 i a \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)-3 \int \frac {1}{-2 \tan ^{\frac {2}{3}}(c+d x)-2}d\left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {12 i a \tan ^{\frac {2}{3}}(c+d x)}{d}}{6 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {15 a \left (\frac {1}{6} \left (-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {12 i a \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {12 i a \tan ^{\frac {2}{3}}(c+d x)}{d}}{6 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {12 i a \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}+\frac {15 a \left (\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )\right )}{d}-\frac {12 i a \tan ^{\frac {2}{3}}(c+d x)}{d}}{6 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

Input:

Int[Tan[c + d*x]^(8/3)/(a + I*a*Tan[c + d*x]),x]
 

Output:

(((12*I)*a*(ArcTan[(-1 + 2*Tan[c + d*x]^(2/3))/Sqrt[3]]/Sqrt[3] + Log[1 + 
Tan[c + d*x]^(2/3)]/3))/d + (15*a*(ArcTan[Tan[c + d*x]^(1/3)]/3 + (-ArcTan 
[Sqrt[3] - 2*Tan[c + d*x]^(1/3)] + (Sqrt[3]*Log[1 - Sqrt[3]*Tan[c + d*x]^( 
1/3) + Tan[c + d*x]^(2/3)])/2)/6 + (ArcTan[Sqrt[3] + 2*Tan[c + d*x]^(1/3)] 
 - (Sqrt[3]*Log[1 + Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)])/2)/6 
))/d - ((12*I)*a*Tan[c + d*x]^(2/3))/d)/(6*a^2) - Tan[c + d*x]^(5/3)/(2*d* 
(a + I*a*Tan[c + d*x]))
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 750

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2)   Int[1/ 
(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]^2)   Int[(2*Rt[a, 3] - 
 Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x] /; 
 FreeQ[{a, b}, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 824

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator 
[Rt[a/b, n]], s = Denominator[Rt[a/b, n]], k, u}, Simp[u = Int[(r*Cos[(2*k 
- 1)*m*(Pi/n)] - s*Cos[(2*k - 1)*(m + 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[(2*k - 
 1)*(Pi/n)]*x + s^2*x^2), x] + Int[(r*Cos[(2*k - 1)*m*(Pi/n)] + s*Cos[(2*k 
- 1)*(m + 1)*(Pi/n)]*x)/(r^2 + 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x] 
; 2*(-1)^(m/2)*(r^(m + 2)/(a*n*s^m))   Int[1/(r^2 + s^2*x^2), x] + 2*(r^(m 
+ 1)/(a*n*s^m))   Sum[u, {k, 1, (n - 2)/4}], x]] /; FreeQ[{a, b}, x] && IGt 
Q[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && PosQ[a/b]
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

rule 4011

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 4021

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

rule 4033

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

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.63

method	result	size
derivativedivides	\(\frac {-\frac {3 i \tan \left (d x +c \right )^{\frac {2}{3}}}{2}-\frac {4 \tan \left (d x +c \right )^{\frac {1}{3}}-2 i}{12 \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}-\frac {13 i \ln \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{24}-\frac {13 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{12}+\frac {13 i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}{12}-\frac {1}{6 \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}+\frac {i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}-i\right )}{4}-\frac {i \ln \left (i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{8}+\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{4}}{d a}\)	\(201\)
default	\(\frac {-\frac {3 i \tan \left (d x +c \right )^{\frac {2}{3}}}{2}-\frac {4 \tan \left (d x +c \right )^{\frac {1}{3}}-2 i}{12 \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}-\frac {13 i \ln \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{24}-\frac {13 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{12}+\frac {13 i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}{12}-\frac {1}{6 \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}+\frac {i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}-i\right )}{4}-\frac {i \ln \left (i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{8}+\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{4}}{d a}\)	\(201\)

Input:

int(tan(d*x+c)^(8/3)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/d/a*(-3/2*I*tan(d*x+c)^(2/3)-1/12*(4*tan(d*x+c)^(1/3)-2*I)/(-I*tan(d*x+c 
)^(1/3)+tan(d*x+c)^(2/3)-1)-13/24*I*ln(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3 
)-1)-13/12*3^(1/2)*arctanh(1/3*(-I+2*tan(d*x+c)^(1/3))*3^(1/2))+13/12*I*ln 
(tan(d*x+c)^(1/3)+I)-1/6/(tan(d*x+c)^(1/3)+I)+1/4*I*ln(tan(d*x+c)^(1/3)-I) 
-1/8*I*ln(I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)+1/4*3^(1/2)*arctanh(1/3*( 
I+2*tan(d*x+c)^(1/3))*3^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 494, normalized size of antiderivative = 1.55 \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx =\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(8/3)/(a+I*a*tan(d*x+c)),x, algorithm="fricas")
 

Output:

1/24*(3*(sqrt(3)*a*d*sqrt(1/(a^2*d^2))*e^(2*I*d*x + 2*I*c) - I*e^(2*I*d*x 
+ 2*I*c))*log(1/2*sqrt(3)*a*d*sqrt(1/(a^2*d^2)) + ((-I*e^(2*I*d*x + 2*I*c) 
 + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) + 1/2*I) - 3*(sqrt(3)*a*d*sqrt(1/(a 
^2*d^2))*e^(2*I*d*x + 2*I*c) + I*e^(2*I*d*x + 2*I*c))*log(-1/2*sqrt(3)*a*d 
*sqrt(1/(a^2*d^2)) + ((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 
1))^(1/3) + 1/2*I) - 13*(3*sqrt(1/3)*a*d*sqrt(1/(a^2*d^2))*e^(2*I*d*x + 2* 
I*c) + I*e^(2*I*d*x + 2*I*c))*log(3/2*sqrt(1/3)*a*d*sqrt(1/(a^2*d^2)) + (( 
-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) - 1/2*I) + 13 
*(3*sqrt(1/3)*a*d*sqrt(1/(a^2*d^2))*e^(2*I*d*x + 2*I*c) - I*e^(2*I*d*x + 2 
*I*c))*log(-3/2*sqrt(1/3)*a*d*sqrt(1/(a^2*d^2)) + ((-I*e^(2*I*d*x + 2*I*c) 
 + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) - 1/2*I) + 26*I*e^(2*I*d*x + 2*I*c) 
*log(((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) + I) + 
 6*I*e^(2*I*d*x + 2*I*c)*log(((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2 
*I*c) + 1))^(1/3) - I) - 6*((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I 
*c) + 1))^(2/3)*(7*I*e^(2*I*d*x + 2*I*c) + I))*e^(-2*I*d*x - 2*I*c)/(a*d)
 

Sympy [F]

\[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=- \frac {i \int \frac {\tan ^{\frac {8}{3}}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx}{a} \] Input:

integrate(tan(d*x+c)**(8/3)/(a+I*a*tan(d*x+c)),x)
 

Output:

-I*Integral(tan(c + d*x)**(8/3)/(tan(c + d*x) - I), x)/a
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(tan(d*x+c)^(8/3)/(a+I*a*tan(d*x+c)),x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.60 \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {i \, {\left (-13 i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}\right ) + 3 i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}\right ) - 36 \, \tan \left (d x + c\right )^{\frac {2}{3}} + \frac {12 i \, \tan \left (d x + c\right )^{\frac {2}{3}}}{\tan \left (d x + c\right ) - i} - 3 \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right ) - 13 \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right ) + 26 \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right ) + 6 \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )\right )}}{24 \, a d} \] Input:

integrate(tan(d*x+c)^(8/3)/(a+I*a*tan(d*x+c)),x, algorithm="giac")
 

Output:

1/24*I*(-13*I*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1/3) + I)/(sqrt(3) + 
 2*tan(d*x + c)^(1/3) - I)) + 3*I*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + c)^( 
1/3) - I)/(sqrt(3) + 2*tan(d*x + c)^(1/3) + I)) - 36*tan(d*x + c)^(2/3) + 
12*I*tan(d*x + c)^(2/3)/(tan(d*x + c) - I) - 3*log(tan(d*x + c)^(2/3) + I* 
tan(d*x + c)^(1/3) - 1) - 13*log(tan(d*x + c)^(2/3) - I*tan(d*x + c)^(1/3) 
 - 1) + 26*log(tan(d*x + c)^(1/3) + I) + 6*log(tan(d*x + c)^(1/3) - I))/(a 
*d)
 

Mupad [B] (verification not implemented)

Time = 2.81 (sec) , antiderivative size = 616, normalized size of antiderivative = 1.93 \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx =\text {Too large to display} \] Input:

int(tan(c + d*x)^(8/3)/(a + a*tan(c + d*x)*1i),x)
 

Output:

log((a^3*d^3*312480i - a^4*d^4*tan(c + d*x)^(1/3)*(-1i/(64*a^3*d^3))^(1/3) 
*307584i)*(-1i/(64*a^3*d^3))^(2/3) + 24336*a*d*tan(c + d*x)^(1/3))*(-1i/(6 
4*a^3*d^3))^(1/3) + log((a^3*d^3*312480i - a^4*d^4*tan(c + d*x)^(1/3)*(-21 
97i/(1728*a^3*d^3))^(1/3)*307584i)*(-2197i/(1728*a^3*d^3))^(2/3) + 24336*a 
*d*tan(c + d*x)^(1/3))*(-2197i/(1728*a^3*d^3))^(1/3) - (tan(c + d*x)^(2/3) 
*3i)/(2*a*d) + (log(((3^(1/2)*1i - 1)^2*(a^3*d^3*312480i - a^4*d^4*tan(c + 
 d*x)^(1/3)*(3^(1/2)*1i - 1)*(-1i/(64*a^3*d^3))^(1/3)*153792i)*(-1i/(64*a^ 
3*d^3))^(2/3))/4 + 24336*a*d*tan(c + d*x)^(1/3))*(3^(1/2)*1i - 1)*(-1i/(64 
*a^3*d^3))^(1/3))/2 - (log(((3^(1/2)*1i + 1)^2*(a^3*d^3*312480i + a^4*d^4* 
tan(c + d*x)^(1/3)*(3^(1/2)*1i + 1)*(-1i/(64*a^3*d^3))^(1/3)*153792i)*(-1i 
/(64*a^3*d^3))^(2/3))/4 + 24336*a*d*tan(c + d*x)^(1/3))*(3^(1/2)*1i + 1)*( 
-1i/(64*a^3*d^3))^(1/3))/2 + (log(((3^(1/2)*1i - 1)^2*(a^3*d^3*312480i - a 
^4*d^4*tan(c + d*x)^(1/3)*(3^(1/2)*1i - 1)*(-2197i/(1728*a^3*d^3))^(1/3)*1 
53792i)*(-2197i/(1728*a^3*d^3))^(2/3))/4 + 24336*a*d*tan(c + d*x)^(1/3))*( 
3^(1/2)*1i - 1)*(-2197i/(1728*a^3*d^3))^(1/3))/2 - (log(((3^(1/2)*1i + 1)^ 
2*(a^3*d^3*312480i + a^4*d^4*tan(c + d*x)^(1/3)*(3^(1/2)*1i + 1)*(-2197i/( 
1728*a^3*d^3))^(1/3)*153792i)*(-2197i/(1728*a^3*d^3))^(2/3))/4 + 24336*a*d 
*tan(c + d*x)^(1/3))*(3^(1/2)*1i + 1)*(-2197i/(1728*a^3*d^3))^(1/3))/2 - ( 
tan(c + d*x)^(2/3)*1i)/(2*a*d*(tan(c + d*x)*1i + 1))
 

Reduce [F]

\[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {\int \frac {\tan \left (d x +c \right )^{\frac {8}{3}}}{\tan \left (d x +c \right ) i +1}d x}{a} \] Input:

int(tan(d*x+c)^(8/3)/(a+I*a*tan(d*x+c)),x)
 

Output:

int((tan(c + d*x)**(2/3)*tan(c + d*x)**2)/(tan(c + d*x)*i + 1),x)/a