\(\int \frac {\tan ^{\frac {8}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [242]

Optimal result

Integrand size = 26, antiderivative size = 337 \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {25 \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {25 \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {2 i \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}-\frac {25 \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {2 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {25 \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}+\frac {25 \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}+\frac {i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{9 a^2 d}+\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \] Output:

-25/72*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))/a^2/d-25/72*arctan(3^(1/2)+2*ta 
n(d*x+c)^(1/3))/a^2/d+2/9*I*arctan(1/3*(1-2*tan(d*x+c)^(2/3))*3^(1/2))*3^( 
1/2)/a^2/d-25/36*arctan(tan(d*x+c)^(1/3))/a^2/d-2/9*I*ln(1+tan(d*x+c)^(2/3 
))/a^2/d-25/144*ln(1-3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*3^(1/2)/a^ 
2/d+25/144*ln(1+3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*3^(1/2)/a^2/d+1 
/9*I*ln(1-tan(d*x+c)^(2/3)+tan(d*x+c)^(4/3))/a^2/d+2/3*I*tan(d*x+c)^(2/3)/ 
a^2/d/(1+I*tan(d*x+c))-1/4*tan(d*x+c)^(5/3)/d/(a+I*a*tan(d*x+c))^2
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 2.16 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.86 \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {27 i \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},i \tan (c+d x)\right ) \tan ^{\frac {2}{3}}(c+d x)+\frac {\sec ^2(c+d x) \left (82 (-1)^{5/6} \sqrt {3} \arctan \left (\frac {-1+2 (-1)^{5/6} \sqrt [3]{\tan (c+d x)}}{\sqrt {3}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+123 \sqrt [3]{-1} \log \left (\sqrt [6]{-1}-\sqrt [3]{\tan (c+d x)}\right ) \sin (2 (c+d x))-41 \sqrt [3]{-1} \log (i-\tan (c+d x)) \sin (2 (c+d x))-48 i \tan ^{\frac {2}{3}}(c+d x)+66 \tan ^{\frac {5}{3}}(c+d x)+\cos (2 (c+d x)) \left (41 (-1)^{5/6} \left (-3 \log \left (\sqrt [6]{-1}-\sqrt [3]{\tan (c+d x)}\right )+\log (i-\tan (c+d x))\right )-48 i \tan ^{\frac {2}{3}}(c+d x)+66 \tan ^{\frac {5}{3}}(c+d x)\right )\right )}{(-i+\tan (c+d x))^2}}{144 a^2 d} \] Input:

Integrate[Tan[c + d*x]^(8/3)/(a + I*a*Tan[c + d*x])^2,x]
 

Output:

((27*I)*Hypergeometric2F1[2/3, 1, 5/3, I*Tan[c + d*x]]*Tan[c + d*x]^(2/3) 
+ (Sec[c + d*x]^2*(82*(-1)^(5/6)*Sqrt[3]*ArcTan[(-1 + 2*(-1)^(5/6)*Tan[c + 
 d*x]^(1/3))/Sqrt[3]]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) + 123*(-1)^( 
1/3)*Log[(-1)^(1/6) - Tan[c + d*x]^(1/3)]*Sin[2*(c + d*x)] - 41*(-1)^(1/3) 
*Log[I - Tan[c + d*x]]*Sin[2*(c + d*x)] - (48*I)*Tan[c + d*x]^(2/3) + 66*T 
an[c + d*x]^(5/3) + Cos[2*(c + d*x)]*(41*(-1)^(5/6)*(-3*Log[(-1)^(1/6) - T 
an[c + d*x]^(1/3)] + Log[I - Tan[c + d*x]]) - (48*I)*Tan[c + d*x]^(2/3) + 
66*Tan[c + d*x]^(5/3))))/(-I + Tan[c + d*x])^2)/(144*a^2*d)
 

Rubi [A] (warning: unable to verify)

Time = 0.96 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.85, number of steps used = 23, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {3042, 4041, 27, 3042, 4078, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4041

\(\displaystyle -\frac {\int -\frac {\tan ^{\frac {2}{3}}(c+d x) (5 a-11 i a \tan (c+d x))}{3 (i \tan (c+d x) a+a)}dx}{4 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\tan ^{\frac {2}{3}}(c+d x) (5 a-11 i a \tan (c+d x))}{i \tan (c+d x) a+a}dx}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\tan (c+d x)^{2/3} (5 a-11 i a \tan (c+d x))}{i \tan (c+d x) a+a}dx}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\frac {8 i \tan ^{\frac {2}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int \frac {2 \left (25 \tan (c+d x) a^2+16 i a^2\right )}{3 \sqrt [3]{\tan (c+d x)}}dx}{2 a^2}}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {8 i \tan ^{\frac {2}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int \frac {25 \tan (c+d x) a^2+16 i a^2}{\sqrt [3]{\tan (c+d x)}}dx}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {8 i \tan ^{\frac {2}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int \frac {25 \tan (c+d x) a^2+16 i a^2}{\sqrt [3]{\tan (c+d x)}}dx}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4021

\(\displaystyle \frac {\frac {8 i \tan ^{\frac {2}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {25 a^2 \int \tan ^{\frac {2}{3}}(c+d x)dx+16 i a^2 \int \frac {1}{\sqrt [3]{\tan (c+d x)}}dx}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {8 i \tan ^{\frac {2}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {25 a^2 \int \tan (c+d x)^{2/3}dx+16 i a^2 \int \frac {1}{\sqrt [3]{\tan (c+d x)}}dx}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {\frac {8 i \tan ^{\frac {2}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {25 a^2 \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}+\frac {16 i a^2 \int \frac {1}{\sqrt [3]{\tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {\frac {8 i \tan ^{\frac {2}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {75 a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {48 i a^2 \int \frac {\sqrt [3]{\tan (c+d x)}}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {\frac {8 i \tan ^{\frac {2}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {75 a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {24 i a^2 \int \frac {1}{\tan (c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 750

\(\displaystyle \frac {\frac {8 i \tan ^{\frac {2}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {75 a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {24 i a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)\right )}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {\frac {8 i \tan ^{\frac {2}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {75 a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {24 i a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 824

\(\displaystyle \frac {\frac {8 i \tan ^{\frac {2}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {75 a^2 \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int -\frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{2 \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int -\frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{2 \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}\right )}{d}+\frac {24 i a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {8 i \tan ^{\frac {2}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {75 a^2 \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )}{d}+\frac {24 i a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {8 i \tan ^{\frac {2}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {75 a^2 \left (-\frac {1}{6} \int \frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {24 i a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {\frac {8 i \tan ^{\frac {2}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {75 a^2 \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int -\frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {24 i a^2 \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{2} \int \left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )d\tan ^{\frac {2}{3}}(c+d x)\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {8 i \tan ^{\frac {2}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {75 a^2 \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {24 i a^2 \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {8 i \tan ^{\frac {2}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {75 a^2 \left (\frac {1}{6} \left (-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}-\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {24 i a^2 \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)-3 \int \frac {1}{-2 \tan ^{\frac {2}{3}}(c+d x)-2}d\left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {8 i \tan ^{\frac {2}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {75 a^2 \left (\frac {1}{6} \left (-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {24 i a^2 \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {8 i \tan ^{\frac {2}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {75 a^2 \left (\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )\right )}{d}+\frac {24 i a^2 \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

Input:

Int[Tan[c + d*x]^(8/3)/(a + I*a*Tan[c + d*x])^2,x]
 

Output:

(-1/3*(((24*I)*a^2*(ArcTan[(-1 + 2*Tan[c + d*x]^(2/3))/Sqrt[3]]/Sqrt[3] + 
Log[1 + Tan[c + d*x]^(2/3)]/3))/d + (75*a^2*(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)/a^2 + ((8*I)*Tan[c + d*x]^(2/3))/(d*(1 + I*Tan[c + d*x]))) 
/(12*a^2) - Tan[c + d*x]^(5/3)/(4*d*(a + I*a*Tan[c + d*x])^2)
 

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 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 4041

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m* 
((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Simp[1/(2*a^2*m)   Int[(a + 
b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n 
 - 1)) - d*(b*c*m + a*d*(n - 1)) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*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] && LtQ[m, 0] && GtQ[n, 1] && (In 
tegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4078

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

Maple [A] (verified)

Time = 1.15 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.67

method	result	size
derivativedivides	\(\frac {\frac {i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}-i\right )}{8}+\frac {44 \tan \left (d x +c \right )-64 i \tan \left (d x +c \right )^{\frac {2}{3}}-58 \tan \left (d x +c \right )^{\frac {1}{3}}+20 i}{72 \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )^{2}}+\frac {41 i \ln \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{144}+\frac {41 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{72}-\frac {41 i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}{72}-\frac {i}{36 \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )^{2}}+\frac {11}{36 \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}-\frac {i \ln \left (i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{16}+\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{8}}{d \,a^{2}}\)	\(225\)
default	\(\frac {\frac {i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}-i\right )}{8}+\frac {44 \tan \left (d x +c \right )-64 i \tan \left (d x +c \right )^{\frac {2}{3}}-58 \tan \left (d x +c \right )^{\frac {1}{3}}+20 i}{72 \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )^{2}}+\frac {41 i \ln \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{144}+\frac {41 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{72}-\frac {41 i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}{72}-\frac {i}{36 \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )^{2}}+\frac {11}{36 \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}-\frac {i \ln \left (i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{16}+\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{8}}{d \,a^{2}}\)	\(225\)

Input:

int(tan(d*x+c)^(8/3)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/d/a^2*(1/8*I*ln(tan(d*x+c)^(1/3)-I)+1/72*(44*tan(d*x+c)-64*I*tan(d*x+c)^ 
(2/3)-58*tan(d*x+c)^(1/3)+20*I)/(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)^2 
+41/144*I*ln(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)+41/72*3^(1/2)*arctanh 
(1/3*(-I+2*tan(d*x+c)^(1/3))*3^(1/2))-41/72*I*ln(tan(d*x+c)^(1/3)+I)-1/36* 
I/(tan(d*x+c)^(1/3)+I)^2+11/36/(tan(d*x+c)^(1/3)+I)-1/16*I*ln(I*tan(d*x+c) 
^(1/3)+tan(d*x+c)^(2/3)-1)+1/8*3^(1/2)*arctanh(1/3*(I+2*tan(d*x+c)^(1/3))* 
3^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 521, normalized size of antiderivative = 1.55 \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx =\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(8/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")
 

Output:

1/144*(9*(sqrt(3)*a^2*d*sqrt(1/(a^4*d^2))*e^(4*I*d*x + 4*I*c) - I*e^(4*I*d 
*x + 4*I*c))*log(1/2*sqrt(3)*a^2*d*sqrt(1/(a^4*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) - 9*(sqrt(3)*a^2*d*sq 
rt(1/(a^4*d^2))*e^(4*I*d*x + 4*I*c) + I*e^(4*I*d*x + 4*I*c))*log(-1/2*sqrt 
(3)*a^2*d*sqrt(1/(a^4*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) + 41*(3*sqrt(1/3)*a^2*d*sqrt(1/(a^4*d^2))*e^(4 
*I*d*x + 4*I*c) + I*e^(4*I*d*x + 4*I*c))*log(3/2*sqrt(1/3)*a^2*d*sqrt(1/(a 
^4*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) - 41*(3*sqrt(1/3)*a^2*d*sqrt(1/(a^4*d^2))*e^(4*I*d*x + 4*I*c) - I 
*e^(4*I*d*x + 4*I*c))*log(-3/2*sqrt(1/3)*a^2*d*sqrt(1/(a^4*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) - 82*I*e^ 
(4*I*d*x + 4*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) + 18*I*e^(4*I*d*x + 4*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) - 3*((-I*e^(2*I*d*x + 2*I*c) + I) 
/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(-19*I*e^(4*I*d*x + 4*I*c) - 16*I*e^(2*I 
*d*x + 2*I*c) + 3*I))*e^(-4*I*d*x - 4*I*c)/(a^2*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate(tan(d*x+c)**(8/3)/(a+I*a*tan(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(tan(d*x+c)^(8/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.58 \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {41 \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}\right ) + 9 \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}\right ) - \frac {12 \, {\left (11 \, \tan \left (d x + c\right )^{\frac {5}{3}} - 8 i \, \tan \left (d x + c\right )^{\frac {2}{3}}\right )}}{{\left (\tan \left (d x + c\right ) - i\right )}^{2}} + 9 i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right ) - 41 i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right ) + 82 i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right ) - 18 i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )}{144 \, a^{2} d} \] Input:

integrate(tan(d*x+c)^(8/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")
 

Output:

-1/144*(41*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1/3) + I)/(sqrt(3) + 2* 
tan(d*x + c)^(1/3) - I)) + 9*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1/3) 
- I)/(sqrt(3) + 2*tan(d*x + c)^(1/3) + I)) - 12*(11*tan(d*x + c)^(5/3) - 8 
*I*tan(d*x + c)^(2/3))/(tan(d*x + c) - I)^2 + 9*I*log(tan(d*x + c)^(2/3) + 
 I*tan(d*x + c)^(1/3) - 1) - 41*I*log(tan(d*x + c)^(2/3) - I*tan(d*x + c)^ 
(1/3) - 1) + 82*I*log(tan(d*x + c)^(1/3) + I) - 18*I*log(tan(d*x + c)^(1/3 
) - I))/(a^2*d)
 

Mupad [B] (verification not implemented)

Time = 2.47 (sec) , antiderivative size = 642, normalized size of antiderivative = 1.91 \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Too large to display} \] Input:

int(tan(c + d*x)^(8/3)/(a + a*tan(c + d*x)*1i)^2,x)
 

Output:

log(- ((a^6*d^3*8915200i)/3 + a^8*d^4*tan(c + d*x)^(1/3)*(-1i/(512*a^6*d^3 
))^(1/3)*5412864i)*(-1i/(512*a^6*d^3))^(2/3) - (107584*a^2*d*tan(c + d*x)^ 
(1/3))/3)*(-1i/(512*a^6*d^3))^(1/3) + log(- ((a^6*d^3*8915200i)/3 + a^8*d^ 
4*tan(c + d*x)^(1/3)*(68921i/(373248*a^6*d^3))^(1/3)*5412864i)*(68921i/(37 
3248*a^6*d^3))^(2/3) - (107584*a^2*d*tan(c + d*x)^(1/3))/3)*(68921i/(37324 
8*a^6*d^3))^(1/3) + ((2*tan(c + d*x)^(2/3))/(3*a^2*d) + (tan(c + d*x)^(5/3 
)*11i)/(12*a^2*d))/(2*tan(c + d*x) + tan(c + d*x)^2*1i - 1i) + (log(- (107 
584*a^2*d*tan(c + d*x)^(1/3))/3 - ((3^(1/2)*1i - 1)^2*((a^6*d^3*8915200i)/ 
3 + a^8*d^4*tan(c + d*x)^(1/3)*(3^(1/2)*1i - 1)*(-1i/(512*a^6*d^3))^(1/3)* 
2706432i)*(-1i/(512*a^6*d^3))^(2/3))/4)*(3^(1/2)*1i - 1)*(-1i/(512*a^6*d^3 
))^(1/3))/2 - (log(- (107584*a^2*d*tan(c + d*x)^(1/3))/3 - ((3^(1/2)*1i + 
1)^2*((a^6*d^3*8915200i)/3 - a^8*d^4*tan(c + d*x)^(1/3)*(3^(1/2)*1i + 1)*( 
-1i/(512*a^6*d^3))^(1/3)*2706432i)*(-1i/(512*a^6*d^3))^(2/3))/4)*(3^(1/2)* 
1i + 1)*(-1i/(512*a^6*d^3))^(1/3))/2 + (log(- (107584*a^2*d*tan(c + d*x)^( 
1/3))/3 - ((3^(1/2)*1i - 1)^2*((a^6*d^3*8915200i)/3 + a^8*d^4*tan(c + d*x) 
^(1/3)*(3^(1/2)*1i - 1)*(68921i/(373248*a^6*d^3))^(1/3)*2706432i)*(68921i/ 
(373248*a^6*d^3))^(2/3))/4)*(3^(1/2)*1i - 1)*(68921i/(373248*a^6*d^3))^(1/ 
3))/2 - (log(- (107584*a^2*d*tan(c + d*x)^(1/3))/3 - ((3^(1/2)*1i + 1)^2*( 
(a^6*d^3*8915200i)/3 - a^8*d^4*tan(c + d*x)^(1/3)*(3^(1/2)*1i + 1)*(68921i 
/(373248*a^6*d^3))^(1/3)*2706432i)*(68921i/(373248*a^6*d^3))^(2/3))/4)*...
 

Reduce [F]

\[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\int \frac {\tan \left (d x +c \right )^{\frac {8}{3}}}{\tan \left (d x +c \right )^{2}-2 \tan \left (d x +c \right ) i -1}d x}{a^{2}} \] Input:

int(tan(d*x+c)^(8/3)/(a+I*a*tan(d*x+c))^2,x)
 

Output:

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