\(\int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))} \, dx\) [238]

Optimal result

Integrand size = 26, antiderivative size = 321 \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {5 i \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {5 i \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {2 \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{\sqrt {3} a d}-\frac {5 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {2 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}+\frac {5 i \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {5 i \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {\log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{3 a d}-\frac {2}{a d \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{2 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 6.21 (sec) , antiderivative size = 465, normalized size of antiderivative = 1.45 \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {1}{2 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))}-\frac {\frac {5 i a \sqrt [3]{\tan (c+d x)} \left (\frac {i \log \left (1-i \sqrt [6]{\tan ^2(c+d x)}\right )}{6 \sqrt [6]{\tan ^2(c+d x)}}-\frac {i \log \left (1+i \sqrt [6]{\tan ^2(c+d x)}\right )}{6 \sqrt [6]{\tan ^2(c+d x)}}-\frac {\sqrt [6]{-1} \log \left (1-e^{-\frac {i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right )}{6 \sqrt [6]{\tan ^2(c+d x)}}+\frac {(-1)^{5/6} \log \left (1-e^{\frac {i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right )}{6 \sqrt [6]{\tan ^2(c+d x)}}-\frac {(-1)^{5/6} \log \left (1-e^{-\frac {5 i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right )}{6 \sqrt [6]{\tan ^2(c+d x)}}+\frac {\sqrt [6]{-1} \log \left (1-e^{\frac {5 i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right )}{6 \sqrt [6]{\tan ^2(c+d x)}}\right )}{d}+\frac {4 a \left (1-\frac {1}{3} \log \left (1+\sqrt [3]{\tan ^2(c+d x)}\right ) \sqrt [3]{\tan ^2(c+d x)}-\frac {1}{3} (-1)^{2/3} \log \left (1-e^{-\frac {i \pi }{3}} \sqrt [3]{\tan ^2(c+d x)}\right ) \sqrt [3]{\tan ^2(c+d x)}+\frac {1}{3} \sqrt [3]{-1} \log \left (1-e^{\frac {i \pi }{3}} \sqrt [3]{\tan ^2(c+d x)}\right ) \sqrt [3]{\tan ^2(c+d x)}\right )}{d \tan ^{\frac {2}{3}}(c+d x)}}{2 a^2} \] Input:

Integrate[1/(Tan[c + d*x]^(5/3)*(a + I*a*Tan[c + d*x])),x]
 

Output:

1/(2*d*Tan[c + d*x]^(2/3)*(a + I*a*Tan[c + d*x])) - (((5*I)*a*Tan[c + d*x] 
^(1/3)*(((I/6)*Log[1 - I*(Tan[c + d*x]^2)^(1/6)])/(Tan[c + d*x]^2)^(1/6) - 
 ((I/6)*Log[1 + I*(Tan[c + d*x]^2)^(1/6)])/(Tan[c + d*x]^2)^(1/6) - ((-1)^ 
(1/6)*Log[1 - (Tan[c + d*x]^2)^(1/6)/E^((I/6)*Pi)])/(6*(Tan[c + d*x]^2)^(1 
/6)) + ((-1)^(5/6)*Log[1 - E^((I/6)*Pi)*(Tan[c + d*x]^2)^(1/6)])/(6*(Tan[c 
 + d*x]^2)^(1/6)) - ((-1)^(5/6)*Log[1 - (Tan[c + d*x]^2)^(1/6)/E^(((5*I)/6 
)*Pi)])/(6*(Tan[c + d*x]^2)^(1/6)) + ((-1)^(1/6)*Log[1 - E^(((5*I)/6)*Pi)* 
(Tan[c + d*x]^2)^(1/6)])/(6*(Tan[c + d*x]^2)^(1/6))))/d + (4*a*(1 - (Log[1 
 + (Tan[c + d*x]^2)^(1/3)]*(Tan[c + d*x]^2)^(1/3))/3 - ((-1)^(2/3)*Log[1 - 
 (Tan[c + d*x]^2)^(1/3)/E^((I/3)*Pi)]*(Tan[c + d*x]^2)^(1/3))/3 + ((-1)^(1 
/3)*Log[1 - E^((I/3)*Pi)*(Tan[c + d*x]^2)^(1/3)]*(Tan[c + d*x]^2)^(1/3))/3 
))/(d*Tan[c + d*x]^(2/3)))/(2*a^2)
 

Rubi [A] (warning: unable to verify)

Time = 0.92 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.81, number of steps used = 23, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {3042, 4035, 27, 3042, 4012, 25, 3042, 4021, 3042, 3957, 266, 753, 27, 216, 807, 821, 16, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4035

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4021

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 753

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 821

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

Input:

Int[1/(Tan[c + d*x]^(5/3)*(a + I*a*Tan[c + d*x])),x]
 

Output:

((-12*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*I)*a*(ArcTan[Tan[c + d*x]^(1/3)]/3 + (-ArcTa 
n[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*a)/(d*Tan[c + d*x]^(2/3)))/(6*a^2) + 1/(2*d*Tan[c + d*x]^(2/3) 
*(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 753

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

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 821

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Simp[-(3*Rt[a, 3]*Rt[b, 3])^(- 
1)   Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]*Rt[b, 3]) 
 Int[(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 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 4012

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

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 4035

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

Maple [A] (verified)

Time = 0.97 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.62

method	result	size
derivativedivides	\(\frac {-\frac {3}{2 \tan \left (d x +c \right )^{\frac {2}{3}}}+\frac {2 i \tan \left (d x +c \right )^{\frac {1}{3}}-2}{-12 i \tan \left (d x +c \right )^{\frac {1}{3}}+12 \tan \left (d x +c \right )^{\frac {2}{3}}-12}-\frac {13 \ln \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{24}-\frac {13 i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{12}+\frac {\ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}-i\right )}{4}-\frac {\ln \left (i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{8}+\frac {i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{4}-\frac {i}{6 \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}+\frac {13 \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}{12}}{d a}\)	\(199\)
default	\(\frac {-\frac {3}{2 \tan \left (d x +c \right )^{\frac {2}{3}}}+\frac {2 i \tan \left (d x +c \right )^{\frac {1}{3}}-2}{-12 i \tan \left (d x +c \right )^{\frac {1}{3}}+12 \tan \left (d x +c \right )^{\frac {2}{3}}-12}-\frac {13 \ln \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{24}-\frac {13 i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{12}+\frac {\ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}-i\right )}{4}-\frac {\ln \left (i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{8}+\frac {i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{4}-\frac {i}{6 \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}+\frac {13 \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}{12}}{d a}\)	\(199\)

Input:

int(1/tan(d*x+c)^(5/3)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 642 vs. \(2 (254) = 508\).

Time = 0.10 (sec) , antiderivative size = 642, normalized size of antiderivative = 2.00 \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))} \, dx =\text {Too large to display} \] Input:

integrate(1/tan(d*x+c)^(5/3)/(a+I*a*tan(d*x+c)),x, algorithm="fricas")
 

Output:

-1/24*(3*(sqrt(3)*(-I*a*d*e^(4*I*d*x + 4*I*c) + I*a*d*e^(2*I*d*x + 2*I*c)) 
*sqrt(1/(a^2*d^2)) + e^(4*I*d*x + 4*I*c) - e^(2*I*d*x + 2*I*c))*log(1/2*sq 
rt(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)*(I*a*d*e^(4*I*d*x + 4*I*c) - I*a* 
d*e^(2*I*d*x + 2*I*c))*sqrt(1/(a^2*d^2)) + e^(4*I*d*x + 4*I*c) - 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)*(I*a* 
d*e^(4*I*d*x + 4*I*c) - I*a*d*e^(2*I*d*x + 2*I*c))*sqrt(1/(a^2*d^2)) + e^( 
4*I*d*x + 4*I*c) - 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)*(-I*a*d*e^(4*I*d*x + 4*I*c) + I*a*d*e^(2*I*d*x + 2 
*I*c))*sqrt(1/(a^2*d^2)) + e^(4*I*d*x + 4*I*c) - 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*(e^(4*I*d*x + 4*I*c) - 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*(e^(4*I*d*x + 4*I*c) - 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))^(1/3)*(7*I*e^(4*I*d*x + 4*I*c) 
+ 6*I*e^(2*I*d*x + 2*I*c) - I))/(a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x 
+ 2*I*c))
 

Sympy [F]

\[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))} \, dx=- \frac {i \int \frac {1}{\tan ^{\frac {8}{3}}{\left (c + d x \right )} - i \tan ^{\frac {5}{3}}{\left (c + d x \right )}}\, dx}{a} \] Input:

integrate(1/tan(d*x+c)**(5/3)/(a+I*a*tan(d*x+c)),x)
 

Output:

-I*Integral(1/(tan(c + d*x)**(8/3) - I*tan(c + d*x)**(5/3)), x)/a
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/tan(d*x+c)^(5/3)/(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 = 0.19 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.60 \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {i \, {\left (13 \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}\right ) - 3 \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}\right ) + \frac {36 i}{\tan \left (d x + c\right )^{\frac {2}{3}}} + \frac {12 i \, \tan \left (d x + c\right )^{\frac {1}{3}}}{\tan \left (d x + c\right ) - i} + 3 i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right ) + 13 i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right ) - 26 i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right ) - 6 i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )\right )}}{24 \, a d} \] Input:

integrate(1/tan(d*x+c)^(5/3)/(a+I*a*tan(d*x+c)),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 2.63 (sec) , antiderivative size = 630, normalized size of antiderivative = 1.96 \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))} \, dx =\text {Too large to display} \] Input:

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

Output:

log((a^3*d^3*312480i + 331776*a^5*d^5*tan(c + d*x)^(1/3)*(1/(64*a^3*d^3))^ 
(2/3))*(1/(64*a^3*d^3))^(1/3) - 83304*a^2*d^2*tan(c + d*x)^(1/3))*(1/(64*a 
^3*d^3))^(1/3) - (3/(2*a*d) + (tan(c + d*x)*2i)/(a*d))/(tan(c + d*x)^(2/3) 
 + tan(c + d*x)^(5/3)*1i) + (13*log((13*(a^3*d^3*312480i + 389376*a^5*d^5* 
tan(c + d*x)^(1/3)*(1/(a^3*d^3))^(2/3))*(1/(a^3*d^3))^(1/3))/12 - 83304*a^ 
2*d^2*tan(c + d*x)^(1/3))*(1/(a^3*d^3))^(1/3))/12 + (13*log((13*(3^(1/2)*1 
i - 1)*(a^3*d^3*312480i + 97344*a^5*d^5*tan(c + d*x)^(1/3)*(3^(1/2)*1i - 1 
)^2*(1/(a^3*d^3))^(2/3))*(1/(a^3*d^3))^(1/3))/24 - 83304*a^2*d^2*tan(c + d 
*x)^(1/3))*(3^(1/2)*1i - 1)*(1/(a^3*d^3))^(1/3))/24 - (13*log((13*(3^(1/2) 
*1i + 1)*(a^3*d^3*312480i + 97344*a^5*d^5*tan(c + d*x)^(1/3)*(3^(1/2)*1i + 
 1)^2*(1/(a^3*d^3))^(2/3))*(1/(a^3*d^3))^(1/3))/24 + 83304*a^2*d^2*tan(c + 
 d*x)^(1/3))*(3^(1/2)*1i + 1)*(1/(a^3*d^3))^(1/3))/24 + log(((3^(1/2)*1i)/ 
2 - 1/2)*(a^3*d^3*312480i + 331776*a^5*d^5*tan(c + d*x)^(1/3)*((3^(1/2)*1i 
)/2 - 1/2)^2*(1/(64*a^3*d^3))^(2/3))*(1/(64*a^3*d^3))^(1/3) - 83304*a^2*d^ 
2*tan(c + d*x)^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(1/(64*a^3*d^3))^(1/3) - log( 
((3^(1/2)*1i)/2 + 1/2)*(a^3*d^3*312480i + 331776*a^5*d^5*tan(c + d*x)^(1/3 
)*((3^(1/2)*1i)/2 + 1/2)^2*(1/(64*a^3*d^3))^(2/3))*(1/(64*a^3*d^3))^(1/3) 
+ 83304*a^2*d^2*tan(c + d*x)^(1/3))*((3^(1/2)*1i)/2 + 1/2)*(1/(64*a^3*d^3) 
)^(1/3)
 

Reduce [F]

\[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {\int \frac {1}{\tan \left (d x +c \right )^{\frac {8}{3}} i +\tan \left (d x +c \right )^{\frac {5}{3}}}d x}{a} \] Input:

int(1/tan(d*x+c)^(5/3)/(a+I*a*tan(d*x+c)),x)
 

Output:

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