\(\int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx\) [235]

Optimal result

Integrand size = 26, antiderivative size = 299 \[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=-\frac {\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {\arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {i \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{\sqrt {3} a d}+\frac {\arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}-\frac {\log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}+\frac {\log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{6 a d}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 4.60 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.08 \[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {\sqrt [3]{\tan (c+d x)} \left (\frac {6 i \tan (c+d x)}{a}+\frac {6 \tan ^2(c+d x)}{a+i a \tan (c+d x)}+\frac {-6+\frac {i \log \left (1-i \sqrt [6]{\tan ^2(c+d x)}\right )-i \log \left (1+i \sqrt [6]{\tan ^2(c+d x)}\right )+\sqrt [6]{-1} \left ((-1)^{2/3} \log \left (1-\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )-(-1)^{2/3} \log \left (1+\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )+\log \left (1-(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )-\log \left (1+(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )\right )}{\sqrt [6]{\tan ^2(c+d x)}}+4 i \cot (c+d x) \left (\log \left (1+\sqrt [3]{\tan ^2(c+d x)}\right )-\sqrt [3]{-1} \log \left (1-\sqrt [3]{-1} \sqrt [3]{\tan ^2(c+d x)}\right )+(-1)^{2/3} \log \left (1+(-1)^{2/3} \sqrt [3]{\tan ^2(c+d x)}\right )\right ) \sqrt [3]{\tan ^2(c+d x)}}{a}\right )}{12 d} \] Input:

Integrate[Tan[c + d*x]^(4/3)/(a + I*a*Tan[c + d*x]),x]
 

Output:

(Tan[c + d*x]^(1/3)*(((6*I)*Tan[c + d*x])/a + (6*Tan[c + d*x]^2)/(a + I*a* 
Tan[c + d*x]) + (-6 + (I*Log[1 - I*(Tan[c + d*x]^2)^(1/6)] - I*Log[1 + I*( 
Tan[c + d*x]^2)^(1/6)] + (-1)^(1/6)*((-1)^(2/3)*Log[1 - (-1)^(1/6)*(Tan[c 
+ d*x]^2)^(1/6)] - (-1)^(2/3)*Log[1 + (-1)^(1/6)*(Tan[c + d*x]^2)^(1/6)] + 
 Log[1 - (-1)^(5/6)*(Tan[c + d*x]^2)^(1/6)] - Log[1 + (-1)^(5/6)*(Tan[c + 
d*x]^2)^(1/6)]))/(Tan[c + d*x]^2)^(1/6) + (4*I)*Cot[c + d*x]*(Log[1 + (Tan 
[c + d*x]^2)^(1/3)] - (-1)^(1/3)*Log[1 - (-1)^(1/3)*(Tan[c + d*x]^2)^(1/3) 
] + (-1)^(2/3)*Log[1 + (-1)^(2/3)*(Tan[c + d*x]^2)^(1/3)])*(Tan[c + d*x]^2 
)^(1/3))/a))/(12*d)
 

Rubi [A] (warning: unable to verify)

Time = 0.77 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.81, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.731, Rules used = {3042, 4033, 27, 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.

\(\displaystyle \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4033

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4021

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 753

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 821

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

Input:

Int[Tan[c + d*x]^(4/3)/(a + I*a*Tan[c + d*x]),x]
 

Output:

(((-6*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 + (3*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)/(6*a^2) - Tan[c + d*x]^(1/3)/(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 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 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 = 1.01 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.64

method	result	size
derivativedivides	\(\frac {\frac {5 i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}{12}-\frac {1}{6 \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}-\frac {i \ln \left (i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{8}-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{4}+\frac {i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}-i\right )}{4}-\frac {-2 \tan \left (d x +c \right )^{\frac {1}{3}}-2 i}{12 \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}-\frac {5 i \ln \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{24}+\frac {5 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{12}}{d a}\)	\(190\)
default	\(\frac {\frac {5 i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}{12}-\frac {1}{6 \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}-\frac {i \ln \left (i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{8}-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{4}+\frac {i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}-i\right )}{4}-\frac {-2 \tan \left (d x +c \right )^{\frac {1}{3}}-2 i}{12 \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}-\frac {5 i \ln \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{24}+\frac {5 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{12}}{d a}\)	\(190\)

Input:

int(tan(d*x+c)^(4/3)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/d/a*(5/12*I*ln(tan(d*x+c)^(1/3)+I)-1/6/(tan(d*x+c)^(1/3)+I)-1/8*I*ln(I*t 
an(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))+1/4*I*ln(tan(d*x+c)^(1/3)-I)-1/12*(-2*tan(d*x+c)^(1/3)-2 
*I)/(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)-5/24*I*ln(-I*tan(d*x+c)^(1/3) 
+tan(d*x+c)^(2/3)-1)+5/12*3^(1/2)*arctanh(1/3*(-I+2*tan(d*x+c)^(1/3))*3^(1 
/2)))
 

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. 492 vs. \(2 (238) = 476\).

Time = 0.09 (sec) , antiderivative size = 492, normalized size of antiderivative = 1.65 \[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx =\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(4/3)/(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) - 5*(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) + 5* 
(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) - 10*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))^(1/3)*(e^(2*I*d*x + 2*I*c) + 1))*e^(-2*I*d*x - 2*I*c)/(a*d)
 

Sympy [F]

\[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=- \frac {i \int \frac {\tan ^{\frac {4}{3}}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx}{a} \] Input:

integrate(tan(d*x+c)**(4/3)/(a+I*a*tan(d*x+c)),x)
 

Output:

-I*Integral(tan(c + d*x)**(4/3)/(tan(c + d*x) - I), x)/a
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(tan(d*x+c)^(4/3)/(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 = 0.16 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.61 \[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {i \, {\left (5 i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}\right ) - 3 i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}\right ) + \frac {12 \, \tan \left (d x + c\right )^{\frac {1}{3}}}{\tan \left (d x + c\right ) - i} - 3 \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right ) - 5 \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right ) + 10 \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right ) + 6 \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )\right )}}{24 \, a d} \] Input:

integrate(tan(d*x+c)^(4/3)/(a+I*a*tan(d*x+c)),x, algorithm="giac")
 

Output:

1/24*I*(5*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)) + 12*tan(d*x + c)^(1/3)/(tan 
(d*x + c) - I) - 3*log(tan(d*x + c)^(2/3) + I*tan(d*x + c)^(1/3) - 1) - 5* 
log(tan(d*x + c)^(2/3) - I*tan(d*x + c)^(1/3) - 1) + 10*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 = 2.51 (sec) , antiderivative size = 622, normalized size of antiderivative = 2.08 \[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx =\text {Too large to display} \] Input:

int(tan(c + d*x)^(4/3)/(a + a*tan(c + d*x)*1i),x)
                                                                                    
                                                                                    
 

Output:

log((a^3*d^3*14112i - 165888*a^5*d^5*tan(c + d*x)^(1/3)*(-1i/(64*a^3*d^3)) 
^(2/3))*(-1i/(64*a^3*d^3))^(1/3) - a^2*d^2*tan(c + d*x)^(1/3)*6120i)*(-1i/ 
(64*a^3*d^3))^(1/3) + log((a^3*d^3*14112i - 165888*a^5*d^5*tan(c + d*x)^(1 
/3)*(-125i/(1728*a^3*d^3))^(2/3))*(-125i/(1728*a^3*d^3))^(1/3) - a^2*d^2*t 
an(c + d*x)^(1/3)*6120i)*(-125i/(1728*a^3*d^3))^(1/3) + (log(((3^(1/2)*1i 
- 1)*(a^3*d^3*14112i - 41472*a^5*d^5*tan(c + d*x)^(1/3)*(3^(1/2)*1i - 1)^2 
*(-1i/(64*a^3*d^3))^(2/3))*(-1i/(64*a^3*d^3))^(1/3))/2 - a^2*d^2*tan(c + d 
*x)^(1/3)*6120i)*(3^(1/2)*1i - 1)*(-1i/(64*a^3*d^3))^(1/3))/2 - (log(((3^( 
1/2)*1i + 1)*(a^3*d^3*14112i - 41472*a^5*d^5*tan(c + d*x)^(1/3)*(3^(1/2)*1 
i + 1)^2*(-1i/(64*a^3*d^3))^(2/3))*(-1i/(64*a^3*d^3))^(1/3))/2 + a^2*d^2*t 
an(c + d*x)^(1/3)*6120i)*(3^(1/2)*1i + 1)*(-1i/(64*a^3*d^3))^(1/3))/2 + (l 
og(((3^(1/2)*1i - 1)*(a^3*d^3*14112i - 41472*a^5*d^5*tan(c + d*x)^(1/3)*(3 
^(1/2)*1i - 1)^2*(-125i/(1728*a^3*d^3))^(2/3))*(-125i/(1728*a^3*d^3))^(1/3 
))/2 - a^2*d^2*tan(c + d*x)^(1/3)*6120i)*(3^(1/2)*1i - 1)*(-125i/(1728*a^3 
*d^3))^(1/3))/2 - (log(((3^(1/2)*1i + 1)*(a^3*d^3*14112i - 41472*a^5*d^5*t 
an(c + d*x)^(1/3)*(3^(1/2)*1i + 1)^2*(-125i/(1728*a^3*d^3))^(2/3))*(-125i/ 
(1728*a^3*d^3))^(1/3))/2 + a^2*d^2*tan(c + d*x)^(1/3)*6120i)*(3^(1/2)*1i + 
 1)*(-125i/(1728*a^3*d^3))^(1/3))/2 - tan(c + d*x)^(1/3)/(2*a*d*(tan(c + d 
*x)*1i + 1))
 

Reduce [F]

\[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {\int \frac {\tan \left (d x +c \right )^{\frac {4}{3}}}{\tan \left (d x +c \right ) i +1}d x}{a} \] Input:

int(tan(d*x+c)^(4/3)/(a+I*a*tan(d*x+c)),x)
 

Output:

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