\(\int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx\) [233]

Optimal result

Integrand size = 26, antiderivative size = 343 \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {7 \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {7 \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {5 i \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 \sqrt {3} a d}-\frac {7 \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {5 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}+\frac {7 \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {7 \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}+\frac {5 i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{12 a d}+\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))} \] Output:

-7/12*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))/a/d-7/12*arctan(3^(1/2)+2*tan(d* 
x+c)^(1/3))/a/d-5/6*I*arctan(1/3*(1-2*tan(d*x+c)^(2/3))*3^(1/2))*3^(1/2)/a 
/d-7/6*arctan(tan(d*x+c)^(1/3))/a/d-5/6*I*ln(1+tan(d*x+c)^(2/3))/a/d+7/24* 
ln(1-3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*3^(1/2)/a/d-7/24*ln(1+3^(1 
/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*3^(1/2)/a/d+5/12*I*ln(1-tan(d*x+c)^ 
(2/3)+tan(d*x+c)^(4/3))/a/d+7/2*tan(d*x+c)^(1/3)/a/d-5/4*I*tan(d*x+c)^(4/3 
)/a/d-1/2*tan(d*x+c)^(7/3)/d/(a+I*a*tan(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 5.68 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.03 \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {\sqrt [3]{\tan (c+d x)} \left (-\frac {15 i \tan (c+d x)}{a}-\frac {6 \tan ^2(c+d x)}{a}+\frac {6 i \tan ^3(c+d x)}{a}+\frac {6 \tan ^4(c+d x)}{a+i a \tan (c+d x)}+\frac {42+\frac {-7 i \log \left (1-i \sqrt [6]{\tan ^2(c+d x)}\right )+7 i \log \left (1+i \sqrt [6]{\tan ^2(c+d x)}\right )+7 \sqrt [6]{-1} \left (-(-1)^{2/3} \log \left (1-\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )+(-1)^{2/3} \log \left (1+\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )-\log \left (1-(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )+\log \left (1+(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )\right )}{\sqrt [6]{\tan ^2(c+d x)}}-10 i \cot (c+d x) \left (\log \left (1+\sqrt [3]{\tan ^2(c+d x)}\right )-\sqrt [3]{-1} \log \left (1-\sqrt [3]{-1} \sqrt [3]{\tan ^2(c+d x)}\right )+(-1)^{2/3} \log \left (1+(-1)^{2/3} \sqrt [3]{\tan ^2(c+d x)}\right )\right ) \sqrt [3]{\tan ^2(c+d x)}}{a}\right )}{12 d} \] Input:

Integrate[Tan[c + d*x]^(10/3)/(a + I*a*Tan[c + d*x]),x]
 

Output:

(Tan[c + d*x]^(1/3)*(((-15*I)*Tan[c + d*x])/a - (6*Tan[c + d*x]^2)/a + ((6 
*I)*Tan[c + d*x]^3)/a + (6*Tan[c + d*x]^4)/(a + I*a*Tan[c + d*x]) + (42 + 
((-7*I)*Log[1 - I*(Tan[c + d*x]^2)^(1/6)] + (7*I)*Log[1 + I*(Tan[c + d*x]^ 
2)^(1/6)] + 7*(-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) - (10*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 = 1.00 (sec) , antiderivative size = 279, normalized size of antiderivative = 0.81, number of steps used = 24, number of rules used = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.885, Rules used = {3042, 4033, 27, 3042, 4011, 3042, 4011, 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.

\(\displaystyle \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^{10/3}}{a+i a \tan (c+d x)}dx\)

\(\Big \downarrow \) 4033

\(\displaystyle \frac {\int \frac {1}{3} \tan ^{\frac {4}{3}}(c+d x) (7 a-10 i a \tan (c+d x))dx}{2 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \tan ^{\frac {4}{3}}(c+d x) (7 a-10 i a \tan (c+d x))dx}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \tan (c+d x)^{4/3} (7 a-10 i a \tan (c+d x))dx}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {\int \sqrt [3]{\tan (c+d x)} (7 \tan (c+d x) a+10 i a)dx-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sqrt [3]{\tan (c+d x)} (7 \tan (c+d x) a+10 i a)dx-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {\int \frac {10 i a \tan (c+d x)-7 a}{\tan ^{\frac {2}{3}}(c+d x)}dx-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {10 i a \tan (c+d x)-7 a}{\tan (c+d x)^{2/3}}dx-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 4021

\(\displaystyle \frac {-7 a \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)}dx+10 i a \int \sqrt [3]{\tan (c+d x)}dx-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-7 a \int \frac {1}{\tan (c+d x)^{2/3}}dx+10 i a \int \sqrt [3]{\tan (c+d x)}dx-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {\frac {10 i a \int \frac {\sqrt [3]{\tan (c+d x)}}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}-\frac {7 a \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x) \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {-\frac {21 a \int \frac {1}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {30 i a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 753

\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{2 \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{2 \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}\right )}{d}+\frac {30 i a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )}{d}+\frac {30 i a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {30 i a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {15 i a \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{\tan (c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 821

\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {15 i a \left (\frac {1}{3} \int \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)\right )}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {15 i a \left (\frac {1}{3} \int \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int -\frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {15 i a \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{2} \int \left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {15 i a \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}-\sqrt {3}\right )\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {15 i a \left (\frac {1}{3} \left (-3 \int \frac {1}{-2 \tan ^{\frac {2}{3}}(c+d x)-2}d\left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )-\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {15 i a \left (\frac {1}{3} \left (\sqrt {3} \arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )-\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {15 i a \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {21 a \left (\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )-\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )+\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )\right )}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\)

Input:

Int[Tan[c + d*x]^(10/3)/(a + I*a*Tan[c + d*x]),x]
 

Output:

-1/2*Tan[c + d*x]^(7/3)/(d*(a + I*a*Tan[c + d*x])) + (((15*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 - (21*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 + S 
qrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)])/2)/6))/d + (21*a*Tan[c + 
d*x]^(1/3))/d - (((15*I)/2)*a*Tan[c + d*x]^(4/3))/d)/(6*a^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 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 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 = 1.07 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.62

method	result	size
derivativedivides	\(\frac {3 \tan \left (d x +c \right )^{\frac {1}{3}}-\frac {3 i \tan \left (d x +c \right )^{\frac {4}{3}}}{4}+\frac {i \ln \left (i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{8}+\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{4}-\frac {i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}-i\right )}{4}+\frac {-2 \tan \left (d x +c \right )^{\frac {1}{3}}-2 i}{-12 i \tan \left (d x +c \right )^{\frac {1}{3}}+12 \tan \left (d x +c \right )^{\frac {2}{3}}-12}+\frac {17 i \ln \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{24}-\frac {17 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{12}-\frac {17 i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}{12}+\frac {1}{6 \tan \left (d x +c \right )^{\frac {1}{3}}+6 i}}{d a}\)	\(211\)
default	\(\frac {3 \tan \left (d x +c \right )^{\frac {1}{3}}-\frac {3 i \tan \left (d x +c \right )^{\frac {4}{3}}}{4}+\frac {i \ln \left (i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{8}+\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{4}-\frac {i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}-i\right )}{4}+\frac {-2 \tan \left (d x +c \right )^{\frac {1}{3}}-2 i}{-12 i \tan \left (d x +c \right )^{\frac {1}{3}}+12 \tan \left (d x +c \right )^{\frac {2}{3}}-12}+\frac {17 i \ln \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{24}-\frac {17 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{12}-\frac {17 i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}{12}+\frac {1}{6 \tan \left (d x +c \right )^{\frac {1}{3}}+6 i}}{d a}\)	\(211\)

Input:

int(tan(d*x+c)^(10/3)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/d/a*(3*tan(d*x+c)^(1/3)-3/4*I*tan(d*x+c)^(4/3)+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))-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)+17/24*I*ln(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^ 
(2/3)-1)-17/12*3^(1/2)*arctanh(1/3*(-I+2*tan(d*x+c)^(1/3))*3^(1/2))-17/12* 
I*ln(tan(d*x+c)^(1/3)+I)+1/6/(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. 645 vs. \(2 (270) = 540\).

Time = 0.09 (sec) , antiderivative size = 645, normalized size of antiderivative = 1.88 \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx =\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(10/3)/(a+I*a*tan(d*x+c)),x, algorithm="fricas")
 

Output:

1/24*(3*(sqrt(3)*(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt( 
1/(a^2*d^2)) + I*e^(4*I*d*x + 4*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*e^(4*I*d*x + 4*I*c) + a*d*e^(2 
*I*d*x + 2*I*c))*sqrt(1/(a^2*d^2)) - I*e^(4*I*d*x + 4*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) - 17*(3*sqrt(1/3)*(a*d*e^ 
(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt(1/(a^2*d^2)) - I*e^(4*I* 
d*x + 4*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) + 17*(3*sqrt(1/3)*(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))* 
sqrt(1/(a^2*d^2)) + I*e^(4*I*d*x + 4*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) - 34*(I*e^(4*I*d*x + 4*I*c) + 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^(4*I*d*x + 4*I*c) + 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)*(10*e^(4*I*d*x + 4*I 
*c) + 17*e^(2*I*d*x + 2*I*c) + 1))/(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d 
*x + 2*I*c))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\text {Timed out} \] Input:

integrate(tan(d*x+c)**(10/3)/(a+I*a*tan(d*x+c)),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(tan(d*x+c)^(10/3)/(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 = 201, normalized size of antiderivative = 0.59 \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {i \, {\left (-17 i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}\right ) + 3 i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}\right ) - 18 \, \tan \left (d x + c\right )^{\frac {4}{3}} - 72 i \, \tan \left (d x + c\right )^{\frac {1}{3}} - \frac {12 \, \tan \left (d x + c\right )^{\frac {1}{3}}}{\tan \left (d x + c\right ) - i} + 3 \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right ) + 17 \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right ) - 34 \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right ) - 6 \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )\right )}}{24 \, a d} \] Input:

integrate(tan(d*x+c)^(10/3)/(a+I*a*tan(d*x+c)),x, algorithm="giac")
 

Output:

1/24*I*(-17*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)) - 18*tan(d*x + c)^(4/3) - 
72*I*tan(d*x + c)^(1/3) - 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) + 17*log(tan(d*x + c)^(2/3 
) - I*tan(d*x + c)^(1/3) - 1) - 34*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 = 3.09 (sec) , antiderivative size = 655, normalized size of antiderivative = 1.91 \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\text {Too large to display} \] Input:

int(tan(c + d*x)^(10/3)/(a + a*tan(c + d*x)*1i),x)
 

Output:

log((a^3*d^3*703584i - 414720*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)*182376i)*(1i/ 
(64*a^3*d^3))^(1/3) + log((a^3*d^3*703584i - 414720*a^5*d^5*tan(c + d*x)^( 
1/3)*(4913i/(1728*a^3*d^3))^(2/3))*(4913i/(1728*a^3*d^3))^(1/3) + a^2*d^2* 
tan(c + d*x)^(1/3)*182376i)*(4913i/(1728*a^3*d^3))^(1/3) + (3*tan(c + d*x) 
^(1/3))/(a*d) - (tan(c + d*x)^(4/3)*3i)/(4*a*d) + (log(((3^(1/2)*1i - 1)*( 
a^3*d^3*703584i - 103680*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)*182376i)*(3^(1/2)*1i - 1)*(1i/(64*a^3*d^3))^(1/3))/2 - (log(((3^(1/2)* 
1i + 1)*(a^3*d^3*703584i - 103680*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)*182376i)*(3^(1/2)*1i + 1)*(1i/(64*a^3*d^3))^(1/3))/2 + (log(( 
(3^(1/2)*1i - 1)*(a^3*d^3*703584i - 103680*a^5*d^5*tan(c + d*x)^(1/3)*(3^( 
1/2)*1i - 1)^2*(4913i/(1728*a^3*d^3))^(2/3))*(4913i/(1728*a^3*d^3))^(1/3)) 
/2 + a^2*d^2*tan(c + d*x)^(1/3)*182376i)*(3^(1/2)*1i - 1)*(4913i/(1728*a^3 
*d^3))^(1/3))/2 - (log(((3^(1/2)*1i + 1)*(a^3*d^3*703584i - 103680*a^5*d^5 
*tan(c + d*x)^(1/3)*(3^(1/2)*1i + 1)^2*(4913i/(1728*a^3*d^3))^(2/3))*(4913 
i/(1728*a^3*d^3))^(1/3))/2 - a^2*d^2*tan(c + d*x)^(1/3)*182376i)*(3^(1/2)* 
1i + 1)*(4913i/(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]

\[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {\int \frac {\tan \left (d x +c \right )^{\frac {10}{3}}}{\tan \left (d x +c \right ) i +1}d x}{a} \] Input:

int(tan(d*x+c)^(10/3)/(a+I*a*tan(d*x+c)),x)
 

Output:

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