\(\int \frac {1}{\tan ^{\frac {7}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx\) [247]

Optimal result

Integrand size = 26, antiderivative size = 381 \[ \int \frac {1}{\tan ^{\frac {7}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=-\frac {91 i \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {91 i \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {25 \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{6 \sqrt {3} a^2 d}+\frac {91 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {25 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{18 a^2 d}+\frac {91 i \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {91 i \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}+\frac {25 \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{36 a^2 d}-\frac {25}{12 a^2 d \tan ^{\frac {4}{3}}(c+d x)}+\frac {13}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}+\frac {91 i}{12 a^2 d \sqrt [3]{\tan (c+d x)}}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2} \] Output:

91/72*I*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))/a^2/d+91/72*I*arctan(3^(1/2)+2 
*tan(d*x+c)^(1/3))/a^2/d+25/18*arctan(1/3*(1-2*tan(d*x+c)^(2/3))*3^(1/2))* 
3^(1/2)/a^2/d+91/36*I*arctan(tan(d*x+c)^(1/3))/a^2/d-25/18*ln(1+tan(d*x+c) 
^(2/3))/a^2/d+91/144*I*ln(1-3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*3^( 
1/2)/a^2/d-91/144*I*ln(1+3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*3^(1/2 
)/a^2/d+25/36*ln(1-tan(d*x+c)^(2/3)+tan(d*x+c)^(4/3))/a^2/d-25/12/a^2/d/ta 
n(d*x+c)^(4/3)+13/12/a^2/d/(1+I*tan(d*x+c))/tan(d*x+c)^(4/3)+91/12*I/a^2/d 
/tan(d*x+c)^(1/3)+1/4/d/tan(d*x+c)^(4/3)/(a+I*a*tan(d*x+c))^2
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 6.30 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.13 \[ \int \frac {1}{\tan ^{\frac {7}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\frac {3 a^2+\frac {1}{6} a (a+i a \tan (c+d x)) \left (78+(-i+\tan (c+d x)) \left (91 \tan (c+d x) \left (-6-i \log \left (1-i \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}+i \log \left (1+i \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}-\sqrt [6]{-1} \log \left (1-\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}+\sqrt [6]{-1} \log \left (1+\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}-(-1)^{5/6} \log \left (1-(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}+(-1)^{5/6} \log \left (1+(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}\right )+50 \left (-3 i-2 i \log \left (1+\sqrt [3]{\tan ^2(c+d x)}\right ) \tan ^2(c+d x)^{2/3}+2 \sqrt [6]{-1} \log \left (1-\sqrt [3]{-1} \sqrt [3]{\tan ^2(c+d x)}\right ) \tan ^2(c+d x)^{2/3}+2 (-1)^{5/6} \log \left (1+(-1)^{2/3} \sqrt [3]{\tan ^2(c+d x)}\right ) \tan ^2(c+d x)^{2/3}\right )\right )\right )}{12 a^2 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2} \] Input:

Integrate[1/(Tan[c + d*x]^(7/3)*(a + I*a*Tan[c + d*x])^2),x]
 

Output:

(3*a^2 + (a*(a + I*a*Tan[c + d*x])*(78 + (-I + Tan[c + d*x])*(91*Tan[c + d 
*x]*(-6 - I*Log[1 - I*(Tan[c + d*x]^2)^(1/6)]*(Tan[c + d*x]^2)^(1/6) + I*L 
og[1 + I*(Tan[c + d*x]^2)^(1/6)]*(Tan[c + d*x]^2)^(1/6) - (-1)^(1/6)*Log[1 
 - (-1)^(1/6)*(Tan[c + d*x]^2)^(1/6)]*(Tan[c + d*x]^2)^(1/6) + (-1)^(1/6)* 
Log[1 + (-1)^(1/6)*(Tan[c + d*x]^2)^(1/6)]*(Tan[c + d*x]^2)^(1/6) - (-1)^( 
5/6)*Log[1 - (-1)^(5/6)*(Tan[c + d*x]^2)^(1/6)]*(Tan[c + d*x]^2)^(1/6) + ( 
-1)^(5/6)*Log[1 + (-1)^(5/6)*(Tan[c + d*x]^2)^(1/6)]*(Tan[c + d*x]^2)^(1/6 
)) + 50*(-3*I - (2*I)*Log[1 + (Tan[c + d*x]^2)^(1/3)]*(Tan[c + d*x]^2)^(2/ 
3) + 2*(-1)^(1/6)*Log[1 - (-1)^(1/3)*(Tan[c + d*x]^2)^(1/3)]*(Tan[c + d*x] 
^2)^(2/3) + 2*(-1)^(5/6)*Log[1 + (-1)^(2/3)*(Tan[c + d*x]^2)^(1/3)]*(Tan[c 
 + d*x]^2)^(2/3)))))/6)/(12*a^2*d*Tan[c + d*x]^(4/3)*(a + I*a*Tan[c + d*x] 
)^2)
 

Rubi [A] (warning: unable to verify)

Time = 1.33 (sec) , antiderivative size = 324, normalized size of antiderivative = 0.85, number of steps used = 28, number of rules used = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.038, Rules used = {3042, 4042, 27, 3042, 4079, 27, 3042, 4012, 25, 3042, 4012, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4042

\(\displaystyle \frac {\int \frac {2 (8 a-5 i a \tan (c+d x))}{3 \tan ^{\frac {7}{3}}(c+d x) (i \tan (c+d x) a+a)}dx}{4 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {8 a-5 i a \tan (c+d x)}{\tan ^{\frac {7}{3}}(c+d x) (i \tan (c+d x) a+a)}dx}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {8 a-5 i a \tan (c+d x)}{\tan (c+d x)^{7/3} (i \tan (c+d x) a+a)}dx}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\int \frac {100 a^2-91 i a^2 \tan (c+d x)}{3 \tan ^{\frac {7}{3}}(c+d x)}dx}{2 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {100 a^2-91 i a^2 \tan (c+d x)}{\tan ^{\frac {7}{3}}(c+d x)}dx}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {100 a^2-91 i a^2 \tan (c+d x)}{\tan (c+d x)^{7/3}}dx}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4012

\(\displaystyle \frac {\frac {-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}+\int -\frac {100 \tan (c+d x) a^2+91 i a^2}{\tan ^{\frac {4}{3}}(c+d x)}dx}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}-\int \frac {100 \tan (c+d x) a^2+91 i a^2}{\tan ^{\frac {4}{3}}(c+d x)}dx}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}-\int \frac {100 \tan (c+d x) a^2+91 i a^2}{\tan (c+d x)^{4/3}}dx}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4012

\(\displaystyle \frac {\frac {-\int \frac {100 a^2-91 i a^2 \tan (c+d x)}{\sqrt [3]{\tan (c+d x)}}dx-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}+\frac {273 i a^2}{d \sqrt [3]{\tan (c+d x)}}}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\int \frac {100 a^2-91 i a^2 \tan (c+d x)}{\sqrt [3]{\tan (c+d x)}}dx-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}+\frac {273 i a^2}{d \sqrt [3]{\tan (c+d x)}}}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4021

\(\displaystyle \frac {\frac {91 i a^2 \int \tan ^{\frac {2}{3}}(c+d x)dx-100 a^2 \int \frac {1}{\sqrt [3]{\tan (c+d x)}}dx-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}+\frac {273 i a^2}{d \sqrt [3]{\tan (c+d x)}}}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-100 a^2 \int \frac {1}{\sqrt [3]{\tan (c+d x)}}dx+91 i a^2 \int \tan (c+d x)^{2/3}dx-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}+\frac {273 i a^2}{d \sqrt [3]{\tan (c+d x)}}}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {\frac {-\frac {100 a^2 \int \frac {1}{\sqrt [3]{\tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}+\frac {91 i a^2 \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}+\frac {273 i a^2}{d \sqrt [3]{\tan (c+d x)}}}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {\frac {-\frac {300 a^2 \int \frac {\sqrt [3]{\tan (c+d x)}}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {273 i a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}+\frac {273 i a^2}{d \sqrt [3]{\tan (c+d x)}}}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {\frac {-\frac {150 a^2 \int \frac {1}{\tan (c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)}{d}+\frac {273 i a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}+\frac {273 i a^2}{d \sqrt [3]{\tan (c+d x)}}}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 750

\(\displaystyle \frac {\frac {-\frac {150 a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)\right )}{d}+\frac {273 i a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}+\frac {273 i a^2}{d \sqrt [3]{\tan (c+d x)}}}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {\frac {\frac {273 i a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {150 a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}+\frac {273 i a^2}{d \sqrt [3]{\tan (c+d x)}}}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 824

\(\displaystyle \frac {\frac {\frac {273 i a^2 \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int -\frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{2 \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int -\frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{2 \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}\right )}{d}-\frac {150 a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}+\frac {273 i a^2}{d \sqrt [3]{\tan (c+d x)}}}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {273 i a^2 \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )}{d}-\frac {150 a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}+\frac {273 i a^2}{d \sqrt [3]{\tan (c+d x)}}}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {273 i a^2 \left (-\frac {1}{6} \int \frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {150 a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}+\frac {273 i a^2}{d \sqrt [3]{\tan (c+d x)}}}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {\frac {\frac {273 i a^2 \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int -\frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {150 a^2 \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{2} \int \left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )d\tan ^{\frac {2}{3}}(c+d x)\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}+\frac {273 i a^2}{d \sqrt [3]{\tan (c+d x)}}}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {273 i a^2 \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {150 a^2 \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}+\frac {273 i a^2}{d \sqrt [3]{\tan (c+d x)}}}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {\frac {273 i a^2 \left (\frac {1}{6} \left (-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}-\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {150 a^2 \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)-3 \int \frac {1}{-2 \tan ^{\frac {2}{3}}(c+d x)-2}d\left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}+\frac {273 i a^2}{d \sqrt [3]{\tan (c+d x)}}}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {\frac {273 i a^2 \left (\frac {1}{6} \left (-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {150 a^2 \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}+\frac {273 i a^2}{d \sqrt [3]{\tan (c+d x)}}}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {-\frac {150 a^2 \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}+\frac {273 i a^2 \left (\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )\right )}{d}-\frac {75 a^2}{d \tan ^{\frac {4}{3}}(c+d x)}+\frac {273 i a^2}{d \sqrt [3]{\tan (c+d x)}}}{6 a^2}+\frac {13}{2 d (1+i \tan (c+d x)) \tan ^{\frac {4}{3}}(c+d x)}}{6 a^2}+\frac {1}{4 d \tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^2}\)

Input:

Int[1/(Tan[c + d*x]^(7/3)*(a + I*a*Tan[c + d*x])^2),x]
 

Output:

(((-150*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 + ((273*I)*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 - (75*a^2)/(d*Tan[c + d*x]^(4/3)) + ((273*I)*a^2)/(d*Tan[c + 
d*x]^(1/3)))/(6*a^2) + 13/(2*d*(1 + I*Tan[c + d*x])*Tan[c + d*x]^(4/3)))/( 
6*a^2) + 1/(4*d*Tan[c + d*x]^(4/3)*(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 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 4042

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

rule 4079

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*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( 
b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   Int[(a + b*Tan[e + f*x])^(m 
 + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m 
- b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free 
Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] 
 && LtQ[m, 0] &&  !GtQ[n, 0]
 

Maple [A] (verified)

Time = 1.08 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.64

method	result	size
derivativedivides	\(\frac {-\frac {-76 i \tan \left (d x +c \right )-116 \tan \left (d x +c \right )^{\frac {2}{3}}+122 i \tan \left (d x +c \right )^{\frac {1}{3}}+40}{72 \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )^{2}}+\frac {191 \ln \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{144}-\frac {191 i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{72}-\frac {3}{4 \tan \left (d x +c \right )^{\frac {4}{3}}}+\frac {6 i}{\tan \left (d x +c \right )^{\frac {1}{3}}}+\frac {19 i}{36 \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}-\frac {1}{36 \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )^{2}}-\frac {191 \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}{72}+\frac {\ln \left (i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{16}+\frac {i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{8}-\frac {\ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}-i\right )}{8}}{d \,a^{2}}\)	\(244\)
default	\(\frac {-\frac {-76 i \tan \left (d x +c \right )-116 \tan \left (d x +c \right )^{\frac {2}{3}}+122 i \tan \left (d x +c \right )^{\frac {1}{3}}+40}{72 \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )^{2}}+\frac {191 \ln \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{144}-\frac {191 i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{72}-\frac {3}{4 \tan \left (d x +c \right )^{\frac {4}{3}}}+\frac {6 i}{\tan \left (d x +c \right )^{\frac {1}{3}}}+\frac {19 i}{36 \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}-\frac {1}{36 \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )^{2}}-\frac {191 \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}{72}+\frac {\ln \left (i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{16}+\frac {i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{8}-\frac {\ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}-i\right )}{8}}{d \,a^{2}}\)	\(244\)

Input:

int(1/tan(d*x+c)^(7/3)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/d/a^2*(-1/72*(-76*I*tan(d*x+c)-116*tan(d*x+c)^(2/3)+122*I*tan(d*x+c)^(1/ 
3)+40)/(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)^2+191/144*ln(-I*tan(d*x+c) 
^(1/3)+tan(d*x+c)^(2/3)-1)-191/72*I*3^(1/2)*arctanh(1/3*(-I+2*tan(d*x+c)^( 
1/3))*3^(1/2))-3/4/tan(d*x+c)^(4/3)+6*I/tan(d*x+c)^(1/3)+19/36*I/(tan(d*x+ 
c)^(1/3)+I)-1/36/(tan(d*x+c)^(1/3)+I)^2-191/72*ln(tan(d*x+c)^(1/3)+I)+1/16 
*ln(I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)+1/8*I*3^(1/2)*arctanh(1/3*(I+2* 
tan(d*x+c)^(1/3))*3^(1/2))-1/8*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. 836 vs. \(2 (298) = 596\).

Time = 0.11 (sec) , antiderivative size = 836, normalized size of antiderivative = 2.19 \[ \int \frac {1}{\tan ^{\frac {7}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\text {Too large to display} \] Input:

integrate(1/tan(d*x+c)^(7/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")
 

Output:

-1/144*(9*(sqrt(3)*(-I*a^2*d*e^(8*I*d*x + 8*I*c) + 2*I*a^2*d*e^(6*I*d*x + 
6*I*c) - I*a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/(a^4*d^2)) - e^(8*I*d*x + 8*I 
*c) + 2*e^(6*I*d*x + 6*I*c) - e^(4*I*d*x + 4*I*c))*log(1/2*sqrt(3)*a^2*d*s 
qrt(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)*(I*a^2*d*e^(8*I*d*x + 8*I*c) - 2*I*a^2*d*e^( 
6*I*d*x + 6*I*c) + I*a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/(a^4*d^2)) - e^(8*I 
*d*x + 8*I*c) + 2*e^(6*I*d*x + 6*I*c) - 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) + 191*(3*sqrt(1/3)*(I*a^2*d*e^(8*I*d*x + 8*I*c 
) - 2*I*a^2*d*e^(6*I*d*x + 6*I*c) + I*a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/(a 
^4*d^2)) - e^(8*I*d*x + 8*I*c) + 2*e^(6*I*d*x + 6*I*c) - 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) + 191*(3*sqrt(1/3)*(-I*a^2*d 
*e^(8*I*d*x + 8*I*c) + 2*I*a^2*d*e^(6*I*d*x + 6*I*c) - I*a^2*d*e^(4*I*d*x 
+ 4*I*c))*sqrt(1/(a^4*d^2)) - e^(8*I*d*x + 8*I*c) + 2*e^(6*I*d*x + 6*I*c) 
- 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) + 382*(e 
^(8*I*d*x + 8*I*c) - 2*e^(6*I*d*x + 6*I*c) + 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*(e^(8 
*I*d*x + 8*I*c) - 2*e^(6*I*d*x + 6*I*c) + e^(4*I*d*x + 4*I*c))*log(((-I...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\tan ^{\frac {7}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate(1/tan(d*x+c)**(7/3)/(a+I*a*tan(d*x+c))**2,x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\tan ^{\frac {7}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/tan(d*x+c)^(7/3)/(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 = 0.20 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.56 \[ \int \frac {1}{\tan ^{\frac {7}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\frac {191 i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}\right ) - 9 i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}\right ) + \frac {108 \, {\left (8 i \, \tan \left (d x + c\right ) - 1\right )}}{\tan \left (d x + c\right )^{\frac {4}{3}}} + \frac {12 \, {\left (19 i \, \tan \left (d x + c\right )^{\frac {5}{3}} + 22 \, \tan \left (d x + c\right )^{\frac {2}{3}}\right )}}{{\left (\tan \left (d x + c\right ) - i\right )}^{2}} + 9 \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right ) + 191 \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right ) - 382 \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right ) - 18 \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )}{144 \, a^{2} d} \] Input:

integrate(1/tan(d*x+c)^(7/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")
 

Output:

1/144*(191*I*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1/3) + I)/(sqrt(3) + 
2*tan(d*x + c)^(1/3) - I)) - 9*I*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1 
/3) - I)/(sqrt(3) + 2*tan(d*x + c)^(1/3) + I)) + 108*(8*I*tan(d*x + c) - 1 
)/tan(d*x + c)^(4/3) + 12*(19*I*tan(d*x + c)^(5/3) + 22*tan(d*x + c)^(2/3) 
)/(tan(d*x + c) - I)^2 + 9*log(tan(d*x + c)^(2/3) + I*tan(d*x + c)^(1/3) - 
 1) + 191*log(tan(d*x + c)^(2/3) - I*tan(d*x + c)^(1/3) - 1) - 382*log(tan 
(d*x + c)^(1/3) + I) - 18*log(tan(d*x + c)^(1/3) - I))/(a^2*d)
 

Mupad [B] (verification not implemented)

Time = 2.78 (sec) , antiderivative size = 652, normalized size of antiderivative = 1.71 \[ \int \frac {1}{\tan ^{\frac {7}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\text {Too large to display} \] Input:

int(1/(tan(c + d*x)^(7/3)*(a + a*tan(c + d*x)*1i)^2),x)
 

Output:

(191*log((14592400*a^2*d*tan(c + d*x)^(1/3))/3 - (36481*((a^6*d^3*89179417 
6i)/3 - (893867776*a^8*d^4*tan(c + d*x)^(1/3)*(-1/(a^6*d^3))^(1/3))/3)*(-1 
/(a^6*d^3))^(2/3))/5184)*(-1/(a^6*d^3))^(1/3))/72 + (3i/(4*a^2*d) + (9*tan 
(c + d*x))/(2*a^2*d) + (tan(c + d*x)^2*157i)/(12*a^2*d) - (91*tan(c + d*x) 
^3)/(12*a^2*d))/(2*tan(c + d*x)^(7/3) - tan(c + d*x)^(4/3)*1i + tan(c + d* 
x)^(10/3)*1i) + log((14592400*a^2*d*tan(c + d*x)^(1/3))/3 - ((a^6*d^3*8917 
94176i)/3 - 112318464*a^8*d^4*tan(c + d*x)^(1/3)*(-1/(512*a^6*d^3))^(1/3)) 
*(-1/(512*a^6*d^3))^(2/3))*(-1/(512*a^6*d^3))^(1/3) + log((14592400*a^2*d* 
tan(c + d*x)^(1/3))/3 - ((3^(1/2)*1i)/2 - 1/2)^2*((a^6*d^3*891794176i)/3 - 
 112318464*a^8*d^4*tan(c + d*x)^(1/3)*((3^(1/2)*1i)/2 - 1/2)*(-1/(512*a^6* 
d^3))^(1/3))*(-1/(512*a^6*d^3))^(2/3))*((3^(1/2)*1i)/2 - 1/2)*(-1/(512*a^6 
*d^3))^(1/3) - log((14592400*a^2*d*tan(c + d*x)^(1/3))/3 - ((3^(1/2)*1i)/2 
 + 1/2)^2*((a^6*d^3*891794176i)/3 + 112318464*a^8*d^4*tan(c + d*x)^(1/3)*( 
(3^(1/2)*1i)/2 + 1/2)*(-1/(512*a^6*d^3))^(1/3))*(-1/(512*a^6*d^3))^(2/3))* 
((3^(1/2)*1i)/2 + 1/2)*(-1/(512*a^6*d^3))^(1/3) + (191*log((14592400*a^2*d 
*tan(c + d*x)^(1/3))/3 - (36481*(3^(1/2)*1i - 1)^2*((a^6*d^3*891794176i)/3 
 - (446933888*a^8*d^4*tan(c + d*x)^(1/3)*(3^(1/2)*1i - 1)*(-1/(a^6*d^3))^( 
1/3))/3)*(-1/(a^6*d^3))^(2/3))/20736)*(3^(1/2)*1i - 1)*(-1/(a^6*d^3))^(1/3 
))/144 - (191*log((14592400*a^2*d*tan(c + d*x)^(1/3))/3 - (36481*(3^(1/2)* 
1i + 1)^2*((a^6*d^3*891794176i)/3 + (446933888*a^8*d^4*tan(c + d*x)^(1/...
 

Reduce [F]

\[ \int \frac {1}{\tan ^{\frac {7}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=-\frac {\int \frac {1}{\tan \left (d x +c \right )^{\frac {13}{3}}-2 \tan \left (d x +c \right )^{\frac {10}{3}} i -\tan \left (d x +c \right )^{\frac {7}{3}}}d x}{a^{2}} \] Input:

int(1/tan(d*x+c)^(7/3)/(a+I*a*tan(d*x+c))^2,x)
 

Output:

( - int(1/(tan(c + d*x)**(1/3)*tan(c + d*x)**4 - 2*tan(c + d*x)**(1/3)*tan 
(c + d*x)**3*i - tan(c + d*x)**(1/3)*tan(c + d*x)**2),x))/a**2