\(\int \frac {\tan ^{\frac {14}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [240]

Optimal result

Integrand size = 26, antiderivative size = 379 \[ \int \frac {\tan ^{\frac {14}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {121 \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {121 \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {14 i \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}+\frac {121 \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {14 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}+\frac {121 \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {121 \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {7 i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{9 a^2 d}-\frac {14 i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d}-\frac {121 \tan ^{\frac {5}{3}}(c+d x)}{60 a^2 d}+\frac {7 i \tan ^{\frac {8}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \] Output:

121/72*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))/a^2/d+121/72*arctan(3^(1/2)+2*t 
an(d*x+c)^(1/3))/a^2/d-14/9*I*arctan(1/3*(1-2*tan(d*x+c)^(2/3))*3^(1/2))*3 
^(1/2)/a^2/d+121/36*arctan(tan(d*x+c)^(1/3))/a^2/d+14/9*I*ln(1+tan(d*x+c)^ 
(2/3))/a^2/d+121/144*ln(1-3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*3^(1/ 
2)/a^2/d-121/144*ln(1+3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*3^(1/2)/a 
^2/d-7/9*I*ln(1-tan(d*x+c)^(2/3)+tan(d*x+c)^(4/3))/a^2/d-14/3*I*tan(d*x+c) 
^(2/3)/a^2/d-121/60*tan(d*x+c)^(5/3)/a^2/d+7/6*I*tan(d*x+c)^(8/3)/a^2/d/(1 
+I*tan(d*x+c))-1/4*tan(d*x+c)^(11/3)/d/(a+I*a*tan(d*x+c))^2
 

Mathematica [C] (warning: unable to verify)

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

Time = 3.57 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.55 \[ \int \frac {\tan ^{\frac {14}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {\sec ^2(c+d x) \left (90 i \sqrt [3]{2} e^{2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {2}{3},\frac {5}{3},\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right )+4 \left (344 i+776 i \cos (2 (c+d x))+1165 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},-\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (-i \cos (2 (c+d x))+\sin (2 (c+d x)))-403 \sec (c+d x) \sin (3 (c+d x))-547 \tan (c+d x)\right )\right ) \tan ^{\frac {2}{3}}(c+d x)}{960 a^2 d (-i+\tan (c+d x))^2} \] Input:

Integrate[Tan[c + d*x]^(14/3)/(a + I*a*Tan[c + d*x])^2,x]
 

Output:

(Sec[c + d*x]^2*((90*I)*2^(1/3)*E^((2*I)*(c + d*x))*(1 + E^((2*I)*(c + d*x 
)))^(2/3)*Hypergeometric2F1[2/3, 2/3, 5/3, (1 - E^((2*I)*(c + d*x)))/2] + 
4*(344*I + (776*I)*Cos[2*(c + d*x)] + 1165*Hypergeometric2F1[2/3, 1, 5/3, 
-((-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x))))]*((-I)*Cos[2*(c + 
d*x)] + Sin[2*(c + d*x)]) - 403*Sec[c + d*x]*Sin[3*(c + d*x)] - 547*Tan[c 
+ d*x]))*Tan[c + d*x]^(2/3))/(960*a^2*d*(-I + Tan[c + d*x])^2)
 

Rubi [A] (warning: unable to verify)

Time = 1.32 (sec) , antiderivative size = 326, normalized size of antiderivative = 0.86, number of steps used = 27, number of rules used = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 4041, 27, 3042, 4078, 27, 3042, 4011, 3042, 4011, 3042, 4021, 3042, 3957, 266, 807, 750, 16, 824, 27, 216, 1142, 25, 1083, 217, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\tan ^{\frac {14}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^{14/3}}{(a+i a \tan (c+d x))^2}dx\)

\(\Big \downarrow \) 4041

\(\displaystyle -\frac {\int -\frac {\tan ^{\frac {8}{3}}(c+d x) (11 a-17 i a \tan (c+d x))}{3 (i \tan (c+d x) a+a)}dx}{4 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\tan ^{\frac {8}{3}}(c+d x) (11 a-17 i a \tan (c+d x))}{i \tan (c+d x) a+a}dx}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\tan (c+d x)^{8/3} (11 a-17 i a \tan (c+d x))}{i \tan (c+d x) a+a}dx}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int \frac {2}{3} \tan ^{\frac {5}{3}}(c+d x) \left (121 \tan (c+d x) a^2+112 i a^2\right )dx}{2 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int \tan ^{\frac {5}{3}}(c+d x) \left (121 \tan (c+d x) a^2+112 i a^2\right )dx}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int \tan (c+d x)^{5/3} \left (121 \tan (c+d x) a^2+112 i a^2\right )dx}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\int \tan ^{\frac {2}{3}}(c+d x) \left (112 i a^2 \tan (c+d x)-121 a^2\right )dx}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\int \tan (c+d x)^{2/3} \left (112 i a^2 \tan (c+d x)-121 a^2\right )dx}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int \frac {-121 \tan (c+d x) a^2-112 i a^2}{\sqrt [3]{\tan (c+d x)}}dx+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int \frac {-121 \tan (c+d x) a^2-112 i a^2}{\sqrt [3]{\tan (c+d x)}}dx+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4021

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-121 a^2 \int \tan ^{\frac {2}{3}}(c+d x)dx-112 i a^2 \int \frac {1}{\sqrt [3]{\tan (c+d x)}}dx+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-112 i a^2 \int \frac {1}{\sqrt [3]{\tan (c+d x)}}dx-121 a^2 \int \tan (c+d x)^{2/3}dx+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {112 i a^2 \int \frac {1}{\sqrt [3]{\tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}-\frac {121 a^2 \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {336 i a^2 \int \frac {\sqrt [3]{\tan (c+d x)}}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {363 a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {168 i a^2 \int \frac {1}{\tan (c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)}{d}-\frac {363 a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 750

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {168 i a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)\right )}{d}-\frac {363 a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {363 a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {168 i a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 824

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {363 a^2 \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int -\frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{2 \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int -\frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{2 \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}\right )}{d}-\frac {168 i a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {363 a^2 \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )}{d}-\frac {168 i a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {363 a^2 \left (-\frac {1}{6} \int \frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {168 i a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {363 a^2 \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int -\frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {168 i a^2 \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{2} \int \left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )d\tan ^{\frac {2}{3}}(c+d x)\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {363 a^2 \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {168 i a^2 \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {363 a^2 \left (\frac {1}{6} \left (-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}-\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {168 i a^2 \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)-3 \int \frac {1}{-2 \tan ^{\frac {2}{3}}(c+d x)-2}d\left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {363 a^2 \left (\frac {1}{6} \left (-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {168 i a^2 \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {168 i a^2 \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {363 a^2 \left (\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )\right )}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

Input:

Int[Tan[c + d*x]^(14/3)/(a + I*a*Tan[c + d*x])^2,x]
 

Output:

-1/4*Tan[c + d*x]^(11/3)/(d*(a + I*a*Tan[c + d*x])^2) + (((14*I)*Tan[c + d 
*x]^(8/3))/(d*(1 + I*Tan[c + d*x])) - (((-168*I)*a^2*(ArcTan[(-1 + 2*Tan[c 
 + d*x]^(2/3))/Sqrt[3]]/Sqrt[3] + Log[1 + Tan[c + d*x]^(2/3)]/3))/d - (363 
*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]*T 
an[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)])/2)/6))/d + ((168*I)*a^2*Tan[c + d 
*x]^(2/3))/d + (363*a^2*Tan[c + d*x]^(5/3))/(5*d))/(3*a^2))/(12*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 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 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 4041

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

rule 4078

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

Maple [A] (verified)

Time = 1.10 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.65

method	result	size
derivativedivides	\(\frac {-\frac {3 \tan \left (d x +c \right )^{\frac {5}{3}}}{5}-3 i \tan \left (d x +c \right )^{\frac {2}{3}}+\frac {i \ln \left (i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{16}-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{8}+\frac {i}{36 \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )^{2}}+\frac {233 i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}{72}-\frac {23}{36 \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}-\frac {i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}-i\right )}{8}-\frac {92 \tan \left (d x +c \right )-136 i \tan \left (d x +c \right )^{\frac {2}{3}}-130 \tan \left (d x +c \right )^{\frac {1}{3}}+44 i}{72 \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )^{2}}-\frac {233 i \ln \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{144}-\frac {233 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{72}}{d \,a^{2}}\)	\(246\)
default	\(\frac {-\frac {3 \tan \left (d x +c \right )^{\frac {5}{3}}}{5}-3 i \tan \left (d x +c \right )^{\frac {2}{3}}+\frac {i \ln \left (i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{16}-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{8}+\frac {i}{36 \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )^{2}}+\frac {233 i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}{72}-\frac {23}{36 \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}-\frac {i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}-i\right )}{8}-\frac {92 \tan \left (d x +c \right )-136 i \tan \left (d x +c \right )^{\frac {2}{3}}-130 \tan \left (d x +c \right )^{\frac {1}{3}}+44 i}{72 \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )^{2}}-\frac {233 i \ln \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{144}-\frac {233 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{72}}{d \,a^{2}}\)	\(246\)

Input:

int(tan(d*x+c)^(14/3)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/d/a^2*(-3/5*tan(d*x+c)^(5/3)-3*I*tan(d*x+c)^(2/3)+1/16*I*ln(I*tan(d*x+c) 
^(1/3)+tan(d*x+c)^(2/3)-1)-1/8*3^(1/2)*arctanh(1/3*(I+2*tan(d*x+c)^(1/3))* 
3^(1/2))+1/36*I/(tan(d*x+c)^(1/3)+I)^2+233/72*I*ln(tan(d*x+c)^(1/3)+I)-23/ 
36/(tan(d*x+c)^(1/3)+I)-1/8*I*ln(tan(d*x+c)^(1/3)-I)-1/72*(92*tan(d*x+c)-1 
36*I*tan(d*x+c)^(2/3)-130*tan(d*x+c)^(1/3)+44*I)/(-I*tan(d*x+c)^(1/3)+tan( 
d*x+c)^(2/3)-1)^2-233/144*I*ln(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)-233 
/72*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. 684 vs. \(2 (298) = 596\).

Time = 0.09 (sec) , antiderivative size = 684, normalized size of antiderivative = 1.80 \[ \int \frac {\tan ^{\frac {14}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx =\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(14/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")
 

Output:

-1/720*(45*(sqrt(3)*(a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4*I*d*x + 4*I*c) 
)*sqrt(1/(a^4*d^2)) - I*e^(6*I*d*x + 6*I*c) - I*e^(4*I*d*x + 4*I*c))*log(1 
/2*sqrt(3)*a^2*d*sqrt(1/(a^4*d^2)) + ((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I 
*d*x + 2*I*c) + 1))^(1/3) + 1/2*I) - 45*(sqrt(3)*(a^2*d*e^(6*I*d*x + 6*I*c 
) + a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/(a^4*d^2)) + I*e^(6*I*d*x + 6*I*c) + 
 I*e^(4*I*d*x + 4*I*c))*log(-1/2*sqrt(3)*a^2*d*sqrt(1/(a^4*d^2)) + ((-I*e^ 
(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) + 1/2*I) + 1165*(3 
*sqrt(1/3)*(a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/ 
(a^4*d^2)) + I*e^(6*I*d*x + 6*I*c) + I*e^(4*I*d*x + 4*I*c))*log(3/2*sqrt(1 
/3)*a^2*d*sqrt(1/(a^4*d^2)) + ((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 
2*I*c) + 1))^(1/3) - 1/2*I) - 1165*(3*sqrt(1/3)*(a^2*d*e^(6*I*d*x + 6*I*c) 
 + a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/(a^4*d^2)) - I*e^(6*I*d*x + 6*I*c) - 
I*e^(4*I*d*x + 4*I*c))*log(-3/2*sqrt(1/3)*a^2*d*sqrt(1/(a^4*d^2)) + ((-I*e 
^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) - 1/2*I) + 2330*( 
-I*e^(6*I*d*x + 6*I*c) - I*e^(4*I*d*x + 4*I*c))*log(((-I*e^(2*I*d*x + 2*I* 
c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) + I) + 90*(I*e^(6*I*d*x + 6*I*c) 
+ I*e^(4*I*d*x + 4*I*c))*log(((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2 
*I*c) + 1))^(1/3) - I) + 3*((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I 
*c) + 1))^(2/3)*(791*I*e^(6*I*d*x + 6*I*c) + 1279*I*e^(4*I*d*x + 4*I*c) + 
185*I*e^(2*I*d*x + 2*I*c) - 15*I))/(a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {14}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate(tan(d*x+c)**(14/3)/(a+I*a*tan(d*x+c))**2,x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^{\frac {14}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(tan(d*x+c)^(14/3)/(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 \[ \int \frac {\tan ^{\frac {14}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {432 \, \tan \left (d x + c\right )^{\frac {5}{3}} - 1165 \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}\right ) - 45 \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}\right ) + 2160 i \, \tan \left (d x + c\right )^{\frac {2}{3}} + \frac {60 \, {\left (23 \, \tan \left (d x + c\right )^{\frac {5}{3}} - 20 i \, \tan \left (d x + c\right )^{\frac {2}{3}}\right )}}{{\left (\tan \left (d x + c\right ) - i\right )}^{2}} - 45 i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right ) + 1165 i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right ) - 2330 i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right ) + 90 i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )}{720 \, a^{2} d} \] Input:

integrate(tan(d*x+c)^(14/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")
 

Output:

-1/720*(432*tan(d*x + c)^(5/3) - 1165*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + 
c)^(1/3) + I)/(sqrt(3) + 2*tan(d*x + c)^(1/3) - I)) - 45*sqrt(3)*log(-(sqr 
t(3) - 2*tan(d*x + c)^(1/3) - I)/(sqrt(3) + 2*tan(d*x + c)^(1/3) + I)) + 2 
160*I*tan(d*x + c)^(2/3) + 60*(23*tan(d*x + c)^(5/3) - 20*I*tan(d*x + c)^( 
2/3))/(tan(d*x + c) - I)^2 - 45*I*log(tan(d*x + c)^(2/3) + I*tan(d*x + c)^ 
(1/3) - 1) + 1165*I*log(tan(d*x + c)^(2/3) - I*tan(d*x + c)^(1/3) - 1) - 2 
330*I*log(tan(d*x + c)^(1/3) + I) + 90*I*log(tan(d*x + c)^(1/3) - I))/(a^2 
*d)
 

Mupad [B] (verification not implemented)

Time = 2.75 (sec) , antiderivative size = 674, normalized size of antiderivative = 1.78 \[ \int \frac {\tan ^{\frac {14}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Too large to display} \] Input:

int(tan(c + d*x)^(14/3)/(a + a*tan(c + d*x)*1i)^2,x)
 

Output:

log(((a^6*d^3*1619208448i)/3 - a^8*d^4*tan(c + d*x)^(1/3)*(1i/(512*a^6*d^3 
))^(1/3)*167024640i)*(1i/(512*a^6*d^3))^(2/3) + (24321472*a^2*d*tan(c + d* 
x)^(1/3))/3)*(1i/(512*a^6*d^3))^(1/3) + log(((a^6*d^3*1619208448i)/3 - a^8 
*d^4*tan(c + d*x)^(1/3)*(-12649337i/(373248*a^6*d^3))^(1/3)*167024640i)*(- 
12649337i/(373248*a^6*d^3))^(2/3) + (24321472*a^2*d*tan(c + d*x)^(1/3))/3) 
*(-12649337i/(373248*a^6*d^3))^(1/3) - ((5*tan(c + d*x)^(2/3))/(3*a^2*d) + 
 (tan(c + d*x)^(5/3)*23i)/(12*a^2*d))/(2*tan(c + d*x) + tan(c + d*x)^2*1i 
- 1i) - (tan(c + d*x)^(2/3)*3i)/(a^2*d) - (3*tan(c + d*x)^(5/3))/(5*a^2*d) 
 + (log((24321472*a^2*d*tan(c + d*x)^(1/3))/3 + ((3^(1/2)*1i - 1)^2*((a^6* 
d^3*1619208448i)/3 - a^8*d^4*tan(c + d*x)^(1/3)*(3^(1/2)*1i - 1)*(1i/(512* 
a^6*d^3))^(1/3)*83512320i)*(1i/(512*a^6*d^3))^(2/3))/4)*(3^(1/2)*1i - 1)*( 
1i/(512*a^6*d^3))^(1/3))/2 - (log((24321472*a^2*d*tan(c + d*x)^(1/3))/3 + 
((3^(1/2)*1i + 1)^2*((a^6*d^3*1619208448i)/3 + a^8*d^4*tan(c + d*x)^(1/3)* 
(3^(1/2)*1i + 1)*(1i/(512*a^6*d^3))^(1/3)*83512320i)*(1i/(512*a^6*d^3))^(2 
/3))/4)*(3^(1/2)*1i + 1)*(1i/(512*a^6*d^3))^(1/3))/2 + (log((24321472*a^2* 
d*tan(c + d*x)^(1/3))/3 + ((3^(1/2)*1i - 1)^2*((a^6*d^3*1619208448i)/3 - a 
^8*d^4*tan(c + d*x)^(1/3)*(3^(1/2)*1i - 1)*(-12649337i/(373248*a^6*d^3))^( 
1/3)*83512320i)*(-12649337i/(373248*a^6*d^3))^(2/3))/4)*(3^(1/2)*1i - 1)*( 
-12649337i/(373248*a^6*d^3))^(1/3))/2 - (log((24321472*a^2*d*tan(c + d*x)^ 
(1/3))/3 + ((3^(1/2)*1i + 1)^2*((a^6*d^3*1619208448i)/3 + a^8*d^4*tan(c...
 

Reduce [F]

\[ \int \frac {\tan ^{\frac {14}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\int \frac {\tan \left (d x +c \right )^{\frac {14}{3}}}{\tan \left (d x +c \right )^{2}-2 \tan \left (d x +c \right ) i -1}d x}{a^{2}} \] Input:

int(tan(d*x+c)^(14/3)/(a+I*a*tan(d*x+c))^2,x)
 

Output:

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