Integrand size = 26, antiderivative size = 339 \[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\frac {7 i \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {7 i \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {2 \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}-\frac {7 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {2 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {7 i \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}+\frac {7 i \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {\log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{9 a^2 d}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{12 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
-7/72*I*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))/a^2/d-7/72*I*arctan(3^(1/2)+2* tan(d*x+c)^(1/3))/a^2/d-7/36*I*arctan(tan(d*x+c)^(1/3))/a^2/d+2/9*ln(1+tan (d*x+c)^(2/3))/a^2/d-1/9*ln(1-tan(d*x+c)^(2/3)+tan(d*x+c)^(4/3))/a^2/d-2/9 *arctan(1/3*(1-2*tan(d*x+c)^(2/3))*3^(1/2))/a^2/d*3^(1/2)-7/144*I*ln(1-3^( 1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))/a^2/d*3^(1/2)+7/144*I*ln(1+3^(1/2) *tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))/a^2/d*3^(1/2)+7/12*tan(d*x+c)^(2/3)/a^ 2/d/(1+I*tan(d*x+c))+1/4*tan(d*x+c)^(2/3)/d/(a+I*a*tan(d*x+c))^2
Time = 4.35 (sec) , antiderivative size = 352, normalized size of antiderivative = 1.04 \[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\frac {18 \tan ^{\frac {2}{3}}(c+d x)+\frac {(a+i a \tan (c+d x)) \left (42 a \tan ^{\frac {2}{3}}(c+d x) \tan ^2(c+d x)^{5/6}-(a+i a \tan (c+d x)) \left (-7 \left (\log \left (1-i \sqrt [6]{\tan ^2(c+d x)}\right )-\log \left (1+i \sqrt [6]{\tan ^2(c+d x)}\right )+\sqrt [3]{-1} \left (-\sqrt [3]{-1} \log \left (1-\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )+\sqrt [3]{-1} \log \left (1+\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )+\log \left (1-(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )-\log \left (1+(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )\right )\right ) \tan ^{\frac {5}{3}}(c+d x)-8 \left (2 \sqrt {3} \arctan \left (\frac {-1+2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )+2 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )-\log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )\right ) \tan ^2(c+d x)^{5/6}\right )\right )}{a^2 \tan ^2(c+d x)^{5/6}}}{72 d (a+i a \tan (c+d x))^2} \]
(18*Tan[c + d*x]^(2/3) + ((a + I*a*Tan[c + d*x])*(42*a*Tan[c + d*x]^(2/3)* (Tan[c + d*x]^2)^(5/6) - (a + I*a*Tan[c + d*x])*(-7*(Log[1 - I*(Tan[c + d* x]^2)^(1/6)] - Log[1 + I*(Tan[c + d*x]^2)^(1/6)] + (-1)^(1/3)*(-((-1)^(1/3 )*Log[1 - (-1)^(1/6)*(Tan[c + d*x]^2)^(1/6)]) + (-1)^(1/3)*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]^(5/3) - 8*(2*Sq rt[3]*ArcTan[(-1 + 2*Tan[c + d*x]^(2/3))/Sqrt[3]] + 2*Log[1 + Tan[c + d*x] ^(2/3)] - Log[1 - Tan[c + d*x]^(2/3) + Tan[c + d*x]^(4/3)])*(Tan[c + d*x]^ 2)^(5/6))))/(a^2*(Tan[c + d*x]^2)^(5/6)))/(72*d*(a + I*a*Tan[c + d*x])^2)
Time = 0.92 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.84, number of steps used = 23, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {3042, 4042, 27, 3042, 4079, 27, 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.
\(\displaystyle \int \frac {1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^2}dx\) |
\(\Big \downarrow \) 4042 |
\(\displaystyle \frac {\int \frac {2 (5 a-2 i a \tan (c+d x))}{3 \sqrt [3]{\tan (c+d x)} (i \tan (c+d x) a+a)}dx}{4 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {5 a-2 i a \tan (c+d x)}{\sqrt [3]{\tan (c+d x)} (i \tan (c+d x) a+a)}dx}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {5 a-2 i a \tan (c+d x)}{\sqrt [3]{\tan (c+d x)} (i \tan (c+d x) a+a)}dx}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 4079 |
\(\displaystyle \frac {\frac {\int \frac {16 a^2-7 i a^2 \tan (c+d x)}{3 \sqrt [3]{\tan (c+d x)}}dx}{2 a^2}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{2 d (1+i \tan (c+d x))}}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {16 a^2-7 i a^2 \tan (c+d x)}{\sqrt [3]{\tan (c+d x)}}dx}{6 a^2}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{2 d (1+i \tan (c+d x))}}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {16 a^2-7 i a^2 \tan (c+d x)}{\sqrt [3]{\tan (c+d x)}}dx}{6 a^2}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{2 d (1+i \tan (c+d x))}}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 4021 |
\(\displaystyle \frac {\frac {16 a^2 \int \frac {1}{\sqrt [3]{\tan (c+d x)}}dx-7 i a^2 \int \tan ^{\frac {2}{3}}(c+d x)dx}{6 a^2}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{2 d (1+i \tan (c+d x))}}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {16 a^2 \int \frac {1}{\sqrt [3]{\tan (c+d x)}}dx-7 i a^2 \int \tan (c+d x)^{2/3}dx}{6 a^2}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{2 d (1+i \tan (c+d x))}}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 3957 |
\(\displaystyle \frac {\frac {\frac {16 a^2 \int \frac {1}{\sqrt [3]{\tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}-\frac {7 i a^2 \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}}{6 a^2}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{2 d (1+i \tan (c+d x))}}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {\frac {\frac {48 a^2 \int \frac {\sqrt [3]{\tan (c+d x)}}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {21 i a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{2 d (1+i \tan (c+d x))}}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {\frac {\frac {24 a^2 \int \frac {1}{\tan (c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)}{d}-\frac {21 i a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{2 d (1+i \tan (c+d x))}}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 750 |
\(\displaystyle \frac {\frac {\frac {24 a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)\right )}{d}-\frac {21 i a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{2 d (1+i \tan (c+d x))}}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {\frac {\frac {24 a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {21 i a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{2 d (1+i \tan (c+d x))}}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 824 |
\(\displaystyle \frac {\frac {\frac {24 a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {21 i a^2 \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int -\frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{2 \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int -\frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{2 \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}\right )}{d}}{6 a^2}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{2 d (1+i \tan (c+d x))}}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {24 a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {21 i a^2 \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )}{d}}{6 a^2}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{2 d (1+i \tan (c+d x))}}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {\frac {24 a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {21 i a^2 \left (-\frac {1}{6} \int \frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}}{6 a^2}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{2 d (1+i \tan (c+d x))}}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {\frac {\frac {24 a^2 \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{2} \int \left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )d\tan ^{\frac {2}{3}}(c+d x)\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {21 i a^2 \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int -\frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}}{6 a^2}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{2 d (1+i \tan (c+d x))}}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\frac {24 a^2 \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {21 i a^2 \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}}{6 a^2}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{2 d (1+i \tan (c+d x))}}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {\frac {\frac {24 a^2 \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)-3 \int \frac {1}{-2 \tan ^{\frac {2}{3}}(c+d x)-2}d\left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {21 i a^2 \left (\frac {1}{6} \left (-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}-\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}}{6 a^2}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{2 d (1+i \tan (c+d x))}}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {\frac {\frac {24 a^2 \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {21 i a^2 \left (\frac {1}{6} \left (-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}}{6 a^2}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{2 d (1+i \tan (c+d x))}}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {\frac {\frac {24 a^2 \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {21 i a^2 \left (\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )\right )}{d}}{6 a^2}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{2 d (1+i \tan (c+d x))}}{6 a^2}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
(((24*a^2*(ArcTan[(-1 + 2*Tan[c + d*x]^(2/3))/Sqrt[3]]/Sqrt[3] + Log[1 + T an[c + d*x]^(2/3)]/3))/d - ((21*I)*a^2*(ArcTan[Tan[c + d*x]^(1/3)]/3 + (-A rcTan[Sqrt[3] - 2*Tan[c + d*x]^(1/3)] + (Sqrt[3]*Log[1 - Sqrt[3]*Tan[c + d *x]^(1/3) + Tan[c + d*x]^(2/3)])/2)/6 + (ArcTan[Sqrt[3] + 2*Tan[c + d*x]^( 1/3)] - (Sqrt[3]*Log[1 + Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)]) /2)/6))/d)/(6*a^2) + (7*Tan[c + d*x]^(2/3))/(2*d*(1 + I*Tan[c + d*x])))/(6 *a^2) + Tan[c + d*x]^(2/3)/(4*d*(a + I*a*Tan[c + d*x])^2)
3.3.45.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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])
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]
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]
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]
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]
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]
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]
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]
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]
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]
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])
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]
Time = 0.68 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.66
method | result | size |
derivativedivides | \(\frac {-\frac {7 i}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}+\frac {1}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )^{2}}+\frac {23 \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{72}-\frac {\ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{16}-\frac {i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{8}+\frac {\ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{8}+\frac {-28 i \tan \left (d x +c \right )-44 \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )+50 i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+16}{72 {\left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}^{2}}-\frac {23 \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{144}+\frac {23 i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{72}}{d \,a^{2}}\) | \(223\) |
default | \(\frac {-\frac {7 i}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}+\frac {1}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )^{2}}+\frac {23 \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{72}-\frac {\ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{16}-\frac {i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{8}+\frac {\ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{8}+\frac {-28 i \tan \left (d x +c \right )-44 \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )+50 i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+16}{72 {\left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}^{2}}-\frac {23 \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{144}+\frac {23 i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{72}}{d \,a^{2}}\) | \(223\) |
1/d/a^2*(-7/36*I/(tan(d*x+c)^(1/3)+I)+1/36/(tan(d*x+c)^(1/3)+I)^2+23/72*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)+1/72*(-28*I*tan(d*x+c)-44*tan(d*x+c)^(2/3)+50*I*tan(d*x+c)^(1/3)+16)/ (-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)^2-23/144*ln(-I*tan(d*x+c)^(1/3)+t an(d*x+c)^(2/3)-1)+23/72*I*3^(1/2)*arctanh(1/3*(-I+2*tan(d*x+c)^(1/3))*3^( 1/2)))
Time = 0.27 (sec) , antiderivative size = 515, normalized size of antiderivative = 1.52 \[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx=-\frac {{\left (9 \, {\left (i \, \sqrt {3} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} + e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (\frac {1}{2} \, \sqrt {3} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} + \frac {1}{2} i\right ) + 9 \, {\left (-i \, \sqrt {3} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} + e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (-\frac {1}{2} \, \sqrt {3} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} + \frac {1}{2} i\right ) + 23 \, {\left (-3 i \, \sqrt {\frac {1}{3}} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} + e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (\frac {3}{2} \, \sqrt {\frac {1}{3}} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} - \frac {1}{2} i\right ) + 23 \, {\left (3 i \, \sqrt {\frac {1}{3}} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} + e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (-\frac {3}{2} \, \sqrt {\frac {1}{3}} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} - \frac {1}{2} i\right ) - 46 \, e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} + i\right ) - 18 \, e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} - i\right ) - 3 \, \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (17 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 20 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{144 \, a^{2} d} \]
-1/144*(9*(I*sqrt(3)*a^2*d*sqrt(1/(a^4*d^2))*e^(4*I*d*x + 4*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) + 9*(-I*sqrt(3)*a^2* d*sqrt(1/(a^4*d^2))*e^(4*I*d*x + 4*I*c) + e^(4*I*d*x + 4*I*c))*log(-1/2*sq rt(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) + 23*(-3*I*sqrt(1/3)*a^2*d*sqrt(1/(a^4*d^2)) *e^(4*I*d*x + 4*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) + 23*(3*I*sqrt(1/3)*a^2*d*sqrt(1/(a^4*d^2))*e^(4*I*d*x + 4*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) - 46*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^(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)*(17*e^(4*I*d*x + 4*I*c) + 20*e^(2*I*d*x + 2*I*c) + 3))*e^(-4*I*d*x - 4*I*c)/(a^2*d)
\[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx=- \frac {\int \frac {1}{\tan ^{\frac {7}{3}}{\left (c + d x \right )} - 2 i \tan ^{\frac {4}{3}}{\left (c + d x \right )} - \sqrt [3]{\tan {\left (c + d x \right )}}}\, dx}{a^{2}} \]
Exception generated. \[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.65 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx=-\frac {23 i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}\right )}{144 \, a^{2} d} + \frac {i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}\right )}{16 \, a^{2} d} - \frac {\log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{16 \, a^{2} d} - \frac {23 \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{144 \, a^{2} d} + \frac {23 \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right )}{72 \, a^{2} d} + \frac {\log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )}{8 \, a^{2} d} + \frac {-7 i \, \tan \left (d x + c\right )^{\frac {5}{3}} - 10 \, \tan \left (d x + c\right )^{\frac {2}{3}}}{12 \, a^{2} d {\left (\tan \left (d x + c\right ) - i\right )}^{2}} \]
-23/144*I*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1/3) + I)/(sqrt(3) + 2*t an(d*x + c)^(1/3) - I))/(a^2*d) + 1/16*I*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1/3) - I)/(sqrt(3) + 2*tan(d*x + c)^(1/3) + I))/(a^2*d) - 1/16*log( tan(d*x + c)^(2/3) + I*tan(d*x + c)^(1/3) - 1)/(a^2*d) - 23/144*log(tan(d* x + c)^(2/3) - I*tan(d*x + c)^(1/3) - 1)/(a^2*d) + 23/72*log(tan(d*x + c)^ (1/3) + I)/(a^2*d) + 1/8*log(tan(d*x + c)^(1/3) - I)/(a^2*d) + 1/12*(-7*I* tan(d*x + c)^(5/3) - 10*tan(d*x + c)^(2/3))/(a^2*d*(tan(d*x + c) - I)^2)
Time = 5.81 (sec) , antiderivative size = 630, normalized size of antiderivative = 1.86 \[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\frac {23\,\ln \left (\frac {529\,\left (\frac {1795840\,a^8\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (\frac {1}{a^6\,d^3}\right )}^{1/3}}{3}+\frac {a^6\,d^3\,1464064{}\mathrm {i}}{3}\right )\,{\left (\frac {1}{a^6\,d^3}\right )}^{2/3}}{5184}-\frac {33856\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{3}\right )\,{\left (\frac {1}{a^6\,d^3}\right )}^{1/3}}{72}+\ln \left (\left (1873920\,a^8\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (\frac {1}{512\,a^6\,d^3}\right )}^{1/3}+\frac {a^6\,d^3\,1464064{}\mathrm {i}}{3}\right )\,{\left (\frac {1}{512\,a^6\,d^3}\right )}^{2/3}-\frac {33856\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{3}\right )\,{\left (\frac {1}{512\,a^6\,d^3}\right )}^{1/3}-\frac {-\frac {7\,{\mathrm {tan}\left (c+d\,x\right )}^{5/3}}{12\,a^2\,d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^{2/3}\,5{}\mathrm {i}}{6\,a^2\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}}+\frac {23\,\ln \left (-\frac {33856\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{3}+\frac {529\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (\frac {a^6\,d^3\,1464064{}\mathrm {i}}{3}+\frac {897920\,a^8\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1}{a^6\,d^3}\right )}^{1/3}}{3}\right )\,{\left (\frac {1}{a^6\,d^3}\right )}^{2/3}}{20736}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1}{a^6\,d^3}\right )}^{1/3}}{144}-\frac {23\,\ln \left (-\frac {33856\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{3}+\frac {529\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (\frac {a^6\,d^3\,1464064{}\mathrm {i}}{3}-\frac {897920\,a^8\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1}{a^6\,d^3}\right )}^{1/3}}{3}\right )\,{\left (\frac {1}{a^6\,d^3}\right )}^{2/3}}{20736}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1}{a^6\,d^3}\right )}^{1/3}}{144}+\ln \left (-\frac {33856\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{3}+{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\,\left (\frac {a^6\,d^3\,1464064{}\mathrm {i}}{3}+1873920\,a^8\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (\frac {1}{512\,a^6\,d^3}\right )}^{1/3}\right )\,{\left (\frac {1}{512\,a^6\,d^3}\right )}^{2/3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (\frac {1}{512\,a^6\,d^3}\right )}^{1/3}-\ln \left (-\frac {33856\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{3}+{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\,\left (\frac {a^6\,d^3\,1464064{}\mathrm {i}}{3}-1873920\,a^8\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (\frac {1}{512\,a^6\,d^3}\right )}^{1/3}\right )\,{\left (\frac {1}{512\,a^6\,d^3}\right )}^{2/3}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (\frac {1}{512\,a^6\,d^3}\right )}^{1/3} \]
(23*log((529*((a^6*d^3*1464064i)/3 + (1795840*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 - (33856*a^2*d*tan(c + d* x)^(1/3))/3)*(1/(a^6*d^3))^(1/3))/72 + log(((a^6*d^3*1464064i)/3 + 1873920 *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) - (33856*a^2*d*tan(c + d*x)^(1/3))/3)*(1/(512*a^6*d^3))^(1/3) - ((tan(c + d*x)^(2/3)*5i)/(6*a^2*d) - (7*tan(c + d*x)^(5/3))/(12*a^2*d))/(2*tan(c + d*x) + tan(c + d*x)^2*1i - 1i) + (23*log((529*(3^(1/2)*1i - 1)^2*((a^6*d ^3*1464064i)/3 + (897920*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 - (33856*a^2*d*tan(c + d*x)^ (1/3))/3)*(3^(1/2)*1i - 1)*(1/(a^6*d^3))^(1/3))/144 - (23*log((529*(3^(1/2 )*1i + 1)^2*((a^6*d^3*1464064i)/3 - (897920*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 - (33856* a^2*d*tan(c + d*x)^(1/3))/3)*(3^(1/2)*1i + 1)*(1/(a^6*d^3))^(1/3))/144 + l og(((3^(1/2)*1i)/2 - 1/2)^2*((a^6*d^3*1464064i)/3 + 1873920*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) - (33856*a^2*d*tan(c + d*x)^(1/3))/3)*((3^(1/2)*1i)/2 - 1/2)*(1 /(512*a^6*d^3))^(1/3) - log(((3^(1/2)*1i)/2 + 1/2)^2*((a^6*d^3*1464064i)/3 - 1873920*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) - (33856*a^2*d*tan(c + d*x)^(1/3))/3)* ((3^(1/2)*1i)/2 + 1/2)*(1/(512*a^6*d^3))^(1/3)