Integrand size = 26, antiderivative size = 299 \[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=-\frac {\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {\arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {i \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{\sqrt {3} a d}+\frac {\arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}-\frac {\log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}+\frac {\log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{6 a d}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))} \]
1/12*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))/a/d+1/12*arctan(3^(1/2)+2*tan(d*x +c)^(1/3))/a/d+1/6*arctan(tan(d*x+c)^(1/3))/a/d+1/3*I*ln(1+tan(d*x+c)^(2/3 ))/a/d-1/6*I*ln(1-tan(d*x+c)^(2/3)+tan(d*x+c)^(4/3))/a/d+1/3*I*arctan(1/3* (1-2*tan(d*x+c)^(2/3))*3^(1/2))/a/d*3^(1/2)-1/24*ln(1-3^(1/2)*tan(d*x+c)^( 1/3)+tan(d*x+c)^(2/3))/a/d*3^(1/2)+1/24*ln(1+3^(1/2)*tan(d*x+c)^(1/3)+tan( d*x+c)^(2/3))/a/d*3^(1/2)-1/2*tan(d*x+c)^(1/3)/d/(a+I*a*tan(d*x+c))
Time = 6.36 (sec) , antiderivative size = 502, normalized size of antiderivative = 1.68 \[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\frac {4 i a \left (\frac {3}{4} \tan ^{\frac {4}{3}}(c+d x)-\frac {3}{4} \tan ^{\frac {4}{3}}(c+d x) \left (-\frac {2 \log \left (1+\sqrt [3]{\tan ^2(c+d x)}\right )}{3 \tan ^2(c+d x)^{2/3}}-\frac {2 (-1)^{2/3} \log \left (1-e^{-\frac {i \pi }{3}} \sqrt [3]{\tan ^2(c+d x)}\right )}{3 \tan ^2(c+d x)^{2/3}}+\frac {2 \sqrt [3]{-1} \log \left (1-e^{\frac {i \pi }{3}} \sqrt [3]{\tan ^2(c+d x)}\right )}{3 \tan ^2(c+d x)^{2/3}}\right )\right )}{3 d}-\frac {a \left (3 \sqrt [3]{\tan (c+d x)}-3 \sqrt [3]{\tan (c+d x)} \left (\frac {i \log \left (1-i \sqrt [6]{\tan ^2(c+d x)}\right )}{6 \sqrt [6]{\tan ^2(c+d x)}}-\frac {i \log \left (1+i \sqrt [6]{\tan ^2(c+d x)}\right )}{6 \sqrt [6]{\tan ^2(c+d x)}}-\frac {\sqrt [6]{-1} \log \left (1-e^{-\frac {i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right )}{6 \sqrt [6]{\tan ^2(c+d x)}}+\frac {(-1)^{5/6} \log \left (1-e^{\frac {i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right )}{6 \sqrt [6]{\tan ^2(c+d x)}}-\frac {(-1)^{5/6} \log \left (1-e^{-\frac {5 i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right )}{6 \sqrt [6]{\tan ^2(c+d x)}}+\frac {\sqrt [6]{-1} \log \left (1-e^{\frac {5 i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right )}{6 \sqrt [6]{\tan ^2(c+d x)}}\right )\right )}{3 d}}{2 a^2} \]
Tan[c + d*x]^(7/3)/(2*d*(a + I*a*Tan[c + d*x])) + ((((4*I)/3)*a*((3*Tan[c + d*x]^(4/3))/4 - (3*Tan[c + d*x]^(4/3)*((-2*Log[1 + (Tan[c + d*x]^2)^(1/3 )])/(3*(Tan[c + d*x]^2)^(2/3)) - (2*(-1)^(2/3)*Log[1 - (Tan[c + d*x]^2)^(1 /3)/E^((I/3)*Pi)])/(3*(Tan[c + d*x]^2)^(2/3)) + (2*(-1)^(1/3)*Log[1 - E^(( I/3)*Pi)*(Tan[c + d*x]^2)^(1/3)])/(3*(Tan[c + d*x]^2)^(2/3))))/4))/d - (a* (3*Tan[c + d*x]^(1/3) - 3*Tan[c + d*x]^(1/3)*(((I/6)*Log[1 - I*(Tan[c + d* x]^2)^(1/6)])/(Tan[c + d*x]^2)^(1/6) - ((I/6)*Log[1 + I*(Tan[c + d*x]^2)^( 1/6)])/(Tan[c + d*x]^2)^(1/6) - ((-1)^(1/6)*Log[1 - (Tan[c + d*x]^2)^(1/6) /E^((I/6)*Pi)])/(6*(Tan[c + d*x]^2)^(1/6)) + ((-1)^(5/6)*Log[1 - E^((I/6)* Pi)*(Tan[c + d*x]^2)^(1/6)])/(6*(Tan[c + d*x]^2)^(1/6)) - ((-1)^(5/6)*Log[ 1 - (Tan[c + d*x]^2)^(1/6)/E^(((5*I)/6)*Pi)])/(6*(Tan[c + d*x]^2)^(1/6)) + ((-1)^(1/6)*Log[1 - E^(((5*I)/6)*Pi)*(Tan[c + d*x]^2)^(1/6)])/(6*(Tan[c + d*x]^2)^(1/6)))))/(3*d))/(2*a^2)
Time = 0.72 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.81, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.731, Rules used = {3042, 4033, 27, 3042, 4021, 3042, 3957, 266, 753, 27, 216, 807, 821, 16, 1142, 25, 1083, 217, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^{4/3}}{a+i a \tan (c+d x)}dx\) |
| \(\Big \downarrow \) 4033 |
| \(\displaystyle \frac {\int \frac {a-4 i a \tan (c+d x)}{3 \tan ^{\frac {2}{3}}(c+d x)}dx}{2 a^2}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a-4 i a \tan (c+d x)}{\tan ^{\frac {2}{3}}(c+d x)}dx}{6 a^2}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a-4 i a \tan (c+d x)}{\tan (c+d x)^{2/3}}dx}{6 a^2}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 4021 |
| \(\displaystyle \frac {a \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)}dx-4 i a \int \sqrt [3]{\tan (c+d x)}dx}{6 a^2}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \int \frac {1}{\tan (c+d x)^{2/3}}dx-4 i a \int \sqrt [3]{\tan (c+d x)}dx}{6 a^2}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 3957 |
| \(\displaystyle \frac {\frac {a \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x) \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}-\frac {4 i a \int \frac {\sqrt [3]{\tan (c+d x)}}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}}{6 a^2}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {\frac {3 a \int \frac {1}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {12 i a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 753 |
| \(\displaystyle \frac {\frac {3 a \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{2 \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{2 \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}\right )}{d}-\frac {12 i a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3 a \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )}{d}-\frac {12 i a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {3 a \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {12 i a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 807 |
| \(\displaystyle \frac {\frac {3 a \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {6 i a \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{\tan (c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)}{d}}{6 a^2}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 821 |
| \(\displaystyle \frac {\frac {3 a \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {6 i a \left (\frac {1}{3} \int \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)\right )}{d}}{6 a^2}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {\frac {3 a \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {6 i a \left (\frac {1}{3} \int \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{6 a^2}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {\frac {3 a \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int -\frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {6 i a \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{2} \int \left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{6 a^2}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {3 a \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {6 i a \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{6 a^2}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {\frac {3 a \left (\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}-\sqrt {3}\right )\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {6 i a \left (\frac {1}{3} \left (-3 \int \frac {1}{-2 \tan ^{\frac {2}{3}}(c+d x)-2}d\left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )-\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{6 a^2}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {3 a \left (\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {6 i a \left (\frac {1}{3} \left (\sqrt {3} \arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )-\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{6 a^2}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {\frac {3 a \left (\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )-\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )+\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )\right )}{d}-\frac {6 i a \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{6 a^2}-\frac {\sqrt [3]{\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\) |
(((-6*I)*a*(ArcTan[(-1 + 2*Tan[c + d*x]^(2/3))/Sqrt[3]]/Sqrt[3] - Log[1 + Tan[c + d*x]^(2/3)]/3))/d + (3*a*(ArcTan[Tan[c + d*x]^(1/3)]/3 + (-ArcTan[ Sqrt[3] - 2*Tan[c + d*x]^(1/3)] - (Sqrt[3]*Log[1 - Sqrt[3]*Tan[c + d*x]^(1 /3) + Tan[c + d*x]^(2/3)])/2)/6 + (ArcTan[Sqrt[3] + 2*Tan[c + d*x]^(1/3)] + (Sqrt[3]*Log[1 + Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)])/2)/6) )/d)/(6*a^2) - Tan[c + d*x]^(1/3)/(2*d*(a + I*a*Tan[c + d*x]))
3.3.35.3.1 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_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[a/ b, n]], s = Denominator[Rt[a/b, n]], k, u, v}, Simp[u = Int[(r - s*Cos[(2*k - 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x] + Int[ (r + s*Cos[(2*k - 1)*(Pi/n)]*x)/(r^2 + 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2* x^2), x]; 2*(r^2/(a*n)) Int[1/(r^2 + s^2*x^2), x] + 2*(r/(a*n)) Sum[u, {k, 1, (n - 2)/4}], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && PosQ[a /b]
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_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Simp[-(3*Rt[a, 3]*Rt[b, 3])^(- 1) Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]*Rt[b, 3]) Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2 *x^2), x], x] /; FreeQ[{a, b}, x]
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[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((c + d*Tan[e + f*x])^(n - 1)/( 2*a*f*(a + b*Tan[e + f*x]))), x] + Simp[1/(2*a^2) Int[(c + d*Tan[e + f*x] )^(n - 2)*Simp[a*c^2 + a*d^2*(n - 1) - b*c*d*n - d*(a*c*(n - 2) + b*d*n)*Ta n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[n, 1]
Time = 0.56 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.64
| method | result | size |
| derivativedivides | \(\frac {-\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{8}-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{4}+\frac {5 i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{12}-\frac {1}{6 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}-\frac {-2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )-2 i}{12 \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}-\frac {5 i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{24}+\frac {5 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{12}+\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{4}}{d a}\) | \(190\) |
| default | \(\frac {-\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{8}-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{4}+\frac {5 i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{12}-\frac {1}{6 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}-\frac {-2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )-2 i}{12 \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}-\frac {5 i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{24}+\frac {5 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{12}+\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{4}}{d a}\) | \(190\) |
1/d/a*(-1/8*I*ln(I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)-1/4*3^(1/2)*arctan h(1/3*(I+2*tan(d*x+c)^(1/3))*3^(1/2))+5/12*I*ln(tan(d*x+c)^(1/3)+I)-1/6/(t an(d*x+c)^(1/3)+I)-1/12*(-2*tan(d*x+c)^(1/3)-2*I)/(-I*tan(d*x+c)^(1/3)+tan (d*x+c)^(2/3)-1)-5/24*I*ln(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)+5/12*3^ (1/2)*arctanh(1/3*(-I+2*tan(d*x+c)^(1/3))*3^(1/2))+1/4*I*ln(tan(d*x+c)^(1/ 3)-I))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 492 vs. \(2 (238) = 476\).
Time = 0.26 (sec) , antiderivative size = 492, normalized size of antiderivative = 1.65 \[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=-\frac {{\left (3 \, {\left (\sqrt {3} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (\frac {1}{2} \, \sqrt {3} a d \sqrt {\frac {1}{a^{2} 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 ) - 3 \, {\left (\sqrt {3} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (-\frac {1}{2} \, \sqrt {3} a d \sqrt {\frac {1}{a^{2} 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 ) - 5 \, {\left (3 \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (\frac {3}{2} \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} 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 ) + 5 \, {\left (3 \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (-\frac {3}{2} \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} 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 ) - 10 i \, e^{\left (2 i \, d x + 2 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 ) - 6 i \, e^{\left (2 i \, d x + 2 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 ) + 6 \, \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}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{24 \, a d} \]
-1/24*(3*(sqrt(3)*a*d*sqrt(1/(a^2*d^2))*e^(2*I*d*x + 2*I*c) + I*e^(2*I*d*x + 2*I*c))*log(1/2*sqrt(3)*a*d*sqrt(1/(a^2*d^2)) + ((-I*e^(2*I*d*x + 2*I*c ) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) + 1/2*I) - 3*(sqrt(3)*a*d*sqrt(1/( a^2*d^2))*e^(2*I*d*x + 2*I*c) - I*e^(2*I*d*x + 2*I*c))*log(-1/2*sqrt(3)*a* d*sqrt(1/(a^2*d^2)) + ((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) + 1/2*I) - 5*(3*sqrt(1/3)*a*d*sqrt(1/(a^2*d^2))*e^(2*I*d*x + 2* I*c) - I*e^(2*I*d*x + 2*I*c))*log(3/2*sqrt(1/3)*a*d*sqrt(1/(a^2*d^2)) + (( -I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) - 1/2*I) + 5* (3*sqrt(1/3)*a*d*sqrt(1/(a^2*d^2))*e^(2*I*d*x + 2*I*c) + I*e^(2*I*d*x + 2* I*c))*log(-3/2*sqrt(1/3)*a*d*sqrt(1/(a^2*d^2)) + ((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) - 1/2*I) - 10*I*e^(2*I*d*x + 2*I*c)* log(((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) + I) - 6*I*e^(2*I*d*x + 2*I*c)*log(((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2* I*c) + 1))^(1/3) - I) + 6*((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I* c) + 1))^(1/3)*(e^(2*I*d*x + 2*I*c) + 1))*e^(-2*I*d*x - 2*I*c)/(a*d)
\[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=- \frac {i \int \frac {\tan ^{\frac {4}{3}}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx}{a} \]
Exception generated. \[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.46 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.72 \[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=-\frac {5 \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}\right )}{24 \, a d} + \frac {\sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}\right )}{8 \, a d} - \frac {i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{8 \, a d} - \frac {5 i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{24 \, a d} + \frac {5 i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right )}{12 \, a d} + \frac {i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )}{4 \, a d} + \frac {i \, \tan \left (d x + c\right )^{\frac {1}{3}}}{2 \, a d {\left (\tan \left (d x + c\right ) - i\right )}} \]
-5/24*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1/3) + I)/(sqrt(3) + 2*tan(d *x + c)^(1/3) - I))/(a*d) + 1/8*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1/ 3) - I)/(sqrt(3) + 2*tan(d*x + c)^(1/3) + I))/(a*d) - 1/8*I*log(tan(d*x + c)^(2/3) + I*tan(d*x + c)^(1/3) - 1)/(a*d) - 5/24*I*log(tan(d*x + c)^(2/3) - I*tan(d*x + c)^(1/3) - 1)/(a*d) + 5/12*I*log(tan(d*x + c)^(1/3) + I)/(a *d) + 1/4*I*log(tan(d*x + c)^(1/3) - I)/(a*d) + 1/2*I*tan(d*x + c)^(1/3)/( a*d*(tan(d*x + c) - I))
Time = 6.17 (sec) , antiderivative size = 622, normalized size of antiderivative = 2.08 \[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\ln \left (\left (a^3\,d^3\,14112{}\mathrm {i}-165888\,a^5\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}-a^2\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,6120{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}+\ln \left (\left (a^3\,d^3\,14112{}\mathrm {i}-165888\,a^5\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (-\frac {125{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {125{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}-a^2\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,6120{}\mathrm {i}\right )\,{\left (-\frac {125{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}+\frac {\ln \left (\frac {\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (a^3\,d^3\,14112{}\mathrm {i}-41472\,a^5\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}}{2}-a^2\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,6120{}\mathrm {i}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}}{2}-\frac {\ln \left (\frac {\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (a^3\,d^3\,14112{}\mathrm {i}-41472\,a^5\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}}{2}+a^2\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,6120{}\mathrm {i}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}}{2}+\frac {\ln \left (\frac {\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (a^3\,d^3\,14112{}\mathrm {i}-41472\,a^5\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (-\frac {125{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {125{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}}{2}-a^2\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,6120{}\mathrm {i}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {125{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}}{2}-\frac {\ln \left (\frac {\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (a^3\,d^3\,14112{}\mathrm {i}-41472\,a^5\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (-\frac {125{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {125{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}}{2}+a^2\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,6120{}\mathrm {i}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {125{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}}{2}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \]
log((a^3*d^3*14112i - 165888*a^5*d^5*tan(c + d*x)^(1/3)*(-1i/(64*a^3*d^3)) ^(2/3))*(-1i/(64*a^3*d^3))^(1/3) - a^2*d^2*tan(c + d*x)^(1/3)*6120i)*(-1i/ (64*a^3*d^3))^(1/3) + log((a^3*d^3*14112i - 165888*a^5*d^5*tan(c + d*x)^(1 /3)*(-125i/(1728*a^3*d^3))^(2/3))*(-125i/(1728*a^3*d^3))^(1/3) - a^2*d^2*t an(c + d*x)^(1/3)*6120i)*(-125i/(1728*a^3*d^3))^(1/3) + (log(((3^(1/2)*1i - 1)*(a^3*d^3*14112i - 41472*a^5*d^5*tan(c + d*x)^(1/3)*(3^(1/2)*1i - 1)^2 *(-1i/(64*a^3*d^3))^(2/3))*(-1i/(64*a^3*d^3))^(1/3))/2 - a^2*d^2*tan(c + d *x)^(1/3)*6120i)*(3^(1/2)*1i - 1)*(-1i/(64*a^3*d^3))^(1/3))/2 - (log(((3^( 1/2)*1i + 1)*(a^3*d^3*14112i - 41472*a^5*d^5*tan(c + d*x)^(1/3)*(3^(1/2)*1 i + 1)^2*(-1i/(64*a^3*d^3))^(2/3))*(-1i/(64*a^3*d^3))^(1/3))/2 + a^2*d^2*t an(c + d*x)^(1/3)*6120i)*(3^(1/2)*1i + 1)*(-1i/(64*a^3*d^3))^(1/3))/2 + (l og(((3^(1/2)*1i - 1)*(a^3*d^3*14112i - 41472*a^5*d^5*tan(c + d*x)^(1/3)*(3 ^(1/2)*1i - 1)^2*(-125i/(1728*a^3*d^3))^(2/3))*(-125i/(1728*a^3*d^3))^(1/3 ))/2 - a^2*d^2*tan(c + d*x)^(1/3)*6120i)*(3^(1/2)*1i - 1)*(-125i/(1728*a^3 *d^3))^(1/3))/2 - (log(((3^(1/2)*1i + 1)*(a^3*d^3*14112i - 41472*a^5*d^5*t an(c + d*x)^(1/3)*(3^(1/2)*1i + 1)^2*(-125i/(1728*a^3*d^3))^(2/3))*(-125i/ (1728*a^3*d^3))^(1/3))/2 + a^2*d^2*tan(c + d*x)^(1/3)*6120i)*(3^(1/2)*1i + 1)*(-125i/(1728*a^3*d^3))^(1/3))/2 - tan(c + d*x)^(1/3)/(2*a*d*(tan(c + d *x)*1i + 1))