Integrand size = 26, antiderivative size = 303 \[ \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=-\frac {\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {\arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {i \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 \sqrt {3} a d}+\frac {\arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}+\frac {\log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {\log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}+\frac {i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{12 a d}+\frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))} \] Output:
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*I*arctan(1/3*(1-2*tan(d*x+c)^(2/3))*3^(1/2))*3^(1/2)/a/ d+1/6*arctan(tan(d*x+c)^(1/3))/a/d-1/6*I*ln(1+tan(d*x+c)^(2/3))/a/d+1/24*l n(1-3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*3^(1/2)/a/d-1/24*ln(1+3^(1/ 2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*3^(1/2)/a/d+1/12*I*ln(1-tan(d*x+c)^( 2/3)+tan(d*x+c)^(4/3))/a/d+1/2*I*tan(d*x+c)^(2/3)/d/(a+I*a*tan(d*x+c))
Time = 2.91 (sec) , antiderivative size = 300, normalized size of antiderivative = 0.99 \[ \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {\frac {i \left (-2 \sqrt {3} \arctan \left (\frac {-1+2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )-2 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )+\log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )+6 \tan ^{\frac {2}{3}}(c+d x)\right )}{a}+\frac {\left (i \log \left (1-i \sqrt [6]{\tan ^2(c+d x)}\right )-i \log \left (1+i \sqrt [6]{\tan ^2(c+d x)}\right )+\sqrt [6]{-1} \left (\log \left (1-\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )-\log \left (1+\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )+(-1)^{2/3} \left (\log \left (1-(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )-\log \left (1+(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )\right )\right )\right ) \tan ^{\frac {5}{3}}(c+d x)}{a \tan ^2(c+d x)^{5/6}}+\frac {6 \tan ^{\frac {5}{3}}(c+d x)}{a+i a \tan (c+d x)}}{12 d} \] Input:
Integrate[Tan[c + d*x]^(2/3)/(a + I*a*Tan[c + d*x]),x]
Output:
((I*(-2*Sqrt[3]*ArcTan[(-1 + 2*Tan[c + d*x]^(2/3))/Sqrt[3]] - 2*Log[1 + Ta n[c + d*x]^(2/3)] + Log[1 - Tan[c + d*x]^(2/3) + Tan[c + d*x]^(4/3)] + 6*T an[c + d*x]^(2/3)))/a + ((I*Log[1 - I*(Tan[c + d*x]^2)^(1/6)] - I*Log[1 + I*(Tan[c + d*x]^2)^(1/6)] + (-1)^(1/6)*(Log[1 - (-1)^(1/6)*(Tan[c + d*x]^2 )^(1/6)] - Log[1 + (-1)^(1/6)*(Tan[c + d*x]^2)^(1/6)] + (-1)^(2/3)*(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))/(a*(Tan[c + d*x]^2)^(5/6)) + (6*Tan[c + d*x ]^(5/3))/(a + I*a*Tan[c + d*x]))/(12*d)
Time = 0.71 (sec) , antiderivative size = 245, 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, 4032, 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.
\(\displaystyle \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^{2/3}}{a+i a \tan (c+d x)}dx\) |
\(\Big \downarrow \) 4032 |
\(\displaystyle \frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\int \frac {2 i a-a \tan (c+d x)}{3 \sqrt [3]{\tan (c+d x)}}dx}{2 a^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\int \frac {2 i a-a \tan (c+d x)}{\sqrt [3]{\tan (c+d x)}}dx}{6 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\int \frac {2 i a-a \tan (c+d x)}{\sqrt [3]{\tan (c+d x)}}dx}{6 a^2}\) |
\(\Big \downarrow \) 4021 |
\(\displaystyle \frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {-a \int \tan ^{\frac {2}{3}}(c+d x)dx+2 i a \int \frac {1}{\sqrt [3]{\tan (c+d x)}}dx}{6 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {-a \int \tan (c+d x)^{2/3}dx+2 i a \int \frac {1}{\sqrt [3]{\tan (c+d x)}}dx}{6 a^2}\) |
\(\Big \downarrow \) 3957 |
\(\displaystyle \frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {-\frac {a \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}+\frac {2 i a \int \frac {1}{\sqrt [3]{\tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}}{6 a^2}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {-\frac {3 a \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {6 i a \int \frac {\sqrt [3]{\tan (c+d x)}}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {-\frac {3 a \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {3 i a \int \frac {1}{\tan (c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)}{d}}{6 a^2}\) |
\(\Big \downarrow \) 750 |
\(\displaystyle \frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {-\frac {3 a \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {3 i a \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)\right )}{d}}{6 a^2}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {-\frac {3 a \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {3 i a \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{6 a^2}\) |
\(\Big \downarrow \) 824 |
\(\displaystyle \frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {-\frac {3 a \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int -\frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{2 \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int -\frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{2 \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}\right )}{d}+\frac {3 i a \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{6 a^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {-\frac {3 a \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )}{d}+\frac {3 i a \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{6 a^2}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {-\frac {3 a \left (-\frac {1}{6} \int \frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {3 i a \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{6 a^2}\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {-\frac {3 a \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int -\frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {3 i a \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{2} \int \left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )d\tan ^{\frac {2}{3}}(c+d x)\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{6 a^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {-\frac {3 a \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {3 i a \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{6 a^2}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {-\frac {3 a \left (\frac {1}{6} \left (-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}-\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {3 i a \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)-3 \int \frac {1}{-2 \tan ^{\frac {2}{3}}(c+d x)-2}d\left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{6 a^2}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {-\frac {3 a \left (\frac {1}{6} \left (-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {3 i a \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{6 a^2}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {-\frac {3 a \left (\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )\right )}{d}+\frac {3 i a \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}}{6 a^2}\) |
Input:
Int[Tan[c + d*x]^(2/3)/(a + I*a*Tan[c + d*x]),x]
Output:
-1/6*(((3*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 + (-Arc Tan[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)/a^2 + ((I/2)*Tan[c + d*x]^(2/3))/(d*(a + I*a*Tan[c + d*x]))
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[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(a*c + b*d))*((c + d*Tan[e + f*x])^n/(2*( b*c - a*d)*f*(a + b*Tan[e + f*x]))), x] + Simp[1/(2*a*(b*c - a*d)) Int[(c + d*Tan[e + f*x])^(n - 1)*Simp[a*c*d*(n - 1) + b*c^2 + b*d^2*n - d*(b*c - a*d)*(n - 1)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && Ne Q[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[0, n, 1]
Time = 1.04 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.63
method | result | size |
derivativedivides | \(\frac {-\frac {i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}-i\right )}{4}-\frac {i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}{12}+\frac {1}{6 \tan \left (d x +c \right )^{\frac {1}{3}}+6 i}+\frac {i \ln \left (i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{8}-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{4}+\frac {4 \tan \left (d x +c \right )^{\frac {1}{3}}-2 i}{-12 i \tan \left (d x +c \right )^{\frac {1}{3}}+12 \tan \left (d x +c \right )^{\frac {2}{3}}-12}+\frac {i \ln \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{24}+\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{12}}{d a}\) | \(190\) |
default | \(\frac {-\frac {i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}-i\right )}{4}-\frac {i \ln \left (\tan \left (d x +c \right )^{\frac {1}{3}}+i\right )}{12}+\frac {1}{6 \tan \left (d x +c \right )^{\frac {1}{3}}+6 i}+\frac {i \ln \left (i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{8}-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{4}+\frac {4 \tan \left (d x +c \right )^{\frac {1}{3}}-2 i}{-12 i \tan \left (d x +c \right )^{\frac {1}{3}}+12 \tan \left (d x +c \right )^{\frac {2}{3}}-12}+\frac {i \ln \left (-i \tan \left (d x +c \right )^{\frac {1}{3}}+\tan \left (d x +c \right )^{\frac {2}{3}}-1\right )}{24}+\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \tan \left (d x +c \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{12}}{d a}\) | \(190\) |
Input:
int(tan(d*x+c)^(2/3)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-1/4*I*ln(tan(d*x+c)^(1/3)-I)-1/12*I*ln(tan(d*x+c)^(1/3)+I)+1/6/(ta n(d*x+c)^(1/3)+I)+1/8*I*ln(I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)-1/4*3^(1 /2)*arctanh(1/3*(I+2*tan(d*x+c)^(1/3))*3^(1/2))+1/12*(4*tan(d*x+c)^(1/3)-2 *I)/(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)+1/24*I*ln(-I*tan(d*x+c)^(1/3) +tan(d*x+c)^(2/3)-1)+1/12*3^(1/2)*arctanh(1/3*(-I+2*tan(d*x+c)^(1/3))*3^(1 /2)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 493 vs. \(2 (238) = 476\).
Time = 0.09 (sec) , antiderivative size = 493, normalized size of antiderivative = 1.63 \[ \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx =\text {Too large to display} \] Input:
integrate(tan(d*x+c)^(2/3)/(a+I*a*tan(d*x+c)),x, algorithm="fricas")
Output:
-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) - (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) + (3*s qrt(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) + 2*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))^(2/3)*(-I*e^(2*I*d*x + 2*I*c) - I))*e^(-2*I*d*x - 2*I*c)/(a*d)
\[ \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=- \frac {i \int \frac {\tan ^{\frac {2}{3}}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx}{a} \] Input:
integrate(tan(d*x+c)**(2/3)/(a+I*a*tan(d*x+c)),x)
Output:
-I*Integral(tan(c + d*x)**(2/3)/(tan(c + d*x) - I), x)/a
Exception generated. \[ \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(tan(d*x+c)^(2/3)/(a+I*a*tan(d*x+c)),x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
Time = 0.16 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.59 \[ \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {i \, {\left (i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}\right ) - 3 i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}\right ) - \frac {12 i \, \tan \left (d x + c\right )^{\frac {2}{3}}}{\tan \left (d x + c\right ) - i} + 3 \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right ) + \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right ) - 2 \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right ) - 6 \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )\right )}}{24 \, a d} \] Input:
integrate(tan(d*x+c)^(2/3)/(a+I*a*tan(d*x+c)),x, algorithm="giac")
Output:
1/24*I*(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)) - 3*I*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1/3) - I)/(sqrt(3) + 2*tan(d*x + c)^(1/3) + I)) - 12*I*tan(d*x + c)^(2/3)/(tan (d*x + c) - I) + 3*log(tan(d*x + c)^(2/3) + I*tan(d*x + c)^(1/3) - 1) + lo g(tan(d*x + c)^(2/3) - I*tan(d*x + c)^(1/3) - 1) - 2*log(tan(d*x + c)^(1/3 ) + I) - 6*log(tan(d*x + c)^(1/3) - I))/(a*d)
Time = 2.51 (sec) , antiderivative size = 599, normalized size of antiderivative = 1.98 \[ \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx =\text {Too large to display} \] Input:
int(tan(c + d*x)^(2/3)/(a + a*tan(c + d*x)*1i),x)
Output:
log((a^3*d^3*3744i - a^4*d^4*tan(c + d*x)^(1/3)*(1i/(64*a^3*d^3))^(1/3)*17 280i)*(1i/(64*a^3*d^3))^(2/3) - 36*a*d*tan(c + d*x)^(1/3))*(1i/(64*a^3*d^3 ))^(1/3) + log((a^3*d^3*3744i - a^4*d^4*tan(c + d*x)^(1/3)*(1i/(1728*a^3*d ^3))^(1/3)*17280i)*(1i/(1728*a^3*d^3))^(2/3) - 36*a*d*tan(c + d*x)^(1/3))* (1i/(1728*a^3*d^3))^(1/3) + (log(((3^(1/2)*1i - 1)^2*(a^3*d^3*3744i - a^4* d^4*tan(c + d*x)^(1/3)*(3^(1/2)*1i - 1)*(1i/(64*a^3*d^3))^(1/3)*8640i)*(1i /(64*a^3*d^3))^(2/3))/4 - 36*a*d*tan(c + d*x)^(1/3))*(3^(1/2)*1i - 1)*(1i/ (64*a^3*d^3))^(1/3))/2 - (log(((3^(1/2)*1i + 1)^2*(a^3*d^3*3744i + a^4*d^4 *tan(c + d*x)^(1/3)*(3^(1/2)*1i + 1)*(1i/(64*a^3*d^3))^(1/3)*8640i)*(1i/(6 4*a^3*d^3))^(2/3))/4 - 36*a*d*tan(c + d*x)^(1/3))*(3^(1/2)*1i + 1)*(1i/(64 *a^3*d^3))^(1/3))/2 + (log(((3^(1/2)*1i - 1)^2*(a^3*d^3*3744i - a^4*d^4*ta n(c + d*x)^(1/3)*(3^(1/2)*1i - 1)*(1i/(1728*a^3*d^3))^(1/3)*8640i)*(1i/(17 28*a^3*d^3))^(2/3))/4 - 36*a*d*tan(c + d*x)^(1/3))*(3^(1/2)*1i - 1)*(1i/(1 728*a^3*d^3))^(1/3))/2 - (log(((3^(1/2)*1i + 1)^2*(a^3*d^3*3744i + a^4*d^4 *tan(c + d*x)^(1/3)*(3^(1/2)*1i + 1)*(1i/(1728*a^3*d^3))^(1/3)*8640i)*(1i/ (1728*a^3*d^3))^(2/3))/4 - 36*a*d*tan(c + d*x)^(1/3))*(3^(1/2)*1i + 1)*(1i /(1728*a^3*d^3))^(1/3))/2 + (tan(c + d*x)^(2/3)*1i)/(2*a*d*(tan(c + d*x)*1 i + 1))
\[ \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {-3 \tan \left (d x +c \right )^{\frac {2}{3}} i +2 \left (\int \frac {\tan \left (d x +c \right )^{\frac {2}{3}}}{\tan \left (d x +c \right )^{2}-\tan \left (d x +c \right ) i}d x \right ) d +2 \left (\int \frac {\tan \left (d x +c \right )^{\frac {8}{3}}}{\tan \left (d x +c \right )-i}d x \right ) d i +2 \left (\int \frac {\tan \left (d x +c \right )^{\frac {5}{3}}}{\tan \left (d x +c \right )-i}d x \right ) d}{2 a d} \] Input:
int(tan(d*x+c)^(2/3)/(a+I*a*tan(d*x+c)),x)
Output:
( - 3*tan(c + d*x)**(2/3)*i + 2*int(tan(c + d*x)**(2/3)/(tan(c + d*x)**2 - tan(c + d*x)*i),x)*d + 2*int((tan(c + d*x)**(2/3)*tan(c + d*x)**2)/(tan(c + d*x) - i),x)*d*i + 2*int((tan(c + d*x)**(2/3)*tan(c + d*x))/(tan(c + d* x) - i),x)*d)/(2*a*d)