Integrand size = 26, antiderivative size = 343 \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {7 \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {7 \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {5 i \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 \sqrt {3} a d}-\frac {7 \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {5 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}+\frac {7 \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {7 \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}+\frac {5 i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{12 a d}+\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))} \]
-7/12*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))/a/d-7/12*arctan(3^(1/2)+2*tan(d* x+c)^(1/3))/a/d-7/6*arctan(tan(d*x+c)^(1/3))/a/d-5/6*I*ln(1+tan(d*x+c)^(2/ 3))/a/d+5/12*I*ln(1-tan(d*x+c)^(2/3)+tan(d*x+c)^(4/3))/a/d-5/6*I*arctan(1/ 3*(1-2*tan(d*x+c)^(2/3))*3^(1/2))/a/d*3^(1/2)+7/24*ln(1-3^(1/2)*tan(d*x+c) ^(1/3)+tan(d*x+c)^(2/3))/a/d*3^(1/2)-7/24*ln(1+3^(1/2)*tan(d*x+c)^(1/3)+ta n(d*x+c)^(2/3))/a/d*3^(1/2)+7/2*tan(d*x+c)^(1/3)/a/d-5/4*I*tan(d*x+c)^(4/3 )/a/d-1/2*tan(d*x+c)^(7/3)/d/(a+I*a*tan(d*x+c))
Time = 6.62 (sec) , antiderivative size = 530, normalized size of antiderivative = 1.55 \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {\tan ^{\frac {13}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\frac {10 i a \left (-\frac {3}{4} \tan ^{\frac {4}{3}}(c+d x)+\frac {3}{10} \tan ^{\frac {10}{3}}(c+d x)+\frac {3}{4} \tan ^{\frac {4}{3}}(c+d x) \left (-\frac {2 \log \left (1+\sqrt [3]{\tan ^2(c+d x)}\right )}{3 \tan ^2(c+d x)^{2/3}}-\frac {2 (-1)^{2/3} \log \left (1-e^{-\frac {i \pi }{3}} \sqrt [3]{\tan ^2(c+d x)}\right )}{3 \tan ^2(c+d x)^{2/3}}+\frac {2 \sqrt [3]{-1} \log \left (1-e^{\frac {i \pi }{3}} \sqrt [3]{\tan ^2(c+d x)}\right )}{3 \tan ^2(c+d x)^{2/3}}\right )\right )}{3 d}-\frac {7 a \left (-3 \sqrt [3]{\tan (c+d x)}+\frac {3}{7} \tan ^{\frac {7}{3}}(c+d x)+3 \sqrt [3]{\tan (c+d x)} \left (\frac {i \log \left (1-i \sqrt [6]{\tan ^2(c+d x)}\right )}{6 \sqrt [6]{\tan ^2(c+d x)}}-\frac {i \log \left (1+i \sqrt [6]{\tan ^2(c+d x)}\right )}{6 \sqrt [6]{\tan ^2(c+d x)}}-\frac {\sqrt [6]{-1} \log \left (1-e^{-\frac {i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right )}{6 \sqrt [6]{\tan ^2(c+d x)}}+\frac {(-1)^{5/6} \log \left (1-e^{\frac {i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right )}{6 \sqrt [6]{\tan ^2(c+d x)}}-\frac {(-1)^{5/6} \log \left (1-e^{-\frac {5 i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right )}{6 \sqrt [6]{\tan ^2(c+d x)}}+\frac {\sqrt [6]{-1} \log \left (1-e^{\frac {5 i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right )}{6 \sqrt [6]{\tan ^2(c+d x)}}\right )\right )}{3 d}}{2 a^2} \]
Tan[c + d*x]^(13/3)/(2*d*(a + I*a*Tan[c + d*x])) + ((((10*I)/3)*a*((-3*Tan [c + d*x]^(4/3))/4 + (3*Tan[c + d*x]^(10/3))/10 + (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 - (7*a*(-3*Tan[c + d*x]^(1/3) + (3*Tan[c + d*x] ^(7/3))/7 + 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 = 0.95 (sec) , antiderivative size = 279, normalized size of antiderivative = 0.81, number of steps used = 24, number of rules used = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.885, Rules used = {3042, 4033, 27, 3042, 4011, 3042, 4011, 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.
\(\displaystyle \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^{10/3}}{a+i a \tan (c+d x)}dx\) |
\(\Big \downarrow \) 4033 |
\(\displaystyle \frac {\int \frac {1}{3} \tan ^{\frac {4}{3}}(c+d x) (7 a-10 i a \tan (c+d x))dx}{2 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \tan ^{\frac {4}{3}}(c+d x) (7 a-10 i a \tan (c+d x))dx}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \tan (c+d x)^{4/3} (7 a-10 i a \tan (c+d x))dx}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \frac {\int \sqrt [3]{\tan (c+d x)} (7 \tan (c+d x) a+10 i a)dx-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sqrt [3]{\tan (c+d x)} (7 \tan (c+d x) a+10 i a)dx-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \frac {\int \frac {10 i a \tan (c+d x)-7 a}{\tan ^{\frac {2}{3}}(c+d x)}dx-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {10 i a \tan (c+d x)-7 a}{\tan (c+d x)^{2/3}}dx-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 4021 |
\(\displaystyle \frac {-7 a \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)}dx+10 i a \int \sqrt [3]{\tan (c+d x)}dx-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-7 a \int \frac {1}{\tan (c+d x)^{2/3}}dx+10 i a \int \sqrt [3]{\tan (c+d x)}dx-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 3957 |
\(\displaystyle \frac {\frac {10 i a \int \frac {\sqrt [3]{\tan (c+d x)}}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}-\frac {7 a \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x) \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {-\frac {21 a \int \frac {1}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {30 i a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 753 |
\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{2 \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{2 \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}\right )}{d}+\frac {30 i a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )}{d}+\frac {30 i a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {30 i a \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {15 i a \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{\tan (c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 821 |
\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {15 i a \left (\frac {1}{3} \int \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)\right )}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{6} \int \frac {2-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+2}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {15 i a \left (\frac {1}{3} \int \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int -\frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {15 i a \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{2} \int \left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {15 i a \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}-\sqrt {3}\right )\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {15 i a \left (\frac {1}{3} \left (-3 \int \frac {1}{-2 \tan ^{\frac {2}{3}}(c+d x)-2}d\left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )-\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {-\frac {21 a \left (\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}+\frac {15 i a \left (\frac {1}{3} \left (\sqrt {3} \arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )-\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {\frac {15 i a \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {21 a \left (\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )-\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )+\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )\right )}{d}-\frac {15 i a \tan ^{\frac {4}{3}}(c+d x)}{2 d}+\frac {21 a \sqrt [3]{\tan (c+d x)}}{d}}{6 a^2}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\) |
-1/2*Tan[c + d*x]^(7/3)/(d*(a + I*a*Tan[c + d*x])) + (((15*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 - (21*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 + S qrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)])/2)/6))/d + (21*a*Tan[c + d*x]^(1/3))/d - (((15*I)/2)*a*Tan[c + d*x]^(4/3))/d)/(6*a^2)
3.3.33.3.1 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[((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]
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 = 0.66 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.62
method | result | size |
derivativedivides | \(\frac {3 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )-\frac {3 i \left (\tan ^{\frac {4}{3}}\left (d x +c \right )\right )}{4}-\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{4}+\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{8}+\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{4}-\frac {17 i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{12}+\frac {1}{6 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+6 i}+\frac {-2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )-2 i}{-12 i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+12 \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )-12}+\frac {17 i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{24}-\frac {17 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{12}}{d a}\) | \(211\) |
default | \(\frac {3 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )-\frac {3 i \left (\tan ^{\frac {4}{3}}\left (d x +c \right )\right )}{4}-\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{4}+\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{8}+\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{4}-\frac {17 i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{12}+\frac {1}{6 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+6 i}+\frac {-2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )-2 i}{-12 i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+12 \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )-12}+\frac {17 i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{24}-\frac {17 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{12}}{d a}\) | \(211\) |
1/d/a*(3*tan(d*x+c)^(1/3)-3/4*I*tan(d*x+c)^(4/3)-1/4*I*ln(tan(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))-17/12*I*ln(tan(d*x+c)^(1/3)+I)+1/6/(tan( 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)+17/24*I*ln(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)-17/12*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. 645 vs. \(2 (270) = 540\).
Time = 0.26 (sec) , antiderivative size = 645, normalized size of antiderivative = 1.88 \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {3 \, {\left (\sqrt {3} {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {1}{a^{2} d^{2}}} + i \, e^{\left (4 i \, d x + 4 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} {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {1}{a^{2} d^{2}}} - i \, e^{\left (4 i \, d x + 4 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 ) - 17 \, {\left (3 \, \sqrt {\frac {1}{3}} {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {1}{a^{2} d^{2}}} - i \, e^{\left (4 i \, d x + 4 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 ) + 17 \, {\left (3 \, \sqrt {\frac {1}{3}} {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {1}{a^{2} d^{2}}} + i \, e^{\left (4 i \, d x + 4 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 ) - 34 \, {\left (i \, e^{\left (4 i \, d x + 4 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\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 (i \, e^{\left (4 i \, d x + 4 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\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 (10 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 17 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}}{24 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]
1/24*(3*(sqrt(3)*(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt( 1/(a^2*d^2)) + I*e^(4*I*d*x + 4*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*e^(4*I*d*x + 4*I*c) + a*d*e^(2 *I*d*x + 2*I*c))*sqrt(1/(a^2*d^2)) - I*e^(4*I*d*x + 4*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) - 17*(3*sqrt(1/3)*(a*d*e^ (4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt(1/(a^2*d^2)) - I*e^(4*I* d*x + 4*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) + 17*(3*sqrt(1/3)*(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))* sqrt(1/(a^2*d^2)) + I*e^(4*I*d*x + 4*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) - 34*(I*e^(4*I*d*x + 4*I*c) + 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^(4*I*d*x + 4*I*c) + 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)*(10*e^(4*I*d*x + 4*I *c) + 17*e^(2*I*d*x + 2*I*c) + 1))/(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d *x + 2*I*c))
\[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=- \frac {i \int \frac {\tan ^{\frac {10}{3}}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx}{a} \]
Exception generated. \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.48 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.75 \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {17 \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}\right )}{24 \, a d} - \frac {\sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}\right )}{8 \, a d} + \frac {i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{8 \, a d} + \frac {17 i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{24 \, a d} - \frac {17 i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right )}{12 \, a d} - \frac {i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )}{4 \, a d} - \frac {i \, \tan \left (d x + c\right )^{\frac {1}{3}}}{2 \, a d {\left (\tan \left (d x + c\right ) - i\right )}} - \frac {3 \, {\left (i \, a^{3} d^{3} \tan \left (d x + c\right )^{\frac {4}{3}} - 4 \, a^{3} d^{3} \tan \left (d x + c\right )^{\frac {1}{3}}\right )}}{4 \, a^{4} d^{4}} \]
17/24*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) + 17/24*I*log(tan(d*x + c)^(2/3 ) - I*tan(d*x + c)^(1/3) - 1)/(a*d) - 17/12*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)) - 3/4*(I*a^3*d^3*tan(d*x + c)^(4/3) - 4*a^3*d^3* tan(d*x + c)^(1/3))/(a^4*d^4)
Time = 6.50 (sec) , antiderivative size = 655, normalized size of antiderivative = 1.91 \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\text {Too large to display} \]
log((a^3*d^3*703584i - 414720*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)*182376i)*(1i/ (64*a^3*d^3))^(1/3) + log((a^3*d^3*703584i - 414720*a^5*d^5*tan(c + d*x)^( 1/3)*(4913i/(1728*a^3*d^3))^(2/3))*(4913i/(1728*a^3*d^3))^(1/3) + a^2*d^2* tan(c + d*x)^(1/3)*182376i)*(4913i/(1728*a^3*d^3))^(1/3) + (3*tan(c + d*x) ^(1/3))/(a*d) - (tan(c + d*x)^(4/3)*3i)/(4*a*d) + (log(((3^(1/2)*1i - 1)*( a^3*d^3*703584i - 103680*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)*182376i)*(3^(1/2)*1i - 1)*(1i/(64*a^3*d^3))^(1/3))/2 - (log(((3^(1/2)* 1i + 1)*(a^3*d^3*703584i - 103680*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)*182376i)*(3^(1/2)*1i + 1)*(1i/(64*a^3*d^3))^(1/3))/2 + (log(( (3^(1/2)*1i - 1)*(a^3*d^3*703584i - 103680*a^5*d^5*tan(c + d*x)^(1/3)*(3^( 1/2)*1i - 1)^2*(4913i/(1728*a^3*d^3))^(2/3))*(4913i/(1728*a^3*d^3))^(1/3)) /2 + a^2*d^2*tan(c + d*x)^(1/3)*182376i)*(3^(1/2)*1i - 1)*(4913i/(1728*a^3 *d^3))^(1/3))/2 - (log(((3^(1/2)*1i + 1)*(a^3*d^3*703584i - 103680*a^5*d^5 *tan(c + d*x)^(1/3)*(3^(1/2)*1i + 1)^2*(4913i/(1728*a^3*d^3))^(2/3))*(4913 i/(1728*a^3*d^3))^(1/3))/2 - a^2*d^2*tan(c + d*x)^(1/3)*182376i)*(3^(1/2)* 1i + 1)*(4913i/(1728*a^3*d^3))^(1/3))/2 + tan(c + d*x)^(1/3)/(2*a*d*(tan(c + d*x)*1i + 1))