Integrand size = 26, antiderivative size = 379 \[ \int \frac {\tan ^{\frac {14}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {121 \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {121 \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {14 i \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}+\frac {121 \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {14 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}+\frac {121 \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {121 \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {7 i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{9 a^2 d}-\frac {14 i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d}-\frac {121 \tan ^{\frac {5}{3}}(c+d x)}{60 a^2 d}+\frac {7 i \tan ^{\frac {8}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
121/72*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))/a^2/d+121/72*arctan(3^(1/2)+2*t an(d*x+c)^(1/3))/a^2/d+121/36*arctan(tan(d*x+c)^(1/3))/a^2/d+14/9*I*ln(1+t an(d*x+c)^(2/3))/a^2/d-7/9*I*ln(1-tan(d*x+c)^(2/3)+tan(d*x+c)^(4/3))/a^2/d -14/9*I*arctan(1/3*(1-2*tan(d*x+c)^(2/3))*3^(1/2))/a^2/d*3^(1/2)+121/144*l n(1-3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))/a^2/d*3^(1/2)-121/144*ln(1+ 3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))/a^2/d*3^(1/2)-14/3*I*tan(d*x+c) ^(2/3)/a^2/d-121/60*tan(d*x+c)^(5/3)/a^2/d+7/6*I*tan(d*x+c)^(8/3)/a^2/d/(1 +I*tan(d*x+c))-1/4*tan(d*x+c)^(11/3)/d/(a+I*a*tan(d*x+c))^2
Time = 6.80 (sec) , antiderivative size = 639, normalized size of antiderivative = 1.69 \[ \int \frac {\tan ^{\frac {14}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {\tan ^{\frac {17}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\frac {56 i \arctan \left (\frac {-1+2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} d}+\frac {56 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 d}-\frac {28 i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{9 d}+\frac {121 i \log \left (1-i \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}}{18 d \sqrt [3]{\tan (c+d x)}}-\frac {121 i \log \left (1+i \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}}{18 d \sqrt [3]{\tan (c+d x)}}-\frac {121 (-1)^{5/6} \log \left (1-e^{-\frac {i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}}{18 d \sqrt [3]{\tan (c+d x)}}+\frac {121 \sqrt [6]{-1} \log \left (1-e^{\frac {i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}}{18 d \sqrt [3]{\tan (c+d x)}}-\frac {121 \sqrt [6]{-1} \log \left (1-e^{-\frac {5 i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}}{18 d \sqrt [3]{\tan (c+d x)}}+\frac {121 (-1)^{5/6} \log \left (1-e^{\frac {5 i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right ) \sqrt [6]{\tan ^2(c+d x)}}{18 d \sqrt [3]{\tan (c+d x)}}-\frac {56 i a \tan ^{\frac {2}{3}}(c+d x)}{3 d (a+i a \tan (c+d x))}+\frac {53 a \tan ^{\frac {5}{3}}(c+d x)}{5 d (a+i a \tan (c+d x))}-\frac {17 i a \tan ^{\frac {8}{3}}(c+d x)}{5 d (a+i a \tan (c+d x))}-\frac {a \tan ^{\frac {11}{3}}(c+d x)}{d (a+i a \tan (c+d x))}+\frac {i a \tan ^{\frac {14}{3}}(c+d x)}{d (a+i a \tan (c+d x))}}{4 a^2} \]
Tan[c + d*x]^(17/3)/(4*d*(a + I*a*Tan[c + d*x])^2) + ((((56*I)/3)*ArcTan[( -1 + 2*Tan[c + d*x]^(2/3))/Sqrt[3]])/(Sqrt[3]*d) + (((56*I)/9)*Log[1 + Tan [c + d*x]^(2/3)])/d - (((28*I)/9)*Log[1 - Tan[c + d*x]^(2/3) + Tan[c + d*x ]^(4/3)])/d + (((121*I)/18)*Log[1 - I*(Tan[c + d*x]^2)^(1/6)]*(Tan[c + d*x ]^2)^(1/6))/(d*Tan[c + d*x]^(1/3)) - (((121*I)/18)*Log[1 + I*(Tan[c + d*x] ^2)^(1/6)]*(Tan[c + d*x]^2)^(1/6))/(d*Tan[c + d*x]^(1/3)) - (121*(-1)^(5/6 )*Log[1 - (Tan[c + d*x]^2)^(1/6)/E^((I/6)*Pi)]*(Tan[c + d*x]^2)^(1/6))/(18 *d*Tan[c + d*x]^(1/3)) + (121*(-1)^(1/6)*Log[1 - E^((I/6)*Pi)*(Tan[c + d*x ]^2)^(1/6)]*(Tan[c + d*x]^2)^(1/6))/(18*d*Tan[c + d*x]^(1/3)) - (121*(-1)^ (1/6)*Log[1 - (Tan[c + d*x]^2)^(1/6)/E^(((5*I)/6)*Pi)]*(Tan[c + d*x]^2)^(1 /6))/(18*d*Tan[c + d*x]^(1/3)) + (121*(-1)^(5/6)*Log[1 - E^(((5*I)/6)*Pi)* (Tan[c + d*x]^2)^(1/6)]*(Tan[c + d*x]^2)^(1/6))/(18*d*Tan[c + d*x]^(1/3)) - (((56*I)/3)*a*Tan[c + d*x]^(2/3))/(d*(a + I*a*Tan[c + d*x])) + (53*a*Tan [c + d*x]^(5/3))/(5*d*(a + I*a*Tan[c + d*x])) - (((17*I)/5)*a*Tan[c + d*x] ^(8/3))/(d*(a + I*a*Tan[c + d*x])) - (a*Tan[c + d*x]^(11/3))/(d*(a + I*a*T an[c + d*x])) + (I*a*Tan[c + d*x]^(14/3))/(d*(a + I*a*Tan[c + d*x])))/(4*a ^2)
Time = 1.30 (sec) , antiderivative size = 326, normalized size of antiderivative = 0.86, number of steps used = 27, number of rules used = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 4041, 27, 3042, 4078, 27, 3042, 4011, 3042, 4011, 3042, 4021, 3042, 3957, 266, 807, 750, 16, 824, 27, 216, 1142, 25, 1083, 217, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\tan ^{\frac {14}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^{14/3}}{(a+i a \tan (c+d x))^2}dx\) |
| \(\Big \downarrow \) 4041 |
| \(\displaystyle -\frac {\int -\frac {\tan ^{\frac {8}{3}}(c+d x) (11 a-17 i a \tan (c+d x))}{3 (i \tan (c+d x) a+a)}dx}{4 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\tan ^{\frac {8}{3}}(c+d x) (11 a-17 i a \tan (c+d x))}{i \tan (c+d x) a+a}dx}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\tan (c+d x)^{8/3} (11 a-17 i a \tan (c+d x))}{i \tan (c+d x) a+a}dx}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int \frac {2}{3} \tan ^{\frac {5}{3}}(c+d x) \left (121 \tan (c+d x) a^2+112 i a^2\right )dx}{2 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int \tan ^{\frac {5}{3}}(c+d x) \left (121 \tan (c+d x) a^2+112 i a^2\right )dx}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int \tan (c+d x)^{5/3} \left (121 \tan (c+d x) a^2+112 i a^2\right )dx}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\int \tan ^{\frac {2}{3}}(c+d x) \left (112 i a^2 \tan (c+d x)-121 a^2\right )dx}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\int \tan (c+d x)^{2/3} \left (112 i a^2 \tan (c+d x)-121 a^2\right )dx}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int \frac {-121 \tan (c+d x) a^2-112 i a^2}{\sqrt [3]{\tan (c+d x)}}dx+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int \frac {-121 \tan (c+d x) a^2-112 i a^2}{\sqrt [3]{\tan (c+d x)}}dx+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 4021 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-121 a^2 \int \tan ^{\frac {2}{3}}(c+d x)dx-112 i a^2 \int \frac {1}{\sqrt [3]{\tan (c+d x)}}dx+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-112 i a^2 \int \frac {1}{\sqrt [3]{\tan (c+d x)}}dx-121 a^2 \int \tan (c+d x)^{2/3}dx+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 3957 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {112 i a^2 \int \frac {1}{\sqrt [3]{\tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}-\frac {121 a^2 \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {336 i a^2 \int \frac {\sqrt [3]{\tan (c+d x)}}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {363 a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 807 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {168 i a^2 \int \frac {1}{\tan (c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)}{d}-\frac {363 a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 750 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {168 i a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\tan ^{\frac {2}{3}}(c+d x)\right )}{d}-\frac {363 a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {363 a^2 \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\tan ^2(c+d x)+1}d\sqrt [3]{\tan (c+d x)}}{d}-\frac {168 i a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 824 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {363 a^2 \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int -\frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{2 \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \int -\frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{2 \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}d\sqrt [3]{\tan (c+d x)}\right )}{d}-\frac {168 i a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {363 a^2 \left (\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )}{d}-\frac {168 i a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {363 a^2 \left (-\frac {1}{6} \int \frac {1-\sqrt {3} \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{6} \int \frac {\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {168 i a^2 \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {363 a^2 \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}+\frac {1}{2} \sqrt {3} \int -\frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {168 i a^2 \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)-\frac {1}{2} \int \left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )d\tan ^{\frac {2}{3}}(c+d x)\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {363 a^2 \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {168 i a^2 \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(c+d x)+\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {363 a^2 \left (\frac {1}{6} \left (-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}-\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (-\int \frac {1}{-\tan ^{\frac {2}{3}}(c+d x)-1}d\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {168 i a^2 \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)-3 \int \frac {1}{-2 \tan ^{\frac {2}{3}}(c+d x)-2}d\left (2 \tan ^{\frac {2}{3}}(c+d x)-1\right )\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {363 a^2 \left (\frac {1}{6} \left (-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}}{\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}}{\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1}d\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )\right )}{d}-\frac {168 i a^2 \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(c+d x)\right )d\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {\frac {14 i \tan ^{\frac {8}{3}}(c+d x)}{d (1+i \tan (c+d x))}-\frac {-\frac {168 i a^2 \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(c+d x)-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )\right )}{d}-\frac {363 a^2 \left (\frac {1}{3} \arctan \left (\sqrt [3]{\tan (c+d x)}\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )-\arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )\right )+\frac {1}{6} \left (\arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )\right )\right )}{d}+\frac {363 a^2 \tan ^{\frac {5}{3}}(c+d x)}{5 d}+\frac {168 i a^2 \tan ^{\frac {2}{3}}(c+d x)}{d}}{3 a^2}}{12 a^2}-\frac {\tan ^{\frac {11}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
-1/4*Tan[c + d*x]^(11/3)/(d*(a + I*a*Tan[c + d*x])^2) + (((14*I)*Tan[c + d *x]^(8/3))/(d*(1 + I*Tan[c + d*x])) - (((-168*I)*a^2*(ArcTan[(-1 + 2*Tan[c + d*x]^(2/3))/Sqrt[3]]/Sqrt[3] + Log[1 + Tan[c + d*x]^(2/3)]/3))/d - (363 *a^2*(ArcTan[Tan[c + d*x]^(1/3)]/3 + (-ArcTan[Sqrt[3] - 2*Tan[c + d*x]^(1/ 3)] + (Sqrt[3]*Log[1 - Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)])/2 )/6 + (ArcTan[Sqrt[3] + 2*Tan[c + d*x]^(1/3)] - (Sqrt[3]*Log[1 + Sqrt[3]*T an[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)])/2)/6))/d + ((168*I)*a^2*Tan[c + d *x]^(2/3))/d + (363*a^2*Tan[c + d*x]^(5/3))/(5*d))/(3*a^2))/(12*a^2)
3.3.40.3.1 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[((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[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m* ((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Ta n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && GtQ[n, 1] && (In tegerQ[m] || IntegersQ[2*m, 2*n])
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-(A*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f* x])^(n - 1)*Simp[A*(a*c*m + b*d*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a *A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]
Time = 0.57 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.65
| method | result | size |
| derivativedivides | \(\frac {-\frac {3 \left (\tan ^{\frac {5}{3}}\left (d x +c \right )\right )}{5}-3 i \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )+\frac {i}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )^{2}}+\frac {233 i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{72}-\frac {23}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}+\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{16}-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{8}-\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{8}-\frac {92 \tan \left (d x +c \right )-136 i \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )-130 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+44 i}{72 {\left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}^{2}}-\frac {233 i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{144}-\frac {233 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{72}}{d \,a^{2}}\) | \(246\) |
| default | \(\frac {-\frac {3 \left (\tan ^{\frac {5}{3}}\left (d x +c \right )\right )}{5}-3 i \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )+\frac {i}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )^{2}}+\frac {233 i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{72}-\frac {23}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}+\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{16}-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{8}-\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{8}-\frac {92 \tan \left (d x +c \right )-136 i \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )-130 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+44 i}{72 {\left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}^{2}}-\frac {233 i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{144}-\frac {233 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{72}}{d \,a^{2}}\) | \(246\) |
1/d/a^2*(-3/5*tan(d*x+c)^(5/3)-3*I*tan(d*x+c)^(2/3)+1/36*I/(tan(d*x+c)^(1/ 3)+I)^2+233/72*I*ln(tan(d*x+c)^(1/3)+I)-23/36/(tan(d*x+c)^(1/3)+I)+1/16*I* ln(I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)-1/8*3^(1/2)*arctanh(1/3*(I+2*tan (d*x+c)^(1/3))*3^(1/2))-1/8*I*ln(tan(d*x+c)^(1/3)-I)-1/72*(92*tan(d*x+c)-1 36*I*tan(d*x+c)^(2/3)-130*tan(d*x+c)^(1/3)+44*I)/(-I*tan(d*x+c)^(1/3)+tan( d*x+c)^(2/3)-1)^2-233/144*I*ln(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)-233 /72*3^(1/2)*arctanh(1/3*(-I+2*tan(d*x+c)^(1/3))*3^(1/2)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 684 vs. \(2 (298) = 596\).
Time = 0.28 (sec) , antiderivative size = 684, normalized size of antiderivative = 1.80 \[ \int \frac {\tan ^{\frac {14}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {45 \, {\left (\sqrt {3} {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt {\frac {1}{a^{4} d^{2}}} - i \, e^{\left (6 i \, d x + 6 i \, c\right )} - i \, 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 ) - 45 \, {\left (\sqrt {3} {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt {\frac {1}{a^{4} d^{2}}} + i \, e^{\left (6 i \, d x + 6 i \, c\right )} + i \, 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 ) + 1165 \, {\left (3 \, \sqrt {\frac {1}{3}} {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt {\frac {1}{a^{4} d^{2}}} + i \, e^{\left (6 i \, d x + 6 i \, c\right )} + i \, 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 ) - 1165 \, {\left (3 \, \sqrt {\frac {1}{3}} {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt {\frac {1}{a^{4} d^{2}}} - i \, e^{\left (6 i \, d x + 6 i \, c\right )} - i \, 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 ) + 2330 \, {\left (-i \, e^{\left (6 i \, d x + 6 i \, c\right )} - i \, e^{\left (4 i \, d x + 4 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 ) + 90 \, {\left (i \, e^{\left (6 i \, d x + 6 i \, c\right )} + i \, e^{\left (4 i \, d x + 4 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 ) + 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 (791 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 1279 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 185 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 15 i\right )}}{720 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \]
-1/720*(45*(sqrt(3)*(a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4*I*d*x + 4*I*c) )*sqrt(1/(a^4*d^2)) - I*e^(6*I*d*x + 6*I*c) - I*e^(4*I*d*x + 4*I*c))*log(1 /2*sqrt(3)*a^2*d*sqrt(1/(a^4*d^2)) + ((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I *d*x + 2*I*c) + 1))^(1/3) + 1/2*I) - 45*(sqrt(3)*(a^2*d*e^(6*I*d*x + 6*I*c ) + a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/(a^4*d^2)) + I*e^(6*I*d*x + 6*I*c) + I*e^(4*I*d*x + 4*I*c))*log(-1/2*sqrt(3)*a^2*d*sqrt(1/(a^4*d^2)) + ((-I*e^ (2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) + 1/2*I) + 1165*(3 *sqrt(1/3)*(a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/ (a^4*d^2)) + I*e^(6*I*d*x + 6*I*c) + I*e^(4*I*d*x + 4*I*c))*log(3/2*sqrt(1 /3)*a^2*d*sqrt(1/(a^4*d^2)) + ((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) - 1/2*I) - 1165*(3*sqrt(1/3)*(a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/(a^4*d^2)) - I*e^(6*I*d*x + 6*I*c) - I*e^(4*I*d*x + 4*I*c))*log(-3/2*sqrt(1/3)*a^2*d*sqrt(1/(a^4*d^2)) + ((-I*e ^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) - 1/2*I) + 2330*( -I*e^(6*I*d*x + 6*I*c) - I*e^(4*I*d*x + 4*I*c))*log(((-I*e^(2*I*d*x + 2*I* c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) + I) + 90*(I*e^(6*I*d*x + 6*I*c) + I*e^(4*I*d*x + 4*I*c))*log(((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2 *I*c) + 1))^(1/3) - I) + 3*((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I *c) + 1))^(2/3)*(791*I*e^(6*I*d*x + 6*I*c) + 1279*I*e^(4*I*d*x + 4*I*c) + 185*I*e^(2*I*d*x + 2*I*c) - 15*I))/(a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e...
Timed out. \[ \int \frac {\tan ^{\frac {14}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\tan ^{\frac {14}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.75 (sec) , antiderivative size = 268, normalized size of antiderivative = 0.71 \[ \int \frac {\tan ^{\frac {14}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {233 \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}\right )}{144 \, a^{2} d} + \frac {\sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}\right )}{16 \, a^{2} d} + \frac {i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{16 \, a^{2} d} - \frac {233 i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{144 \, a^{2} d} + \frac {233 i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right )}{72 \, a^{2} d} - \frac {i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )}{8 \, a^{2} d} - \frac {23 \, \tan \left (d x + c\right )^{\frac {5}{3}} - 20 i \, \tan \left (d x + c\right )^{\frac {2}{3}}}{12 \, a^{2} d {\left (\tan \left (d x + c\right ) - i\right )}^{2}} - \frac {3 \, {\left (a^{8} d^{4} \tan \left (d x + c\right )^{\frac {5}{3}} + 5 i \, a^{8} d^{4} \tan \left (d x + c\right )^{\frac {2}{3}}\right )}}{5 \, a^{10} d^{5}} \]
233/144*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*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*I*log(ta n(d*x + c)^(2/3) + I*tan(d*x + c)^(1/3) - 1)/(a^2*d) - 233/144*I*log(tan(d *x + c)^(2/3) - I*tan(d*x + c)^(1/3) - 1)/(a^2*d) + 233/72*I*log(tan(d*x + c)^(1/3) + I)/(a^2*d) - 1/8*I*log(tan(d*x + c)^(1/3) - I)/(a^2*d) - 1/12* (23*tan(d*x + c)^(5/3) - 20*I*tan(d*x + c)^(2/3))/(a^2*d*(tan(d*x + c) - I )^2) - 3/5*(a^8*d^4*tan(d*x + c)^(5/3) + 5*I*a^8*d^4*tan(d*x + c)^(2/3))/( a^10*d^5)
Time = 6.37 (sec) , antiderivative size = 674, normalized size of antiderivative = 1.78 \[ \int \frac {\tan ^{\frac {14}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Too large to display} \]
log(((a^6*d^3*1619208448i)/3 - a^8*d^4*tan(c + d*x)^(1/3)*(1i/(512*a^6*d^3 ))^(1/3)*167024640i)*(1i/(512*a^6*d^3))^(2/3) + (24321472*a^2*d*tan(c + d* x)^(1/3))/3)*(1i/(512*a^6*d^3))^(1/3) + log(((a^6*d^3*1619208448i)/3 - a^8 *d^4*tan(c + d*x)^(1/3)*(-12649337i/(373248*a^6*d^3))^(1/3)*167024640i)*(- 12649337i/(373248*a^6*d^3))^(2/3) + (24321472*a^2*d*tan(c + d*x)^(1/3))/3) *(-12649337i/(373248*a^6*d^3))^(1/3) - ((5*tan(c + d*x)^(2/3))/(3*a^2*d) + (tan(c + d*x)^(5/3)*23i)/(12*a^2*d))/(2*tan(c + d*x) + tan(c + d*x)^2*1i - 1i) - (tan(c + d*x)^(2/3)*3i)/(a^2*d) - (3*tan(c + d*x)^(5/3))/(5*a^2*d) + (log((24321472*a^2*d*tan(c + d*x)^(1/3))/3 + ((3^(1/2)*1i - 1)^2*((a^6* d^3*1619208448i)/3 - a^8*d^4*tan(c + d*x)^(1/3)*(3^(1/2)*1i - 1)*(1i/(512* a^6*d^3))^(1/3)*83512320i)*(1i/(512*a^6*d^3))^(2/3))/4)*(3^(1/2)*1i - 1)*( 1i/(512*a^6*d^3))^(1/3))/2 - (log((24321472*a^2*d*tan(c + d*x)^(1/3))/3 + ((3^(1/2)*1i + 1)^2*((a^6*d^3*1619208448i)/3 + a^8*d^4*tan(c + d*x)^(1/3)* (3^(1/2)*1i + 1)*(1i/(512*a^6*d^3))^(1/3)*83512320i)*(1i/(512*a^6*d^3))^(2 /3))/4)*(3^(1/2)*1i + 1)*(1i/(512*a^6*d^3))^(1/3))/2 + (log((24321472*a^2* d*tan(c + d*x)^(1/3))/3 + ((3^(1/2)*1i - 1)^2*((a^6*d^3*1619208448i)/3 - a ^8*d^4*tan(c + d*x)^(1/3)*(3^(1/2)*1i - 1)*(-12649337i/(373248*a^6*d^3))^( 1/3)*83512320i)*(-12649337i/(373248*a^6*d^3))^(2/3))/4)*(3^(1/2)*1i - 1)*( -12649337i/(373248*a^6*d^3))^(1/3))/2 - (log((24321472*a^2*d*tan(c + d*x)^ (1/3))/3 + ((3^(1/2)*1i + 1)^2*((a^6*d^3*1619208448i)/3 + a^8*d^4*tan(c...