Integrand size = 28, antiderivative size = 353 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^2} \, dx=\frac {\left (\frac {49}{16}-\frac {45 i}{16}\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 d^{5/2} f}-\frac {\left (\frac {49}{16}-\frac {45 i}{16}\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 d^{5/2} f}+\frac {\left (\frac {49}{32}+\frac {45 i}{32}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 d^{5/2} f}-\frac {\left (\frac {49}{32}+\frac {45 i}{32}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 d^{5/2} f}-\frac {49}{24 a^2 d f (d \tan (e+f x))^{3/2}}+\frac {9}{8 a^2 d f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}+\frac {45 i}{8 a^2 d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{4 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2} \]
(49/32-45/32*I)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^2/d^(5/2) /f*2^(1/2)+(-49/32+45/32*I)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2)) /a^2/d^(5/2)/f*2^(1/2)+(49/64+45/64*I)*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e))^( 1/2)+d^(1/2)*tan(f*x+e))/a^2/d^(5/2)/f*2^(1/2)-(49/64+45/64*I)*ln(d^(1/2)+ 2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/a^2/d^(5/2)/f*2^(1/2)+45/ 8*I/a^2/d^2/f/(d*tan(f*x+e))^(1/2)-49/24/a^2/d/f/(d*tan(f*x+e))^(3/2)+9/8/ a^2/d/f/(1+I*tan(f*x+e))/(d*tan(f*x+e))^(3/2)+1/4/d/f/(d*tan(f*x+e))^(3/2) /(a+I*a*tan(f*x+e))^2
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.71 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.44 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^2} \, dx=\frac {\sec ^2(e+f x) \left (94 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-i \tan (e+f x)\right ) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+4 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},i \tan (e+f x)\right ) (\cos (2 (e+f x))+i \sin (2 (e+f x)))-3 (11+11 \cos (2 (e+f x))+9 i \sin (2 (e+f x)))\right )}{48 a^2 d f (d \tan (e+f x))^{3/2} (-i+\tan (e+f x))^2} \]
(Sec[e + f*x]^2*(94*Hypergeometric2F1[-3/2, 1, -1/2, (-I)*Tan[e + f*x]]*(C os[2*(e + f*x)] + I*Sin[2*(e + f*x)]) + 4*Hypergeometric2F1[-3/2, 1, -1/2, I*Tan[e + f*x]]*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]) - 3*(11 + 11*Cos[ 2*(e + f*x)] + (9*I)*Sin[2*(e + f*x)])))/(48*a^2*d*f*(d*Tan[e + f*x])^(3/2 )*(-I + Tan[e + f*x])^2)
Time = 1.22 (sec) , antiderivative size = 348, normalized size of antiderivative = 0.99, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 4042, 27, 3042, 4079, 3042, 4012, 25, 3042, 4012, 3042, 4017, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^2 (d \tan (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^2 (d \tan (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4042 |
| \(\displaystyle \frac {\int \frac {11 a d-7 i a d \tan (e+f x)}{2 (d \tan (e+f x))^{5/2} (i \tan (e+f x) a+a)}dx}{4 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {11 a d-7 i a d \tan (e+f x)}{(d \tan (e+f x))^{5/2} (i \tan (e+f x) a+a)}dx}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {11 a d-7 i a d \tan (e+f x)}{(d \tan (e+f x))^{5/2} (i \tan (e+f x) a+a)}dx}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\frac {\int \frac {49 a^2 d^2-45 i a^2 d^2 \tan (e+f x)}{(d \tan (e+f x))^{5/2}}dx}{2 a^2 d}+\frac {9}{f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {49 a^2 d^2-45 i a^2 d^2 \tan (e+f x)}{(d \tan (e+f x))^{5/2}}dx}{2 a^2 d}+\frac {9}{f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {\frac {-\frac {98 a^2 d}{3 f (d \tan (e+f x))^{3/2}}+\frac {\int -\frac {45 i a^2 d^3+49 a^2 \tan (e+f x) d^3}{(d \tan (e+f x))^{3/2}}dx}{d^2}}{2 a^2 d}+\frac {9}{f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {-\frac {98 a^2 d}{3 f (d \tan (e+f x))^{3/2}}-\frac {\int \frac {45 i a^2 d^3+49 a^2 \tan (e+f x) d^3}{(d \tan (e+f x))^{3/2}}dx}{d^2}}{2 a^2 d}+\frac {9}{f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {-\frac {98 a^2 d}{3 f (d \tan (e+f x))^{3/2}}-\frac {\int \frac {45 i a^2 d^3+49 a^2 \tan (e+f x) d^3}{(d \tan (e+f x))^{3/2}}dx}{d^2}}{2 a^2 d}+\frac {9}{f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {\frac {-\frac {98 a^2 d}{3 f (d \tan (e+f x))^{3/2}}-\frac {\frac {\int \frac {49 a^2 d^4-45 i a^2 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{d^2}-\frac {90 i a^2 d^2}{f \sqrt {d \tan (e+f x)}}}{d^2}}{2 a^2 d}+\frac {9}{f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {-\frac {98 a^2 d}{3 f (d \tan (e+f x))^{3/2}}-\frac {\frac {\int \frac {49 a^2 d^4-45 i a^2 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{d^2}-\frac {90 i a^2 d^2}{f \sqrt {d \tan (e+f x)}}}{d^2}}{2 a^2 d}+\frac {9}{f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {\frac {-\frac {98 a^2 d}{3 f (d \tan (e+f x))^{3/2}}-\frac {\frac {2 \int \frac {a^2 d^4 (49 d-45 i d \tan (e+f x))}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{d^2 f}-\frac {90 i a^2 d^2}{f \sqrt {d \tan (e+f x)}}}{d^2}}{2 a^2 d}+\frac {9}{f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {-\frac {98 a^2 d}{3 f (d \tan (e+f x))^{3/2}}-\frac {\frac {2 a^2 d^2 \int \frac {49 d-45 i d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}-\frac {90 i a^2 d^2}{f \sqrt {d \tan (e+f x)}}}{d^2}}{2 a^2 d}+\frac {9}{f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {\frac {-\frac {98 a^2 d}{3 f (d \tan (e+f x))^{3/2}}-\frac {\frac {2 a^2 d^2 \left (\left (\frac {49}{2}+\frac {45 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {49}{2}-\frac {45 i}{2}\right ) \int \frac {\tan (e+f x) d+d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}-\frac {90 i a^2 d^2}{f \sqrt {d \tan (e+f x)}}}{d^2}}{2 a^2 d}+\frac {9}{f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {\frac {-\frac {98 a^2 d}{3 f (d \tan (e+f x))^{3/2}}-\frac {\frac {2 a^2 d^2 \left (\left (\frac {49}{2}+\frac {45 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {49}{2}-\frac {45 i}{2}\right ) \left (\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}\right )\right )}{f}-\frac {90 i a^2 d^2}{f \sqrt {d \tan (e+f x)}}}{d^2}}{2 a^2 d}+\frac {9}{f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {\frac {-\frac {98 a^2 d}{3 f (d \tan (e+f x))^{3/2}}-\frac {\frac {2 a^2 d^2 \left (\left (\frac {49}{2}+\frac {45 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {49}{2}-\frac {45 i}{2}\right ) \left (\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}-\frac {90 i a^2 d^2}{f \sqrt {d \tan (e+f x)}}}{d^2}}{2 a^2 d}+\frac {9}{f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {-\frac {98 a^2 d}{3 f (d \tan (e+f x))^{3/2}}-\frac {\frac {2 a^2 d^2 \left (\left (\frac {49}{2}+\frac {45 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {49}{2}-\frac {45 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}-\frac {90 i a^2 d^2}{f \sqrt {d \tan (e+f x)}}}{d^2}}{2 a^2 d}+\frac {9}{f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {\frac {-\frac {98 a^2 d}{3 f (d \tan (e+f x))^{3/2}}-\frac {\frac {2 a^2 d^2 \left (\left (\frac {49}{2}+\frac {45 i}{2}\right ) \left (-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\left (\frac {49}{2}-\frac {45 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}-\frac {90 i a^2 d^2}{f \sqrt {d \tan (e+f x)}}}{d^2}}{2 a^2 d}+\frac {9}{f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {-\frac {98 a^2 d}{3 f (d \tan (e+f x))^{3/2}}-\frac {\frac {2 a^2 d^2 \left (\left (\frac {49}{2}+\frac {45 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\left (\frac {49}{2}-\frac {45 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}-\frac {90 i a^2 d^2}{f \sqrt {d \tan (e+f x)}}}{d^2}}{2 a^2 d}+\frac {9}{f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {-\frac {98 a^2 d}{3 f (d \tan (e+f x))^{3/2}}-\frac {\frac {2 a^2 d^2 \left (\left (\frac {49}{2}+\frac {45 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}\right )+\left (\frac {49}{2}-\frac {45 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}-\frac {90 i a^2 d^2}{f \sqrt {d \tan (e+f x)}}}{d^2}}{2 a^2 d}+\frac {9}{f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {\frac {-\frac {98 a^2 d}{3 f (d \tan (e+f x))^{3/2}}-\frac {\frac {2 a^2 d^2 \left (\left (\frac {49}{2}-\frac {45 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )+\left (\frac {49}{2}+\frac {45 i}{2}\right ) \left (\frac {\log \left (d \tan (e+f x)+\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (d \tan (e+f x)-\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}-\frac {90 i a^2 d^2}{f \sqrt {d \tan (e+f x)}}}{d^2}}{2 a^2 d}+\frac {9}{f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\) |
1/(4*d*f*(d*Tan[e + f*x])^(3/2)*(a + I*a*Tan[e + f*x])^2) + (9/(f*(1 + I*T an[e + f*x])*(d*Tan[e + f*x])^(3/2)) + ((-98*a^2*d)/(3*f*(d*Tan[e + f*x])^ (3/2)) - ((2*a^2*d^2*((49/2 - (45*I)/2)*(-(ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[ e + f*x]])/Sqrt[d]]/(Sqrt[2]*Sqrt[d])) + ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2]*Sqrt[d])) + (49/2 + (45*I)/2)*(-1/2*Log[d + d*T an[e + f*x] - Sqrt[2]*Sqrt[d]*Sqrt[d*Tan[e + f*x]]]/(Sqrt[2]*Sqrt[d]) + Lo g[d + d*Tan[e + f*x] + Sqrt[2]*Sqrt[d]*Sqrt[d*Tan[e + f*x]]]/(2*Sqrt[2]*Sq rt[d]))))/f - ((90*I)*a^2*d^2)/(f*Sqrt[d*Tan[e + f*x]]))/d^2)/(2*a^2*d))/( 8*a^2*d)
3.2.78.3.1 Defintions of rubi rules used
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[(-(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[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{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_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ a*c, 2]}, Simp[(d*q + a*e)/(2*a*c) Int[(q + c*x^2)/(a + c*x^4), x], x] + Simp[(d*q - a*e)/(2*a*c) Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a , c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- a)*c]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ (f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x] )^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ )]], x_Symbol] :> Simp[2/f Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & & NeQ[c^2 + d^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*(b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d)) In t[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && !GtQ[n, 0]
Time = 0.88 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.42
| method | result | size |
| derivativedivides | \(\frac {2 d^{3} \left (-\frac {1}{3 d^{4} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {2 i}{d^{5} \sqrt {d \tan \left (f x +e \right )}}-\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 d^{5} \sqrt {i d}}+\frac {\frac {-\frac {13 i \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2}-\frac {15 d \sqrt {d \tan \left (f x +e \right )}}{2}}{\left (i d \tan \left (f x +e \right )+d \right )^{2}}+\frac {47 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{2 \sqrt {-i d}}}{8 d^{5}}\right )}{f \,a^{2}}\) | \(149\) |
| default | \(\frac {2 d^{3} \left (-\frac {1}{3 d^{4} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {2 i}{d^{5} \sqrt {d \tan \left (f x +e \right )}}-\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 d^{5} \sqrt {i d}}+\frac {\frac {-\frac {13 i \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2}-\frac {15 d \sqrt {d \tan \left (f x +e \right )}}{2}}{\left (i d \tan \left (f x +e \right )+d \right )^{2}}+\frac {47 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{2 \sqrt {-i d}}}{8 d^{5}}\right )}{f \,a^{2}}\) | \(149\) |
2/f/a^2*d^3*(-1/3/d^4/(d*tan(f*x+e))^(3/2)+2*I/d^5/(d*tan(f*x+e))^(1/2)-1/ 8*I/d^5/(I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(I*d)^(1/2))+1/8/d^5*((-13 /2*I*(d*tan(f*x+e))^(3/2)-15/2*d*(d*tan(f*x+e))^(1/2))/(I*d*tan(f*x+e)+d)^ 2+47/2*I/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2))))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 798 vs. \(2 (263) = 526\).
Time = 0.27 (sec) , antiderivative size = 798, normalized size of antiderivative = 2.26 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^2} \, dx=\frac {12 \, {\left (a^{2} d^{3} f e^{\left (8 i \, f x + 8 i \, e\right )} - 2 \, a^{2} d^{3} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {-\frac {i}{16 \, a^{4} d^{5} f^{2}}} \log \left (-2 \, {\left (4 \, {\left (a^{2} d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} d^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {i}{16 \, a^{4} d^{5} f^{2}}} + i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - 12 \, {\left (a^{2} d^{3} f e^{\left (8 i \, f x + 8 i \, e\right )} - 2 \, a^{2} d^{3} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {-\frac {i}{16 \, a^{4} d^{5} f^{2}}} \log \left (2 \, {\left (4 \, {\left (a^{2} d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} d^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {i}{16 \, a^{4} d^{5} f^{2}}} - i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - 12 \, {\left (a^{2} d^{3} f e^{\left (8 i \, f x + 8 i \, e\right )} - 2 \, a^{2} d^{3} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {\frac {2209 i}{64 \, a^{4} d^{5} f^{2}}} \log \left (-\frac {{\left (8 \, {\left (a^{2} d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} d^{2} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {2209 i}{64 \, a^{4} d^{5} f^{2}}} + 47 i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{2} d^{2} f}\right ) + 12 \, {\left (a^{2} d^{3} f e^{\left (8 i \, f x + 8 i \, e\right )} - 2 \, a^{2} d^{3} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {\frac {2209 i}{64 \, a^{4} d^{5} f^{2}}} \log \left (\frac {{\left (8 \, {\left (a^{2} d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} d^{2} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {2209 i}{64 \, a^{4} d^{5} f^{2}}} - 47 i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{2} d^{2} f}\right ) - \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (202 \, e^{\left (8 i \, f x + 8 i \, e\right )} - 103 \, e^{\left (6 i \, f x + 6 i \, e\right )} - 269 \, e^{\left (4 i \, f x + 4 i \, e\right )} + 39 \, e^{\left (2 i \, f x + 2 i \, e\right )} + 3\right )}}{48 \, {\left (a^{2} d^{3} f e^{\left (8 i \, f x + 8 i \, e\right )} - 2 \, a^{2} d^{3} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )}} \]
1/48*(12*(a^2*d^3*f*e^(8*I*f*x + 8*I*e) - 2*a^2*d^3*f*e^(6*I*f*x + 6*I*e) + a^2*d^3*f*e^(4*I*f*x + 4*I*e))*sqrt(-1/16*I/(a^4*d^5*f^2))*log(-2*(4*(a^ 2*d^3*f*e^(2*I*f*x + 2*I*e) + a^2*d^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-1/16*I/(a^4*d^5*f^2)) + I*d*e^(2*I*f *x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) - 12*(a^2*d^3*f*e^(8*I*f*x + 8*I*e) - 2 *a^2*d^3*f*e^(6*I*f*x + 6*I*e) + a^2*d^3*f*e^(4*I*f*x + 4*I*e))*sqrt(-1/16 *I/(a^4*d^5*f^2))*log(2*(4*(a^2*d^3*f*e^(2*I*f*x + 2*I*e) + a^2*d^3*f)*sqr t((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-1/16*I /(a^4*d^5*f^2)) - I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) - 12*(a^2 *d^3*f*e^(8*I*f*x + 8*I*e) - 2*a^2*d^3*f*e^(6*I*f*x + 6*I*e) + a^2*d^3*f*e ^(4*I*f*x + 4*I*e))*sqrt(2209/64*I/(a^4*d^5*f^2))*log(-1/8*(8*(a^2*d^2*f*e ^(2*I*f*x + 2*I*e) + a^2*d^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^( 2*I*f*x + 2*I*e) + 1))*sqrt(2209/64*I/(a^4*d^5*f^2)) + 47*I)*e^(-2*I*f*x - 2*I*e)/(a^2*d^2*f)) + 12*(a^2*d^3*f*e^(8*I*f*x + 8*I*e) - 2*a^2*d^3*f*e^( 6*I*f*x + 6*I*e) + a^2*d^3*f*e^(4*I*f*x + 4*I*e))*sqrt(2209/64*I/(a^4*d^5* f^2))*log(1/8*(8*(a^2*d^2*f*e^(2*I*f*x + 2*I*e) + a^2*d^2*f)*sqrt((-I*d*e^ (2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(2209/64*I/(a^4*d^ 5*f^2)) - 47*I)*e^(-2*I*f*x - 2*I*e)/(a^2*d^2*f)) - sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*(202*e^(8*I*f*x + 8*I*e) - 103* e^(6*I*f*x + 6*I*e) - 269*e^(4*I*f*x + 4*I*e) + 39*e^(2*I*f*x + 2*I*e) ...
\[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^2} \, dx=- \frac {\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan ^{2}{\left (e + f x \right )} - 2 i \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan {\left (e + f x \right )} - \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx}{a^{2}} \]
-Integral(1/((d*tan(e + f*x))**(5/2)*tan(e + f*x)**2 - 2*I*(d*tan(e + f*x) )**(5/2)*tan(e + f*x) - (d*tan(e + f*x))**(5/2)), x)/a**2
Exception generated. \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \]
Time = 1.28 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.68 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^2} \, dx=-\frac {1}{24} \, d^{3} {\left (\frac {6 i \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{2} d^{\frac {11}{2}} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {141 i \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{2} d^{\frac {11}{2}} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {3 \, {\left (-13 i \, \sqrt {d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) - 15 \, \sqrt {d \tan \left (f x + e\right )} d\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} a^{2} d^{5} f} + \frac {16 \, {\left (-6 i \, d \tan \left (f x + e\right ) + d\right )}}{\sqrt {d \tan \left (f x + e\right )} a^{2} d^{6} f \tan \left (f x + e\right )}\right )} \]
-1/24*d^3*(6*I*sqrt(2)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(4*I*sqrt(2 )*d^(3/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(d)))/(a^2*d^(11/2)*f*(I*d/sqrt(d^2) + 1)) - 141*I*sqrt(2)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(-4*I*sqrt(2) *d^(3/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(d)))/(a^2*d^(11/2)*f*(-I*d/sqrt(d^2) + 1)) + 3*(-13*I*sqrt(d*tan(f*x + e))*d*tan(f*x + e) - 15*sqrt(d*tan(f*x + e))*d)/((d*tan(f*x + e) - I*d)^2*a^2*d^5*f) + 16*(-6*I*d*tan(f*x + e) + d) /(sqrt(d*tan(f*x + e))*a^2*d^6*f*tan(f*x + e)))
Time = 7.60 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.59 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^2} \, dx=-\mathrm {atan}\left (8\,a^2\,d^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {1{}\mathrm {i}}{64\,a^4\,d^5\,f^2}}\right )\,\sqrt {-\frac {1{}\mathrm {i}}{64\,a^4\,d^5\,f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {16\,a^2\,d^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {2209{}\mathrm {i}}{256\,a^4\,d^5\,f^2}}}{47}\right )\,\sqrt {\frac {2209{}\mathrm {i}}{256\,a^4\,d^5\,f^2}}\,2{}\mathrm {i}-\frac {\frac {2\,d}{3\,a^2\,f}+\frac {221\,d\,{\mathrm {tan}\left (e+f\,x\right )}^2}{24\,a^2\,f}+\frac {d\,{\mathrm {tan}\left (e+f\,x\right )}^3\,45{}\mathrm {i}}{8\,a^2\,f}-\frac {d\,\mathrm {tan}\left (e+f\,x\right )\,8{}\mathrm {i}}{3\,a^2\,f}}{d^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}-{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}+d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,2{}\mathrm {i}} \]
atan((16*a^2*d^2*f*(d*tan(e + f*x))^(1/2)*(2209i/(256*a^4*d^5*f^2))^(1/2)) /47)*(2209i/(256*a^4*d^5*f^2))^(1/2)*2i - atan(8*a^2*d^2*f*(d*tan(e + f*x) )^(1/2)*(-1i/(64*a^4*d^5*f^2))^(1/2))*(-1i/(64*a^4*d^5*f^2))^(1/2)*2i - (( 2*d)/(3*a^2*f) + (221*d*tan(e + f*x)^2)/(24*a^2*f) + (d*tan(e + f*x)^3*45i )/(8*a^2*f) - (d*tan(e + f*x)*8i)/(3*a^2*f))/(d*(d*tan(e + f*x))^(5/2)*2i - (d*tan(e + f*x))^(7/2) + d^2*(d*tan(e + f*x))^(3/2))