Integrand size = 28, antiderivative size = 368 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^3} \, dx=\frac {\left (\frac {15}{8}+\frac {7 i}{4}\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 d^{3/2} f}-\frac {\left (\frac {15}{8}+\frac {7 i}{4}\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 d^{3/2} f}-\frac {\left (\frac {15}{16}-\frac {7 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 d^{3/2} f}+\frac {\left (\frac {15}{16}-\frac {7 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 d^{3/2} f}-\frac {15}{4 a^3 d f \sqrt {d \tan (e+f x)}}+\frac {1}{6 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3}+\frac {5}{12 a d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2}+\frac {7}{6 d f \sqrt {d \tan (e+f x)} \left (a^3+i a^3 \tan (e+f x)\right )} \]
(15/16+7/8*I)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/d^(3/2)/f *2^(1/2)-(15/16+7/8*I)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/ d^(3/2)/f*2^(1/2)+(-15/32+7/16*I)*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e))^(1/2)+ d^(1/2)*tan(f*x+e))/a^3/d^(3/2)/f*2^(1/2)+(15/32-7/16*I)*ln(d^(1/2)+2^(1/2 )*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/a^3/d^(3/2)/f*2^(1/2)-15/4/a^3/ d/f/(d*tan(f*x+e))^(1/2)+1/6/d/f/(d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^3 +5/12/a/d/f/(d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^2+7/6/d/f/(d*tan(f*x+e ))^(1/2)/(a^3+I*a^3*tan(f*x+e))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.71 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.46 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^3} \, dx=-\frac {i \sec ^3(e+f x) \left (-49 \cos (e+f x)-35 \cos (3 (e+f x))-33 i \sin (e+f x)+174 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-i \tan (e+f x)\right ) (\cos (3 (e+f x))+i \sin (3 (e+f x)))+6 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},i \tan (e+f x)\right ) (\cos (3 (e+f x))+i \sin (3 (e+f x)))-33 i \sin (3 (e+f x))\right )}{48 a^3 d f \sqrt {d \tan (e+f x)} (-i+\tan (e+f x))^3} \]
((-1/48*I)*Sec[e + f*x]^3*(-49*Cos[e + f*x] - 35*Cos[3*(e + f*x)] - (33*I) *Sin[e + f*x] + 174*Hypergeometric2F1[-1/2, 1, 1/2, (-I)*Tan[e + f*x]]*(Co s[3*(e + f*x)] + I*Sin[3*(e + f*x)]) + 6*Hypergeometric2F1[-1/2, 1, 1/2, I *Tan[e + f*x]]*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)]) - (33*I)*Sin[3*(e + f*x)]))/(a^3*d*f*Sqrt[d*Tan[e + f*x]]*(-I + Tan[e + f*x])^3)
Time = 1.45 (sec) , antiderivative size = 360, normalized size of antiderivative = 0.98, number of steps used = 24, number of rules used = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.821, Rules used = {3042, 4042, 27, 3042, 4079, 27, 3042, 4079, 27, 3042, 4012, 25, 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.
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^3 (d \tan (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^3 (d \tan (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4042 |
| \(\displaystyle \frac {\int \frac {13 a d-7 i a d \tan (e+f x)}{2 (d \tan (e+f x))^{3/2} (i \tan (e+f x) a+a)^2}dx}{6 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {13 a d-7 i a d \tan (e+f x)}{(d \tan (e+f x))^{3/2} (i \tan (e+f x) a+a)^2}dx}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {13 a d-7 i a d \tan (e+f x)}{(d \tan (e+f x))^{3/2} (i \tan (e+f x) a+a)^2}dx}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\frac {\int \frac {2 \left (31 a^2 d^2-25 i a^2 d^2 \tan (e+f x)\right )}{(d \tan (e+f x))^{3/2} (i \tan (e+f x) a+a)}dx}{4 a^2 d}+\frac {5 a}{f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {31 a^2 d^2-25 i a^2 d^2 \tan (e+f x)}{(d \tan (e+f x))^{3/2} (i \tan (e+f x) a+a)}dx}{2 a^2 d}+\frac {5 a}{f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {31 a^2 d^2-25 i a^2 d^2 \tan (e+f x)}{(d \tan (e+f x))^{3/2} (i \tan (e+f x) a+a)}dx}{2 a^2 d}+\frac {5 a}{f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {6 \left (15 a^3 d^3-14 i a^3 d^3 \tan (e+f x)\right )}{(d \tan (e+f x))^{3/2}}dx}{2 a^2 d}+\frac {28 a^2 d}{f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{2 a^2 d}+\frac {5 a}{f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3 \int \frac {15 a^3 d^3-14 i a^3 d^3 \tan (e+f x)}{(d \tan (e+f x))^{3/2}}dx}{a^2 d}+\frac {28 a^2 d}{f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{2 a^2 d}+\frac {5 a}{f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3 \int \frac {15 a^3 d^3-14 i a^3 d^3 \tan (e+f x)}{(d \tan (e+f x))^{3/2}}dx}{a^2 d}+\frac {28 a^2 d}{f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{2 a^2 d}+\frac {5 a}{f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {30 a^3 d^2}{f \sqrt {d \tan (e+f x)}}+\frac {\int -\frac {14 i a^3 d^4+15 a^3 \tan (e+f x) d^4}{\sqrt {d \tan (e+f x)}}dx}{d^2}\right )}{a^2 d}+\frac {28 a^2 d}{f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{2 a^2 d}+\frac {5 a}{f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {30 a^3 d^2}{f \sqrt {d \tan (e+f x)}}-\frac {\int \frac {14 i a^3 d^4+15 a^3 \tan (e+f x) d^4}{\sqrt {d \tan (e+f x)}}dx}{d^2}\right )}{a^2 d}+\frac {28 a^2 d}{f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{2 a^2 d}+\frac {5 a}{f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {30 a^3 d^2}{f \sqrt {d \tan (e+f x)}}-\frac {\int \frac {14 i a^3 d^4+15 a^3 \tan (e+f x) d^4}{\sqrt {d \tan (e+f x)}}dx}{d^2}\right )}{a^2 d}+\frac {28 a^2 d}{f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{2 a^2 d}+\frac {5 a}{f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {30 a^3 d^2}{f \sqrt {d \tan (e+f x)}}-\frac {2 \int \frac {a^3 d^4 (15 \tan (e+f x) d+14 i d)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{d^2 f}\right )}{a^2 d}+\frac {28 a^2 d}{f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{2 a^2 d}+\frac {5 a}{f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {30 a^3 d^2}{f \sqrt {d \tan (e+f x)}}-\frac {2 a^3 d^2 \int \frac {15 \tan (e+f x) d+14 i d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}\right )}{a^2 d}+\frac {28 a^2 d}{f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{2 a^2 d}+\frac {5 a}{f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {30 a^3 d^2}{f \sqrt {d \tan (e+f x)}}-\frac {2 a^3 d^2 \left (\left (\frac {15}{2}+7 i\right ) \int \frac {\tan (e+f x) d+d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}-\left (\frac {15}{2}-7 i\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}\right )}{a^2 d}+\frac {28 a^2 d}{f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{2 a^2 d}+\frac {5 a}{f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {30 a^3 d^2}{f \sqrt {d \tan (e+f x)}}-\frac {2 a^3 d^2 \left (\left (\frac {15}{2}+7 i\right ) \left (\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}\right )-\left (\frac {15}{2}-7 i\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}\right )}{a^2 d}+\frac {28 a^2 d}{f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{2 a^2 d}+\frac {5 a}{f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {30 a^3 d^2}{f \sqrt {d \tan (e+f x)}}-\frac {2 a^3 d^2 \left (\left (\frac {15}{2}+7 i\right ) \left (\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {15}{2}-7 i\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}\right )}{a^2 d}+\frac {28 a^2 d}{f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{2 a^2 d}+\frac {5 a}{f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {30 a^3 d^2}{f \sqrt {d \tan (e+f x)}}-\frac {2 a^3 d^2 \left (\left (\frac {15}{2}+7 i\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {15}{2}-7 i\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}\right )}{a^2 d}+\frac {28 a^2 d}{f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{2 a^2 d}+\frac {5 a}{f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {30 a^3 d^2}{f \sqrt {d \tan (e+f x)}}-\frac {2 a^3 d^2 \left (\left (\frac {15}{2}+7 i\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {15}{2}-7 i\right ) \left (-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}\right )}{a^2 d}+\frac {28 a^2 d}{f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{2 a^2 d}+\frac {5 a}{f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {30 a^3 d^2}{f \sqrt {d \tan (e+f x)}}-\frac {2 a^3 d^2 \left (\left (\frac {15}{2}+7 i\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {15}{2}-7 i\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}\right )}{a^2 d}+\frac {28 a^2 d}{f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{2 a^2 d}+\frac {5 a}{f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {30 a^3 d^2}{f \sqrt {d \tan (e+f x)}}-\frac {2 a^3 d^2 \left (\left (\frac {15}{2}+7 i\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {15}{2}-7 i\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}\right )\right )}{f}\right )}{a^2 d}+\frac {28 a^2 d}{f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{2 a^2 d}+\frac {5 a}{f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {\frac {\frac {28 a^2 d}{f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}+\frac {3 \left (-\frac {30 a^3 d^2}{f \sqrt {d \tan (e+f x)}}-\frac {2 a^3 d^2 \left (\left (\frac {15}{2}+7 i\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {15}{2}-7 i\right ) \left (\frac {\log \left (d \tan (e+f x)+\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (d \tan (e+f x)-\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}\right )}{a^2 d}}{2 a^2 d}+\frac {5 a}{f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}}{12 a^2 d}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}}\) |
1/(6*d*f*Sqrt[d*Tan[e + f*x]]*(a + I*a*Tan[e + f*x])^3) + ((5*a)/(f*Sqrt[d *Tan[e + f*x]]*(a + I*a*Tan[e + f*x])^2) + ((28*a^2*d)/(f*Sqrt[d*Tan[e + f *x]]*(a + I*a*Tan[e + f*x])) + (3*((-2*a^3*d^2*((15/2 + 7*I)*(-(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])) - (15/2 - 7*I)*( -1/2*Log[d + d*Tan[e + f*x] - Sqrt[2]*Sqrt[d]*Sqrt[d*Tan[e + f*x]]]/(Sqrt[ 2]*Sqrt[d]) + Log[d + d*Tan[e + f*x] + Sqrt[2]*Sqrt[d]*Sqrt[d*Tan[e + f*x] ]]/(2*Sqrt[2]*Sqrt[d]))))/f - (30*a^3*d^2)/(f*Sqrt[d*Tan[e + f*x]])))/(a^2 *d))/(2*a^2*d))/(12*a^2*d)
3.2.85.3.1 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 = 0.92 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.40
| method | result | size |
| derivativedivides | \(\frac {2 d^{4} \left (-\frac {\frac {14 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}-\frac {98 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-20 d^{2} \sqrt {d \tan \left (f x +e \right )}}{\left (d \tan \left (f x +e \right )-i d \right )^{3}}+\frac {29 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d^{5}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d^{5} \sqrt {i d}}-\frac {1}{d^{5} \sqrt {d \tan \left (f x +e \right )}}\right )}{f \,a^{3}}\) | \(147\) |
| default | \(\frac {2 d^{4} \left (-\frac {\frac {14 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}-\frac {98 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-20 d^{2} \sqrt {d \tan \left (f x +e \right )}}{\left (d \tan \left (f x +e \right )-i d \right )^{3}}+\frac {29 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d^{5}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d^{5} \sqrt {i d}}-\frac {1}{d^{5} \sqrt {d \tan \left (f x +e \right )}}\right )}{f \,a^{3}}\) | \(147\) |
2/f/a^3*d^4*(-1/16/d^5*((14*(d*tan(f*x+e))^(5/2)-98/3*I*d*(d*tan(f*x+e))^( 3/2)-20*d^2*(d*tan(f*x+e))^(1/2))/(d*tan(f*x+e)-I*d)^3+29/(-I*d)^(1/2)*arc tan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2)))-1/16/d^5/(I*d)^(1/2)*arctan((d*tan (f*x+e))^(1/2)/(I*d)^(1/2))-1/d^5/(d*tan(f*x+e))^(1/2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 708 vs. \(2 (278) = 556\).
Time = 0.26 (sec) , antiderivative size = 708, normalized size of antiderivative = 1.92 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^3} \, dx=\frac {12 \, {\left (a^{3} d^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} - a^{3} d^{2} f e^{\left (6 i \, f x + 6 i \, e\right )}\right )} \sqrt {\frac {i}{64 \, a^{6} d^{3} f^{2}}} \log \left (-2 \, {\left (8 \, {\left (i \, a^{3} d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{3} 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 {i}{64 \, a^{6} d^{3} 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^{3} d^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} - a^{3} d^{2} f e^{\left (6 i \, f x + 6 i \, e\right )}\right )} \sqrt {\frac {i}{64 \, a^{6} d^{3} f^{2}}} \log \left (-2 \, {\left (8 \, {\left (-i \, a^{3} d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{3} 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 {i}{64 \, a^{6} d^{3} 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^{3} d^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} - a^{3} d^{2} f e^{\left (6 i \, f x + 6 i \, e\right )}\right )} \sqrt {-\frac {841 i}{64 \, a^{6} d^{3} f^{2}}} \log \left (\frac {{\left (8 \, {\left (a^{3} d f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} d 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 {841 i}{64 \, a^{6} d^{3} f^{2}}} + 29\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{3} d f}\right ) - 12 \, {\left (a^{3} d^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} - a^{3} d^{2} f e^{\left (6 i \, f x + 6 i \, e\right )}\right )} \sqrt {-\frac {841 i}{64 \, a^{6} d^{3} f^{2}}} \log \left (-\frac {{\left (8 \, {\left (a^{3} d f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} d 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 {841 i}{64 \, a^{6} d^{3} f^{2}}} - 29\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{3} d 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 (-146 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 105 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 49 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 9 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )}}{48 \, {\left (a^{3} d^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} - a^{3} d^{2} f e^{\left (6 i \, f x + 6 i \, e\right )}\right )}} \]
1/48*(12*(a^3*d^2*f*e^(8*I*f*x + 8*I*e) - a^3*d^2*f*e^(6*I*f*x + 6*I*e))*s qrt(1/64*I/(a^6*d^3*f^2))*log(-2*(8*(I*a^3*d^2*f*e^(2*I*f*x + 2*I*e) + I*a ^3*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(1/64*I/(a^6*d^3*f^2)) + I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e )) - 12*(a^3*d^2*f*e^(8*I*f*x + 8*I*e) - a^3*d^2*f*e^(6*I*f*x + 6*I*e))*sq rt(1/64*I/(a^6*d^3*f^2))*log(-2*(8*(-I*a^3*d^2*f*e^(2*I*f*x + 2*I*e) - I*a ^3*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(1/64*I/(a^6*d^3*f^2)) + I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e )) + 12*(a^3*d^2*f*e^(8*I*f*x + 8*I*e) - a^3*d^2*f*e^(6*I*f*x + 6*I*e))*sq rt(-841/64*I/(a^6*d^3*f^2))*log(1/8*(8*(a^3*d*f*e^(2*I*f*x + 2*I*e) + a^3* d*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt (-841/64*I/(a^6*d^3*f^2)) + 29)*e^(-2*I*f*x - 2*I*e)/(a^3*d*f)) - 12*(a^3* d^2*f*e^(8*I*f*x + 8*I*e) - a^3*d^2*f*e^(6*I*f*x + 6*I*e))*sqrt(-841/64*I/ (a^6*d^3*f^2))*log(-1/8*(8*(a^3*d*f*e^(2*I*f*x + 2*I*e) + a^3*d*f)*sqrt((- I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-841/64*I/( a^6*d^3*f^2)) - 29)*e^(-2*I*f*x - 2*I*e)/(a^3*d*f)) + sqrt((-I*d*e^(2*I*f* x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*(-146*I*e^(8*I*f*x + 8*I*e) - 105*I*e^(6*I*f*x + 6*I*e) + 49*I*e^(4*I*f*x + 4*I*e) + 9*I*e^(2*I*f*x + 2 *I*e) + I))/(a^3*d^2*f*e^(8*I*f*x + 8*I*e) - a^3*d^2*f*e^(6*I*f*x + 6*I*e) )
\[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^3} \, dx=\frac {i \int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{3}{\left (e + f x \right )} - 3 i \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{2}{\left (e + f x \right )} - 3 \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )} + i \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx}{a^{3}} \]
I*Integral(1/((d*tan(e + f*x))**(3/2)*tan(e + f*x)**3 - 3*I*(d*tan(e + f*x ))**(3/2)*tan(e + f*x)**2 - 3*(d*tan(e + f*x))**(3/2)*tan(e + f*x) + I*(d* tan(e + f*x))**(3/2)), x)/a**3
Exception generated. \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \]
Time = 1.38 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.68 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^3} \, dx=-\frac {1}{24} \, d^{4} {\left (\frac {3 \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} d^{\frac {11}{2}} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {87 \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} d^{\frac {11}{2}} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {48}{\sqrt {d \tan \left (f x + e\right )} a^{3} d^{5} f} + \frac {2 \, {\left (21 i \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right )^{2} + 49 \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right ) - 30 i \, \sqrt {d \tan \left (f x + e\right )} d^{2}\right )}}{{\left (-i \, d \tan \left (f x + e\right ) - d\right )}^{3} a^{3} d^{5} f}\right )} \]
-1/24*d^4*(3*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^3*d^(11/2)*f*(I*d/sqrt(d^2) + 1 )) + 87*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^3*d^(11/2)*f*(-I*d/sqrt(d^2) + 1)) + 48/(sqrt(d*tan(f*x + e))*a^3*d^5*f) + 2*(21*I*sqrt(d*tan(f*x + e))*d^2*t an(f*x + e)^2 + 49*sqrt(d*tan(f*x + e))*d^2*tan(f*x + e) - 30*I*sqrt(d*tan (f*x + e))*d^2)/((-I*d*tan(f*x + e) - d)^3*a^3*d^5*f))
Time = 7.16 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.62 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^3} \, dx=\frac {\frac {15\,d^2\,{\mathrm {tan}\left (e+f\,x\right )}^3}{4\,a^3\,f}-\frac {17\,d^2\,\mathrm {tan}\left (e+f\,x\right )}{2\,a^3\,f}+\frac {d^2\,2{}\mathrm {i}}{a^3\,f}-\frac {d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2\,121{}\mathrm {i}}{12\,a^3\,f}}{3\,d^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}-{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}+d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,3{}\mathrm {i}-d^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}+2\,\mathrm {atanh}\left (16\,a^3\,d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {1{}\mathrm {i}}{256\,a^6\,d^3\,f^2}}\right )\,\sqrt {\frac {1{}\mathrm {i}}{256\,a^6\,d^3\,f^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^3\,d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {841{}\mathrm {i}}{256\,a^6\,d^3\,f^2}}}{29}\right )\,\sqrt {-\frac {841{}\mathrm {i}}{256\,a^6\,d^3\,f^2}} \]
((d^2*2i)/(a^3*f) - (17*d^2*tan(e + f*x))/(2*a^3*f) - (d^2*tan(e + f*x)^2* 121i)/(12*a^3*f) + (15*d^2*tan(e + f*x)^3)/(4*a^3*f))/(d*(d*tan(e + f*x))^ (5/2)*3i - (d*tan(e + f*x))^(7/2) - d^3*(d*tan(e + f*x))^(1/2)*1i + 3*d^2* (d*tan(e + f*x))^(3/2)) + 2*atanh(16*a^3*d*f*(d*tan(e + f*x))^(1/2)*(1i/(2 56*a^6*d^3*f^2))^(1/2))*(1i/(256*a^6*d^3*f^2))^(1/2) + 2*atanh((16*a^3*d*f *(d*tan(e + f*x))^(1/2)*(-841i/(256*a^6*d^3*f^2))^(1/2))/29)*(-841i/(256*a ^6*d^3*f^2))^(1/2)