Integrand size = 28, antiderivative size = 343 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) d^{7/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) d^{7/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}+\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 f}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )} \]
(5/32-7/32*I)*d^(7/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/f *2^(1/2)+(-5/32+7/32*I)*d^(7/2)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1 /2))/a^3/f*2^(1/2)+(5/64+7/64*I)*d^(7/2)*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e)) ^(1/2)+d^(1/2)*tan(f*x+e))/a^3/f*2^(1/2)-(5/64+7/64*I)*d^(7/2)*ln(d^(1/2)+ 2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/a^3/f*2^(1/2)-1/6*d*(d*ta n(f*x+e))^(5/2)/f/(a+I*a*tan(f*x+e))^3+1/3*I*d^2*(d*tan(f*x+e))^(3/2)/a/f/ (a+I*a*tan(f*x+e))^2+5/8*d^3*(d*tan(f*x+e))^(1/2)/f/(a^3+I*a^3*tan(f*x+e))
Time = 1.71 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.58 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {d^3 \sec ^3(e+f x) \left (6 (-1)^{3/4} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right ) (\cos (3 (e+f x))+i \sin (3 (e+f x)))+36 \sqrt [4]{-1} \sqrt {d} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right ) (-i \cos (3 (e+f x))+\sin (3 (e+f x)))+\left (-9 i \cos (e+f x)-21 i \cos (3 (e+f x))+76 \cos ^2(e+f x) \sin (e+f x)\right ) \sqrt {d \tan (e+f x)}\right )}{48 a^3 f (-i+\tan (e+f x))^3} \]
-1/48*(d^3*Sec[e + f*x]^3*(6*(-1)^(3/4)*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d* Tan[e + f*x]])/Sqrt[d]]*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)]) + 36*(-1)^ (1/4)*Sqrt[d]*ArcTanh[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]*((-I)*Cos [3*(e + f*x)] + Sin[3*(e + f*x)]) + ((-9*I)*Cos[e + f*x] - (21*I)*Cos[3*(e + f*x)] + 76*Cos[e + f*x]^2*Sin[e + f*x])*Sqrt[d*Tan[e + f*x]]))/(a^3*f*( -I + Tan[e + f*x])^3)
Time = 1.13 (sec) , antiderivative size = 331, normalized size of antiderivative = 0.97, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 4041, 27, 3042, 4078, 27, 3042, 4078, 27, 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.
| \(\displaystyle \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3}dx\) |
| \(\Big \downarrow \) 4041 |
| \(\displaystyle -\frac {\int -\frac {(d \tan (e+f x))^{3/2} \left (5 a d^2-11 i a d^2 \tan (e+f x)\right )}{2 (i \tan (e+f x) a+a)^2}dx}{6 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(d \tan (e+f x))^{3/2} \left (5 a d^2-11 i a d^2 \tan (e+f x)\right )}{(i \tan (e+f x) a+a)^2}dx}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(d \tan (e+f x))^{3/2} \left (5 a d^2-11 i a d^2 \tan (e+f x)\right )}{(i \tan (e+f x) a+a)^2}dx}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {\int \frac {12 \sqrt {d \tan (e+f x)} \left (2 i a^2 d^3+3 a^2 \tan (e+f x) d^3\right )}{i \tan (e+f x) a+a}dx}{4 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \int \frac {\sqrt {d \tan (e+f x)} \left (2 i a^2 d^3+3 a^2 \tan (e+f x) d^3\right )}{i \tan (e+f x) a+a}dx}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \int \frac {\sqrt {d \tan (e+f x)} \left (2 i a^2 d^3+3 a^2 \tan (e+f x) d^3\right )}{i \tan (e+f x) a+a}dx}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (-\frac {\int -\frac {5 a^3 d^4-7 i a^3 d^4 \tan (e+f x)}{2 \sqrt {d \tan (e+f x)}}dx}{2 a^2}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {\int \frac {5 a^3 d^4-7 i a^3 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{4 a^2}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {\int \frac {5 a^3 d^4-7 i a^3 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{4 a^2}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {\int \frac {a^3 d^4 (5 d-7 i d \tan (e+f x))}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{2 a^2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {a d^4 \int \frac {5 d-7 i d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \int \frac {\tan (e+f x) d+d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}\right )\right )}{2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \left (-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {a d^4 \left (\left (\frac {5}{2}+\frac {7 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}\right )+\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {\frac {4 i a d^2 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))^2}-\frac {3 \left (\frac {a d^4 \left (\left (\frac {5}{2}-\frac {7 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )+\left (\frac {5}{2}+\frac {7 i}{2}\right ) \left (\frac {\log \left (d \tan (e+f x)+\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (d \tan (e+f x)-\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{2 f}-\frac {5 a^2 d^3 \sqrt {d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\right )}{a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}\) |
-1/6*(d*(d*Tan[e + f*x])^(5/2))/(f*(a + I*a*Tan[e + f*x])^3) + (((4*I)*a*d ^2*(d*Tan[e + f*x])^(3/2))/(f*(a + I*a*Tan[e + f*x])^2) - (3*((a*d^4*((5/2 - (7*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]*S qrt[d])) + (5/2 + (7*I)/2)*(-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]))))/(2*f) - (5*a^2*d^3*S qrt[d*Tan[e + f*x]])/(2*f*(a + I*a*Tan[e + f*x]))))/a^2)/(12*a^2)
3.2.80.3.1 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[((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[(-(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 = 0.75 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.36
| method | result | size |
| derivativedivides | \(\frac {2 d^{4} \left (-\frac {9 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}-\frac {38 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-5 d^{2} \sqrt {d \tan \left (f x +e \right )}}{16 \left (i d \tan \left (f x +e \right )+d \right )^{3}}+\frac {3 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{8 \sqrt {-i d}}+\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 \sqrt {i d}}\right )}{f \,a^{3}}\) | \(125\) |
| default | \(\frac {2 d^{4} \left (-\frac {9 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}-\frac {38 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-5 d^{2} \sqrt {d \tan \left (f x +e \right )}}{16 \left (i d \tan \left (f x +e \right )+d \right )^{3}}+\frac {3 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{8 \sqrt {-i d}}+\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 \sqrt {i d}}\right )}{f \,a^{3}}\) | \(125\) |
2/f/a^3*d^4*(-1/16*(9*(d*tan(f*x+e))^(5/2)-38/3*I*d*(d*tan(f*x+e))^(3/2)-5 *d^2*(d*tan(f*x+e))^(1/2))/(I*d*tan(f*x+e)+d)^3+3/8*I/(-I*d)^(1/2)*arctan( (d*tan(f*x+e))^(1/2)/(-I*d)^(1/2))+1/16*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. 586 vs. \(2 (255) = 510\).
Time = 0.26 (sec) , antiderivative size = 586, normalized size of antiderivative = 1.71 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {{\left (12 \, a^{3} f \sqrt {-\frac {i \, d^{7}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{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 \, d^{7}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{3}}\right ) - 12 \, a^{3} f \sqrt {-\frac {i \, d^{7}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{4} e^{\left (2 i \, f x + 2 i \, e\right )} - 8 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{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 \, d^{7}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{3}}\right ) - 12 \, a^{3} f \sqrt {\frac {9 i \, d^{7}}{16 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (-3 i \, d^{4} + 4 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{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 {9 i \, d^{7}}{16 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3} f}\right ) + 12 \, a^{3} f \sqrt {\frac {9 i \, d^{7}}{16 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (-3 i \, d^{4} - 4 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{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 {9 i \, d^{7}}{16 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3} f}\right ) - {\left (20 \, d^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + 14 \, d^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 5 \, d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3}\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}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{48 \, a^{3} f} \]
-1/48*(12*a^3*f*sqrt(-1/64*I*d^7/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(-2*(I* d^4*e^(2*I*f*x + 2*I*e) + 8*(a^3*f*e^(2*I*f*x + 2*I*e) + a^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(-1/64*I*d^7/(a ^6*f^2)))*e^(-2*I*f*x - 2*I*e)/d^3) - 12*a^3*f*sqrt(-1/64*I*d^7/(a^6*f^2)) *e^(6*I*f*x + 6*I*e)*log(-2*(I*d^4*e^(2*I*f*x + 2*I*e) - 8*(a^3*f*e^(2*I*f *x + 2*I*e) + a^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(-1/64*I*d^7/(a^6*f^2)))*e^(-2*I*f*x - 2*I*e)/d^3) - 12*a^ 3*f*sqrt(9/16*I*d^7/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/4*(-3*I*d^4 + 4*( a^3*f*e^(2*I*f*x + 2*I*e) + a^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(9/16*I*d^7/(a^6*f^2)))*e^(-2*I*f*x - 2*I*e) /(a^3*f)) + 12*a^3*f*sqrt(9/16*I*d^7/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/ 4*(-3*I*d^4 - 4*(a^3*f*e^(2*I*f*x + 2*I*e) + a^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(9/16*I*d^7/(a^6*f^2)))*e^( -2*I*f*x - 2*I*e)/(a^3*f)) - (20*d^3*e^(6*I*f*x + 6*I*e) + 14*d^3*e^(4*I*f *x + 4*I*e) - 5*d^3*e^(2*I*f*x + 2*I*e) + d^3)*sqrt((-I*d*e^(2*I*f*x + 2*I *e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-6*I*f*x - 6*I*e)/(a^3*f)
\[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {i \int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx}{a^{3}} \]
I*Integral((d*tan(e + f*x))**(7/2)/(tan(e + f*x)**3 - 3*I*tan(e + f*x)**2 - 3*tan(e + f*x) + I), x)/a**3
Exception generated. \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.89 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.65 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {1}{24} \, d^{4} {\left (-\frac {3 i \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} \sqrt {d} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {18 i \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} \sqrt {d} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {27 i \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right )^{2} + 38 \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right ) - 15 i \, \sqrt {d \tan \left (f x + e\right )} d^{2}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} a^{3} f}\right )} \]
-1/24*d^4*(-3*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^3*sqrt(d)*f*(I*d/sqrt(d^2) + 1)) - 18*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^3*sqrt(d)*f*(-I*d/sqrt(d^2) + 1 )) + (27*I*sqrt(d*tan(f*x + e))*d^2*tan(f*x + e)^2 + 38*sqrt(d*tan(f*x + e ))*d^2*tan(f*x + e) - 15*I*sqrt(d*tan(f*x + e))*d^2)/((d*tan(f*x + e) - I* d)^3*a^3*f))
Time = 6.35 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.63 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=\mathrm {atan}\left (\frac {8\,a^3\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^7\,9{}\mathrm {i}}{64\,a^6\,f^2}}}{3\,d^4}\right )\,\sqrt {\frac {d^7\,9{}\mathrm {i}}{64\,a^6\,f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {16\,a^3\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^7\,1{}\mathrm {i}}{256\,a^6\,f^2}}}{d^4}\right )\,\sqrt {-\frac {d^7\,1{}\mathrm {i}}{256\,a^6\,f^2}}\,2{}\mathrm {i}+\frac {\frac {19\,d^5\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{12\,a^3\,f}-\frac {d^6\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,5{}\mathrm {i}}{8\,a^3\,f}+\frac {d^4\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,9{}\mathrm {i}}{8\,a^3\,f}}{-d^3\,{\mathrm {tan}\left (e+f\,x\right )}^3+d^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,d^3\,\mathrm {tan}\left (e+f\,x\right )-d^3\,1{}\mathrm {i}} \]
atan((8*a^3*f*(d*tan(e + f*x))^(1/2)*((d^7*9i)/(64*a^6*f^2))^(1/2))/(3*d^4 ))*((d^7*9i)/(64*a^6*f^2))^(1/2)*2i + atan((16*a^3*f*(d*tan(e + f*x))^(1/2 )*(-(d^7*1i)/(256*a^6*f^2))^(1/2))/d^4)*(-(d^7*1i)/(256*a^6*f^2))^(1/2)*2i + ((19*d^5*(d*tan(e + f*x))^(3/2))/(12*a^3*f) - (d^6*(d*tan(e + f*x))^(1/ 2)*5i)/(8*a^3*f) + (d^4*(d*tan(e + f*x))^(5/2)*9i)/(8*a^3*f))/(3*d^3*tan(e + f*x) - d^3*1i + d^3*tan(e + f*x)^2*3i - d^3*tan(e + f*x)^3)