\(\int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx\) [179]

Optimal result

Integrand size = 28, antiderivative size = 310 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {\left (\frac {7}{4}+\frac {15 i}{8}\right ) d^{9/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {7}{4}+\frac {15 i}{8}\right ) d^{9/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}+\frac {\left (\frac {7}{4}-\frac {15 i}{8}\right ) d^{9/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}+\sqrt {d} \tan (e+f x)}\right )}{\sqrt {2} a^3 f}+\frac {15 i d^4 \sqrt {d \tan (e+f x)}}{4 a^3 f}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}+\frac {7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )} \] Output:

(7/8+15/16*I)*d^(9/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*2^(1/ 
2)/a^3/f-(7/8+15/16*I)*d^(9/2)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/ 
2))*2^(1/2)/a^3/f+(7/8-15/16*I)*d^(9/2)*arctanh(2^(1/2)*(d*tan(f*x+e))^(1/ 
2)/(d^(1/2)+d^(1/2)*tan(f*x+e)))*2^(1/2)/a^3/f+15/4*I*d^4*(d*tan(f*x+e))^( 
1/2)/a^3/f-1/6*d*(d*tan(f*x+e))^(7/2)/f/(a+I*a*tan(f*x+e))^3+5/12*I*d^2*(d 
*tan(f*x+e))^(5/2)/a/f/(a+I*a*tan(f*x+e))^2+7/6*d^3*(d*tan(f*x+e))^(3/2)/f 
/(a^3+I*a^3*tan(f*x+e))
 

Mathematica [A] (warning: unable to verify)

Time = 3.83 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.76 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {i d^4 e^{-6 i (e+f x)} \left (-1+9 e^{2 i (e+f x)}-49 e^{4 i (e+f x)}-105 e^{6 i (e+f x)}+146 e^{8 i (e+f x)}-87 e^{6 i (e+f x)} \sqrt {-1+e^{4 i (e+f x)}} \arctan \left (\sqrt {-1+e^{4 i (e+f x)}}\right )-6 e^{6 i (e+f x)} \sqrt {-1+e^{2 i (e+f x)}} \sqrt {1+e^{2 i (e+f x)}} \text {arctanh}\left (\sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )\right ) \sqrt {d \tan (e+f x)}}{48 a^3 \left (-1+e^{2 i (e+f x)}\right ) f} \] Input:

Integrate[(d*Tan[e + f*x])^(9/2)/(a + I*a*Tan[e + f*x])^3,x]
 

Output:

((I/48)*d^4*(-1 + 9*E^((2*I)*(e + f*x)) - 49*E^((4*I)*(e + f*x)) - 105*E^( 
(6*I)*(e + f*x)) + 146*E^((8*I)*(e + f*x)) - 87*E^((6*I)*(e + f*x))*Sqrt[- 
1 + E^((4*I)*(e + f*x))]*ArcTan[Sqrt[-1 + E^((4*I)*(e + f*x))]] - 6*E^((6* 
I)*(e + f*x))*Sqrt[-1 + E^((2*I)*(e + f*x))]*Sqrt[1 + E^((2*I)*(e + f*x))] 
*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]])*Sqrt 
[d*Tan[e + f*x]])/(a^3*E^((6*I)*(e + f*x))*(-1 + E^((2*I)*(e + f*x)))*f)
 

Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.15, number of steps used = 23, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {3042, 4041, 27, 3042, 4078, 27, 3042, 4078, 27, 3042, 4011, 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.

\(\displaystyle \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3}dx\)

\(\Big \downarrow \) 4041

\(\displaystyle -\frac {\int -\frac {(d \tan (e+f x))^{5/2} \left (7 a d^2-13 i a d^2 \tan (e+f x)\right )}{2 (i \tan (e+f x) a+a)^2}dx}{6 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(d \tan (e+f x))^{5/2} \left (7 a d^2-13 i a d^2 \tan (e+f x)\right )}{(i \tan (e+f x) a+a)^2}dx}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(d \tan (e+f x))^{5/2} \left (7 a d^2-13 i a d^2 \tan (e+f x)\right )}{(i \tan (e+f x) a+a)^2}dx}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\int \frac {2 (d \tan (e+f x))^{3/2} \left (25 i a^2 d^3+31 a^2 \tan (e+f x) d^3\right )}{i \tan (e+f x) a+a}dx}{4 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\int \frac {(d \tan (e+f x))^{3/2} \left (25 i a^2 d^3+31 a^2 \tan (e+f x) d^3\right )}{i \tan (e+f x) a+a}dx}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\int \frac {(d \tan (e+f x))^{3/2} \left (25 i a^2 d^3+31 a^2 \tan (e+f x) d^3\right )}{i \tan (e+f x) a+a}dx}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {-\frac {\int -6 \sqrt {d \tan (e+f x)} \left (14 a^3 d^4-15 i a^3 d^4 \tan (e+f x)\right )dx}{2 a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \int \sqrt {d \tan (e+f x)} \left (14 a^3 d^4-15 i a^3 d^4 \tan (e+f x)\right )dx}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \int \sqrt {d \tan (e+f x)} \left (14 a^3 d^4-15 i a^3 d^4 \tan (e+f x)\right )dx}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\int \frac {15 i a^3 d^5+14 a^3 \tan (e+f x) d^5}{\sqrt {d \tan (e+f x)}}dx-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\int \frac {15 i a^3 d^5+14 a^3 \tan (e+f x) d^5}{\sqrt {d \tan (e+f x)}}dx-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 4017

\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 \int \frac {a^3 d^5 (14 \tan (e+f x) d+15 i d)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 a^3 d^5 \int \frac {14 \tan (e+f x) d+15 i d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 1482

\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 a^3 d^5 \left (\left (7+\frac {15 i}{2}\right ) \int \frac {\tan (e+f x) d+d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}-\left (7-\frac {15 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 a^3 d^5 \left (\left (7+\frac {15 i}{2}\right ) \left (\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}\right )-\left (7-\frac {15 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 a^3 d^5 \left (\left (7+\frac {15 i}{2}\right ) \left (\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}\right )-\left (7-\frac {15 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 a^3 d^5 \left (\left (7+\frac {15 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (7-\frac {15 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 a^3 d^5 \left (\left (7+\frac {15 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (7-\frac {15 i}{2}\right ) \left (-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 a^3 d^5 \left (\left (7+\frac {15 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (7-\frac {15 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 a^3 d^5 \left (\left (7+\frac {15 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (7-\frac {15 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}\right )\right )}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 a^3 d^5 \left (\left (7+\frac {15 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (7-\frac {15 i}{2}\right ) \left (\frac {\log \left (d \tan (e+f x)+\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (d \tan (e+f x)-\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\)

Input:

Int[(d*Tan[e + f*x])^(9/2)/(a + I*a*Tan[e + f*x])^3,x]
 

Output:

-1/6*(d*(d*Tan[e + f*x])^(7/2))/(f*(a + I*a*Tan[e + f*x])^3) + (((5*I)*a*d 
^2*(d*Tan[e + f*x])^(5/2))/(f*(a + I*a*Tan[e + f*x])^2) - ((-28*a^2*d^3*(d 
*Tan[e + f*x])^(3/2))/(f*(a + I*a*Tan[e + f*x])) + (3*((2*a^3*d^5*((7 + (1 
5*I)/2)*(-(ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2]*Sqr 
t[d])) + ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2]*Sqrt[ 
d])) - (7 - (15*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]))))/f - ((30*I)*a^3*d^4*Sqrt[ 
d*Tan[e + f*x]])/f))/a^2)/(2*a^2))/(12*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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])
 

rule 1082

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]
 

rule 1103

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]
 

rule 1476

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]
 

rule 1479

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]
 

rule 1482

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4011

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]
 

rule 4017

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]
 

rule 4041

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])
 

rule 4078

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]
 

Maple [A] (verified)

Time = 1.44 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.45

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (i \sqrt {d \tan \left (f x +e \right )}+\frac {d \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 \sqrt {i d}}-\frac {d \left (\frac {-20 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}+\frac {98 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+14 d^{2} \sqrt {d \tan \left (f x +e \right )}}{\left (d \tan \left (f x +e \right )-i d \right )^{3}}+\frac {29 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}\right )}{16}\right )}{f \,a^{3}}\)	\(141\)
default	\(\frac {2 d^{4} \left (i \sqrt {d \tan \left (f x +e \right )}+\frac {d \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 \sqrt {i d}}-\frac {d \left (\frac {-20 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}+\frac {98 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+14 d^{2} \sqrt {d \tan \left (f x +e \right )}}{\left (d \tan \left (f x +e \right )-i d \right )^{3}}+\frac {29 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}\right )}{16}\right )}{f \,a^{3}}\)	\(141\)

Input:

int((d*tan(f*x+e))^(9/2)/(a+I*a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)
 

Output:

2/f/a^3*d^4*(I*(d*tan(f*x+e))^(1/2)+1/16*d/(I*d)^(1/2)*arctan((d*tan(f*x+e 
))^(1/2)/(I*d)^(1/2))-1/16*d*((-20*(d*tan(f*x+e))^(5/2)+98/3*I*d*(d*tan(f* 
x+e))^(3/2)+14*d^2*(d*tan(f*x+e))^(1/2))/(d*tan(f*x+e)-I*d)^3+29/(-I*d)^(1 
/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2))))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 592 vs. \(2 (234) = 468\).

Time = 0.10 (sec) , antiderivative size = 592, normalized size of antiderivative = 1.91 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx =\text {Too large to display} \] Input:

integrate((d*tan(f*x+e))^(9/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")
 

Output:

-1/48*(12*a^3*sqrt(1/64*I*d^9/(a^6*f^2))*f*e^(6*I*f*x + 6*I*e)*log(-2*(I*d 
^5*e^(2*I*f*x + 2*I*e) + 8*(I*a^3*f*e^(2*I*f*x + 2*I*e) + I*a^3*f)*sqrt(1/ 
64*I*d^9/(a^6*f^2))*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2* 
I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/d^4) - 12*a^3*sqrt(1/64*I*d^9/(a^6*f^2))* 
f*e^(6*I*f*x + 6*I*e)*log(-2*(I*d^5*e^(2*I*f*x + 2*I*e) + 8*(-I*a^3*f*e^(2 
*I*f*x + 2*I*e) - I*a^3*f)*sqrt(1/64*I*d^9/(a^6*f^2))*sqrt((-I*d*e^(2*I*f* 
x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/d^4) - 
12*a^3*sqrt(-841/64*I*d^9/(a^6*f^2))*f*e^(6*I*f*x + 6*I*e)*log(1/8*(29*d^5 
 + 8*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt(-841/64*I*d^9/(a^6*f^2))*sqr 
t((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x 
 - 2*I*e)/(a^3*f)) + 12*a^3*sqrt(-841/64*I*d^9/(a^6*f^2))*f*e^(6*I*f*x + 6 
*I*e)*log(1/8*(29*d^5 - 8*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt(-841/64 
*I*d^9/(a^6*f^2))*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I* 
e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a^3*f)) - (146*I*d^4*e^(6*I*f*x + 6*I*e) + 
 41*I*d^4*e^(4*I*f*x + 4*I*e) - 8*I*d^4*e^(2*I*f*x + 2*I*e) + I*d^4)*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)
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=\text {Timed out} \] Input:

integrate((d*tan(f*x+e))**(9/2)/(a+I*a*tan(f*x+e))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((d*tan(f*x+e))^(9/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.67 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {i \, d^{4} {\left (\frac {3 i \, \sqrt {2} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{\frac {i \, d}{{\left | d \right |}} + 1} - \frac {87 i \, \sqrt {2} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{-i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{-\frac {i \, d}{{\left | d \right |}} + 1} - 48 \, \sqrt {d \tan \left (f x + e\right )} - \frac {2 \, {\left (-30 i \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )^{2} - 49 \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right ) + 21 i \, \sqrt {d \tan \left (f x + e\right )} d^{3}\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3}}\right )}}{24 \, a^{3} f} \] Input:

integrate((d*tan(f*x+e))^(9/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")
 

Output:

-1/24*I*d^4*(3*I*sqrt(2)*sqrt(d)*arctan(2*sqrt(d*tan(f*x + e))*abs(d)/(I*s 
qrt(2)*d^(3/2) + sqrt(2)*sqrt(d)*abs(d)))/(I*d/abs(d) + 1) - 87*I*sqrt(2)* 
sqrt(d)*arctan(2*sqrt(d*tan(f*x + e))*abs(d)/(-I*sqrt(2)*d^(3/2) + sqrt(2) 
*sqrt(d)*abs(d)))/(-I*d/abs(d) + 1) - 48*sqrt(d*tan(f*x + e)) - 2*(-30*I*s 
qrt(d*tan(f*x + e))*d^3*tan(f*x + e)^2 - 49*sqrt(d*tan(f*x + e))*d^3*tan(f 
*x + e) + 21*I*sqrt(d*tan(f*x + e))*d^3)/(d*tan(f*x + e) - I*d)^3)/(a^3*f)
 

Mupad [B] (verification not implemented)

Time = 2.55 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.77 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=\mathrm {atan}\left (\frac {a^3\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^9\,1{}\mathrm {i}}{256\,a^6\,f^2}}\,16{}\mathrm {i}}{d^5}\right )\,\sqrt {\frac {d^9\,1{}\mathrm {i}}{256\,a^6\,f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {a^3\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^9\,841{}\mathrm {i}}{256\,a^6\,f^2}}\,16{}\mathrm {i}}{29\,d^5}\right )\,\sqrt {-\frac {d^9\,841{}\mathrm {i}}{256\,a^6\,f^2}}\,2{}\mathrm {i}+\frac {\frac {7\,d^7\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{4\,a^3\,f}-\frac {5\,d^5\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{2\,a^3\,f}+\frac {d^6\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,49{}\mathrm {i}}{12\,a^3\,f}}{-d^3\,{\mathrm {tan}\left (e+f\,x\right )}^3+d^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,d^3\,\mathrm {tan}\left (e+f\,x\right )-d^3\,1{}\mathrm {i}}+\frac {d^4\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,2{}\mathrm {i}}{a^3\,f} \] Input:

int((d*tan(e + f*x))^(9/2)/(a + a*tan(e + f*x)*1i)^3,x)
 

Output:

atan((a^3*f*(d*tan(e + f*x))^(1/2)*((d^9*1i)/(256*a^6*f^2))^(1/2)*16i)/d^5 
)*((d^9*1i)/(256*a^6*f^2))^(1/2)*2i - atan((a^3*f*(d*tan(e + f*x))^(1/2)*( 
-(d^9*841i)/(256*a^6*f^2))^(1/2)*16i)/(29*d^5))*(-(d^9*841i)/(256*a^6*f^2) 
)^(1/2)*2i + ((7*d^7*(d*tan(e + f*x))^(1/2))/(4*a^3*f) + (d^6*(d*tan(e + f 
*x))^(3/2)*49i)/(12*a^3*f) - (5*d^5*(d*tan(e + f*x))^(5/2))/(2*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) + 
(d^4*(d*tan(e + f*x))^(1/2)*2i)/(a^3*f)
 

Reduce [F]

\[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{4}}{\tan \left (f x +e \right )^{3} i +3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i -1}d x \right ) d^{4}}{a^{3}} \] Input:

int((d*tan(f*x+e))^(9/2)/(a+I*a*tan(f*x+e))^3,x)
 

Output:

( - sqrt(d)*int((sqrt(tan(e + f*x))*tan(e + f*x)**4)/(tan(e + f*x)**3*i + 
3*tan(e + f*x)**2 - 3*tan(e + f*x)*i - 1),x)*d**4)/a**3