\(\int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^2} \, dx\) [178]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F(-2)]
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 28, antiderivative size = 293 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^2} \, dx=\frac {\left (\frac {49}{16}-\frac {45 i}{16}\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 d^{5/2} f}-\frac {\left (\frac {49}{16}-\frac {45 i}{16}\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 d^{5/2} f}-\frac {\left (\frac {49}{16}+\frac {45 i}{16}\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}+\sqrt {d} \tan (e+f x)}\right )}{\sqrt {2} a^2 d^{5/2} f}-\frac {49}{24 a^2 d f (d \tan (e+f x))^{3/2}}+\frac {9}{8 a^2 d f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}+\frac {45 i}{8 a^2 d^2 f \sqrt {d \tan (e+f x)}}+\frac {1}{4 d f (d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2} \] Output:

(49/32-45/32*I)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*2^(1/2)/a^2 
/d^(5/2)/f+(-49/32+45/32*I)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2)) 
*2^(1/2)/a^2/d^(5/2)/f-(49/32+45/32*I)*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^2/d^(5/2)/f-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)+45/8*I/a^ 
2/d^2/f/(d*tan(f*x+e))^(1/2)+1/4/d/f/(d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e 
))^2
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.45 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.53 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^2} \, dx=\frac {\sec ^2(e+f x) \left (94 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-i \tan (e+f x)\right ) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+4 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},i \tan (e+f x)\right ) (\cos (2 (e+f x))+i \sin (2 (e+f x)))-3 (11+11 \cos (2 (e+f x))+9 i \sin (2 (e+f x)))\right )}{48 a^2 d f (d \tan (e+f x))^{3/2} (-i+\tan (e+f x))^2} \] Input:

Integrate[1/((d*Tan[e + f*x])^(5/2)*(a + I*a*Tan[e + f*x])^2),x]
 

Output:

(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)
 

Rubi [A] (verified)

Time = 1.25 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.19, 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

\(\displaystyle \frac {\frac {-\frac {98 a^2 d}{3 f (d \tan (e+f x))^{3/2}}-\frac {\frac {2 a^2 d^2 \left (\left (\frac {49}{2}-\frac {45 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )+\left (\frac {49}{2}+\frac {45 i}{2}\right ) \left (\frac {\log \left (d \tan (e+f x)+\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (d \tan (e+f x)-\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}-\frac {90 i a^2 d^2}{f \sqrt {d \tan (e+f x)}}}{d^2}}{2 a^2 d}+\frac {9}{f (1+i \tan (e+f x)) (d \tan (e+f x))^{3/2}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}\)

Input:

Int[1/((d*Tan[e + f*x])^(5/2)*(a + I*a*Tan[e + f*x])^2),x]
 

Output:

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)
 

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 4012
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 
]
 

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 4042
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])
 

rule 4079
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]
 
Maple [A] (verified)

Time = 1.44 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.51

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 \(\frac {2 d^{3} \left (-\frac {1}{3 d^{4} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {2 i}{d^{5} \sqrt {d \tan \left (f x +e \right )}}-\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 d^{5} \sqrt {i d}}+\frac {\frac {-\frac {13 i \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2}-\frac {15 d \sqrt {d \tan \left (f x +e \right )}}{2}}{\left (i d \tan \left (f x +e \right )+d \right )^{2}}+\frac {47 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{2 \sqrt {-i d}}}{8 d^{5}}\right )}{f \,a^{2}}\) \(149\)

Input:

int(1/(d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^2,x,method=_RETURNVERBOSE)
 

Output:

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))))
 

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. 798 vs. \(2 (221) = 442\).

Time = 0.12 (sec) , antiderivative size = 798, normalized size of antiderivative = 2.72 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^2} \, dx =\text {Too large to display} \] Input:

integrate(1/(d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas 
")
 

Output:

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) ...
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^2} \, dx=- \frac {\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan ^{2}{\left (e + f x \right )} - 2 i \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan {\left (e + f x \right )} - \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx}{a^{2}} \] Input:

integrate(1/(d*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e))**2,x)
 

Output:

-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
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/(d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="maxima 
")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.70 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^2} \, dx=\frac {-\frac {6 i \, \sqrt {2} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{d^{\frac {5}{2}} {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )}} + \frac {141 i \, \sqrt {2} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{-i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{d^{\frac {5}{2}} {\left (-\frac {i \, d}{{\left | d \right |}} + 1\right )}} + \frac {3 \, {\left (13 i \, \sqrt {d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) + 15 \, \sqrt {d \tan \left (f x + e\right )} d\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} d^{2}} + \frac {16 \, {\left (6 i \, d \tan \left (f x + e\right ) - d\right )}}{\sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )}}{24 \, a^{2} f} \] Input:

integrate(1/(d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")
 

Output:

1/24*(-6*I*sqrt(2)*arctan(2*sqrt(d*tan(f*x + e))*abs(d)/(I*sqrt(2)*d^(3/2) 
 + sqrt(2)*sqrt(d)*abs(d)))/(d^(5/2)*(I*d/abs(d) + 1)) + 141*I*sqrt(2)*arc 
tan(2*sqrt(d*tan(f*x + e))*abs(d)/(-I*sqrt(2)*d^(3/2) + sqrt(2)*sqrt(d)*ab 
s(d)))/(d^(5/2)*(-I*d/abs(d) + 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*d^2) + 16*(6 
*I*d*tan(f*x + e) - d)/(sqrt(d*tan(f*x + e))*d^3*tan(f*x + e)))/(a^2*f)
 

Mupad [B] (verification not implemented)

Time = 3.39 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^2} \, dx=-\mathrm {atan}\left (8\,a^2\,d^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {1{}\mathrm {i}}{64\,a^4\,d^5\,f^2}}\right )\,\sqrt {-\frac {1{}\mathrm {i}}{64\,a^4\,d^5\,f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {16\,a^2\,d^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {2209{}\mathrm {i}}{256\,a^4\,d^5\,f^2}}}{47}\right )\,\sqrt {\frac {2209{}\mathrm {i}}{256\,a^4\,d^5\,f^2}}\,2{}\mathrm {i}-\frac {\frac {2\,d}{3\,a^2\,f}+\frac {221\,d\,{\mathrm {tan}\left (e+f\,x\right )}^2}{24\,a^2\,f}+\frac {d\,{\mathrm {tan}\left (e+f\,x\right )}^3\,45{}\mathrm {i}}{8\,a^2\,f}-\frac {d\,\mathrm {tan}\left (e+f\,x\right )\,8{}\mathrm {i}}{3\,a^2\,f}}{d^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}-{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}+d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,2{}\mathrm {i}} \] Input:

int(1/((d*tan(e + f*x))^(5/2)*(a + a*tan(e + f*x)*1i)^2),x)
 

Output:

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))
 

Reduce [F]

\[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^2} \, dx=-\frac {\int \frac {1}{\sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{4}-2 \sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{3} i -\sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{2}}d x}{\sqrt {d}\, a^{2} d^{2}} \] Input:

int(1/(d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^2,x)
 

Output:

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