\(\int \frac {1}{(a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2}} \, dx\) [1130]

Optimal result

Integrand size = 30, antiderivative size = 368 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{8 a^3 (c-i d)^{3/2} f}+\frac {\left (2 i c^3-12 c^2 d-33 i c d^2+58 d^3\right ) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{16 a^3 (c+i d)^{9/2} f}+\frac {d \left (2 c^3+9 i c^2 d-17 c d^2+60 i d^3\right )}{16 a^3 (c-i d) (c+i d)^4 f \sqrt {c+d \tan (e+f x)}}-\frac {1}{6 (i c-d) f (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}}+\frac {3 i c-10 d}{24 a (c+i d)^2 f (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}+\frac {6 c^2+27 i c d-56 d^2}{48 (i c-d)^3 f \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}} \] Output:

-1/8*I*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/a^3/(c-I*d)^(3/2)/f+1 
/16*(2*I*c^3-12*c^2*d-33*I*c*d^2+58*d^3)*arctanh((c+d*tan(f*x+e))^(1/2)/(c 
+I*d)^(1/2))/a^3/(c+I*d)^(9/2)/f+1/16*d*(2*c^3+9*I*c^2*d-17*c*d^2+60*I*d^3 
)/a^3/(c-I*d)/(c+I*d)^4/f/(c+d*tan(f*x+e))^(1/2)-1/6/(I*c-d)/f/(a+I*a*tan( 
f*x+e))^3/(c+d*tan(f*x+e))^(1/2)+1/24*(3*I*c-10*d)/a/(c+I*d)^2/f/(a+I*a*ta 
n(f*x+e))^2/(c+d*tan(f*x+e))^(1/2)+1/48*(6*c^2+27*I*c*d-56*d^2)/(I*c-d)^3/ 
f/(a^3+I*a^3*tan(f*x+e))/(c+d*tan(f*x+e))^(1/2)
 

Mathematica [C] (verified)

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

Time = 1.30 (sec) , antiderivative size = 294, normalized size of antiderivative = 0.80 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {8 a^6 (c-i d) (c+i d)^3+2 a^6 (c+i d)^2 (3 c+10 i d) (i c+d) (-i+\tan (e+f x))-a^6 (c-i d) (c+i d) \left (6 c^2+27 i c d-56 d^2\right ) (-i+\tan (e+f x))^2+3 i a^3 \left (-2 i (c+i d)^4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c-i d}\right )+(i c+d) \left (2 c^3+12 i c^2 d-33 c d^2-58 i d^3\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c+i d}\right )\right ) (a+i a \tan (e+f x))^3}{48 a^6 (c+i d)^4 (i c+d) f (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}} \] Input:

Integrate[1/((a + I*a*Tan[e + f*x])^3*(c + d*Tan[e + f*x])^(3/2)),x]
 

Output:

-1/48*(8*a^6*(c - I*d)*(c + I*d)^3 + 2*a^6*(c + I*d)^2*(3*c + (10*I)*d)*(I 
*c + d)*(-I + Tan[e + f*x]) - a^6*(c - I*d)*(c + I*d)*(6*c^2 + (27*I)*c*d 
- 56*d^2)*(-I + Tan[e + f*x])^2 + (3*I)*a^3*((-2*I)*(c + I*d)^4*Hypergeome 
tric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c - I*d)] + (I*c + d)*(2*c^3 + 
 (12*I)*c^2*d - 33*c*d^2 - (58*I)*d^3)*Hypergeometric2F1[-1/2, 1, 1/2, (c 
+ d*Tan[e + f*x])/(c + I*d)])*(a + I*a*Tan[e + f*x])^3)/(a^6*(c + I*d)^4*( 
I*c + d)*f*(a + I*a*Tan[e + f*x])^3*Sqrt[c + d*Tan[e + f*x]])
 

Rubi [A] (warning: unable to verify)

Time = 2.20 (sec) , antiderivative size = 415, normalized size of antiderivative = 1.13, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.567, Rules used = {3042, 4042, 27, 3042, 4079, 3042, 4079, 27, 3042, 4012, 3042, 4022, 3042, 4020, 25, 73, 221}

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.

\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 4042

\(\displaystyle -\frac {\int -\frac {a (6 i c-13 d)+7 i a d \tan (e+f x)}{2 (i \tan (e+f x) a+a)^2 (c+d \tan (e+f x))^{3/2}}dx}{6 a^2 (-d+i c)}-\frac {1}{6 f (-d+i c) (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (6 i c-13 d)+7 i a d \tan (e+f x)}{(i \tan (e+f x) a+a)^2 (c+d \tan (e+f x))^{3/2}}dx}{12 a^2 (-d+i c)}-\frac {1}{6 f (-d+i c) (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (6 i c-13 d)+7 i a d \tan (e+f x)}{(i \tan (e+f x) a+a)^2 (c+d \tan (e+f x))^{3/2}}dx}{12 a^2 (-d+i c)}-\frac {1}{6 f (-d+i c) (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {-\frac {\int \frac {\left (12 c^2+39 i d c-62 d^2\right ) a^2+5 (3 c+10 i d) d \tan (e+f x) a^2}{(i \tan (e+f x) a+a) (c+d \tan (e+f x))^{3/2}}dx}{4 a^2 (-d+i c)}-\frac {a (3 c+10 i d)}{2 f (c+i d) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}}{12 a^2 (-d+i c)}-\frac {1}{6 f (-d+i c) (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {\left (12 c^2+39 i d c-62 d^2\right ) a^2+5 (3 c+10 i d) d \tan (e+f x) a^2}{(i \tan (e+f x) a+a) (c+d \tan (e+f x))^{3/2}}dx}{4 a^2 (-d+i c)}-\frac {a (3 c+10 i d)}{2 f (c+i d) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}}{12 a^2 (-d+i c)}-\frac {1}{6 f (-d+i c) (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {-\frac {\frac {a^2 \left (6 i c^2-27 c d-56 i d^2\right )}{f (c+i d) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}-\frac {\int -\frac {3 \left (\left (4 i c^3-18 d c^2-39 i d^2 c+60 d^3\right ) a^3+d \left (6 i c^2-27 d c-56 i d^2\right ) \tan (e+f x) a^3\right )}{(c+d \tan (e+f x))^{3/2}}dx}{2 a^2 (-d+i c)}}{4 a^2 (-d+i c)}-\frac {a (3 c+10 i d)}{2 f (c+i d) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}}{12 a^2 (-d+i c)}-\frac {1}{6 f (-d+i c) (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\frac {3 \int \frac {\left (4 i c^3-18 d c^2-39 i d^2 c+60 d^3\right ) a^3+d \left (6 i c^2-27 d c-56 i d^2\right ) \tan (e+f x) a^3}{(c+d \tan (e+f x))^{3/2}}dx}{2 a^2 (-d+i c)}+\frac {a^2 \left (6 i c^2-27 c d-56 i d^2\right )}{f (c+i d) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}}{4 a^2 (-d+i c)}-\frac {a (3 c+10 i d)}{2 f (c+i d) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}}{12 a^2 (-d+i c)}-\frac {1}{6 f (-d+i c) (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {3 \int \frac {\left (4 i c^3-18 d c^2-39 i d^2 c+60 d^3\right ) a^3+d \left (6 i c^2-27 d c-56 i d^2\right ) \tan (e+f x) a^3}{(c+d \tan (e+f x))^{3/2}}dx}{2 a^2 (-d+i c)}+\frac {a^2 \left (6 i c^2-27 c d-56 i d^2\right )}{f (c+i d) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}}{4 a^2 (-d+i c)}-\frac {a (3 c+10 i d)}{2 f (c+i d) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}}{12 a^2 (-d+i c)}-\frac {1}{6 f (-d+i c) (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4012

\(\displaystyle \frac {-\frac {\frac {3 \left (\frac {\int \frac {\left (4 i c^4-18 d c^3-33 i d^2 c^2+33 d^3 c-56 i d^4\right ) a^3+d \left (2 i c^3-9 d c^2-17 i d^2 c-60 d^3\right ) \tan (e+f x) a^3}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a^3 d \left (2 i c^3-9 c^2 d-17 i c d^2-60 d^3\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\right )}{2 a^2 (-d+i c)}+\frac {a^2 \left (6 i c^2-27 c d-56 i d^2\right )}{f (c+i d) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}}{4 a^2 (-d+i c)}-\frac {a (3 c+10 i d)}{2 f (c+i d) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}}{12 a^2 (-d+i c)}-\frac {1}{6 f (-d+i c) (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {3 \left (\frac {\int \frac {\left (4 i c^4-18 d c^3-33 i d^2 c^2+33 d^3 c-56 i d^4\right ) a^3+d \left (2 i c^3-9 d c^2-17 i d^2 c-60 d^3\right ) \tan (e+f x) a^3}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a^3 d \left (2 i c^3-9 c^2 d-17 i c d^2-60 d^3\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\right )}{2 a^2 (-d+i c)}+\frac {a^2 \left (6 i c^2-27 c d-56 i d^2\right )}{f (c+i d) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}}{4 a^2 (-d+i c)}-\frac {a (3 c+10 i d)}{2 f (c+i d) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}}{12 a^2 (-d+i c)}-\frac {1}{6 f (-d+i c) (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {-\frac {\frac {3 \left (\frac {a^3 (c-i d) \left (2 i c^3-12 c^2 d-33 i c d^2+58 d^3\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+2 i a^3 (c+i d)^4 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a^3 d \left (2 i c^3-9 c^2 d-17 i c d^2-60 d^3\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\right )}{2 a^2 (-d+i c)}+\frac {a^2 \left (6 i c^2-27 c d-56 i d^2\right )}{f (c+i d) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}}{4 a^2 (-d+i c)}-\frac {a (3 c+10 i d)}{2 f (c+i d) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}}{12 a^2 (-d+i c)}-\frac {1}{6 f (-d+i c) (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {3 \left (\frac {a^3 (c-i d) \left (2 i c^3-12 c^2 d-33 i c d^2+58 d^3\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+2 i a^3 (c+i d)^4 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a^3 d \left (2 i c^3-9 c^2 d-17 i c d^2-60 d^3\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\right )}{2 a^2 (-d+i c)}+\frac {a^2 \left (6 i c^2-27 c d-56 i d^2\right )}{f (c+i d) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}}{4 a^2 (-d+i c)}-\frac {a (3 c+10 i d)}{2 f (c+i d) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}}{12 a^2 (-d+i c)}-\frac {1}{6 f (-d+i c) (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {-\frac {\frac {3 \left (\frac {-\frac {i a^3 (c-i d) \left (2 i c^3-12 c^2 d-33 i c d^2+58 d^3\right ) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{f}-\frac {2 a^3 (c+i d)^4 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{f}}{c^2+d^2}+\frac {2 a^3 d \left (2 i c^3-9 c^2 d-17 i c d^2-60 d^3\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\right )}{2 a^2 (-d+i c)}+\frac {a^2 \left (6 i c^2-27 c d-56 i d^2\right )}{f (c+i d) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}}{4 a^2 (-d+i c)}-\frac {a (3 c+10 i d)}{2 f (c+i d) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}}{12 a^2 (-d+i c)}-\frac {1}{6 f (-d+i c) (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {\frac {3 \left (\frac {\frac {i a^3 (c-i d) \left (2 i c^3-12 c^2 d-33 i c d^2+58 d^3\right ) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{f}+\frac {2 a^3 (c+i d)^4 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{f}}{c^2+d^2}+\frac {2 a^3 d \left (2 i c^3-9 c^2 d-17 i c d^2-60 d^3\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\right )}{2 a^2 (-d+i c)}+\frac {a^2 \left (6 i c^2-27 c d-56 i d^2\right )}{f (c+i d) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}}{4 a^2 (-d+i c)}-\frac {a (3 c+10 i d)}{2 f (c+i d) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}}{12 a^2 (-d+i c)}-\frac {1}{6 f (-d+i c) (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {-\frac {\frac {3 \left (\frac {\frac {2 a^3 (c-i d) \left (2 i c^3-12 c^2 d-33 i c d^2+58 d^3\right ) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {4 i a^3 (c+i d)^4 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}}{c^2+d^2}+\frac {2 a^3 d \left (2 i c^3-9 c^2 d-17 i c d^2-60 d^3\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\right )}{2 a^2 (-d+i c)}+\frac {a^2 \left (6 i c^2-27 c d-56 i d^2\right )}{f (c+i d) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}}{4 a^2 (-d+i c)}-\frac {a (3 c+10 i d)}{2 f (c+i d) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}}{12 a^2 (-d+i c)}-\frac {1}{6 f (-d+i c) (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {-\frac {\frac {a^2 \left (6 i c^2-27 c d-56 i d^2\right )}{f (c+i d) (a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}+\frac {3 \left (\frac {\frac {2 a^3 (c-i d) \left (2 i c^3-12 c^2 d-33 i c d^2+58 d^3\right ) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}+\frac {4 i a^3 (c+i d)^4 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}}{c^2+d^2}+\frac {2 a^3 d \left (2 i c^3-9 c^2 d-17 i c d^2-60 d^3\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\right )}{2 a^2 (-d+i c)}}{4 a^2 (-d+i c)}-\frac {a (3 c+10 i d)}{2 f (c+i d) (a+i a \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}}{12 a^2 (-d+i c)}-\frac {1}{6 f (-d+i c) (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}}\)

Input:

Int[1/((a + I*a*Tan[e + f*x])^3*(c + d*Tan[e + f*x])^(3/2)),x]
 

Output:

-1/6*1/((I*c - d)*f*(a + I*a*Tan[e + f*x])^3*Sqrt[c + d*Tan[e + f*x]]) + ( 
-1/2*(a*(3*c + (10*I)*d))/((c + I*d)*f*(a + I*a*Tan[e + f*x])^2*Sqrt[c + d 
*Tan[e + f*x]]) - ((a^2*((6*I)*c^2 - 27*c*d - (56*I)*d^2))/((c + I*d)*f*(a 
 + I*a*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]]) + (3*((((4*I)*a^3*(c + I*d) 
^4*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) + (2*a^3*(c - I*d 
)*((2*I)*c^3 - 12*c^2*d - (33*I)*c*d^2 + 58*d^3)*ArcTan[Tan[e + f*x]/Sqrt[ 
c + I*d]])/(Sqrt[c + I*d]*f))/(c^2 + d^2) + (2*a^3*d*((2*I)*c^3 - 9*c^2*d 
- (17*I)*c*d^2 - 60*d^3))/((c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]])))/(2*a^ 
2*(I*c - d)))/(4*a^2*(I*c - d)))/(12*a^2*(I*c - 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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

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 4020

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f)   Subst[Int[(a + (b/d)*x)^m/(d^2 + 
c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
 

rule 4022

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2   Int[(a + b*Tan[e + f*x])^m*( 
1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2   Int[(a + b*Tan[e + f*x])^m 
*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c 
 - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]
 

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 = 0.38 (sec) , antiderivative size = 596, normalized size of antiderivative = 1.62

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (-\frac {i}{\left (i c -d \right ) \left (i c +d \right ) \left (i d +c \right )^{3} \sqrt {c +d \tan \left (f x +e \right )}}-\frac {i \left (\frac {\frac {d \left (2 i c^{6}-50 i c^{4} d^{2}-24 i c^{2} d^{4}+28 i d^{6}-15 c^{5} d +52 c^{3} d^{3}+67 c \,d^{5}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{6 i c^{2} d -2 i d^{3}+2 c^{3}-6 c \,d^{2}}-\frac {2 d \left (3 i c^{7}-109 i c^{5} d^{2}+53 i c^{3} d^{4}+165 i c \,d^{6}-27 c^{6} d +177 c^{4} d^{3}+155 c^{2} d^{5}-49 d^{7}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right )}+\frac {d \left (2 i c^{8}-102 i c^{6} d^{2}+190 i c^{4} d^{4}+254 i c^{2} d^{6}-40 i d^{8}-21 d \,c^{7}+225 d^{3} c^{5}+73 c^{3} d^{5}-173 c \,d^{7}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{6 i c^{2} d -2 i d^{3}+2 c^{3}-6 c \,d^{2}}}{\left (-d \tan \left (f x +e \right )+i d \right )^{3}}-\frac {\left (16 i c^{6} d -120 i c^{4} d^{3}-78 i c^{2} d^{5}+58 i d^{7}+2 c^{7}-57 c^{5} d^{2}+90 c^{3} d^{4}+149 c \,d^{6}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right ) \sqrt {-i d -c}}\right )}{16 d^{4} \left (i d +c \right )^{4} \left (i d -c \right )}+\frac {\left (-i c^{4}+6 i c^{2} d^{2}-i d^{4}+4 c^{3} d -4 c \,d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{16 \left (i d -c \right )^{\frac {3}{2}} \left (i d +c \right )^{4} d^{4}}\right )}{f \,a^{3}}\)	\(596\)
default	\(\frac {2 d^{4} \left (-\frac {i}{\left (i c -d \right ) \left (i c +d \right ) \left (i d +c \right )^{3} \sqrt {c +d \tan \left (f x +e \right )}}-\frac {i \left (\frac {\frac {d \left (2 i c^{6}-50 i c^{4} d^{2}-24 i c^{2} d^{4}+28 i d^{6}-15 c^{5} d +52 c^{3} d^{3}+67 c \,d^{5}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{6 i c^{2} d -2 i d^{3}+2 c^{3}-6 c \,d^{2}}-\frac {2 d \left (3 i c^{7}-109 i c^{5} d^{2}+53 i c^{3} d^{4}+165 i c \,d^{6}-27 c^{6} d +177 c^{4} d^{3}+155 c^{2} d^{5}-49 d^{7}\right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right )}+\frac {d \left (2 i c^{8}-102 i c^{6} d^{2}+190 i c^{4} d^{4}+254 i c^{2} d^{6}-40 i d^{8}-21 d \,c^{7}+225 d^{3} c^{5}+73 c^{3} d^{5}-173 c \,d^{7}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{6 i c^{2} d -2 i d^{3}+2 c^{3}-6 c \,d^{2}}}{\left (-d \tan \left (f x +e \right )+i d \right )^{3}}-\frac {\left (16 i c^{6} d -120 i c^{4} d^{3}-78 i c^{2} d^{5}+58 i d^{7}+2 c^{7}-57 c^{5} d^{2}+90 c^{3} d^{4}+149 c \,d^{6}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right ) \sqrt {-i d -c}}\right )}{16 d^{4} \left (i d +c \right )^{4} \left (i d -c \right )}+\frac {\left (-i c^{4}+6 i c^{2} d^{2}-i d^{4}+4 c^{3} d -4 c \,d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{16 \left (i d -c \right )^{\frac {3}{2}} \left (i d +c \right )^{4} d^{4}}\right )}{f \,a^{3}}\)	\(596\)

Input:

int(1/(a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2/f/a^3*d^4*(-I/(I*c-d)/(I*c+d)/(c+I*d)^3/(c+d*tan(f*x+e))^(1/2)-1/16*I/d^ 
4/(c+I*d)^4/(I*d-c)*((1/2*d*(2*I*c^6-50*I*c^4*d^2-24*I*c^2*d^4+28*I*d^6-15 
*c^5*d+52*c^3*d^3+67*c*d^5)/(3*I*c^2*d-I*d^3+c^3-3*c*d^2)*(c+d*tan(f*x+e)) 
^(5/2)-2/3*d*(-27*c^6*d+177*c^4*d^3+155*c^2*d^5-49*d^7+3*I*c^7-109*I*c^5*d 
^2+53*I*c^3*d^4+165*I*c*d^6)/(3*I*c^2*d-I*d^3+c^3-3*c*d^2)*(c+d*tan(f*x+e) 
)^(3/2)+1/2*d*(2*I*c^8-102*I*c^6*d^2+190*I*c^4*d^4+254*I*c^2*d^6-40*I*d^8- 
21*d*c^7+225*d^3*c^5+73*c^3*d^5-173*c*d^7)/(3*I*c^2*d-I*d^3+c^3-3*c*d^2)*( 
c+d*tan(f*x+e))^(1/2))/(-d*tan(f*x+e)+I*d)^3-1/2*(-57*c^5*d^2+90*c^3*d^4+1 
49*c*d^6+16*I*c^6*d-120*I*c^4*d^3-78*I*c^2*d^5+58*I*d^7+2*c^7)/(3*I*c^2*d- 
I*d^3+c^3-3*c*d^2)/(-c-I*d)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(-c-I*d)^( 
1/2)))+1/16*(-I*c^4+6*I*c^2*d^2-I*d^4+4*c^3*d-4*c*d^3)/(I*d-c)^(3/2)/(c+I* 
d)^4/d^4*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^(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. 2638 vs. \(2 (298) = 596\).

Time = 2.66 (sec) , antiderivative size = 2638, normalized size of antiderivative = 7.17 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e))^(3/2),x, algorithm="fric 
as")
 

Output:

-1/192*(48*((a^3*c^6 + 2*I*a^3*c^5*d + a^3*c^4*d^2 + 4*I*a^3*c^3*d^3 - a^3 
*c^2*d^4 + 2*I*a^3*c*d^5 - a^3*d^6)*f*e^(8*I*f*x + 8*I*e) + (a^3*c^6 + 4*I 
*a^3*c^5*d - 5*a^3*c^4*d^2 - 5*a^3*c^2*d^4 - 4*I*a^3*c*d^5 + a^3*d^6)*f*e^ 
(6*I*f*x + 6*I*e))*sqrt(-1/64*I/((I*a^6*c^3 + 3*a^6*c^2*d - 3*I*a^6*c*d^2 
- a^6*d^3)*f^2))*log(-2*(8*((I*a^3*c^2 + 2*a^3*c*d - I*a^3*d^2)*f*e^(2*I*f 
*x + 2*I*e) + (I*a^3*c^2 + 2*a^3*c*d - I*a^3*d^2)*f)*sqrt(((c - I*d)*e^(2* 
I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-1/64*I/((I*a^6* 
c^3 + 3*a^6*c^2*d - 3*I*a^6*c*d^2 - a^6*d^3)*f^2)) - (c - I*d)*e^(2*I*f*x 
+ 2*I*e) - c)*e^(-2*I*f*x - 2*I*e)) - 48*((a^3*c^6 + 2*I*a^3*c^5*d + a^3*c 
^4*d^2 + 4*I*a^3*c^3*d^3 - a^3*c^2*d^4 + 2*I*a^3*c*d^5 - a^3*d^6)*f*e^(8*I 
*f*x + 8*I*e) + (a^3*c^6 + 4*I*a^3*c^5*d - 5*a^3*c^4*d^2 - 5*a^3*c^2*d^4 - 
 4*I*a^3*c*d^5 + a^3*d^6)*f*e^(6*I*f*x + 6*I*e))*sqrt(-1/64*I/((I*a^6*c^3 
+ 3*a^6*c^2*d - 3*I*a^6*c*d^2 - a^6*d^3)*f^2))*log(-2*(8*((-I*a^3*c^2 - 2* 
a^3*c*d + I*a^3*d^2)*f*e^(2*I*f*x + 2*I*e) + (-I*a^3*c^2 - 2*a^3*c*d + I*a 
^3*d^2)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2* 
I*e) + 1))*sqrt(-1/64*I/((I*a^6*c^3 + 3*a^6*c^2*d - 3*I*a^6*c*d^2 - a^6*d^ 
3)*f^2)) - (c - I*d)*e^(2*I*f*x + 2*I*e) - c)*e^(-2*I*f*x - 2*I*e)) - 3*(( 
a^3*c^6 + 2*I*a^3*c^5*d + a^3*c^4*d^2 + 4*I*a^3*c^3*d^3 - a^3*c^2*d^4 + 2* 
I*a^3*c*d^5 - a^3*d^6)*f*e^(8*I*f*x + 8*I*e) + (a^3*c^6 + 4*I*a^3*c^5*d - 
5*a^3*c^4*d^2 - 5*a^3*c^2*d^4 - 4*I*a^3*c*d^5 + a^3*d^6)*f*e^(6*I*f*x +...
 

Sympy [F]

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

integrate(1/(a+I*a*tan(f*x+e))**3/(c+d*tan(f*x+e))**(3/2),x)
 

Output:

I*Integral(1/(c*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**3 - 3*I*c*sqrt(c + 
d*tan(e + f*x))*tan(e + f*x)**2 - 3*c*sqrt(c + d*tan(e + f*x))*tan(e + f*x 
) + I*c*sqrt(c + d*tan(e + f*x)) + d*sqrt(c + d*tan(e + f*x))*tan(e + f*x) 
**4 - 3*I*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**3 - 3*d*sqrt(c + d*tan( 
e + f*x))*tan(e + f*x)**2 + I*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)), x) 
/a**3
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e))^(3/2),x, algorithm="maxi 
ma")
 

Output:

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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 708 vs. \(2 (298) = 596\).

Time = 0.81 (sec) , antiderivative size = 708, normalized size of antiderivative = 1.92 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e))^(3/2),x, algorithm="giac 
")
 

Output:

1/8*I*(16*d^4/((c^5 + 3*I*c^4*d - 2*c^3*d^2 + 2*I*c^2*d^3 - 3*c*d^4 - I*d^ 
5)*sqrt(d*tan(f*x + e) + c)) + 8*sqrt(2)*(-2*I*c^3 + 12*c^2*d + 33*I*c*d^2 
 - 58*d^3)*arctan(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*t 
an(f*x + e) + c))/(sqrt(2)*c*sqrt(-c + sqrt(c^2 + d^2)) + I*sqrt(2)*sqrt(- 
c + sqrt(c^2 + d^2))*d - sqrt(2)*sqrt(c^2 + d^2)*sqrt(-c + sqrt(c^2 + d^2) 
)))/((16*I*c^4 - 64*c^3*d - 96*I*c^2*d^2 + 64*c*d^3 + 16*I*d^4)*sqrt(-c + 
sqrt(c^2 + d^2))*(I*d/(c - sqrt(c^2 + d^2)) + 1)) + sqrt(2)*arctan(2*(sqrt 
(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(sqrt(2 
)*c*sqrt(-c + sqrt(c^2 + d^2)) - I*sqrt(2)*sqrt(-c + sqrt(c^2 + d^2))*d - 
sqrt(2)*sqrt(c^2 + d^2)*sqrt(-c + sqrt(c^2 + d^2))))/((c - I*d)*sqrt(-c + 
sqrt(c^2 + d^2))*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) + 16*(6*(d*tan(f*x + e) 
 + c)^(5/2)*c^2*d - 12*(d*tan(f*x + e) + c)^(3/2)*c^3*d + 6*sqrt(d*tan(f*x 
 + e) + c)*c^4*d + 33*I*(d*tan(f*x + e) + c)^(5/2)*c*d^2 - 84*I*(d*tan(f*x 
 + e) + c)^(3/2)*c^2*d^2 + 51*I*sqrt(d*tan(f*x + e) + c)*c^3*d^2 - 84*(d*t 
an(f*x + e) + c)^(5/2)*d^3 + 268*(d*tan(f*x + e) + c)^(3/2)*c*d^3 - 204*sq 
rt(d*tan(f*x + e) + c)*c^2*d^3 + 196*I*(d*tan(f*x + e) + c)^(3/2)*d^4 - 27 
9*I*sqrt(d*tan(f*x + e) + c)*c*d^4 + 120*sqrt(d*tan(f*x + e) + c)*d^5)/((9 
6*I*c^4 - 384*c^3*d - 576*I*c^2*d^2 + 384*c*d^3 + 96*I*d^4)*(d*tan(f*x + e 
) - I*d)^3))/(a^3*f)
 

Mupad [B] (verification not implemented)

Time = 10.55 (sec) , antiderivative size = 106340, normalized size of antiderivative = 288.97 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

log(6960*a^3*d^15*f - ((-(c*d^12*6300i + 3360*d^13 - 945*c^2*d^11 + c^3*d^ 
10*1365i + 315*c^4*d^9 + c^5*d^8*693i + 672*c^6*d^7 - c^7*d^6*288i - 72*c^ 
8*d^5 + c^9*d^4*8i - a^6*c^10*f^2*(4*(((29973*c^2*d^14)/1024 - (841*d^16)/ 
256 - (5247*c^4*d^12)/1024 + (987*c^6*d^10)/1024 + (423*c^8*d^8)/1024 + (3 
*c^10*d^6)/64 - (c^12*d^4)/256)/(a^12*c^16*f^4 + a^12*d^16*f^4 + 8*a^12*c^ 
2*d^14*f^4 + 28*a^12*c^4*d^12*f^4 + 56*a^12*c^6*d^10*f^4 + 70*a^12*c^8*d^8 
*f^4 + 56*a^12*c^10*d^6*f^4 + 28*a^12*c^12*d^4*f^4 + 8*a^12*c^14*d^2*f^4) 
+ (((11861*c^3*d^13)/512 - (4089*c*d^15)/256 + (171*c^5*d^11)/128 + (261*c 
^7*d^9)/512 + (c^9*d^7)/128 - (3*c^11*d^5)/128)*1i)/(a^12*c^16*f^4 + a^12* 
d^16*f^4 + 8*a^12*c^2*d^14*f^4 + 28*a^12*c^4*d^12*f^4 + 56*a^12*c^6*d^10*f 
^4 + 70*a^12*c^8*d^8*f^4 + 56*a^12*c^10*d^6*f^4 + 28*a^12*c^12*d^4*f^4 + 8 
*a^12*c^14*d^2*f^4))*(256*d^6 + 256*c^2*d^4) + (((840*d^19 - (89145*c^2*d^ 
17)/4 + 45675*c^4*d^15 - (18123*c^6*d^13)/2 + 729*c^8*d^11 + (879*c^10*d^9 
)/4 + 34*c^12*d^7 + 6*c^14*d^5)*1i)/(a^6*c^16*f^2 + a^6*d^16*f^2 + 8*a^6*c 
^2*d^14*f^2 + 28*a^6*c^4*d^12*f^2 + 56*a^6*c^6*d^10*f^2 + 70*a^6*c^8*d^8*f 
^2 + 56*a^6*c^10*d^6*f^2 + 28*a^6*c^12*d^4*f^2 + 8*a^6*c^14*d^2*f^2) - (66 
15*c*d^18 - (166005*c^3*d^16)/4 + 28917*c^5*d^14 - (2223*c^7*d^12)/2 + 344 
*c^9*d^10 + (339*c^11*d^8)/4 - 6*c^13*d^6 - 2*c^15*d^4)/(a^6*c^16*f^2 + a^ 
6*d^16*f^2 + 8*a^6*c^2*d^14*f^2 + 28*a^6*c^4*d^12*f^2 + 56*a^6*c^6*d^10*f^ 
2 + 70*a^6*c^8*d^8*f^2 + 56*a^6*c^10*d^6*f^2 + 28*a^6*c^12*d^4*f^2 + 8*...
 

Reduce [F]

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

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

Output:

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