\(\int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2} \, dx\) [368]

Optimal result

Integrand size = 25, antiderivative size = 250 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2} \, dx=-\frac {5 \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 d^{3/2} f}+\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 d^{3/2} f}-\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 d^{3/2} f}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}+\sqrt {d} \tan (e+f x)}\right )}{2 \sqrt {2} a^2 d^{3/2} f}-\frac {5}{2 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \sqrt {d \tan (e+f x)} \left (a^2+a^2 \tan (e+f x)\right )} \] Output:

-5/2*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a^2/d^(3/2)/f+1/4*arctan(1-2^(1/ 
2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*2^(1/2)/a^2/d^(3/2)/f-1/4*arctan(1+2^(1/2 
)*(d*tan(f*x+e))^(1/2)/d^(1/2))*2^(1/2)/a^2/d^(3/2)/f-1/4*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^(3/2)/f-5 
/2/a^2/d/f/(d*tan(f*x+e))^(1/2)+1/2/d/f/(d*tan(f*x+e))^(1/2)/(a^2+a^2*tan( 
f*x+e))
 

Mathematica [A] (verified)

Time = 6.31 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.20 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2} \, dx=\frac {1}{2 a d f \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))}+\frac {-\frac {5 a \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {d} f}+\frac {a \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d} f}-\frac {a \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d} f}+\frac {a \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{2 \sqrt {2} \sqrt {d} f}-\frac {a \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{2 \sqrt {2} \sqrt {d} f}-\frac {5 a \cot (e+f x) \sqrt {d \tan (e+f x)}}{d f}}{2 a^3 d} \] Input:

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

Output:

1/(2*a*d*f*Sqrt[d*Tan[e + f*x]]*(a + a*Tan[e + f*x])) + ((-5*a*ArcTan[Sqrt 
[d*Tan[e + f*x]]/Sqrt[d]])/(Sqrt[d]*f) + (a*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan 
[e + f*x]])/Sqrt[d]])/(Sqrt[2]*Sqrt[d]*f) - (a*ArcTan[1 + (Sqrt[2]*Sqrt[d* 
Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*Sqrt[d]*f) + (a*Log[Sqrt[d] + Sqrt[d]*Ta 
n[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(2*Sqrt[2]*Sqrt[d]*f) - (a*Log 
[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(2*Sqrt[2 
]*Sqrt[d]*f) - (5*a*Cot[e + f*x]*Sqrt[d*Tan[e + f*x]])/(d*f))/(2*a^3*d)
 

Rubi [A] (warning: unable to verify)

Time = 1.48 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.18, number of steps used = 25, number of rules used = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.960, Rules used = {3042, 4052, 27, 3042, 4132, 27, 3042, 4136, 27, 3042, 3957, 266, 755, 1476, 1082, 217, 1479, 25, 27, 1103, 4117, 27, 73, 216}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4052

\(\displaystyle \frac {\int \frac {3 d \tan ^2(e+f x) a^2+5 d a^2-2 d \tan (e+f x) a^2}{2 (d \tan (e+f x))^{3/2} (\tan (e+f x) a+a)}dx}{2 a^3 d}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {3 d \tan ^2(e+f x) a^2+5 d a^2-2 d \tan (e+f x) a^2}{(d \tan (e+f x))^{3/2} (\tan (e+f x) a+a)}dx}{4 a^3 d}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {3 d \tan (e+f x)^2 a^2+5 d a^2-2 d \tan (e+f x) a^2}{(d \tan (e+f x))^{3/2} (\tan (e+f x) a+a)}dx}{4 a^3 d}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 4132

\(\displaystyle \frac {-\frac {2 \int \frac {7 a^3 d^3+5 a^3 \tan ^2(e+f x) d^3+2 a^3 \tan (e+f x) d^3}{2 \sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{a d^3}-\frac {10 a}{f \sqrt {d \tan (e+f x)}}}{4 a^3 d}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {7 a^3 d^3+5 a^3 \tan ^2(e+f x) d^3+2 a^3 \tan (e+f x) d^3}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{a d^3}-\frac {10 a}{f \sqrt {d \tan (e+f x)}}}{4 a^3 d}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {7 a^3 d^3+5 a^3 \tan (e+f x)^2 d^3+2 a^3 \tan (e+f x) d^3}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{a d^3}-\frac {10 a}{f \sqrt {d \tan (e+f x)}}}{4 a^3 d}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 4136

\(\displaystyle \frac {-\frac {5 a^3 d^3 \int \frac {\tan ^2(e+f x)+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {\int \frac {4 a^4 d^3}{\sqrt {d \tan (e+f x)}}dx}{2 a^2}}{a d^3}-\frac {10 a}{f \sqrt {d \tan (e+f x)}}}{4 a^3 d}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {5 a^3 d^3 \int \frac {\tan ^2(e+f x)+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+2 a^2 d^3 \int \frac {1}{\sqrt {d \tan (e+f x)}}dx}{a d^3}-\frac {10 a}{f \sqrt {d \tan (e+f x)}}}{4 a^3 d}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {5 a^3 d^3 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+2 a^2 d^3 \int \frac {1}{\sqrt {d \tan (e+f x)}}dx}{a d^3}-\frac {10 a}{f \sqrt {d \tan (e+f x)}}}{4 a^3 d}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {-\frac {5 a^3 d^3 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {2 a^2 d^4 \int \frac {1}{\sqrt {d \tan (e+f x)} \left (\tan ^2(e+f x) d^2+d^2\right )}d(d \tan (e+f x))}{f}}{a d^3}-\frac {10 a}{f \sqrt {d \tan (e+f x)}}}{4 a^3 d}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {-\frac {5 a^3 d^3 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {4 a^2 d^4 \int \frac {1}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{f}}{a d^3}-\frac {10 a}{f \sqrt {d \tan (e+f x)}}}{4 a^3 d}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {-\frac {5 a^3 d^3 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {4 a^2 d^4 \left (\frac {\int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{2 d}+\frac {\int \frac {d^2 \tan ^2(e+f x)+d}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{2 d}\right )}{f}}{a d^3}-\frac {10 a}{f \sqrt {d \tan (e+f x)}}}{4 a^3 d}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

\(\displaystyle \frac {-\frac {5 a^3 d^3 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {4 a^2 d^4 \left (\frac {\frac {\int \frac {1}{-d^2 \tan ^2(e+f x)-1}d\left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d^2 \tan ^2(e+f x)-1}d\left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{2 d}\right )}{f}}{a d^3}-\frac {10 a}{f \sqrt {d \tan (e+f x)}}}{4 a^3 d}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {-\frac {5 a^3 d^3 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {4 a^2 d^4 \left (\frac {\int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{f}}{a d^3}-\frac {10 a}{f \sqrt {d \tan (e+f x)}}}{4 a^3 d}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

\(\displaystyle \frac {-\frac {5 a^3 d^3 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {4 a^2 d^4 \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\log \left (\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (-\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}}{2 d}\right )}{f}}{a d^3}-\frac {10 a}{f \sqrt {d \tan (e+f x)}}}{4 a^3 d}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {-\frac {\frac {5 a^3 d^3 \int \frac {1}{a \sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{f}+\frac {4 a^2 d^4 \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\log \left (\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (-\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}}{2 d}\right )}{f}}{a d^3}-\frac {10 a}{f \sqrt {d \tan (e+f x)}}}{4 a^3 d}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\frac {5 a^2 d^3 \int \frac {1}{\sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{f}+\frac {4 a^2 d^4 \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\log \left (\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (-\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}}{2 d}\right )}{f}}{a d^3}-\frac {10 a}{f \sqrt {d \tan (e+f x)}}}{4 a^3 d}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {-\frac {\frac {10 a^2 d^2 \int \frac {1}{\tan (e+f x)+1}d\sqrt {d \tan (e+f x)}}{f}+\frac {4 a^2 d^4 \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\log \left (\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (-\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}}{2 d}\right )}{f}}{a d^3}-\frac {10 a}{f \sqrt {d \tan (e+f x)}}}{4 a^3 d}+\frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{2 d f \left (a^2 \tan (e+f x)+a^2\right ) \sqrt {d \tan (e+f x)}}+\frac {-\frac {\frac {10 a^2 d^{5/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {4 a^2 d^4 \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\log \left (\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (-\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}}{2 d}\right )}{f}}{a d^3}-\frac {10 a}{f \sqrt {d \tan (e+f x)}}}{4 a^3 d}\)

Input:

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

Output:

1/(2*d*f*Sqrt[d*Tan[e + f*x]]*(a^2 + a^2*Tan[e + f*x])) + (-(((10*a^2*d^(5 
/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/f + (4*a^2*d^4*((-(ArcTan[1 - Sq 
rt[2]*Sqrt[d]*Tan[e + f*x]]/(Sqrt[2]*Sqrt[d])) + ArcTan[1 + Sqrt[2]*Sqrt[d 
]*Tan[e + f*x]]/(Sqrt[2]*Sqrt[d]))/(2*d) + (-1/2*Log[d - Sqrt[2]*d^(3/2)*T 
an[e + f*x] + d^2*Tan[e + f*x]^2]/(Sqrt[2]*Sqrt[d]) + Log[d + Sqrt[2]*d^(3 
/2)*Tan[e + f*x] + d^2*Tan[e + f*x]^2]/(2*Sqrt[2]*Sqrt[d]))/(2*d)))/f)/(a* 
d^3)) - (10*a)/(f*Sqrt[d*Tan[e + f*x]]))/(4*a^3*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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

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 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 755

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[1/(2*r)   Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, 
 b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & 
& AtomQ[SplitProduct[SumBaseQ, b]]))
 

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 3042

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

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

rule 4052

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

rule 4117

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

rule 4132

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Tan[e + 
 f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + 
b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2))   Int[(a + b*Tan[e + 
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* 
(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d 
)*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ 
[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && 
!(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4136

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) 
+ (f_.)*(x_)]), x_Symbol] :> Simp[1/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^ 
n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Simp[ 
(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^n*((1 + Tan[ 
e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, 
 C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] & 
&  !GtQ[n, 0] &&  !LeQ[n, -1]
 

Maple [A] (verified)

Time = 1.38 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {2 d^{3} \left (-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 d^{5}}-\frac {1}{d^{4} \sqrt {d \tan \left (f x +e \right )}}-\frac {\frac {\sqrt {d \tan \left (f x +e \right )}}{2 d \tan \left (f x +e \right )+2 d}+\frac {5 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 \sqrt {d}}}{2 d^{4}}\right )}{f \,a^{2}}\)	\(212\)
default	\(\frac {2 d^{3} \left (-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 d^{5}}-\frac {1}{d^{4} \sqrt {d \tan \left (f x +e \right )}}-\frac {\frac {\sqrt {d \tan \left (f x +e \right )}}{2 d \tan \left (f x +e \right )+2 d}+\frac {5 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 \sqrt {d}}}{2 d^{4}}\right )}{f \,a^{2}}\)	\(212\)

Input:

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

Output:

2/f/a^2*d^3*(-1/16/d^5*(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)+(d^2)^(1/4)*( 
d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)-(d^2)^(1/4)*(d*tan( 
f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f* 
x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))-1/d^ 
4/(d*tan(f*x+e))^(1/2)-1/2/d^4*(1/2*(d*tan(f*x+e))^(1/2)/(d*tan(f*x+e)+d)+ 
5/2/d^(1/2)*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.20 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2} \, dx=\frac {20 \, {\left (\tan \left (f x + e\right )^{2} + \tan \left (f x + e\right )\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d} \tan \left (f x + e\right )}\right ) - \frac {2 \, \sqrt {2} {\left (d \tan \left (f x + e\right )^{2} + d \tan \left (f x + e\right )\right )} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}} + 1\right )}{\sqrt {d}} - \frac {2 \, \sqrt {2} {\left (d \tan \left (f x + e\right )^{2} + d \tan \left (f x + e\right )\right )} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}} - 1\right )}{\sqrt {d}} - \frac {\sqrt {2} {\left (d \tan \left (f x + e\right )^{2} + d \tan \left (f x + e\right )\right )} \log \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}} + \tan \left (f x + e\right ) + 1\right )}{\sqrt {d}} + \frac {\sqrt {2} {\left (d \tan \left (f x + e\right )^{2} + d \tan \left (f x + e\right )\right )} \log \left (-\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}} + \tan \left (f x + e\right ) + 1\right )}{\sqrt {d}} - 4 \, \sqrt {d \tan \left (f x + e\right )} {\left (5 \, \tan \left (f x + e\right ) + 4\right )}}{8 \, {\left (a^{2} d^{2} f \tan \left (f x + e\right )^{2} + a^{2} d^{2} f \tan \left (f x + e\right )\right )}} \] Input:

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

Output:

1/8*(20*(tan(f*x + e)^2 + tan(f*x + e))*sqrt(d)*arctan(sqrt(d*tan(f*x + e) 
)/(sqrt(d)*tan(f*x + e))) - 2*sqrt(2)*(d*tan(f*x + e)^2 + d*tan(f*x + e))* 
arctan(sqrt(2)*sqrt(d*tan(f*x + e))/sqrt(d) + 1)/sqrt(d) - 2*sqrt(2)*(d*ta 
n(f*x + e)^2 + d*tan(f*x + e))*arctan(sqrt(2)*sqrt(d*tan(f*x + e))/sqrt(d) 
 - 1)/sqrt(d) - sqrt(2)*(d*tan(f*x + e)^2 + d*tan(f*x + e))*log(sqrt(2)*sq 
rt(d*tan(f*x + e))/sqrt(d) + tan(f*x + e) + 1)/sqrt(d) + sqrt(2)*(d*tan(f* 
x + e)^2 + d*tan(f*x + e))*log(-sqrt(2)*sqrt(d*tan(f*x + e))/sqrt(d) + tan 
(f*x + e) + 1)/sqrt(d) - 4*sqrt(d*tan(f*x + e))*(5*tan(f*x + e) + 4))/(a^2 
*d^2*f*tan(f*x + e)^2 + a^2*d^2*f*tan(f*x + e))
 

Sympy [F]

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

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

Output:

Integral(1/((d*tan(e + f*x))**(3/2)*tan(e + f*x)**2 + 2*(d*tan(e + f*x))** 
(3/2)*tan(e + f*x) + (d*tan(e + f*x))**(3/2)), x)/a**2
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2} \, dx=-\frac {\frac {4 \, {\left (5 \, d \tan \left (f x + e\right ) + 4 \, d\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{2} + \sqrt {d \tan \left (f x + e\right )} a^{2} d} + \frac {\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}}{a^{2}} + \frac {20 \, \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{2} \sqrt {d}}}{8 \, d f} \] Input:

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

Output:

-1/8*(4*(5*d*tan(f*x + e) + 4*d)/((d*tan(f*x + e))^(3/2)*a^2 + sqrt(d*tan( 
f*x + e))*a^2*d) + (2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt 
(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2 
)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + sqrt(2)*log(d*tan(f 
*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) - sqrt(2)*log( 
d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d))/a^2 + 
20*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^2*sqrt(d)))/(d*f)
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
 

Mupad [B] (verification not implemented)

Time = 1.73 (sec) , antiderivative size = 415, normalized size of antiderivative = 1.66 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2} \, dx=\frac {\mathrm {atan}\left (\frac {2048\,a^{10}\,d^{13}\,f^5\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {1}{a^8\,d^6\,f^4}\right )}^{1/4}}{51200\,a^8\,d^{12}\,f^4-2048\,a^{12}\,d^{15}\,f^6\,\sqrt {-\frac {1}{a^8\,d^6\,f^4}}}+\frac {51200\,a^{14}\,d^{16}\,f^7\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {1}{a^8\,d^6\,f^4}\right )}^{3/4}}{51200\,a^8\,d^{12}\,f^4-2048\,a^{12}\,d^{15}\,f^6\,\sqrt {-\frac {1}{a^8\,d^6\,f^4}}}\right )\,{\left (-\frac {1}{a^8\,d^6\,f^4}\right )}^{1/4}}{2}+\mathrm {atan}\left (\frac {a^{10}\,d^{13}\,f^5\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {1}{256\,a^8\,d^6\,f^4}\right )}^{1/4}\,8192{}\mathrm {i}}{51200\,a^8\,d^{12}\,f^4+32768\,a^{12}\,d^{15}\,f^6\,\sqrt {-\frac {1}{256\,a^8\,d^6\,f^4}}}-\frac {a^{14}\,d^{16}\,f^7\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {1}{256\,a^8\,d^6\,f^4}\right )}^{3/4}\,3276800{}\mathrm {i}}{51200\,a^8\,d^{12}\,f^4+32768\,a^{12}\,d^{15}\,f^6\,\sqrt {-\frac {1}{256\,a^8\,d^6\,f^4}}}\right )\,{\left (-\frac {1}{256\,a^8\,d^6\,f^4}\right )}^{1/4}\,2{}\mathrm {i}-\frac {\frac {5\,\mathrm {tan}\left (e+f\,x\right )}{2}+2}{a^2\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}+a^2\,d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}-\frac {\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-d^3}\,1{}\mathrm {i}}{d^2}\right )\,\sqrt {-d^3}\,5{}\mathrm {i}}{2\,a^2\,d^3\,f} \] Input:

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

Output:

(atan((2048*a^10*d^13*f^5*(d*tan(e + f*x))^(1/2)*(-1/(a^8*d^6*f^4))^(1/4)) 
/(51200*a^8*d^12*f^4 - 2048*a^12*d^15*f^6*(-1/(a^8*d^6*f^4))^(1/2)) + (512 
00*a^14*d^16*f^7*(d*tan(e + f*x))^(1/2)*(-1/(a^8*d^6*f^4))^(3/4))/(51200*a 
^8*d^12*f^4 - 2048*a^12*d^15*f^6*(-1/(a^8*d^6*f^4))^(1/2)))*(-1/(a^8*d^6*f 
^4))^(1/4))/2 + atan((a^10*d^13*f^5*(d*tan(e + f*x))^(1/2)*(-1/(256*a^8*d^ 
6*f^4))^(1/4)*8192i)/(51200*a^8*d^12*f^4 + 32768*a^12*d^15*f^6*(-1/(256*a^ 
8*d^6*f^4))^(1/2)) - (a^14*d^16*f^7*(d*tan(e + f*x))^(1/2)*(-1/(256*a^8*d^ 
6*f^4))^(3/4)*3276800i)/(51200*a^8*d^12*f^4 + 32768*a^12*d^15*f^6*(-1/(256 
*a^8*d^6*f^4))^(1/2)))*(-1/(256*a^8*d^6*f^4))^(1/4)*2i - ((5*tan(e + f*x)) 
/2 + 2)/(a^2*f*(d*tan(e + f*x))^(3/2) + a^2*d*f*(d*tan(e + f*x))^(1/2)) - 
(atan(((d*tan(e + f*x))^(1/2)*(-d^3)^(1/2)*1i)/d^2)*(-d^3)^(1/2)*5i)/(2*a^ 
2*d^3*f)
 

Reduce [F]

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

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

Output:

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