\(\int \frac {(d \tan (e+f x))^{5/2}}{(a+a \tan (e+f x))^2} \, dx\) [364]

Optimal result

Integrand size = 25, antiderivative size = 225 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+a \tan (e+f x))^2} \, dx=\frac {3 d^{5/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 f}+\frac {d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 f}-\frac {d^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 f}-\frac {d^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}+\sqrt {d} \tan (e+f x)}\right )}{2 \sqrt {2} a^2 f}-\frac {d^2 \sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )} \] Output:

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

Mathematica [A] (verified)

Time = 3.73 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.00 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+a \tan (e+f x))^2} \, dx=\frac {\csc (e+f x) (\cos (e+f x)+\sin (e+f x))^2 \left (-\frac {4 \cot (e+f x)}{\cos (e+f x)+\sin (e+f x)}+\frac {\left (2 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )-2 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {\tan (e+f x)}\right )+12 \arctan \left (\sqrt {\tan (e+f x)}\right )+\sqrt {2} \log \left (-1+\sqrt {2} \sqrt {\tan (e+f x)}-\tan (e+f x)\right )-\sqrt {2} \log \left (1+\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right )\right ) \sec (e+f x)}{\tan ^{\frac {3}{2}}(e+f x)}\right ) (d \tan (e+f x))^{5/2}}{8 a^2 f (1+\tan (e+f x))^2} \] Input:

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

Output:

(Csc[e + f*x]*(Cos[e + f*x] + Sin[e + f*x])^2*((-4*Cot[e + f*x])/(Cos[e + 
f*x] + Sin[e + f*x]) + ((2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]] 
- 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]] + 12*ArcTan[Sqrt[Tan[e 
+ f*x]]] + Sqrt[2]*Log[-1 + Sqrt[2]*Sqrt[Tan[e + f*x]] - Tan[e + f*x]] - S 
qrt[2]*Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]])*Sec[e + f*x])/T 
an[e + f*x]^(3/2))*(d*Tan[e + f*x])^(5/2))/(8*a^2*f*(1 + Tan[e + f*x])^2)
 

Rubi [A] (warning: unable to verify)

Time = 1.15 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.16, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.840, Rules used = {3042, 4048, 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.

\(\displaystyle \int \frac {(d \tan (e+f x))^{5/2}}{(a \tan (e+f x)+a)^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \tan (e+f x))^{5/2}}{(a \tan (e+f x)+a)^2}dx\)

\(\Big \downarrow \) 4048

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

\(\Big \downarrow \) 4117

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {6 a d^2 \int \frac {1}{\tan (e+f x)+1}d\sqrt {d \tan (e+f x)}}{f}-\frac {4 a d^4 \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\log \left (\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (-\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}}{2 d}\right )}{f}}{4 a^3}-\frac {d^2 \sqrt {d \tan (e+f x)}}{2 f \left (a^2 \tan (e+f x)+a^2\right )}\)

\(\Big \downarrow \) 216

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

Input:

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

Output:

((6*a*d^(5/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/f - (4*a*d^4*((-(ArcTa 
n[1 - Sqrt[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)*Tan[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)/(4*a^3) - (d^2*Sqrt[d*Tan[e + f*x]])/(2*f*(a^2 + a^2*Tan[e + f*x]))
 

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 4048

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m 
 - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1 
/(d*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + 
f*x])^(n + 1)*Simp[a^2*d*(b*d*(m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c 
*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3*a*b^2*d) 
*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*( 
n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 2] && LtQ 
[n, -1] && IntegerQ[2*m]
 

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 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 = 1.48 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {2 d^{3} \left (-\frac {\sqrt {d \tan \left (f x +e \right )}}{4 \left (d \tan \left (f x +e \right )+d \right )}+\frac {3 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 d}\right )}{f \,a^{2}}\)	\(191\)
default	\(\frac {2 d^{3} \left (-\frac {\sqrt {d \tan \left (f x +e \right )}}{4 \left (d \tan \left (f x +e \right )+d \right )}+\frac {3 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 d}\right )}{f \,a^{2}}\)	\(191\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.21 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+a \tan (e+f x))^2} \, dx=-\frac {2 \, \sqrt {2} {\left (d^{2} \tan \left (f x + e\right ) + d^{2}\right )} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d}{d}\right ) + 2 \, \sqrt {2} {\left (d^{2} \tan \left (f x + e\right ) + d^{2}\right )} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} - d}{d}\right ) + \sqrt {2} {\left (d^{2} \tan \left (f x + e\right ) + d^{2}\right )} \sqrt {d} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right ) - \sqrt {2} {\left (d^{2} \tan \left (f x + e\right ) + d^{2}\right )} \sqrt {d} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right ) + 4 \, \sqrt {d \tan \left (f x + e\right )} d^{2} + 12 \, {\left (d^{2} \tan \left (f x + e\right ) + d^{2}\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d} \tan \left (f x + e\right )}\right )}{8 \, {\left (a^{2} f \tan \left (f x + e\right ) + a^{2} f\right )}} \] Input:

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

Output:

-1/8*(2*sqrt(2)*(d^2*tan(f*x + e) + d^2)*sqrt(d)*arctan((sqrt(2)*sqrt(d*ta 
n(f*x + e))*sqrt(d) + d)/d) + 2*sqrt(2)*(d^2*tan(f*x + e) + d^2)*sqrt(d)*a 
rctan((sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) - d)/d) + sqrt(2)*(d^2*tan(f*x 
 + e) + d^2)*sqrt(d)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqr 
t(d) + d) - sqrt(2)*(d^2*tan(f*x + e) + d^2)*sqrt(d)*log(d*tan(f*x + e) - 
sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d) + 4*sqrt(d*tan(f*x + e))*d^2 + 1 
2*(d^2*tan(f*x + e) + d^2)*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/(sqrt(d)*ta 
n(f*x + e))))/(a^2*f*tan(f*x + e) + a^2*f)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.96 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+a \tan (e+f x))^2} \, dx=-\frac {\frac {4 \, \sqrt {d \tan \left (f x + e\right )} d^{4}}{a^{2} d \tan \left (f x + e\right ) + a^{2} d} - \frac {12 \, d^{\frac {7}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{2}} + \frac {2 \, \sqrt {2} d^{\frac {7}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right ) + 2 \, \sqrt {2} d^{\frac {7}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right ) + \sqrt {2} d^{\frac {7}{2}} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right ) - \sqrt {2} d^{\frac {7}{2}} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{a^{2}}}{8 \, d f} \] Input:

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

Output:

-1/8*(4*sqrt(d*tan(f*x + e))*d^4/(a^2*d*tan(f*x + e) + a^2*d) - 12*d^(7/2) 
*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/a^2 + (2*sqrt(2)*d^(7/2)*arctan(1/2* 
sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d)) + 2*sqrt(2)*d^ 
(7/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt( 
d)) + sqrt(2)*d^(7/2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sq 
rt(d) + d) - sqrt(2)*d^(7/2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + 
 e))*sqrt(d) + d))/a^2)/(d*f)
 

Giac [F(-2)]

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

integrate((d*tan(f*x+e))^(5/2)/(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:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 1.75 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.67 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+a \tan (e+f x))^2} \, dx=\frac {\mathrm {atan}\left (\frac {4\,d^{20}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^{10}}{a^8\,f^4}\right )}^{1/4}}{\frac {36\,d^{23}}{a^2\,f}-4\,a^2\,d^{18}\,f\,\sqrt {-\frac {d^{10}}{a^8\,f^4}}}+\frac {36\,d^{15}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^{10}}{a^8\,f^4}\right )}^{3/4}}{\frac {36\,d^{23}}{a^6\,f^3}-\frac {4\,d^{18}\,\sqrt {-\frac {d^{10}}{a^8\,f^4}}}{a^2\,f}}\right )\,{\left (-\frac {d^{10}}{a^8\,f^4}\right )}^{1/4}}{2}+\mathrm {atan}\left (\frac {d^{20}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^{10}}{256\,a^8\,f^4}\right )}^{1/4}\,16{}\mathrm {i}}{\frac {36\,d^{23}}{a^2\,f}+64\,a^2\,d^{18}\,f\,\sqrt {-\frac {d^{10}}{256\,a^8\,f^4}}}-\frac {d^{15}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^{10}}{256\,a^8\,f^4}\right )}^{3/4}\,2304{}\mathrm {i}}{\frac {36\,d^{23}}{a^6\,f^3}+\frac {64\,d^{18}\,\sqrt {-\frac {d^{10}}{256\,a^8\,f^4}}}{a^2\,f}}\right )\,{\left (-\frac {d^{10}}{256\,a^8\,f^4}\right )}^{1/4}\,2{}\mathrm {i}-\frac {d^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\left (a^2\,d\,f+a^2\,d\,f\,\mathrm {tan}\left (e+f\,x\right )\right )}+\frac {\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-d^5}\,1{}\mathrm {i}}{d^3}\right )\,\sqrt {-d^5}\,3{}\mathrm {i}}{2\,a^2\,f} \] Input:

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

Output:

(atan((4*d^20*(d*tan(e + f*x))^(1/2)*(-d^10/(a^8*f^4))^(1/4))/((36*d^23)/( 
a^2*f) - 4*a^2*d^18*f*(-d^10/(a^8*f^4))^(1/2)) + (36*d^15*(d*tan(e + f*x)) 
^(1/2)*(-d^10/(a^8*f^4))^(3/4))/((36*d^23)/(a^6*f^3) - (4*d^18*(-d^10/(a^8 
*f^4))^(1/2))/(a^2*f)))*(-d^10/(a^8*f^4))^(1/4))/2 + atan((d^20*(d*tan(e + 
 f*x))^(1/2)*(-d^10/(256*a^8*f^4))^(1/4)*16i)/((36*d^23)/(a^2*f) + 64*a^2* 
d^18*f*(-d^10/(256*a^8*f^4))^(1/2)) - (d^15*(d*tan(e + f*x))^(1/2)*(-d^10/ 
(256*a^8*f^4))^(3/4)*2304i)/((36*d^23)/(a^6*f^3) + (64*d^18*(-d^10/(256*a^ 
8*f^4))^(1/2))/(a^2*f)))*(-d^10/(256*a^8*f^4))^(1/4)*2i - (d^3*(d*tan(e + 
f*x))^(1/2))/(2*(a^2*d*f + a^2*d*f*tan(e + f*x))) + (atan(((d*tan(e + f*x) 
)^(1/2)*(-d^5)^(1/2)*1i)/d^3)*(-d^5)^(1/2)*3i)/(2*a^2*f)
 

Reduce [F]

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

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

Output:

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