\(\int \frac {(d \tan (e+f x))^{3/2}}{(a+a \tan (e+f x))^2} \, dx\) [365]

Optimal result

Integrand size = 25, antiderivative size = 223 \[ \int \frac {(d \tan (e+f x))^{3/2}}{(a+a \tan (e+f x))^2} \, dx=-\frac {d^{3/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 f}-\frac {d^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 f}+\frac {d^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 f}-\frac {d^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}+\sqrt {d} \tan (e+f x)}\right )}{2 \sqrt {2} a^2 f}+\frac {d \sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.82 \[ \int \frac {(d \tan (e+f x))^{3/2}}{(a+a \tan (e+f x))^2} \, dx=-\frac {\left (-d^2\right )^{3/4} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt [4]{-d^2}}\right )+\left (-d^2\right )^{3/4} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt [4]{-d^2}}\right ) \tan (e+f x)-d \sqrt {d \tan (e+f x)}+d^{3/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right ) (1+\tan (e+f x))-\left (-d^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt [4]{-d^2}}\right ) (1+\tan (e+f x))}{2 a^2 f (1+\tan (e+f x))} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 1.23 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.12, number of steps used = 23, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.880, Rules used = {3042, 4050, 27, 3042, 4136, 27, 2030, 3042, 3957, 266, 826, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4050

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

\(\Big \downarrow \) 4117

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

-1/4*((2*d^(3/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/f - (4*d^2*((-(ArcT 
an[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 + (Log[d - Sqrt[2]*d^(3/2)*T 
an[e + f*x] + d^2*Tan[e + f*x]^2]/(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))/f)/a^2 + 
 (d*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 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   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 2030

Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

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 4050

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*(a + b*Tan[e + f*x])^(m + 
 1)*((c + d*Tan[e + f*x])^(n - 1)/(f*(m + 1)*(a^2 + b^2))), x] + Simp[1/((m 
 + 1)*(a^2 + b^2))   Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^ 
(n - 2)*Simp[a*c^2*(m + 1) + a*d^2*(n - 1) + b*c*d*(m - n + 2) - (b*c^2 - 2 
*a*c*d - b*d^2)*(m + 1)*Tan[e + f*x] - d*(b*c - a*d)*(m + n)*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] && LtQ[m, -1] && LtQ[1, n, 2] && 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 = 1.41 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {2 d^{3} \left (\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 d \left (d^{2}\right )^{\frac {1}{4}}}-\frac {-\frac {\sqrt {d \tan \left (f x +e \right )}}{2 \left (d \tan \left (f x +e \right )+d \right )}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 \sqrt {d}}}{2 d}\right )}{f \,a^{2}}\)	\(197\)
default	\(\frac {2 d^{3} \left (\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 d \left (d^{2}\right )^{\frac {1}{4}}}-\frac {-\frac {\sqrt {d \tan \left (f x +e \right )}}{2 \left (d \tan \left (f x +e \right )+d \right )}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 \sqrt {d}}}{2 d}\right )}{f \,a^{2}}\)	\(197\)

Input:

int((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/(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)-(d^2)^(1/4)*(d*t 
an(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))-1/2/d*( 
-1/2*(d*tan(f*x+e))^(1/2)/(d*tan(f*x+e)+d)+1/2/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 = 251, normalized size of antiderivative = 1.13 \[ \int \frac {(d \tan (e+f x))^{3/2}}{(a+a \tan (e+f x))^2} \, dx=\frac {2 \, \sqrt {2} {\left (d \tan \left (f x + e\right ) + d\right )} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d}{d}\right ) + 2 \, \sqrt {2} {\left (d \tan \left (f x + e\right ) + d\right )} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} - d}{d}\right ) - \sqrt {2} {\left (d \tan \left (f x + e\right ) + d\right )} \sqrt {d} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right ) + \sqrt {2} {\left (d \tan \left (f x + e\right ) + d\right )} \sqrt {d} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right ) + 4 \, {\left (d \tan \left (f x + e\right ) + d\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d} \tan \left (f x + e\right )}\right ) + 4 \, \sqrt {d \tan \left (f x + e\right )} d}{8 \, {\left (a^{2} f \tan \left (f x + e\right ) + a^{2} f\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.99 \[ \int \frac {(d \tan (e+f x))^{3/2}}{(a+a \tan (e+f x))^2} \, dx=\frac {\frac {4 \, \sqrt {d \tan \left (f x + e\right )} d^{3}}{a^{2} d \tan \left (f x + e\right ) + a^{2} d} + \frac {d^{3} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{a^{2}} - \frac {4 \, d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{2}}}{8 \, d f} \] Input:

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

Output:

1/8*(4*sqrt(d*tan(f*x + e))*d^3/(a^2*d*tan(f*x + e) + a^2*d) + d^3*(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*t 
an(f*x + e))*sqrt(d) + d)/sqrt(d) + sqrt(2)*log(d*tan(f*x + e) - sqrt(2)*s 
qrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d))/a^2 - 4*d^(5/2)*arctan(sqrt(d*ta 
n(f*x + e))/sqrt(d))/a^2)/(d*f)
 

Giac [F(-2)]

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

integrate((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:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 1.66 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.68 \[ \int \frac {(d \tan (e+f x))^{3/2}}{(a+a \tan (e+f x))^2} \, dx=\frac {\mathrm {atan}\left (\frac {4\,d^{16}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^6}{a^8\,f^4}\right )}^{1/4}}{\frac {4\,d^{18}}{a^2\,f}+4\,a^2\,d^{15}\,f\,\sqrt {-\frac {d^6}{a^8\,f^4}}}+\frac {4\,d^{13}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^6}{a^8\,f^4}\right )}^{3/4}}{\frac {4\,d^{18}}{a^6\,f^3}+\frac {4\,d^{15}\,\sqrt {-\frac {d^6}{a^8\,f^4}}}{a^2\,f}}\right )\,{\left (-\frac {d^6}{a^8\,f^4}\right )}^{1/4}}{2}+\mathrm {atan}\left (\frac {d^{16}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^6}{256\,a^8\,f^4}\right )}^{1/4}\,16{}\mathrm {i}}{\frac {4\,d^{18}}{a^2\,f}-64\,a^2\,d^{15}\,f\,\sqrt {-\frac {d^6}{256\,a^8\,f^4}}}-\frac {d^{13}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^6}{256\,a^8\,f^4}\right )}^{3/4}\,256{}\mathrm {i}}{\frac {4\,d^{18}}{a^6\,f^3}-\frac {64\,d^{15}\,\sqrt {-\frac {d^6}{256\,a^8\,f^4}}}{a^2\,f}}\right )\,{\left (-\frac {d^6}{256\,a^8\,f^4}\right )}^{1/4}\,2{}\mathrm {i}+\frac {d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\left (a^2\,d\,f+a^2\,d\,f\,\mathrm {tan}\left (e+f\,x\right )\right )}-\frac {\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-d^3}\,1{}\mathrm {i}}{d^2}\right )\,\sqrt {-d^3}\,1{}\mathrm {i}}{2\,a^2\,f} \] Input:

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

Output:

(atan((4*d^16*(d*tan(e + f*x))^(1/2)*(-d^6/(a^8*f^4))^(1/4))/((4*d^18)/(a^ 
2*f) + 4*a^2*d^15*f*(-d^6/(a^8*f^4))^(1/2)) + (4*d^13*(d*tan(e + f*x))^(1/ 
2)*(-d^6/(a^8*f^4))^(3/4))/((4*d^18)/(a^6*f^3) + (4*d^15*(-d^6/(a^8*f^4))^ 
(1/2))/(a^2*f)))*(-d^6/(a^8*f^4))^(1/4))/2 + atan((d^16*(d*tan(e + f*x))^( 
1/2)*(-d^6/(256*a^8*f^4))^(1/4)*16i)/((4*d^18)/(a^2*f) - 64*a^2*d^15*f*(-d 
^6/(256*a^8*f^4))^(1/2)) - (d^13*(d*tan(e + f*x))^(1/2)*(-d^6/(256*a^8*f^4 
))^(3/4)*256i)/((4*d^18)/(a^6*f^3) - (64*d^15*(-d^6/(256*a^8*f^4))^(1/2))/ 
(a^2*f)))*(-d^6/(256*a^8*f^4))^(1/4)*2i + (d^2*(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^3)^( 
1/2)*1i)/d^2)*(-d^3)^(1/2)*1i)/(2*a^2*f)
 

Reduce [F]

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

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