\(\int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^2} \, dx\) [366]

Optimal result

Integrand size = 25, antiderivative size = 222 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^2} \, dx=-\frac {\sqrt {d} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 a^2 f}-\frac {\sqrt {d} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 f}+\frac {\sqrt {d} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 \sqrt {2} a^2 f}+\frac {\sqrt {d} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}+\sqrt {d} \tan (e+f x)}\right )}{2 \sqrt {2} a^2 f}-\frac {\sqrt {d \tan (e+f x)}}{2 f \left (a^2+a^2 \tan (e+f x)\right )} \] Output:

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

Mathematica [A] (verified)

Time = 1.86 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^2} \, dx=\frac {-4 d^{3/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )-2 \sqrt {2} d^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+2 \sqrt {2} d^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )-\sqrt {2} d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )+\sqrt {2} d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )-4 d \sqrt {d \tan (e+f x)}+\frac {4 (d \tan (e+f x))^{3/2}}{1+\tan (e+f x)}}{8 a^2 d f} \] Input:

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

Output:

(-4*d^(3/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]] - 2*Sqrt[2]*d^(3/2)*ArcTa 
n[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]] + 2*Sqrt[2]*d^(3/2)*ArcTan[1 
 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]] - Sqrt[2]*d^(3/2)*Log[Sqrt[d] + 
 Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]] + Sqrt[2]*d^(3/2)*Lo 
g[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]] - 4*d*Sqr 
t[d*Tan[e + f*x]] + (4*(d*Tan[e + f*x])^(3/2))/(1 + Tan[e + f*x]))/(8*a^2* 
d*f)
 

Rubi [A] (warning: unable to verify)

Time = 1.23 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.15, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.840, Rules used = {3042, 4051, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4051

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

\(\Big \downarrow \) 4117

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

((-2*Sqrt[d]*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/f + (4*d^2*((-(ArcTan[1 
 - Sqrt[2]*Sqrt[d]*Tan[e + f*x]]/(Sqrt[2]*Sqrt[d])) + ArcTan[1 + Sqrt[2]*S 
qrt[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^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 4051

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a + b*Tan[e + f*x])^(m + 1)*((c + 
d*Tan[e + f*x])^n/(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 - 1)*Simp[a*c 
*(m + 1) - b*d*n - (b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(m + 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] && LtQ[m, -1] && GtQ[n, 0] && Int 
egerQ[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 = 1.42 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(\frac {2 d^{3} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 d^{3}}-\frac {\frac {\sqrt {d \tan \left (f x +e \right )}}{2 d \left (d \tan \left (f x +e \right )+d \right )}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 d^{\frac {3}{2}}}}{2 d}\right )}{f \,a^{2}}\)	\(200\)
default	\(\frac {2 d^{3} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 d^{3}}-\frac {\frac {\sqrt {d \tan \left (f x +e \right )}}{2 d \left (d \tan \left (f x +e \right )+d \right )}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 d^{\frac {3}{2}}}}{2 d}\right )}{f \,a^{2}}\)	\(200\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^2} \, dx=\frac {2 \, \sqrt {2} \sqrt {d} {\left (\tan \left (f x + e\right ) + 1\right )} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d}{d}\right ) + 2 \, \sqrt {2} \sqrt {d} {\left (\tan \left (f x + e\right ) + 1\right )} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} - d}{d}\right ) + \sqrt {2} \sqrt {d} {\left (\tan \left (f x + e\right ) + 1\right )} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right ) - \sqrt {2} \sqrt {d} {\left (\tan \left (f x + e\right ) + 1\right )} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right ) + 4 \, \sqrt {d} {\left (\tan \left (f x + e\right ) + 1\right )} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d} \tan \left (f x + e\right )}\right ) - 4 \, \sqrt {d \tan \left (f x + e\right )}}{8 \, {\left (a^{2} f \tan \left (f x + e\right ) + a^{2} f\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral(sqrt(d*tan(e + f*x))/(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 = 218, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^2} \, dx=-\frac {\frac {4 \, \sqrt {d \tan \left (f x + e\right )} d^{2}}{a^{2} d \tan \left (f x + e\right ) + a^{2} d} + \frac {4 \, d^{\frac {3}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{2}} - \frac {2 \, \sqrt {2} d^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right ) + 2 \, \sqrt {2} d^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right ) + \sqrt {2} d^{\frac {3}{2}} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right ) - \sqrt {2} d^{\frac {3}{2}} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{a^{2}}}{8 \, d f} \] Input:

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

Output:

-1/8*(4*sqrt(d*tan(f*x + e))*d^2/(a^2*d*tan(f*x + e) + a^2*d) + 4*d^(3/2)* 
arctan(sqrt(d*tan(f*x + e))/sqrt(d))/a^2 - (2*sqrt(2)*d^(3/2)*arctan(1/2*s 
qrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d)) + 2*sqrt(2)*d^( 
3/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d 
)) + sqrt(2)*d^(3/2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqr 
t(d) + d) - sqrt(2)*d^(3/2)*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. \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((d*tan(f*x+e))^(1/2)/(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 = 1.52 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.65 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^2} \, dx=-\frac {\mathrm {atan}\left (\frac {4\,d^{12}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^2}{a^8\,f^4}\right )}^{1/4}}{\frac {4\,d^{13}}{a^2\,f}-4\,a^2\,d^{12}\,f\,\sqrt {-\frac {d^2}{a^8\,f^4}}}+\frac {4\,d^{11}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^2}{a^8\,f^4}\right )}^{3/4}}{\frac {4\,d^{13}}{a^6\,f^3}-\frac {4\,d^{12}\,\sqrt {-\frac {d^2}{a^8\,f^4}}}{a^2\,f}}\right )\,{\left (-\frac {d^2}{a^8\,f^4}\right )}^{1/4}}{2}-\mathrm {atan}\left (\frac {d^{12}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^2}{256\,a^8\,f^4}\right )}^{1/4}\,16{}\mathrm {i}}{\frac {4\,d^{13}}{a^2\,f}+64\,a^2\,d^{12}\,f\,\sqrt {-\frac {d^2}{256\,a^8\,f^4}}}-\frac {d^{11}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (-\frac {d^2}{256\,a^8\,f^4}\right )}^{3/4}\,256{}\mathrm {i}}{\frac {4\,d^{13}}{a^6\,f^3}+\frac {64\,d^{12}\,\sqrt {-\frac {d^2}{256\,a^8\,f^4}}}{a^2\,f}}\right )\,{\left (-\frac {d^2}{256\,a^8\,f^4}\right )}^{1/4}\,2{}\mathrm {i}-\frac {d\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\left (a^2\,d\,f+a^2\,d\,f\,\mathrm {tan}\left (e+f\,x\right )\right )}+\frac {\sqrt {-d}\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{\sqrt {-d}}\right )\,1{}\mathrm {i}}{2\,a^2\,f} \] Input:

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

Output:

((-d)^(1/2)*atan(((d*tan(e + f*x))^(1/2)*1i)/(-d)^(1/2))*1i)/(2*a^2*f) - a 
tan((d^12*(d*tan(e + f*x))^(1/2)*(-d^2/(256*a^8*f^4))^(1/4)*16i)/((4*d^13) 
/(a^2*f) + 64*a^2*d^12*f*(-d^2/(256*a^8*f^4))^(1/2)) - (d^11*(d*tan(e + f* 
x))^(1/2)*(-d^2/(256*a^8*f^4))^(3/4)*256i)/((4*d^13)/(a^6*f^3) + (64*d^12* 
(-d^2/(256*a^8*f^4))^(1/2))/(a^2*f)))*(-d^2/(256*a^8*f^4))^(1/4)*2i - (d*( 
d*tan(e + f*x))^(1/2))/(2*(a^2*d*f + a^2*d*f*tan(e + f*x))) - (atan((4*d^1 
2*(d*tan(e + f*x))^(1/2)*(-d^2/(a^8*f^4))^(1/4))/((4*d^13)/(a^2*f) - 4*a^2 
*d^12*f*(-d^2/(a^8*f^4))^(1/2)) + (4*d^11*(d*tan(e + f*x))^(1/2)*(-d^2/(a^ 
8*f^4))^(3/4))/((4*d^13)/(a^6*f^3) - (4*d^12*(-d^2/(a^8*f^4))^(1/2))/(a^2* 
f)))*(-d^2/(a^8*f^4))^(1/4))/2
 

Reduce [F]

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

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

Output:

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