\(\int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx\) [374]

Optimal result

Integrand size = 25, antiderivative size = 161 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx=\frac {\sqrt {d} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 f}+\frac {\sqrt {d} \text {arctanh}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 f}-\frac {\sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac {3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+a^3 \tan (e+f x)\right )} \] Output:

1/8*d^(1/2)*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/f+1/4*d^(1/2)*arctanh 
(1/2*(d^(1/2)+d^(1/2)*tan(f*x+e))*2^(1/2)/(d*tan(f*x+e))^(1/2))*2^(1/2)/a^ 
3/f-1/4*(d*tan(f*x+e))^(1/2)/a/f/(a+a*tan(f*x+e))^2-3/8*(d*tan(f*x+e))^(1/ 
2)/f/(a^3+a^3*tan(f*x+e))
 

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.16 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx=\frac {d^{3/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )-\sqrt {2} d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )+\sqrt {2} d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )+\frac {2 (d \tan (e+f x))^{3/2}}{(1+\tan (e+f x))^2}-\frac {5 d \sqrt {d \tan (e+f x)}}{1+\tan (e+f x)}}{8 a^3 d f} \] Input:

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

Output:

(d^(3/2)*ArcTan[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) 
*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]] + (2*( 
d*Tan[e + f*x])^(3/2))/(1 + Tan[e + f*x])^2 - (5*d*Sqrt[d*Tan[e + f*x]])/( 
1 + Tan[e + f*x]))/(8*a^3*d*f)
 

Rubi [A] (verified)

Time = 1.13 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.12, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 4051, 27, 3042, 4132, 3042, 4137, 27, 3042, 4015, 221, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4051

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {-3 a d \tan ^2(e+f x)+4 a d \tan (e+f x)+a d}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)^2}dx}{8 a^2}-\frac {\sqrt {d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {-3 a d \tan (e+f x)^2+4 a d \tan (e+f x)+a d}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)^2}dx}{8 a^2}-\frac {\sqrt {d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 4132

\(\displaystyle \frac {\frac {\int \frac {5 a^3 d^2-3 a^3 d^2 \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{2 a^3 d}-\frac {3 \sqrt {d \tan (e+f x)}}{f (a \tan (e+f x)+a)}}{8 a^2}-\frac {\sqrt {d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {5 a^3 d^2-3 a^3 d^2 \tan (e+f x)^2}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{2 a^3 d}-\frac {3 \sqrt {d \tan (e+f x)}}{f (a \tan (e+f x)+a)}}{8 a^2}-\frac {\sqrt {d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 4137

\(\displaystyle \frac {\frac {a^3 d^2 \int \frac {\tan ^2(e+f x)+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {\int \frac {8 \left (a^4 d^2-a^4 d^2 \tan (e+f x)\right )}{\sqrt {d \tan (e+f x)}}dx}{2 a^2}}{2 a^3 d}-\frac {3 \sqrt {d \tan (e+f x)}}{f (a \tan (e+f x)+a)}}{8 a^2}-\frac {\sqrt {d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {a^3 d^2 \int \frac {\tan ^2(e+f x)+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {4 \int \frac {a^4 d^2-a^4 d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{a^2}}{2 a^3 d}-\frac {3 \sqrt {d \tan (e+f x)}}{f (a \tan (e+f x)+a)}}{8 a^2}-\frac {\sqrt {d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^3 d^2 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {4 \int \frac {a^4 d^2-a^4 d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{a^2}}{2 a^3 d}-\frac {3 \sqrt {d \tan (e+f x)}}{f (a \tan (e+f x)+a)}}{8 a^2}-\frac {\sqrt {d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 4015

\(\displaystyle \frac {\frac {a^3 d^2 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx-\frac {8 a^6 d^4 \int \frac {1}{\cot (e+f x) \left (d^2 a^4+d^2 \tan (e+f x) a^4\right )^2-2 a^8 d^4}d\frac {d^2 a^4+d^2 \tan (e+f x) a^4}{\sqrt {d \tan (e+f x)}}}{f}}{2 a^3 d}-\frac {3 \sqrt {d \tan (e+f x)}}{f (a \tan (e+f x)+a)}}{8 a^2}-\frac {\sqrt {d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {a^3 d^2 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {4 \sqrt {2} a^2 d^{3/2} \text {arctanh}\left (\frac {a^4 d^2 \tan (e+f x)+a^4 d^2}{\sqrt {2} a^4 d^{3/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{2 a^3 d}-\frac {3 \sqrt {d \tan (e+f x)}}{f (a \tan (e+f x)+a)}}{8 a^2}-\frac {\sqrt {d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {\frac {\frac {a^3 d^2 \int \frac {1}{a \sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{f}+\frac {4 \sqrt {2} a^2 d^{3/2} \text {arctanh}\left (\frac {a^4 d^2 \tan (e+f x)+a^4 d^2}{\sqrt {2} a^4 d^{3/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{2 a^3 d}-\frac {3 \sqrt {d \tan (e+f x)}}{f (a \tan (e+f x)+a)}}{8 a^2}-\frac {\sqrt {d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {a^2 d^2 \int \frac {1}{\sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{f}+\frac {4 \sqrt {2} a^2 d^{3/2} \text {arctanh}\left (\frac {a^4 d^2 \tan (e+f x)+a^4 d^2}{\sqrt {2} a^4 d^{3/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{2 a^3 d}-\frac {3 \sqrt {d \tan (e+f x)}}{f (a \tan (e+f x)+a)}}{8 a^2}-\frac {\sqrt {d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\frac {2 a^2 d \int \frac {1}{\tan (e+f x)+1}d\sqrt {d \tan (e+f x)}}{f}+\frac {4 \sqrt {2} a^2 d^{3/2} \text {arctanh}\left (\frac {a^4 d^2 \tan (e+f x)+a^4 d^2}{\sqrt {2} a^4 d^{3/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{2 a^3 d}-\frac {3 \sqrt {d \tan (e+f x)}}{f (a \tan (e+f x)+a)}}{8 a^2}-\frac {\sqrt {d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {2 a^2 d^{3/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {4 \sqrt {2} a^2 d^{3/2} \text {arctanh}\left (\frac {a^4 d^2 \tan (e+f x)+a^4 d^2}{\sqrt {2} a^4 d^{3/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{2 a^3 d}-\frac {3 \sqrt {d \tan (e+f x)}}{f (a \tan (e+f x)+a)}}{8 a^2}-\frac {\sqrt {d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2}\)

Input:

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

Output:

-1/4*Sqrt[d*Tan[e + f*x]]/(a*f*(a + a*Tan[e + f*x])^2) + (((2*a^2*d^(3/2)* 
ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/f + (4*Sqrt[2]*a^2*d^(3/2)*ArcTanh[( 
a^4*d^2 + a^4*d^2*Tan[e + f*x])/(Sqrt[2]*a^4*d^(3/2)*Sqrt[d*Tan[e + f*x]]) 
])/f)/(2*a^3*d) - (3*Sqrt[d*Tan[e + f*x]])/(f*(a + a*Tan[e + f*x])))/(8*a^ 
2)
 

Defintions of rubi rules used

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 221

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

rule 3042

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

rule 4015

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[-2*(d^2/f)   Subst[Int[1/(2*c*d + b*x^2), x], x, (c 
- d*Tan[e + f*x])/Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && 
 EqQ[c^2 - d^2, 0]
 

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 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 4137

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*tan[(e_.) 
+ (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Sim 
p[1/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^n*Simp[a*(A - C) - (A*b - b*C)*T 
an[e + f*x], x], x], x] + Simp[(A*b^2 + 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, 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(348\) vs. \(2(132)=264\).

Time = 1.52 (sec) , antiderivative size = 349, normalized size of antiderivative = 2.17

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (-\frac {\frac {\frac {3 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}+\frac {5 d \sqrt {d \tan \left (f x +e \right )}}{4}}{\left (d \tan \left (f x +e \right )+d \right )^{2}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}}{4 d^{3}}+\frac {\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{4 d^{3}}\right )}{f \,a^{3}}\)	\(349\)
default	\(\frac {2 d^{4} \left (-\frac {\frac {\frac {3 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}+\frac {5 d \sqrt {d \tan \left (f x +e \right )}}{4}}{\left (d \tan \left (f x +e \right )+d \right )^{2}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}}{4 d^{3}}+\frac {\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{4 d^{3}}\right )}{f \,a^{3}}\)	\(349\)

Input:

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

Output:

2/f/a^3*d^4*(-1/4/d^3*((3/4*(d*tan(f*x+e))^(3/2)+5/4*d*(d*tan(f*x+e))^(1/2 
))/(d*tan(f*x+e)+d)^2-1/4/d^(1/2)*arctan((d*tan(f*x+e))^(1/2)/d^(1/2)))+1/ 
4/d^3*(1/8/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))-1/8/(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 = 383, normalized size of antiderivative = 2.38 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx=\left [-\frac {8 \, \sqrt {\frac {1}{2}} {\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) + 1\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {\frac {1}{2}} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) + 1\right )}}{d \tan \left (f x + e\right )}\right ) - {\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) + 1\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) + 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right ) + 2 \, \sqrt {d \tan \left (f x + e\right )} {\left (3 \, \tan \left (f x + e\right ) + 5\right )}}{16 \, {\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}, \frac {2 \, \sqrt {\frac {1}{2}} {\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) + 1\right )} \sqrt {d} \log \left (\frac {d \tan \left (f x + e\right )^{2} + 4 \, \sqrt {\frac {1}{2}} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} {\left (\tan \left (f x + e\right ) + 1\right )} + 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) - {\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) + 1\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d} \tan \left (f x + e\right )}\right ) - \sqrt {d \tan \left (f x + e\right )} {\left (3 \, \tan \left (f x + e\right ) + 5\right )}}{8 \, {\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}\right ] \] Input:

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

Output:

[-1/16*(8*sqrt(1/2)*(tan(f*x + e)^2 + 2*tan(f*x + e) + 1)*sqrt(-d)*arctan( 
sqrt(1/2)*sqrt(d*tan(f*x + e))*sqrt(-d)*(tan(f*x + e) + 1)/(d*tan(f*x + e) 
)) - (tan(f*x + e)^2 + 2*tan(f*x + e) + 1)*sqrt(-d)*log((d*tan(f*x + e) + 
2*sqrt(d*tan(f*x + e))*sqrt(-d) - d)/(tan(f*x + e) + 1)) + 2*sqrt(d*tan(f* 
x + e))*(3*tan(f*x + e) + 5))/(a^3*f*tan(f*x + e)^2 + 2*a^3*f*tan(f*x + e) 
 + a^3*f), 1/8*(2*sqrt(1/2)*(tan(f*x + e)^2 + 2*tan(f*x + e) + 1)*sqrt(d)* 
log((d*tan(f*x + e)^2 + 4*sqrt(1/2)*sqrt(d*tan(f*x + e))*sqrt(d)*(tan(f*x 
+ e) + 1) + 4*d*tan(f*x + e) + d)/(tan(f*x + e)^2 + 1)) - (tan(f*x + e)^2 
+ 2*tan(f*x + e) + 1)*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/(sqrt(d)*tan(f*x 
 + e))) - sqrt(d*tan(f*x + e))*(3*tan(f*x + e) + 5))/(a^3*f*tan(f*x + e)^2 
 + 2*a^3*f*tan(f*x + e) + a^3*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx=-\frac {\frac {3 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d^{2} + 5 \, \sqrt {d \tan \left (f x + e\right )} d^{3}}{a^{3} d^{2} \tan \left (f x + e\right )^{2} + 2 \, a^{3} d^{2} \tan \left (f x + e\right ) + a^{3} d^{2}} - \frac {d^{2} {\left (\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{a^{3}} - \frac {d^{\frac {3}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{3}}}{8 \, d f} \] Input:

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

Output:

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

Giac [F(-2)]

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

integrate((d*tan(f*x+e))^(1/2)/(a+a*tan(f*x+e))^3,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 = 1.67 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx=\frac {\sqrt {d}\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {\frac {3\,d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{8}+\frac {5\,d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{8}}{f\,a^3\,d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+2\,f\,a^3\,d^2\,\mathrm {tan}\left (e+f\,x\right )+f\,a^3\,d^2}+\frac {\sqrt {2}\,\sqrt {d}\,\mathrm {atanh}\left (\frac {9\,\sqrt {2}\,d^{17/2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{32\,\left (\frac {9\,d^9\,\mathrm {tan}\left (e+f\,x\right )}{32}+\frac {9\,d^9}{32}\right )}\right )}{4\,a^3\,f} \] Input:

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

Output:

(d^(1/2)*atan((d*tan(e + f*x))^(1/2)/d^(1/2)))/(8*a^3*f) - ((3*d*(d*tan(e 
+ f*x))^(3/2))/8 + (5*d^2*(d*tan(e + f*x))^(1/2))/8)/(a^3*d^2*f + a^3*d^2* 
f*tan(e + f*x)^2 + 2*a^3*d^2*f*tan(e + f*x)) + (2^(1/2)*d^(1/2)*atanh((9*2 
^(1/2)*d^(17/2)*(d*tan(e + f*x))^(1/2))/(32*((9*d^9*tan(e + f*x))/32 + (9* 
d^9)/32))))/(4*a^3*f)
 

Reduce [F]

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

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

Output:

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