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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3649, 3730, 3735, 3613, 214, 3715, 65, 211} \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx=\frac {\sqrt {d} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 f}+\frac {\sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} \tan (e+f x)+\sqrt {d}}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 f}-\frac {3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3 \tan (e+f x)+a^3\right )}-\frac {\sqrt {d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2} \]

[In]

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

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 211

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

Rule 214

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 3613

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-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 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[b*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^n/(f*(m + 1)*(a^2 + b^2))), x] + Dist[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] && IntegerQ[2*m]

Rule 3715

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[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 3730

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Ta
n[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] + Dist[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)*Tan[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] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3735

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac {\int \frac {-\frac {a d}{2}-2 a d \tan (e+f x)+\frac {3}{2} a d \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2} \, dx}{4 a^2} \\ & = -\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 )}-\frac {\int \frac {-\frac {5}{2} a^3 d^2+\frac {3}{2} a^3 d^2 \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{8 a^5 d} \\ & = -\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 )}-\frac {\int \frac {-4 a^4 d^2+4 a^4 d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{16 a^7 d}+\frac {d \int \frac {1+\tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{16 a^2} \\ & = -\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 )}+\frac {d \text {Subst}\left (\int \frac {1}{\sqrt {d x} (a+a x)} \, dx,x,\tan (e+f x)\right )}{16 a^2 f}+\frac {\left (2 a d^3\right ) \text {Subst}\left (\int \frac {1}{-32 a^8 d^4+d x^2} \, dx,x,\frac {-4 a^4 d^2-4 a^4 d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}\right )}{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 )}+\frac {\text {Subst}\left (\int \frac {1}{a+\frac {a x^2}{d}} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 a^2 f} \\ & = \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 )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.68 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.16 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx=\frac {d^{3/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )-\sqrt {2} d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )+\sqrt {2} d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )+\frac {2 (d \tan (e+f x))^{3/2}}{(1+\tan (e+f x))^2}-\frac {5 d \sqrt {d \tan (e+f x)}}{1+\tan (e+f x)}}{8 a^3 d f} \]

[In]

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

[Out]

(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]*S
qrt[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)

Maple [B] (verified)

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

Time = 0.92 (sec) , antiderivative size = 349, normalized size of antiderivative = 2.17

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 392, normalized size of antiderivative = 2.43 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx=\left [-\frac {4 \, {\left (\sqrt {2} \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} \tan \left (f x + e\right ) + \sqrt {2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )} {\left (\sqrt {2} \tan \left (f x + e\right ) + \sqrt {2}\right )} \sqrt {-d}}{2 \, d \tan \left (f x + e\right )}\right ) - {\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) + 1\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) + 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right ) + 2 \, \sqrt {d \tan \left (f x + e\right )} {\left (3 \, \tan \left (f x + e\right ) + 5\right )}}{16 \, {\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}, \frac {{\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) + 1\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right ) + {\left (\sqrt {2} \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} \tan \left (f x + e\right ) + \sqrt {2}\right )} \sqrt {d} \log \left (\frac {d \tan \left (f x + e\right )^{2} + 2 \, \sqrt {d \tan \left (f x + e\right )} {\left (\sqrt {2} \tan \left (f x + e\right ) + \sqrt {2}\right )} \sqrt {d} + 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) - \sqrt {d \tan \left (f x + e\right )} {\left (3 \, \tan \left (f x + e\right ) + 5\right )}}{8 \, {\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}\right ] \]

[In]

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

[Out]

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

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

[In]

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

[Out]

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)

none

Time = 0.33 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx=-\frac {\frac {3 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d^{2} + 5 \, \sqrt {d \tan \left (f x + e\right )} d^{3}}{a^{3} d^{2} \tan \left (f x + e\right )^{2} + 2 \, a^{3} d^{2} \tan \left (f x + e\right ) + a^{3} d^{2}} - \frac {d^{2} {\left (\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{a^{3}} - \frac {d^{\frac {3}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{3}}}{8 \, d f} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.64 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx=\frac {\sqrt {d}\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {\frac {3\,d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{8}+\frac {5\,d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{8}}{f\,a^3\,d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+2\,f\,a^3\,d^2\,\mathrm {tan}\left (e+f\,x\right )+f\,a^3\,d^2}+\frac {\sqrt {2}\,\sqrt {d}\,\mathrm {atanh}\left (\frac {9\,\sqrt {2}\,d^{17/2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{32\,\left (\frac {9\,d^9\,\mathrm {tan}\left (e+f\,x\right )}{32}+\frac {9\,d^9}{32}\right )}\right )}{4\,a^3\,f} \]

[In]

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

[Out]

(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)