3.464 \(\int \frac{1}{(d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}})^{3/2}} \, dx\)

Optimal. Leaf size=158 \[ -\frac{a f^2 \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{2 d^2 e \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x\right )}-\frac{\frac{a f^2}{d^2}+1}{e \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}+\frac{3 a f^2 \tanh ^{-1}\left (\frac{\sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{d}}\right )}{2 d^{5/2} e} \]

[Out]

-((1 + (a*f^2)/d^2)/(e*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]])) - (a*f^2*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2
)/f^2]])/(2*d^2*e*(e*x + f*Sqrt[a + (e^2*x^2)/f^2])) + (3*a*f^2*ArcTanh[Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^
2]]/Sqrt[d]])/(2*d^(5/2)*e)

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Rubi [A]  time = 0.155884, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {2117, 897, 1259, 453, 206} \[ -\frac{a f^2 \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{2 d^2 e \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x\right )}-\frac{\frac{a f^2}{d^2}+1}{e \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}+\frac{3 a f^2 \tanh ^{-1}\left (\frac{\sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{d}}\right )}{2 d^{5/2} e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 + (a*f^2)/d^2)/(e*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]])) - (a*f^2*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2
)/f^2]])/(2*d^2*e*(e*x + f*Sqrt[a + (e^2*x^2)/f^2])) + (3*a*f^2*ArcTanh[Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^
2]]/Sqrt[d]])/(2*d^(5/2)*e)

Rule 2117

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Dist[1/(2*
e), Subst[Int[((g + h*x^n)^p*(d^2 + a*f^2 - 2*d*x + x^2))/(d - x)^2, x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /
; FreeQ[{a, c, d, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1259

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{\left (d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{d^2+a f^2-2 d x+x^2}{(d-x)^2 x^{3/2}} \, dx,x,d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )}{2 e}\\ &=\frac{\operatorname{Subst}\left (\int \frac{d^2+a f^2-2 d x^2+x^4}{x^2 \left (d-x^2\right )^2} \, dx,x,\sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}\right )}{e}\\ &=-\frac{a f^2 \sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}{2 d^2 e \left (e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )}-\frac{\operatorname{Subst}\left (\int \frac{-2 d \left (d^2+a f^2\right )+\left (2 d^2-a f^2\right ) x^2}{x^2 \left (d-x^2\right )} \, dx,x,\sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}\right )}{2 d^2 e}\\ &=-\frac{d^2+a f^2}{d^2 e \sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}-\frac{a f^2 \sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}{2 d^2 e \left (e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )}+\frac{\left (3 a f^2\right ) \operatorname{Subst}\left (\int \frac{1}{d-x^2} \, dx,x,\sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}\right )}{2 d^2 e}\\ &=-\frac{d^2+a f^2}{d^2 e \sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}-\frac{a f^2 \sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}{2 d^2 e \left (e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )}+\frac{3 a f^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}{\sqrt{d}}\right )}{2 d^{5/2} e}\\ \end{align*}

Mathematica [A]  time = 0.408244, size = 167, normalized size = 1.06 \[ \frac{\frac{-2 d^2 \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x\right )-a f^2 \left (3 f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+3 e x\right )}{d^2 \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x\right ) \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}+\frac{3 a f^2 \tanh ^{-1}\left (\frac{\sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{d}}\right )}{d^{5/2}}}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*d^2*(e*x + f*Sqrt[a + (e^2*x^2)/f^2]) - a*f^2*(d + 3*e*x + 3*f*Sqrt[a + (e^2*x^2)/f^2]))/(d^2*(e*x + f*Sq
rt[a + (e^2*x^2)/f^2])*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]]) + (3*a*f^2*ArcTanh[Sqrt[d + e*x + f*Sqrt[a +
(e^2*x^2)/f^2]]/Sqrt[d]])/d^(5/2))/(2*e)

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Maple [F]  time = 0.01, size = 0, normalized size = 0. \begin{align*} \int \left ( d+ex+f\sqrt{a+{\frac{{e}^{2}{x}^{2}}{{f}^{2}}}} \right ) ^{-{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (e x + \sqrt{\frac{e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + sqrt(e^2*x^2/f^2 + a)*f + d)^(-3/2), x)

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Fricas [A]  time = 1.38619, size = 1037, normalized size = 6.56 \begin{align*} \left [\frac{3 \,{\left (a^{2} f^{4} - 2 \, a d e f^{2} x - a d^{2} f^{2}\right )} \sqrt{d} \log \left (a f^{2} - 2 \, d e x + 2 \, d f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}} - 2 \,{\left (\sqrt{d} e x - \sqrt{d} f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt{e x + f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}} + d}\right ) - 2 \,{\left (2 \, d^{2} e^{2} x^{2} - 2 \, a d^{2} f^{2} - 2 \, d^{4} -{\left (3 \, a d e f^{2} + d^{3} e\right )} x +{\left (3 \, a d f^{3} - 2 \, d^{2} e f x + d^{3} f\right )} \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt{e x + f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{4 \,{\left (a d^{3} e f^{2} - 2 \, d^{4} e^{2} x - d^{5} e\right )}}, -\frac{3 \,{\left (a^{2} f^{4} - 2 \, a d e f^{2} x - a d^{2} f^{2}\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{e x + f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}} + d} \sqrt{-d}}{d}\right ) +{\left (2 \, d^{2} e^{2} x^{2} - 2 \, a d^{2} f^{2} - 2 \, d^{4} -{\left (3 \, a d e f^{2} + d^{3} e\right )} x +{\left (3 \, a d f^{3} - 2 \, d^{2} e f x + d^{3} f\right )} \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt{e x + f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{2 \,{\left (a d^{3} e f^{2} - 2 \, d^{4} e^{2} x - d^{5} e\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d + e x + f \sqrt{a + \frac{e^{2} x^{2}}{f^{2}}}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (e x + \sqrt{\frac{e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + sqrt(e^2*x^2/f^2 + a)*f + d)^(-3/2), x)