3.722 \(\int e^{-3 \coth ^{-1}(a x)} x^2 \sqrt{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=186 \[ \frac{x^3 \sqrt{c-a^2 c x^2}}{4 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{x^2 \sqrt{c-a^2 c x^2}}{a \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{2 x \sqrt{c-a^2 c x^2}}{a^2 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{4 \sqrt{c-a^2 c x^2}}{a^3 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{4 \sqrt{c-a^2 c x^2} \log (a x+1)}{a^4 x \sqrt{1-\frac{1}{a^2 x^2}}} \]

[Out]

(-4*Sqrt[c - a^2*c*x^2])/(a^3*Sqrt[1 - 1/(a^2*x^2)]) + (2*x*Sqrt[c - a^2*c*x^2])/(a^2*Sqrt[1 - 1/(a^2*x^2)]) -
(x^2*Sqrt[c - a^2*c*x^2])/(a*Sqrt[1 - 1/(a^2*x^2)]) + (x^3*Sqrt[c - a^2*c*x^2])/(4*Sqrt[1 - 1/(a^2*x^2)]) + (
4*Sqrt[c - a^2*c*x^2]*Log[1 + a*x])/(a^4*Sqrt[1 - 1/(a^2*x^2)]*x)

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Rubi [A]  time = 0.257982, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {6192, 6193, 88} \[ \frac{x^3 \sqrt{c-a^2 c x^2}}{4 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{x^2 \sqrt{c-a^2 c x^2}}{a \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{2 x \sqrt{c-a^2 c x^2}}{a^2 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{4 \sqrt{c-a^2 c x^2}}{a^3 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{4 \sqrt{c-a^2 c x^2} \log (a x+1)}{a^4 x \sqrt{1-\frac{1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*Sqrt[c - a^2*c*x^2])/E^(3*ArcCoth[a*x]),x]

[Out]

(-4*Sqrt[c - a^2*c*x^2])/(a^3*Sqrt[1 - 1/(a^2*x^2)]) + (2*x*Sqrt[c - a^2*c*x^2])/(a^2*Sqrt[1 - 1/(a^2*x^2)]) -
(x^2*Sqrt[c - a^2*c*x^2])/(a*Sqrt[1 - 1/(a^2*x^2)]) + (x^3*Sqrt[c - a^2*c*x^2])/(4*Sqrt[1 - 1/(a^2*x^2)]) + (
4*Sqrt[c - a^2*c*x^2]*Log[1 + a*x])/(a^4*Sqrt[1 - 1/(a^2*x^2)]*x)

Rule 6192

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c + d*x^2)^p/(x^(2*p)*(
1 - 1/(a^2*x^2))^p), Int[u*x^(2*p)*(1 - 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x]
&& EqQ[a^2*c + d, 0] &&  !IntegerQ[n/2] &&  !IntegerQ[p]

Rule 6193

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[c^p/a^(2*p), Int[(u*(-1
+ a*x)^(p - n/2)*(1 + a*x)^(p + n/2))/x^(2*p), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] &&  !
IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegersQ[2*p, p + n/2]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int e^{-3 \coth ^{-1}(a x)} x^2 \sqrt{c-a^2 c x^2} \, dx &=\frac{\sqrt{c-a^2 c x^2} \int e^{-3 \coth ^{-1}(a x)} \sqrt{1-\frac{1}{a^2 x^2}} x^3 \, dx}{\sqrt{1-\frac{1}{a^2 x^2}} x}\\ &=\frac{\sqrt{c-a^2 c x^2} \int \frac{x^2 (-1+a x)^2}{1+a x} \, dx}{a \sqrt{1-\frac{1}{a^2 x^2}} x}\\ &=\frac{\sqrt{c-a^2 c x^2} \int \left (-\frac{4}{a^2}+\frac{4 x}{a}-3 x^2+a x^3+\frac{4}{a^2 (1+a x)}\right ) \, dx}{a \sqrt{1-\frac{1}{a^2 x^2}} x}\\ &=-\frac{4 \sqrt{c-a^2 c x^2}}{a^3 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{2 x \sqrt{c-a^2 c x^2}}{a^2 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{x^2 \sqrt{c-a^2 c x^2}}{a \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{x^3 \sqrt{c-a^2 c x^2}}{4 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{4 \sqrt{c-a^2 c x^2} \log (1+a x)}{a^4 \sqrt{1-\frac{1}{a^2 x^2}} x}\\ \end{align*}

Mathematica [A]  time = 0.0510912, size = 72, normalized size = 0.39 \[ \frac{\sqrt{c-a^2 c x^2} \left (a x \left (a^3 x^3-4 a^2 x^2+8 a x-16\right )+16 \log (a x+1)\right )}{4 a^4 x \sqrt{1-\frac{1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*Sqrt[c - a^2*c*x^2])/E^(3*ArcCoth[a*x]),x]

[Out]

(Sqrt[c - a^2*c*x^2]*(a*x*(-16 + 8*a*x - 4*a^2*x^2 + a^3*x^3) + 16*Log[1 + a*x]))/(4*a^4*Sqrt[1 - 1/(a^2*x^2)]
*x)

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Maple [A]  time = 0.132, size = 83, normalized size = 0.5 \begin{align*}{\frac{ \left ({x}^{4}{a}^{4}-4\,{x}^{3}{a}^{3}+8\,{a}^{2}{x}^{2}-16\,ax+16\,\ln \left ( ax+1 \right ) \right ) \left ( ax+1 \right ) }{4\,{a}^{3} \left ( ax-1 \right ) ^{2}}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) } \left ({\frac{ax-1}{ax+1}} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(-a^2*c*x^2+c)^(1/2)*((a*x-1)/(a*x+1))^(3/2),x)

[Out]

1/4*(x^4*a^4-4*x^3*a^3+8*a^2*x^2-16*a*x+16*ln(a*x+1))*(-c*(a^2*x^2-1))^(1/2)*(a*x+1)*((a*x-1)/(a*x+1))^(3/2)/a
^3/(a*x-1)^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-a^2*c*x^2+c)^(1/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt(-a^2*c*x^2 + c)*x^2*((a*x - 1)/(a*x + 1))^(3/2), x)

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Fricas [A]  time = 1.56866, size = 112, normalized size = 0.6 \begin{align*} \frac{{\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 8 \, a^{2} x^{2} - 16 \, a x + 16 \, \log \left (a x + 1\right )\right )} \sqrt{-a^{2} c}}{4 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-a^2*c*x^2+c)^(1/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="fricas")

[Out]

1/4*(a^4*x^4 - 4*a^3*x^3 + 8*a^2*x^2 - 16*a*x + 16*log(a*x + 1))*sqrt(-a^2*c)/a^4

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(-a**2*c*x**2+c)**(1/2)*((a*x-1)/(a*x+1))**(3/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-a^2*c*x^2+c)^(1/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt(-a^2*c*x^2 + c)*x^2*((a*x - 1)/(a*x + 1))^(3/2), x)