3.1263 \(\int e^{-3 \tanh ^{-1}(a x)} x^2 \sqrt{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=184 \[ \frac{a x^4 \sqrt{c-a^2 c x^2}}{4 \sqrt{1-a^2 x^2}}-\frac{x^3 \sqrt{c-a^2 c x^2}}{\sqrt{1-a^2 x^2}}+\frac{2 x^2 \sqrt{c-a^2 c x^2}}{a \sqrt{1-a^2 x^2}}-\frac{4 x \sqrt{c-a^2 c x^2}}{a^2 \sqrt{1-a^2 x^2}}+\frac{4 \sqrt{c-a^2 c x^2} \log (a x+1)}{a^3 \sqrt{1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.218766, antiderivative size = 184, 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 = {6153, 6150, 88} \[ \frac{a x^4 \sqrt{c-a^2 c x^2}}{4 \sqrt{1-a^2 x^2}}-\frac{x^3 \sqrt{c-a^2 c x^2}}{\sqrt{1-a^2 x^2}}+\frac{2 x^2 \sqrt{c-a^2 c x^2}}{a \sqrt{1-a^2 x^2}}-\frac{4 x \sqrt{c-a^2 c x^2}}{a^2 \sqrt{1-a^2 x^2}}+\frac{4 \sqrt{c-a^2 c x^2} \log (a x+1)}{a^3 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6153

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

Rule 6150

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

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

Mathematica [A]  time = 0.0466033, size = 69, normalized size = 0.38 \[ \frac{\sqrt{c-a^2 c x^2} \left (-\frac{4 x}{a^2}+\frac{4 \log (a x+1)}{a^3}+\frac{a x^4}{4}+\frac{2 x^2}{a}-x^3\right )}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.085, size = 79, normalized size = 0.4 \begin{align*} -{\frac{{x}^{4}{a}^{4}-4\,{x}^{3}{a}^{3}+8\,{a}^{2}{x}^{2}-16\,ax+16\,\ln \left ( ax+1 \right ) }{ \left ( 4\,{a}^{2}{x}^{2}-4 \right ){a}^{3}}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) }\sqrt{-{a}^{2}{x}^{2}+1}} \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)^3*(-a^2*x^2+1)^(3/2),x)

[Out]

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

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

[Out]

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

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Fricas [A]  time = 2.95694, size = 807, normalized size = 4.39 \begin{align*} \left [\frac{8 \,{\left (a^{2} x^{2} - 1\right )} \sqrt{c} \log \left (\frac{a^{6} c x^{6} + 4 \, a^{5} c x^{5} + 5 \, a^{4} c x^{4} - 4 \, a^{2} c x^{2} - 4 \, a c x -{\left (a^{4} x^{4} + 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} + 4 \, a x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1} \sqrt{c} - 2 \, c}{a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1}\right ) -{\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 8 \, a^{2} x^{2} - 16 \, a x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}}{4 \,{\left (a^{5} x^{2} - a^{3}\right )}}, \frac{16 \,{\left (a^{2} x^{2} - 1\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-a^{2} c x^{2} + c}{\left (a^{2} x^{2} + 2 \, a x + 2\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-c}}{a^{4} c x^{4} + 2 \, a^{3} c x^{3} - a^{2} c x^{2} - 2 \, a c x}\right ) -{\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 8 \, a^{2} x^{2} - 16 \, a x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}}{4 \,{\left (a^{5} x^{2} - a^{3}\right )}}\right ] \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)^3*(-a^2*x^2+1)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(8*(a^2*x^2 - 1)*sqrt(c)*log((a^6*c*x^6 + 4*a^5*c*x^5 + 5*a^4*c*x^4 - 4*a^2*c*x^2 - 4*a*c*x - (a^4*x^4 +
4*a^3*x^3 + 6*a^2*x^2 + 4*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) - 2*c)/(a^4*x^4 + 2*a^3*x^3 - 2
*a*x - 1)) - (a^4*x^4 - 4*a^3*x^3 + 8*a^2*x^2 - 16*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^5*x^2 - a^
3), 1/4*(16*(a^2*x^2 - 1)*sqrt(-c)*arctan(sqrt(-a^2*c*x^2 + c)*(a^2*x^2 + 2*a*x + 2)*sqrt(-a^2*x^2 + 1)*sqrt(-
c)/(a^4*c*x^4 + 2*a^3*c*x^3 - a^2*c*x^2 - 2*a*c*x)) - (a^4*x^4 - 4*a^3*x^3 + 8*a^2*x^2 - 16*a*x)*sqrt(-a^2*c*x
^2 + c)*sqrt(-a^2*x^2 + 1))/(a^5*x^2 - a^3)]

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

[Out]

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

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

[Out]

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