3.35 \(\int e^{-\tanh ^{-1}(a x)} x^2 \, dx\)

Optimal. Leaf size=73 \[ -\frac{\left (1-a^2 x^2\right )^{3/2}}{3 a^3}-\frac{x \sqrt{1-a^2 x^2}}{2 a^2}+\frac{\sqrt{1-a^2 x^2}}{a^3}+\frac{\sin ^{-1}(a x)}{2 a^3} \]

[Out]

Sqrt[1 - a^2*x^2]/a^3 - (x*Sqrt[1 - a^2*x^2])/(2*a^2) - (1 - a^2*x^2)^(3/2)/(3*a^3) + ArcSin[a*x]/(2*a^3)

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Rubi [A]  time = 0.055245, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {6124, 797, 641, 195, 216} \[ -\frac{\left (1-a^2 x^2\right )^{3/2}}{3 a^3}-\frac{x \sqrt{1-a^2 x^2}}{2 a^2}+\frac{\sqrt{1-a^2 x^2}}{a^3}+\frac{\sin ^{-1}(a x)}{2 a^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2/E^ArcTanh[a*x],x]

[Out]

Sqrt[1 - a^2*x^2]/a^3 - (x*Sqrt[1 - a^2*x^2])/(2*a^2) - (1 - a^2*x^2)^(3/2)/(3*a^3) + ArcSin[a*x]/(2*a^3)

Rule 6124

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 + a*x)^((n + 1)/2)/((1 - a*x)^((n - 1)/
2)*Sqrt[1 - a^2*x^2])), x] /; FreeQ[{a, m}, x] && IntegerQ[(n - 1)/2]

Rule 797

Int[(x_)^2*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/c, Int[(f + g*x)*(a + c*x^2)^(p
+ 1), x], x] - Dist[a/c, Int[(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && EqQ[a*g^2 + f^2*
c, 0]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 216

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

Rubi steps

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

Mathematica [A]  time = 0.0316323, size = 43, normalized size = 0.59 \[ \frac{\sqrt{1-a^2 x^2} \left (2 a^2 x^2-3 a x+4\right )+3 \sin ^{-1}(a x)}{6 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/E^ArcTanh[a*x],x]

[Out]

(Sqrt[1 - a^2*x^2]*(4 - 3*a*x + 2*a^2*x^2) + 3*ArcSin[a*x])/(6*a^3)

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Maple [B]  time = 0.038, size = 134, normalized size = 1.8 \begin{align*} -{\frac{1}{3\,{a}^{3}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}-{\frac{x}{2\,{a}^{2}}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{1}{2\,{a}^{2}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{1}{{a}^{3}}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}+{\frac{1}{{a}^{2}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-a^2*x^2+1)^(3/2)/a^3-1/2*x*(-a^2*x^2+1)^(1/2)/a^2-1/2/a^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)
^(1/2))+1/a^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)+1/a^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x
+1/a))^(1/2))

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Maxima [A]  time = 1.43345, size = 82, normalized size = 1.12 \begin{align*} -\frac{\sqrt{-a^{2} x^{2} + 1} x}{2 \, a^{2}} - \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{3 \, a^{3}} + \frac{\arcsin \left (a x\right )}{2 \, a^{3}} + \frac{\sqrt{-a^{2} x^{2} + 1}}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(-a^2*x^2 + 1)*x/a^2 - 1/3*(-a^2*x^2 + 1)^(3/2)/a^3 + 1/2*arcsin(a*x)/a^3 + sqrt(-a^2*x^2 + 1)/a^3

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Fricas [A]  time = 1.85855, size = 131, normalized size = 1.79 \begin{align*} \frac{{\left (2 \, a^{2} x^{2} - 3 \, a x + 4\right )} \sqrt{-a^{2} x^{2} + 1} - 6 \, \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right )}{6 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((2*a^2*x^2 - 3*a*x + 4)*sqrt(-a^2*x^2 + 1) - 6*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)))/a^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17102, size = 68, normalized size = 0.93 \begin{align*} \frac{1}{6} \, \sqrt{-a^{2} x^{2} + 1}{\left (x{\left (\frac{2 \, x}{a} - \frac{3}{a^{2}}\right )} + \frac{4}{a^{3}}\right )} + \frac{\arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{2 \, a^{2}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(-a^2*x^2 + 1)*(x*(2*x/a - 3/a^2) + 4/a^3) + 1/2*arcsin(a*x)*sgn(a)/(a^2*abs(a))