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

Optimal. Leaf size=32 \[ -\frac{2 x}{a^2}+\frac{2 \log (a x+1)}{a^3}+\frac{x^2}{a}-\frac{x^3}{3} \]

[Out]

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

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Rubi [A]  time = 0.0331909, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6126, 77} \[ -\frac{2 x}{a^2}+\frac{2 \log (a x+1)}{a^3}+\frac{x^2}{a}-\frac{x^3}{3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6126

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.0138062, size = 32, normalized size = 1. \[ -\frac{2 x}{a^2}+\frac{2 \log (a x+1)}{a^3}+\frac{x^2}{a}-\frac{x^3}{3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.03, size = 31, normalized size = 1. \begin{align*} -2\,{\frac{x}{{a}^{2}}}+{\frac{{x}^{2}}{a}}-{\frac{{x}^{3}}{3}}+2\,{\frac{\ln \left ( ax+1 \right ) }{{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.974067, size = 46, normalized size = 1.44 \begin{align*} -\frac{a^{2} x^{3} - 3 \, a x^{2} + 6 \, x}{3 \, a^{2}} + \frac{2 \, \log \left (a x + 1\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(a^2*x^3 - 3*a*x^2 + 6*x)/a^2 + 2*log(a*x + 1)/a^3

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Fricas [A]  time = 1.91128, size = 77, normalized size = 2.41 \begin{align*} -\frac{a^{3} x^{3} - 3 \, a^{2} x^{2} + 6 \, a x - 6 \, \log \left (a x + 1\right )}{3 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(a^3*x^3 - 3*a^2*x^2 + 6*a*x - 6*log(a*x + 1))/a^3

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Sympy [A]  time = 0.356007, size = 27, normalized size = 0.84 \begin{align*} - \frac{x^{3}}{3} + \frac{x^{2}}{a} - \frac{2 x}{a^{2}} + \frac{2 \log{\left (a x + 1 \right )}}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**3/3 + x**2/a - 2*x/a**2 + 2*log(a*x + 1)/a**3

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Giac [A]  time = 1.17283, size = 77, normalized size = 2.41 \begin{align*} \frac{{\left (a x + 1\right )}^{3}{\left (\frac{6}{a x + 1} - \frac{15}{{\left (a x + 1\right )}^{2}} - 1\right )}}{3 \, a^{3}} - \frac{2 \, \log \left (\frac{{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2}{\left | a \right |}}\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a*x + 1)^3*(6/(a*x + 1) - 15/(a*x + 1)^2 - 1)/a^3 - 2*log(abs(a*x + 1)/((a*x + 1)^2*abs(a)))/a^3