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

Optimal. Leaf size=34 \[ -\frac{2 x}{a^2}-\frac{2 \log (1-a x)}{a^3}-\frac{x^2}{a}-\frac{x^3}{3} \]

[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 = 0.0339557, antiderivative size = 34, 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} \[ -\frac{2 x}{a^2}-\frac{2 \log (1-a x)}{a^3}-\frac{x^2}{a}-\frac{x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a*x])*x^2,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 = 0.0134466, size = 34, normalized size = 1. \[ -\frac{2 x}{a^2}-\frac{2 \log (1-a x)}{a^3}-\frac{x^2}{a}-\frac{x^3}{3} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTanh[a*x])*x^2,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 = 0.027, size = 32, normalized size = 0.9 \begin{align*} -{\frac{{x}^{3}}{3}}-{\frac{{x}^{2}}{a}}-2\,{\frac{x}{{a}^{2}}}-2\,{\frac{\ln \left ( ax-1 \right ) }{{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.960327, size = 46, normalized size = 1.35 \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((a*x+1)^2/(-a^2*x^2+1)*x^2,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 = 1.68955, size = 77, normalized size = 2.26 \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((a*x+1)^2/(-a^2*x^2+1)*x^2,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 = 0.38483, size = 29, normalized size = 0.85 \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((a*x+1)**2/(-a**2*x**2+1)*x**2,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 = 1.17959, size = 51, normalized size = 1.5 \begin{align*} -\frac{a^{3} x^{3} + 3 \, a^{2} x^{2} + 6 \, a x}{3 \, a^{3}} - \frac{2 \, \log \left ({\left | a x - 1 \right |}\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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