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

Optimal. Leaf size=42 \[ \frac{x^2}{a^2}-\frac{2 x}{a^3}+\frac{2 \log (a x+1)}{a^4}-\frac{2 x^3}{3 a}+\frac{x^4}{4} \]

[Out]

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

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Rubi [A]  time = 0.0570863, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6167, 6126, 77} \[ \frac{x^2}{a^2}-\frac{2 x}{a^3}+\frac{2 \log (a x+1)}{a^4}-\frac{2 x^3}{3 a}+\frac{x^4}{4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

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

Mathematica [A]  time = 0.0193023, size = 42, normalized size = 1. \[ \frac{x^2}{a^2}-\frac{2 x}{a^3}+\frac{2 \log (a x+1)}{a^4}-\frac{2 x^3}{3 a}+\frac{x^4}{4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.039, size = 39, normalized size = 0.9 \begin{align*} -2\,{\frac{x}{{a}^{3}}}+{\frac{{x}^{2}}{{a}^{2}}}-{\frac{2\,{x}^{3}}{3\,a}}+{\frac{{x}^{4}}{4}}+2\,{\frac{\ln \left ( ax+1 \right ) }{{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.985613, size = 58, normalized size = 1.38 \begin{align*} \frac{3 \, a^{3} x^{4} - 8 \, a^{2} x^{3} + 12 \, a x^{2} - 24 \, x}{12 \, a^{3}} + \frac{2 \, \log \left (a x + 1\right )}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a^3*x^4 - 8*a^2*x^3 + 12*a*x^2 - 24*x)/a^3 + 2*log(a*x + 1)/a^4

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Fricas [A]  time = 1.7463, size = 100, normalized size = 2.38 \begin{align*} \frac{3 \, a^{4} x^{4} - 8 \, a^{3} x^{3} + 12 \, a^{2} x^{2} - 24 \, a x + 24 \, \log \left (a x + 1\right )}{12 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a^4*x^4 - 8*a^3*x^3 + 12*a^2*x^2 - 24*a*x + 24*log(a*x + 1))/a^4

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Sympy [A]  time = 0.276682, size = 37, normalized size = 0.88 \begin{align*} \frac{x^{4}}{4} - \frac{2 x^{3}}{3 a} + \frac{x^{2}}{a^{2}} - \frac{2 x}{a^{3}} + \frac{2 \log{\left (a x + 1 \right )}}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**4/4 - 2*x**3/(3*a) + x**2/a**2 - 2*x/a**3 + 2*log(a*x + 1)/a**4

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Giac [A]  time = 1.13709, size = 63, normalized size = 1.5 \begin{align*} \frac{3 \, a^{4} x^{4} - 8 \, a^{3} x^{3} + 12 \, a^{2} x^{2} - 24 \, a x}{12 \, a^{4}} + \frac{2 \, \log \left ({\left | a x + 1 \right |}\right )}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a^4*x^4 - 8*a^3*x^3 + 12*a^2*x^2 - 24*a*x)/a^4 + 2*log(abs(a*x + 1))/a^4