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

Optimal. Leaf size=25 \[ -\frac{2 \log (a x+1)}{a^2}+\frac{2 x}{a}-\frac{x^2}{2} \]

[Out]

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

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Rubi [A]  time = 0.0239714, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6126, 77} \[ -\frac{2 \log (a x+1)}{a^2}+\frac{2 x}{a}-\frac{x^2}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x)/a - x^2/2 - (2*Log[1 + a*x])/a^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 \tanh ^{-1}(a x)} x \, dx &=\int \frac{x (1-a x)}{1+a x} \, dx\\ &=\int \left (\frac{2}{a}-x-\frac{2}{a (1+a x)}\right ) \, dx\\ &=\frac{2 x}{a}-\frac{x^2}{2}-\frac{2 \log (1+a x)}{a^2}\\ \end{align*}

Mathematica [A]  time = 0.0108768, size = 25, normalized size = 1. \[ -\frac{2 \log (a x+1)}{a^2}+\frac{2 x}{a}-\frac{x^2}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.032, size = 24, normalized size = 1. \begin{align*} 2\,{\frac{x}{a}}-{\frac{{x}^{2}}{2}}-2\,{\frac{\ln \left ( ax+1 \right ) }{{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.947156, size = 35, normalized size = 1.4 \begin{align*} -\frac{a x^{2} - 4 \, x}{2 \, a} - \frac{2 \, \log \left (a x + 1\right )}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.84173, size = 61, normalized size = 2.44 \begin{align*} -\frac{a^{2} x^{2} - 4 \, a x + 4 \, \log \left (a x + 1\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.286735, size = 20, normalized size = 0.8 \begin{align*} - \frac{x^{2}}{2} + \frac{2 x}{a} - \frac{2 \log{\left (a x + 1 \right )}}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**2/2 + 2*x/a - 2*log(a*x + 1)/a**2

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Giac [B]  time = 1.16934, size = 70, normalized size = 2.8 \begin{align*} \frac{\frac{{\left (a x + 1\right )}^{2}{\left (\frac{6}{a x + 1} - 1\right )}}{a} + \frac{4 \, \log \left (\frac{{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2}{\left | a \right |}}\right )}{a}}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a*x + 1)^2*(6/(a*x + 1) - 1)/a + 4*log(abs(a*x + 1)/((a*x + 1)^2*abs(a)))/a)/a