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

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Rubi [A]  time = 0.02, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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]))))

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]

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*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.00 \[ -\frac {2 \log (a x+1)}{a^2}+\frac {2 x}{a}-\frac {x^2}{2} \]

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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fricas [A]  time = 0.46, size = 25, normalized size = 1.00 \[ -\frac {a^{2} x^{2} - 4 \, a x + 4 \, \log \left (a x + 1\right )}{2 \, a^{2}} \]

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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giac [B]  time = 0.18, size = 52, normalized size = 2.08 \[ \frac {\frac {{\left (a x + 1\right )}^{2} {\left (\frac {6}{a x + 1} - 1\right )}}{a} + \frac {4 \, \log \left (\frac {{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2} {\left | a \right |}}\right )}{a}}{2 \, a} \]

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

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maple [A]  time = 0.02, size = 24, normalized size = 0.96 \[ \frac {2 x}{a}-\frac {x^{2}}{2}-\frac {2 \ln \left (a x +1\right )}{a^{2}} \]

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 = 0.31, size = 26, normalized size = 1.04 \[ -\frac {a x^{2} - 4 \, x}{2 \, a} - \frac {2 \, \log \left (a x + 1\right )}{a^{2}} \]

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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mupad [B]  time = 0.79, size = 23, normalized size = 0.92 \[ \frac {2\,x}{a}-\frac {2\,\ln \left (a\,x+1\right )}{a^2}-\frac {x^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x*(a^2*x^2 - 1))/(a*x + 1)^2,x)

[Out]

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

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sympy [A]  time = 0.09, size = 20, normalized size = 0.80 \[ - \frac {x^{2}}{2} + \frac {2 x}{a} - \frac {2 \log {\left (a x + 1 \right )}}{a^{2}} \]

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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