3.932 \(\int \frac {e^{\tanh ^{-1}(a x)} x}{(1-a^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=27 \[ \frac {1}{2 a^2 (1-a x)}-\frac {\tanh ^{-1}(a x)}{2 a^2} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6150, 77, 207} \[ \frac {1}{2 a^2 (1-a x)}-\frac {\tanh ^{-1}(a x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x)/(1 - a^2*x^2)^(3/2),x]

[Out]

1/(2*a^2*(1 - a*x)) - ArcTanh[a*x]/(2*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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 22, normalized size = 0.81 \[ \frac {\frac {1}{1-a x}-\tanh ^{-1}(a x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*x)/(1 - a^2*x^2)^(3/2),x]

[Out]

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

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fricas [A]  time = 0.51, size = 42, normalized size = 1.56 \[ -\frac {{\left (a x - 1\right )} \log \left (a x + 1\right ) - {\left (a x - 1\right )} \log \left (a x - 1\right ) + 2}{4 \, {\left (a^{3} x - a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 37, normalized size = 1.37 \[ -\frac {\log \left ({\left | a x + 1 \right |}\right )}{4 \, a^{2}} + \frac {\log \left ({\left | a x - 1 \right |}\right )}{4 \, a^{2}} - \frac {1}{2 \, {\left (a x - 1\right )} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 36, normalized size = 1.33 \[ -\frac {1}{2 a^{2} \left (a x -1\right )}+\frac {\ln \left (a x -1\right )}{4 a^{2}}-\frac {\ln \left (a x +1\right )}{4 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 38, normalized size = 1.41 \[ -\frac {1}{2 \, {\left (a^{3} x - a^{2}\right )}} - \frac {\log \left (a x + 1\right )}{4 \, a^{2}} + \frac {\log \left (a x - 1\right )}{4 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.05, size = 22, normalized size = 0.81 \[ -\frac {1}{2\,a^2\,\left (a\,x-1\right )}-\frac {\mathrm {atanh}\left (a\,x\right )}{2\,a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a*x + 1))/(a^2*x^2 - 1)^2,x)

[Out]

- 1/(2*a^2*(a*x - 1)) - atanh(a*x)/(2*a^2)

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sympy [A]  time = 0.18, size = 32, normalized size = 1.19 \[ - \frac {1}{2 a^{3} x - 2 a^{2}} + \frac {\frac {\log {\left (x - \frac {1}{a} \right )}}{4} - \frac {\log {\left (x + \frac {1}{a} \right )}}{4}}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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