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

Optimal. Leaf size=19 \[ -\frac{\log (1-a x)}{a^2}-\frac{x}{a} \]

[Out]

-(x/a) - Log[1 - a*x]/a^2

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Rubi [A]  time = 0.0622626, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {6150, 43} \[ -\frac{\log (1-a x)}{a^2}-\frac{x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/a) - Log[1 - a*x]/a^2

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])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0129541, size = 19, normalized size = 1. \[ -\frac{\log (1-a x)}{a^2}-\frac{x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/a) - Log[1 - a*x]/a^2

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Maple [A]  time = 0.026, size = 19, normalized size = 1. \begin{align*} -{\frac{x}{a}}-{\frac{\ln \left ( ax-1 \right ) }{{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.944947, size = 24, normalized size = 1.26 \begin{align*} -\frac{x}{a} - \frac{\log \left (a x - 1\right )}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/a - log(a*x - 1)/a^2

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Fricas [A]  time = 1.71153, size = 36, normalized size = 1.89 \begin{align*} -\frac{a x + \log \left (a x - 1\right )}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*x + log(a*x - 1))/a^2

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Sympy [A]  time = 0.247795, size = 14, normalized size = 0.74 \begin{align*} - \frac{x}{a} - \frac{\log{\left (a x - 1 \right )}}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/a - log(a*x - 1)/a**2

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Giac [A]  time = 1.17948, size = 26, normalized size = 1.37 \begin{align*} -\frac{x}{a} - \frac{\log \left ({\left | a x - 1 \right |}\right )}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/a - log(abs(a*x - 1))/a^2