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

Optimal. Leaf size=36 \[ \frac{1}{2 (1-a x)}-\frac{3}{4} \log (1-a x)-\frac{1}{4} \log (a x+1)+\log (x) \]

[Out]

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

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Rubi [A]  time = 0.108868, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {6150, 72} \[ \frac{1}{2 (1-a x)}-\frac{3}{4} \log (1-a x)-\frac{1}{4} \log (a x+1)+\log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

Mathematica [A]  time = 0.0308583, size = 32, normalized size = 0.89 \[ \frac{1}{2-2 a x}-\frac{3}{4} \log (1-a x)-\frac{1}{4} \log (a x+1)+\log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.036, size = 29, normalized size = 0.8 \begin{align*} \ln \left ( x \right ) -{\frac{\ln \left ( ax+1 \right ) }{4}}-{\frac{1}{2\,ax-2}}-{\frac{3\,\ln \left ( ax-1 \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.944727, size = 38, normalized size = 1.06 \begin{align*} -\frac{1}{2 \,{\left (a x - 1\right )}} - \frac{1}{4} \, \log \left (a x + 1\right ) - \frac{3}{4} \, \log \left (a x - 1\right ) + \log \left (x\right ) \end{align*}

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

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Fricas [A]  time = 1.83539, size = 126, normalized size = 3.5 \begin{align*} -\frac{{\left (a x - 1\right )} \log \left (a x + 1\right ) + 3 \,{\left (a x - 1\right )} \log \left (a x - 1\right ) - 4 \,{\left (a x - 1\right )} \log \left (x\right ) + 2}{4 \,{\left (a x - 1\right )}} \end{align*}

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

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Sympy [A]  time = 0.419124, size = 29, normalized size = 0.81 \begin{align*} \log{\left (x \right )} - \frac{3 \log{\left (x - \frac{1}{a} \right )}}{4} - \frac{\log{\left (x + \frac{1}{a} \right )}}{4} - \frac{1}{2 a x - 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13694, size = 42, normalized size = 1.17 \begin{align*} -\frac{1}{2 \,{\left (a x - 1\right )}} - \frac{1}{4} \, \log \left ({\left | a x + 1 \right |}\right ) - \frac{3}{4} \, \log \left ({\left | a x - 1 \right |}\right ) + \log \left ({\left | x \right |}\right ) \end{align*}

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]

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