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

Optimal. Leaf size=32 \[ a^2 \log (x)-a^2 \log (1-a x)-\frac{a}{x}-\frac{1}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.090373, antiderivative size = 32, 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, 44} \[ a^2 \log (x)-a^2 \log (1-a x)-\frac{a}{x}-\frac{1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0136628, size = 32, normalized size = 1. \[ a^2 \log (x)-a^2 \log (1-a x)-\frac{a}{x}-\frac{1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.031, size = 30, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,{x}^{2}}}-{\frac{a}{x}}+{a}^{2}\ln \left ( x \right ) -{a}^{2}\ln \left ( ax-1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)/x^3,x)

[Out]

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

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Maxima [A]  time = 0.955309, size = 39, normalized size = 1.22 \begin{align*} -a^{2} \log \left (a x - 1\right ) + a^{2} \log \left (x\right ) - \frac{2 \, a x + 1}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.67278, size = 89, normalized size = 2.78 \begin{align*} -\frac{2 \, a^{2} x^{2} \log \left (a x - 1\right ) - 2 \, a^{2} x^{2} \log \left (x\right ) + 2 \, a x + 1}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.311324, size = 26, normalized size = 0.81 \begin{align*} - a^{2} \left (- \log{\left (x \right )} + \log{\left (x - \frac{1}{a} \right )}\right ) - \frac{2 a x + 1}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)/x**3,x)

[Out]

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

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Giac [A]  time = 1.1655, size = 42, normalized size = 1.31 \begin{align*} -a^{2} \log \left ({\left | a x - 1 \right |}\right ) + a^{2} \log \left ({\left | x \right |}\right ) - \frac{2 \, a x + 1}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)/x^3,x, algorithm="giac")

[Out]

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