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

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Rubi [A]  time = 0.09, antiderivative size = 32, normalized size of antiderivative = 1.00, 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 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])

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^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*}

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Mathematica [A]  time = 0.01, size = 32, normalized size = 1.00 \[ a^2 \log (x)-a^2 \log (1-a x)-\frac {a}{x}-\frac {1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 35, normalized size = 1.09 \[ -\frac {2 \, a^{2} x^{2} \log \left (a x - 1\right ) - 2 \, a^{2} x^{2} \log \relax (x) + 2 \, a x + 1}{2 \, x^{2}} \]

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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giac [A]  time = 0.14, size = 31, normalized size = 0.97 \[ -a^{2} \log \left ({\left | a x - 1 \right |}\right ) + a^{2} \log \left ({\left | x \right |}\right ) - \frac {2 \, a x + 1}{2 \, x^{2}} \]

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

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maple [A]  time = 0.03, size = 30, normalized size = 0.94 \[ -\frac {1}{2 x^{2}}-\frac {a}{x}+a^{2} \ln \relax (x )-a^{2} \ln \left (a x -1\right ) \]

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 = 0.31, size = 29, normalized size = 0.91 \[ -a^{2} \log \left (a x - 1\right ) + a^{2} \log \relax (x) - \frac {2 \, a x + 1}{2 \, x^{2}} \]

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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mupad [B]  time = 0.05, size = 23, normalized size = 0.72 \[ 2\,a^2\,\mathrm {atanh}\left (2\,a\,x-1\right )-\frac {a\,x+\frac {1}{2}}{x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a*x + 1)/(x^3*(a^2*x^2 - 1)),x)

[Out]

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

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sympy [A]  time = 0.17, size = 26, normalized size = 0.81 \[ - a^{2} \left (- \log {\relax (x )} + \log {\left (x - \frac {1}{a} \right )}\right ) - \frac {2 a x + 1}{2 x^{2}} \]

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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