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

Optimal. Leaf size=46 \[ \frac {a}{2 (1-a x)}+a \log (x)-\frac {5}{4} a \log (1-a x)+\frac {1}{4} a \log (a x+1)-\frac {1}{x} \]

[Out]

-1/x+1/2*a/(-a*x+1)+a*ln(x)-5/4*a*ln(-a*x+1)+1/4*a*ln(a*x+1)

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Rubi [A]  time = 0.11, antiderivative size = 46, 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, 88} \[ \frac {a}{2 (1-a x)}+a \log (x)-\frac {5}{4} a \log (1-a x)+\frac {1}{4} a \log (a x+1)-\frac {1}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x^(-1) + a/(2*(1 - a*x)) + a*Log[x] - (5*a*Log[1 - a*x])/4 + (a*Log[1 + a*x])/4

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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^2 \left (1-a^2 x^2\right )^{3/2}} \, dx &=\int \frac {1}{x^2 (1-a x)^2 (1+a x)} \, dx\\ &=\int \left (\frac {1}{x^2}+\frac {a}{x}+\frac {a^2}{2 (-1+a x)^2}-\frac {5 a^2}{4 (-1+a x)}+\frac {a^2}{4 (1+a x)}\right ) \, dx\\ &=-\frac {1}{x}+\frac {a}{2 (1-a x)}+a \log (x)-\frac {5}{4} a \log (1-a x)+\frac {1}{4} a \log (1+a x)\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 46, normalized size = 1.00 \[ \frac {a}{2 (1-a x)}+a \log (x)-\frac {5}{4} a \log (1-a x)+\frac {1}{4} a \log (a x+1)-\frac {1}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x^(-1) + a/(2*(1 - a*x)) + a*Log[x] - (5*a*Log[1 - a*x])/4 + (a*Log[1 + a*x])/4

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fricas [A]  time = 0.64, size = 75, normalized size = 1.63 \[ -\frac {6 \, a x - {\left (a^{2} x^{2} - a x\right )} \log \left (a x + 1\right ) + 5 \, {\left (a^{2} x^{2} - a x\right )} \log \left (a x - 1\right ) - 4 \, {\left (a^{2} x^{2} - a x\right )} \log \relax (x) - 4}{4 \, {\left (a x^{2} - x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(6*a*x - (a^2*x^2 - a*x)*log(a*x + 1) + 5*(a^2*x^2 - a*x)*log(a*x - 1) - 4*(a^2*x^2 - a*x)*log(x) - 4)/(a
*x^2 - x)

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giac [A]  time = 0.20, size = 44, normalized size = 0.96 \[ \frac {1}{4} \, a \log \left ({\left | a x + 1 \right |}\right ) - \frac {5}{4} \, a \log \left ({\left | a x - 1 \right |}\right ) + a \log \left ({\left | x \right |}\right ) - \frac {3 \, a x - 2}{2 \, {\left (a x - 1\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*log(abs(a*x + 1)) - 5/4*a*log(abs(a*x - 1)) + a*log(abs(x)) - 1/2*(3*a*x - 2)/((a*x - 1)*x)

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maple [A]  time = 0.03, size = 39, normalized size = 0.85 \[ -\frac {1}{x}+a \ln \relax (x )-\frac {a}{2 \left (a x -1\right )}-\frac {5 a \ln \left (a x -1\right )}{4}+\frac {a \ln \left (a x +1\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/x+a*ln(x)-1/2*a/(a*x-1)-5/4*a*ln(a*x-1)+1/4*a*ln(a*x+1)

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maxima [A]  time = 0.32, size = 42, normalized size = 0.91 \[ \frac {1}{4} \, a \log \left (a x + 1\right ) - \frac {5}{4} \, a \log \left (a x - 1\right ) + a \log \relax (x) - \frac {3 \, a x - 2}{2 \, {\left (a x^{2} - x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*log(a*x + 1) - 5/4*a*log(a*x - 1) + a*log(x) - 1/2*(3*a*x - 2)/(a*x^2 - x)

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mupad [B]  time = 0.89, size = 40, normalized size = 0.87 \[ a\,\ln \relax (x)-\frac {5\,a\,\ln \left (a\,x-1\right )}{4}+\frac {a\,\ln \left (a\,x+1\right )}{4}+\frac {\frac {3\,a\,x}{2}-1}{x-a\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*log(x) - (5*a*log(a*x - 1))/4 + (a*log(a*x + 1))/4 + ((3*a*x)/2 - 1)/(x - a*x^2)

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sympy [A]  time = 0.37, size = 42, normalized size = 0.91 \[ a \log {\relax (x )} - \frac {5 a \log {\left (x - \frac {1}{a} \right )}}{4} + \frac {a \log {\left (x + \frac {1}{a} \right )}}{4} + \frac {- 3 a x + 2}{2 a x^{2} - 2 x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*log(x) - 5*a*log(x - 1/a)/4 + a*log(x + 1/a)/4 + (-3*a*x + 2)/(2*a*x**2 - 2*x)

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