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

Optimal. Leaf size=63 \[ \frac{a^2}{2 (1-a x)}+2 a^2 \log (x)-\frac{7}{4} a^2 \log (1-a x)-\frac{1}{4} a^2 \log (a x+1)-\frac{a}{x}-\frac{1}{2 x^2} \]

[Out]

-1/(2*x^2) - a/x + a^2/(2*(1 - a*x)) + 2*a^2*Log[x] - (7*a^2*Log[1 - a*x])/4 - (a^2*Log[1 + a*x])/4

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Rubi [A]  time = 0.118239, antiderivative size = 63, 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, 88} \[ \frac{a^2}{2 (1-a x)}+2 a^2 \log (x)-\frac{7}{4} a^2 \log (1-a x)-\frac{1}{4} a^2 \log (a x+1)-\frac{a}{x}-\frac{1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*x^2) - a/x + a^2/(2*(1 - a*x)) + 2*a^2*Log[x] - (7*a^2*Log[1 - a*x])/4 - (a^2*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 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]))

Rubi steps

\begin{align*} \int \frac{e^{\tanh ^{-1}(a x)}}{x^3 \left (1-a^2 x^2\right )^{3/2}} \, dx &=\int \frac{1}{x^3 (1-a x)^2 (1+a x)} \, dx\\ &=\int \left (\frac{1}{x^3}+\frac{a}{x^2}+\frac{2 a^2}{x}+\frac{a^3}{2 (-1+a x)^2}-\frac{7 a^3}{4 (-1+a x)}-\frac{a^3}{4 (1+a x)}\right ) \, dx\\ &=-\frac{1}{2 x^2}-\frac{a}{x}+\frac{a^2}{2 (1-a x)}+2 a^2 \log (x)-\frac{7}{4} a^2 \log (1-a x)-\frac{1}{4} a^2 \log (1+a x)\\ \end{align*}

Mathematica [A]  time = 0.0483971, size = 59, normalized size = 0.94 \[ \frac{1}{4} \left (\frac{2 a^2}{1-a x}+8 a^2 \log (x)-7 a^2 \log (1-a x)-a^2 \log (a x+1)-\frac{4 a}{x}-\frac{2}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2/x^2 - (4*a)/x + (2*a^2)/(1 - a*x) + 8*a^2*Log[x] - 7*a^2*Log[1 - a*x] - a^2*Log[1 + a*x])/4

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Maple [A]  time = 0.039, size = 54, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,{x}^{2}}}-{\frac{a}{x}}+2\,{a}^{2}\ln \left ( x \right ) -{\frac{{a}^{2}\ln \left ( ax+1 \right ) }{4}}-{\frac{{a}^{2}}{2\,ax-2}}-{\frac{7\,{a}^{2}\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^3,x)

[Out]

-1/2/x^2-a/x+2*a^2*ln(x)-1/4*a^2*ln(a*x+1)-1/2*a^2/(a*x-1)-7/4*a^2*ln(a*x-1)

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Maxima [A]  time = 0.954497, size = 80, normalized size = 1.27 \begin{align*} -\frac{1}{4} \, a^{2} \log \left (a x + 1\right ) - \frac{7}{4} \, a^{2} \log \left (a x - 1\right ) + 2 \, a^{2} \log \left (x\right ) - \frac{3 \, a^{2} x^{2} - a x - 1}{2 \,{\left (a x^{3} - x^{2}\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^3,x, algorithm="maxima")

[Out]

-1/4*a^2*log(a*x + 1) - 7/4*a^2*log(a*x - 1) + 2*a^2*log(x) - 1/2*(3*a^2*x^2 - a*x - 1)/(a*x^3 - x^2)

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Fricas [A]  time = 1.69433, size = 198, normalized size = 3.14 \begin{align*} -\frac{6 \, a^{2} x^{2} - 2 \, a x +{\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (a x + 1\right ) + 7 \,{\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (a x - 1\right ) - 8 \,{\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (x\right ) - 2}{4 \,{\left (a x^{3} - x^{2}\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^3,x, algorithm="fricas")

[Out]

-1/4*(6*a^2*x^2 - 2*a*x + (a^3*x^3 - a^2*x^2)*log(a*x + 1) + 7*(a^3*x^3 - a^2*x^2)*log(a*x - 1) - 8*(a^3*x^3 -
a^2*x^2)*log(x) - 2)/(a*x^3 - x^2)

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Sympy [A]  time = 0.583806, size = 58, normalized size = 0.92 \begin{align*} 2 a^{2} \log{\left (x \right )} - \frac{7 a^{2} \log{\left (x - \frac{1}{a} \right )}}{4} - \frac{a^{2} \log{\left (x + \frac{1}{a} \right )}}{4} - \frac{3 a^{2} x^{2} - a x - 1}{2 a x^{3} - 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)**2/x**3,x)

[Out]

2*a**2*log(x) - 7*a**2*log(x - 1/a)/4 - a**2*log(x + 1/a)/4 - (3*a**2*x**2 - a*x - 1)/(2*a*x**3 - 2*x**2)

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Giac [A]  time = 1.17738, size = 80, normalized size = 1.27 \begin{align*} -\frac{1}{4} \, a^{2} \log \left ({\left | a x + 1 \right |}\right ) - \frac{7}{4} \, a^{2} \log \left ({\left | a x - 1 \right |}\right ) + 2 \, a^{2} \log \left ({\left | x \right |}\right ) - \frac{3 \, a^{2} x^{2} - a x - 1}{2 \,{\left (a x - 1\right )} 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^3,x, algorithm="giac")

[Out]

-1/4*a^2*log(abs(a*x + 1)) - 7/4*a^2*log(abs(a*x - 1)) + 2*a^2*log(abs(x)) - 1/2*(3*a^2*x^2 - a*x - 1)/((a*x -
1)*x^2)