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

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Rubi [A]  time = 0.12, antiderivative size = 63, 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} \[ \frac {a^2}{2 (1-a x)}+2 a^2 \log (x)-\frac {7}{4} a^2 \log (1-a x)-\frac {1}{4} a^2 \log (a x+1)-\frac {a}{x}-\frac {1}{2 x^2} \]

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 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^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*}

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Mathematica [A]  time = 0.05, 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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fricas [A]  time = 0.62, size = 96, normalized size = 1.52 \[ -\frac {6 \, a^{2} x^{2} - 2 \, a x + {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (a x + 1\right ) + 7 \, {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (a x - 1\right ) - 8 \, {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \relax (x) - 2}{4 \, {\left (a x^{3} - x^{2}\right )}} \]

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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giac [A]  time = 0.19, size = 59, normalized size = 0.94 \[ -\frac {1}{4} \, a^{2} \log \left ({\left | a x + 1 \right |}\right ) - \frac {7}{4} \, a^{2} \log \left ({\left | a x - 1 \right |}\right ) + 2 \, a^{2} \log \left ({\left | x \right |}\right ) - \frac {3 \, a^{2} x^{2} - a x - 1}{2 \, {\left (a x - 1\right )} x^{2}} \]

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)

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maple [A]  time = 0.04, size = 54, normalized size = 0.86 \[ -\frac {1}{2 x^{2}}-\frac {a}{x}+2 a^{2} \ln \relax (x )-\frac {a^{2}}{2 \left (a x -1\right )}-\frac {7 a^{2} \ln \left (a x -1\right )}{4}-\frac {a^{2} \ln \left (a x +1\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 59, normalized size = 0.94 \[ -\frac {1}{4} \, a^{2} \log \left (a x + 1\right ) - \frac {7}{4} \, a^{2} \log \left (a x - 1\right ) + 2 \, a^{2} \log \relax (x) - \frac {3 \, a^{2} x^{2} - a x - 1}{2 \, {\left (a x^{3} - x^{2}\right )}} \]

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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mupad [B]  time = 0.91, size = 58, normalized size = 0.92 \[ 2\,a^2\,\ln \relax (x)-\frac {7\,a^2\,\ln \left (a\,x-1\right )}{4}-\frac {a^2\,\ln \left (a\,x+1\right )}{4}+\frac {-\frac {3\,a^2\,x^2}{2}+\frac {a\,x}{2}+\frac {1}{2}}{a\,x^3-x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.40, size = 58, normalized size = 0.92 \[ 2 a^{2} \log {\relax (x )} - \frac {7 a^{2} \log {\left (x - \frac {1}{a} \right )}}{4} - \frac {a^{2} \log {\left (x + \frac {1}{a} \right )}}{4} + \frac {- 3 a^{2} x^{2} + a x + 1}{2 a x^{3} - 2 x^{2}} \]

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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