3.10.35 \(\int \frac {e^{\tanh ^{-1}(a x)}}{x^2 (1-a^2 x^2)^{3/2}} \, dx\) [935]

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

[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 = 0.08, 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 = {6285, 90} \begin {gather*} \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} \end {gather*}

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 90

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 6285

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 = 0.05, size = 46, normalized size = 1.00 \begin {gather*} -\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 {gather*}

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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Maple [A]
time = 0.06, size = 39, normalized size = 0.85

method	result	size
default	\(\frac {a \ln \left (a x +1\right )}{4}-\frac {1}{x}+a \ln \left (x \right )-\frac {a}{2 \left (a x -1\right )}-\frac {5 a \ln \left (a x -1\right )}{4}\)	\(39\)
risch	\(\frac {-\frac {3 a x}{2}+1}{x \left (a x -1\right )}-\frac {5 a \ln \left (-a x +1\right )}{4}+\frac {a \ln \left (a x +1\right )}{4}+a \ln \left (-x \right )\)	\(44\)
norman	\(\frac {1-\frac {1}{2} a^{3} x^{3}-\frac {3}{2} a^{2} x^{2}}{x \left (a^{2} x^{2}-1\right )}+a \ln \left (x \right )-\frac {5 a \ln \left (a x -1\right )}{4}+\frac {a \ln \left (a x +1\right )}{4}\)	\(57\)
meijerg	\(\frac {a \left (\frac {2 a^{2} x^{2}}{-2 a^{2} x^{2}+2}-\ln \left (-a^{2} x^{2}+1\right )+1+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right )}{2}-\frac {a^{2} \left (-\frac {2 \left (-3 a^{2} x^{2}+2\right )}{x \sqrt {-a^{2}}\, \left (-2 a^{2} x^{2}+2\right )}+\frac {3 a \arctanh \left (a x \right )}{\sqrt {-a^{2}}}\right )}{2 \sqrt {-a^{2}}}\)	\(111\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.25, size = 42, normalized size = 0.91 \begin {gather*} \frac {1}{4} \, a \log \left (a x + 1\right ) - \frac {5}{4} \, a \log \left (a x - 1\right ) + a \log \left (x\right ) - \frac {3 \, a x - 2}{2 \, {\left (a x^{2} - x\right )}} \end {gather*}

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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Fricas [A]
time = 0.35, size = 75, normalized size = 1.63 \begin {gather*} -\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 \left (x\right ) - 4}{4 \, {\left (a x^{2} - x\right )}} \end {gather*}

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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Sympy [A]
time = 0.24, size = 42, normalized size = 0.91 \begin {gather*} a \log {\left (x \right )} - \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} \end {gather*}

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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Giac [A]
time = 0.41, size = 44, normalized size = 0.96 \begin {gather*} \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} \end {gather*}

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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Mupad [B]
time = 0.89, size = 40, normalized size = 0.87 \begin {gather*} a\,\ln \left (x\right )-\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} \end {gather*}

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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