3.10.34 \(\int \frac {e^{\tanh ^{-1}(a x)}}{x (1-a^2 x^2)^{3/2}} \, dx\) [934]

Optimal. Leaf size=36 \[ \frac {1}{2 (1-a x)}+\log (x)-\frac {3}{4} \log (1-a x)-\frac {1}{4} \log (1+a x) \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 36, 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, 84} \begin {gather*} \frac {1}{2 (1-a x)}-\frac {3}{4} \log (1-a x)-\frac {1}{4} \log (a x+1)+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1/(2*(1 - a*x)) + Log[x] - (3*Log[1 - a*x])/4 - Log[1 + a*x]/4

Rule 84

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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

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Mathematica [A]
time = 0.05, size = 32, normalized size = 0.89 \begin {gather*} \frac {1}{2-2 a x}+\log (x)-\frac {3}{4} \log (1-a x)-\frac {1}{4} \log (1+a x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2 - 2*a*x)^(-1) + Log[x] - (3*Log[1 - a*x])/4 - Log[1 + a*x]/4

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Maple [A]
time = 0.07, size = 29, normalized size = 0.81

method	result	size
default	\(-\frac {\ln \left (a x +1\right )}{4}+\ln \left (x \right )-\frac {1}{2 \left (a x -1\right )}-\frac {3 \ln \left (a x -1\right )}{4}\)	\(29\)
risch	\(-\frac {1}{2 \left (a x -1\right )}+\ln \left (-x \right )-\frac {\ln \left (a x +1\right )}{4}-\frac {3 \ln \left (-a x +1\right )}{4}\)	\(32\)
norman	\(\frac {-\frac {1}{2} a^{2} x^{2}-\frac {1}{2} a x}{a^{2} x^{2}-1}-\frac {3 \ln \left (a x -1\right )}{4}-\frac {\ln \left (a x +1\right )}{4}+\ln \left (x \right )\)	\(45\)
meijerg	\(\frac {a \left (\frac {2 x \sqrt {-a^{2}}}{-2 a^{2} x^{2}+2}+\frac {\sqrt {-a^{2}}\, \arctanh \left (a x \right )}{a}\right )}{2 \sqrt {-a^{2}}}+\frac {a^{2} x^{2}}{-2 a^{2} x^{2}+2}-\frac {\ln \left (-a^{2} x^{2}+1\right )}{2}+\frac {1}{2}+\ln \left (x \right )+\frac {\ln \left (-a^{2}\right )}{2}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.25, size = 28, normalized size = 0.78 \begin {gather*} -\frac {1}{2 \, {\left (a x - 1\right )}} - \frac {1}{4} \, \log \left (a x + 1\right ) - \frac {3}{4} \, \log \left (a x - 1\right ) + \log \left (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,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.32, size = 45, normalized size = 1.25 \begin {gather*} -\frac {{\left (a x - 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a x - 1\right )} \log \left (a x - 1\right ) - 4 \, {\left (a x - 1\right )} \log \left (x\right ) + 2}{4 \, {\left (a x - 1\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,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.13, size = 29, normalized size = 0.81 \begin {gather*} \log {\left (x \right )} - \frac {3 \log {\left (x - \frac {1}{a} \right )}}{4} - \frac {\log {\left (x + \frac {1}{a} \right )}}{4} - \frac {1}{2 a x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.41, size = 31, normalized size = 0.86 \begin {gather*} -\frac {1}{2 \, {\left (a x - 1\right )}} - \frac {1}{4} \, \log \left ({\left | a x + 1 \right |}\right ) - \frac {3}{4} \, \log \left ({\left | a x - 1 \right |}\right ) + \log \left ({\left | x \right |}\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,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.05, size = 30, normalized size = 0.83 \begin {gather*} \ln \left (x\right )-\frac {3\,\ln \left (1-a\,x\right )}{4}-\frac {\ln \left (a\,x+1\right )}{4}-\frac {1}{2\,\left (a\,x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + 1)/(x*(a^2*x^2 - 1)^2),x)

[Out]

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

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