3.243 \(\int \frac{\log (x)}{1-x^2} \, dx\)

Optimal. Leaf size=22 \[ \frac{1}{2} \text{PolyLog}(2,-x)-\frac{1}{2} \text{PolyLog}(2,x)+\log (x) \tanh ^{-1}(x) \]

[Out]

ArcTanh[x]*Log[x] + PolyLog[2, -x]/2 - PolyLog[2, x]/2

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Rubi [A]  time = 0.0220914, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {206, 2324, 5912} \[ \frac{1}{2} \text{PolyLog}(2,-x)-\frac{1}{2} \text{PolyLog}(2,x)+\log (x) \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[Log[x]/(1 - x^2),x]

[Out]

ArcTanh[x]*Log[x] + PolyLog[2, -x]/2 - PolyLog[2, x]/2

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2324

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> With[{u = IntHide[1/(d + e*x^2),
x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[u/x, x], x]] /; FreeQ[{a, b, c, d, e, n}, x]

Rule 5912

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (-Simp[(b*PolyLog[2, -(c*x)])/2
, x] + Simp[(b*PolyLog[2, c*x])/2, x]) /; FreeQ[{a, b, c}, x]

Rubi steps

\begin{align*} \int \frac{\log (x)}{1-x^2} \, dx &=\tanh ^{-1}(x) \log (x)-\int \frac{\tanh ^{-1}(x)}{x} \, dx\\ &=\tanh ^{-1}(x) \log (x)+\frac{\text{Li}_2(-x)}{2}-\frac{\text{Li}_2(x)}{2}\\ \end{align*}

Mathematica [A]  time = 0.0055929, size = 31, normalized size = 1.41 \[ \frac{1}{2} \text{PolyLog}(2,1-x)+\frac{1}{2} \text{PolyLog}(2,-x)+\frac{1}{2} \log (x) \log (x+1) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[x]/(1 - x^2),x]

[Out]

(Log[x]*Log[1 + x])/2 + PolyLog[2, 1 - x]/2 + PolyLog[2, -x]/2

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Maple [A]  time = 0.041, size = 20, normalized size = 0.9 \begin{align*}{\frac{{\it dilog} \left ( x \right ) }{2}}+{\frac{{\it dilog} \left ( 1+x \right ) }{2}}+{\frac{\ln \left ( x \right ) \ln \left ( 1+x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(x)/(-x^2+1),x)

[Out]

1/2*dilog(x)+1/2*dilog(1+x)+1/2*ln(x)*ln(1+x)

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Maxima [B]  time = 1.17053, size = 65, normalized size = 2.95 \begin{align*} -\frac{1}{2} \, \log \left (-x\right ) \log \left (x + 1\right ) + \frac{1}{2} \,{\left (\log \left (x + 1\right ) - \log \left (x - 1\right )\right )} \log \left (x\right ) + \frac{1}{2} \, \log \left (x - 1\right ) \log \left (x\right ) - \frac{1}{2} \,{\rm Li}_2\left (x + 1\right ) + \frac{1}{2} \,{\rm Li}_2\left (-x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(-x)*log(x + 1) + 1/2*(log(x + 1) - log(x - 1))*log(x) + 1/2*log(x - 1)*log(x) - 1/2*dilog(x + 1) + 1/
2*dilog(-x + 1)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\log \left (x\right )}{x^{2} - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-log(x)/(x^2 - 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{\log{\left (x \right )}}{x^{2} - 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(log(x)/(x**2 - 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{\log \left (x\right )}{x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-log(x)/(x^2 - 1), x)