3.56 \(\int \frac{x \coth ^{-1}(x)}{1-x^2} \, dx\)

Optimal. Leaf size=37 \[ \frac{1}{2} \text{PolyLog}\left (2,\frac{x+1}{x-1}\right )-\frac{1}{2} \coth ^{-1}(x)^2+\log \left (\frac{2}{1-x}\right ) \coth ^{-1}(x) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0592361, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {5985, 5919, 2402, 2315} \[ \frac{1}{2} \text{PolyLog}\left (2,\frac{x+1}{x-1}\right )-\frac{1}{2} \coth ^{-1}(x)^2+\log \left (\frac{2}{1-x}\right ) \coth ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x*ArcCoth[x])/(1 - x^2),x]

[Out]

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

Rule 5985

Int[(((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcCoth[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcCoth[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rule 5919

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcCoth[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcCoth[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 - c^2
*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rubi steps

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

Mathematica [A]  time = 0.0491352, size = 34, normalized size = 0.92 \[ \frac{1}{2} \left (\coth ^{-1}(x) \left (\coth ^{-1}(x)+2 \log \left (1-e^{-2 \coth ^{-1}(x)}\right )\right )-\text{PolyLog}\left (2,e^{-2 \coth ^{-1}(x)}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x*ArcCoth[x])/(1 - x^2),x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.035, size = 75, normalized size = 2. \begin{align*} -{\frac{{\rm arccoth} \left (x\right )\ln \left ( -1+x \right ) }{2}}-{\frac{{\rm arccoth} \left (x\right )\ln \left ( 1+x \right ) }{2}}-{\frac{ \left ( \ln \left ( -1+x \right ) \right ) ^{2}}{8}}+{\frac{1}{2}{\it dilog} \left ({\frac{1}{2}}+{\frac{x}{2}} \right ) }+{\frac{\ln \left ( -1+x \right ) }{4}\ln \left ({\frac{1}{2}}+{\frac{x}{2}} \right ) }-{\frac{1}{4} \left ( \ln \left ( 1+x \right ) -\ln \left ({\frac{1}{2}}+{\frac{x}{2}} \right ) \right ) \ln \left ( -{\frac{x}{2}}+{\frac{1}{2}} \right ) }+{\frac{ \left ( \ln \left ( 1+x \right ) \right ) ^{2}}{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*arccoth(x)*ln(-1+x)-1/2*arccoth(x)*ln(1+x)-1/8*ln(-1+x)^2+1/2*dilog(1/2+1/2*x)+1/4*ln(-1+x)*ln(1/2+1/2*x)
-1/4*(ln(1+x)-ln(1/2+1/2*x))*ln(-1/2*x+1/2)+1/8*ln(1+x)^2

________________________________________________________________________________________

Maxima [B]  time = 0.96325, size = 103, normalized size = 2.78 \begin{align*} \frac{1}{4} \,{\left (\log \left (x + 1\right ) - \log \left (x - 1\right )\right )} \log \left (x^{2} - 1\right ) - \frac{1}{2} \, \operatorname{arcoth}\left (x\right ) \log \left (x^{2} - 1\right ) - \frac{1}{8} \, \log \left (x + 1\right )^{2} - \frac{1}{4} \, \log \left (x + 1\right ) \log \left (x - 1\right ) + \frac{1}{8} \, \log \left (x - 1\right )^{2} + \frac{1}{2} \, \log \left (x - 1\right ) \log \left (\frac{1}{2} \, x + \frac{1}{2}\right ) + \frac{1}{2} \,{\rm Li}_2\left (-\frac{1}{2} \, x + \frac{1}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(log(x + 1) - log(x - 1))*log(x^2 - 1) - 1/2*arccoth(x)*log(x^2 - 1) - 1/8*log(x + 1)^2 - 1/4*log(x + 1)*l
og(x - 1) + 1/8*log(x - 1)^2 + 1/2*log(x - 1)*log(1/2*x + 1/2) + 1/2*dilog(-1/2*x + 1/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-x*arccoth(x)/(x^2 - 1), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(x*acoth(x)/(x**2 - 1), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-x*arccoth(x)/(x^2 - 1), x)