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

Optimal. Leaf size=37 \[ \frac {1}{2} \text {Li}_2\left (\frac {x+1}{x-1}\right )-\frac {1}{2} \coth ^{-1}(x)^2+\log \left (\frac {2}{1-x}\right ) \coth ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 37, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.05, size = 34, normalized size = 0.92 \[ \frac {1}{2} \left (\coth ^{-1}(x) \left (\coth ^{-1}(x)+2 \log \left (1-e^{-2 \coth ^{-1}(x)}\right )\right )-\text {Li}_2\left (e^{-2 \coth ^{-1}(x)}\right )\right ) \]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {x \operatorname {arcoth}\relax (x)}{x^{2} - 1}, x\right ) \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {x \operatorname {arcoth}\relax (x)}{x^{2} - 1}\,{d x} \]

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)

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maple [B]  time = 0.05, size = 75, normalized size = 2.03 \[ -\frac {\mathrm {arccoth}\relax (x ) \ln \left (-1+x \right )}{2}-\frac {\mathrm {arccoth}\relax (x ) \ln \left (1+x \right )}{2}-\frac {\ln \left (-1+x \right )^{2}}{8}+\frac {\dilog \left (\frac {1}{2}+\frac {x}{2}\right )}{2}+\frac {\ln \left (-1+x \right ) \ln \left (\frac {1}{2}+\frac {x}{2}\right )}{4}+\frac {\ln \left (1+x \right )^{2}}{8}-\frac {\left (\ln \left (1+x \right )-\ln \left (\frac {1}{2}+\frac {x}{2}\right )\right ) \ln \left (-\frac {x}{2}+\frac {1}{2}\right )}{4} \]

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

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maxima [B]  time = 0.30, size = 76, normalized size = 2.05 \[ \frac {1}{4} \, {\left (\log \left (x + 1\right ) - \log \left (x - 1\right )\right )} \log \left (x^{2} - 1\right ) - \frac {1}{2} \, \operatorname {arcoth}\relax (x) \log \left (x^{2} - 1\right ) - \frac {1}{8} \, \log \left (x + 1\right )^{2} - \frac {1}{4} \, \log \left (x + 1\right ) \log \left (x - 1\right ) + \frac {1}{8} \, \log \left (x - 1\right )^{2} + \frac {1}{2} \, \log \left (x - 1\right ) \log \left (\frac {1}{2} \, x + \frac {1}{2}\right ) + \frac {1}{2} \, {\rm Li}_2\left (-\frac {1}{2} \, x + \frac {1}{2}\right ) \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ -\int \frac {x\,\mathrm {acoth}\relax (x)}{x^2-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x*acoth(x))/(x^2 - 1),x)

[Out]

-int((x*acoth(x))/(x^2 - 1), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {x \operatorname {acoth}{\relax (x )}}{x^{2} - 1}\, dx \]

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)

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