3.211 \(\int \log (\coth (x)) \, dx\)

Optimal. Leaf size=39 \[ -\frac{1}{2} \text{PolyLog}\left (2,-e^{2 x}\right )+\frac{1}{2} \text{PolyLog}\left (2,e^{2 x}\right )-2 x \tanh ^{-1}\left (e^{2 x}\right )+x \log (\coth (x)) \]

[Out]

-2*x*ArcTanh[E^(2*x)] + x*Log[Coth[x]] - PolyLog[2, -E^(2*x)]/2 + PolyLog[2, E^(2*x)]/2

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Rubi [A]  time = 0.0458674, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 3, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.667, Rules used = {2548, 5461, 4182, 2279, 2391} \[ -\frac{1}{2} \text{PolyLog}\left (2,-e^{2 x}\right )+\frac{1}{2} \text{PolyLog}\left (2,e^{2 x}\right )-2 x \tanh ^{-1}\left (e^{2 x}\right )+x \log (\coth (x)) \]

Antiderivative was successfully verified.

[In]

Int[Log[Coth[x]],x]

[Out]

-2*x*ArcTanh[E^(2*x)] + x*Log[Coth[x]] - PolyLog[2, -E^(2*x)]/2 + PolyLog[2, E^(2*x)]/2

Rule 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

Rule 5461

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

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

Rubi steps

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

Mathematica [A]  time = 0.0067925, size = 35, normalized size = 0.9 \[ \frac{1}{2} \text{PolyLog}(2,1-\coth (x))+\frac{1}{2} \text{PolyLog}(2,-\coth (x))+\frac{1}{2} \log (\coth (x)) \log (\coth (x)+1) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[Coth[x]],x]

[Out]

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

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Maple [A]  time = 0.013, size = 24, normalized size = 0.6 \begin{align*}{\frac{{\it dilog} \left ({\rm coth} \left (x\right ) \right ) }{2}}+{\frac{{\it dilog} \left ({\rm coth} \left (x\right )+1 \right ) }{2}}+{\frac{\ln \left ({\rm coth} \left (x\right ) \right ) \ln \left ({\rm coth} \left (x\right )+1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(coth(x)),x)

[Out]

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

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Maxima [A]  time = 1.5903, size = 66, normalized size = 1.69 \begin{align*} -x \log \left (e^{\left (2 \, x\right )} + 1\right ) + x \log \left (e^{x} + 1\right ) + x \log \left (-e^{x} + 1\right ) + x \log \left (\coth \left (x\right )\right ) - \frac{1}{2} \,{\rm Li}_2\left (-e^{\left (2 \, x\right )}\right ) +{\rm Li}_2\left (-e^{x}\right ) +{\rm Li}_2\left (e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(coth(x)),x, algorithm="maxima")

[Out]

-x*log(e^(2*x) + 1) + x*log(e^x + 1) + x*log(-e^x + 1) + x*log(coth(x)) - 1/2*dilog(-e^(2*x)) + dilog(-e^x) +
dilog(e^x)

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Fricas [C]  time = 1.98114, size = 373, normalized size = 9.56 \begin{align*} x \log \left (\frac{\cosh \left (x\right )}{\sinh \left (x\right )}\right ) + x \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) - x \log \left (i \, \cosh \left (x\right ) + i \, \sinh \left (x\right ) + 1\right ) - x \log \left (-i \, \cosh \left (x\right ) - i \, \sinh \left (x\right ) + 1\right ) + x \log \left (-\cosh \left (x\right ) - \sinh \left (x\right ) + 1\right ) +{\rm Li}_2\left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) -{\rm Li}_2\left (i \, \cosh \left (x\right ) + i \, \sinh \left (x\right )\right ) -{\rm Li}_2\left (-i \, \cosh \left (x\right ) - i \, \sinh \left (x\right )\right ) +{\rm Li}_2\left (-\cosh \left (x\right ) - \sinh \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(coth(x)),x, algorithm="fricas")

[Out]

x*log(cosh(x)/sinh(x)) + x*log(cosh(x) + sinh(x) + 1) - x*log(I*cosh(x) + I*sinh(x) + 1) - x*log(-I*cosh(x) -
I*sinh(x) + 1) + x*log(-cosh(x) - sinh(x) + 1) + dilog(cosh(x) + sinh(x)) - dilog(I*cosh(x) + I*sinh(x)) - dil
og(-I*cosh(x) - I*sinh(x)) + dilog(-cosh(x) - sinh(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(coth(x)),x)

[Out]

Integral(log(coth(x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(coth(x)),x, algorithm="giac")

[Out]

integrate(log(coth(x)), x)