3.284 \(\int x \coth ^{-1}(e^x) \, dx\)

Optimal. Leaf size=51 \[ \frac{1}{2} x \text{PolyLog}\left (2,-e^{-x}\right )-\frac{1}{2} x \text{PolyLog}\left (2,e^{-x}\right )+\frac{1}{2} \text{PolyLog}\left (3,-e^{-x}\right )-\frac{1}{2} \text{PolyLog}\left (3,e^{-x}\right ) \]

[Out]

(x*PolyLog[2, -E^(-x)])/2 - (x*PolyLog[2, E^(-x)])/2 + PolyLog[3, -E^(-x)]/2 - PolyLog[3, E^(-x)]/2

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Rubi [A]  time = 0.0474447, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {6214, 2531, 2282, 6589} \[ \frac{1}{2} x \text{PolyLog}\left (2,-e^{-x}\right )-\frac{1}{2} x \text{PolyLog}\left (2,e^{-x}\right )+\frac{1}{2} \text{PolyLog}\left (3,-e^{-x}\right )-\frac{1}{2} \text{PolyLog}\left (3,e^{-x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCoth[E^x],x]

[Out]

(x*PolyLog[2, -E^(-x)])/2 - (x*PolyLog[2, E^(-x)])/2 + PolyLog[3, -E^(-x)]/2 - PolyLog[3, E^(-x)]/2

Rule 6214

Int[ArcCoth[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] :> Dist[1/2, Int[x^m*Log[1 + 1/(a +
b*f^(c + d*x))], x], x] - Dist[1/2, Int[x^m*Log[1 - 1/(a + b*f^(c + d*x))], x], x] /; FreeQ[{a, b, c, d, f},
x] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

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

Rubi steps

\begin{align*} \int x \coth ^{-1}\left (e^x\right ) \, dx &=-\left (\frac{1}{2} \int x \log \left (1-e^{-x}\right ) \, dx\right )+\frac{1}{2} \int x \log \left (1+e^{-x}\right ) \, dx\\ &=\frac{1}{2} x \text{Li}_2\left (-e^{-x}\right )-\frac{1}{2} x \text{Li}_2\left (e^{-x}\right )-\frac{1}{2} \int \text{Li}_2\left (-e^{-x}\right ) \, dx+\frac{1}{2} \int \text{Li}_2\left (e^{-x}\right ) \, dx\\ &=\frac{1}{2} x \text{Li}_2\left (-e^{-x}\right )-\frac{1}{2} x \text{Li}_2\left (e^{-x}\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{-x}\right )-\frac{1}{2} \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{-x}\right )\\ &=\frac{1}{2} x \text{Li}_2\left (-e^{-x}\right )-\frac{1}{2} x \text{Li}_2\left (e^{-x}\right )+\frac{\text{Li}_3\left (-e^{-x}\right )}{2}-\frac{\text{Li}_3\left (e^{-x}\right )}{2}\\ \end{align*}

Mathematica [A]  time = 0.0223064, size = 71, normalized size = 1.39 \[ \frac{1}{4} \left (-2 x \text{PolyLog}\left (2,-e^x\right )+2 x \text{PolyLog}\left (2,e^x\right )+2 \text{PolyLog}\left (3,-e^x\right )-2 \text{PolyLog}\left (3,e^x\right )+x^2 \log \left (1-e^x\right )-x^2 \log \left (e^x+1\right )+2 x^2 \coth ^{-1}\left (e^x\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCoth[E^x],x]

[Out]

(2*x^2*ArcCoth[E^x] + x^2*Log[1 - E^x] - x^2*Log[1 + E^x] - 2*x*PolyLog[2, -E^x] + 2*x*PolyLog[2, E^x] + 2*Pol
yLog[3, -E^x] - 2*PolyLog[3, E^x])/4

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Maple [A]  time = 0.049, size = 62, normalized size = 1.2 \begin{align*}{\frac{{x}^{2}{\rm arccoth} \left ({{\rm e}^{x}}\right )}{2}}-{\frac{{x}^{2}\ln \left ({{\rm e}^{x}}+1 \right ) }{4}}-{\frac{x{\it polylog} \left ( 2,-{{\rm e}^{x}} \right ) }{2}}+{\frac{{\it polylog} \left ( 3,-{{\rm e}^{x}} \right ) }{2}}+{\frac{{x}^{2}\ln \left ( 1-{{\rm e}^{x}} \right ) }{4}}+{\frac{x{\it polylog} \left ( 2,{{\rm e}^{x}} \right ) }{2}}-{\frac{{\it polylog} \left ( 3,{{\rm e}^{x}} \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccoth(exp(x)),x)

[Out]

1/2*x^2*arccoth(exp(x))-1/4*x^2*ln(exp(x)+1)-1/2*x*polylog(2,-exp(x))+1/2*polylog(3,-exp(x))+1/4*x^2*ln(1-exp(
x))+1/2*x*polylog(2,exp(x))-1/2*polylog(3,exp(x))

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Maxima [A]  time = 1.1231, size = 80, normalized size = 1.57 \begin{align*} \frac{1}{2} \, x^{2} \operatorname{arcoth}\left (e^{x}\right ) - \frac{1}{4} \, x^{2} \log \left (e^{x} + 1\right ) + \frac{1}{4} \, x^{2} \log \left (-e^{x} + 1\right ) - \frac{1}{2} \, x{\rm Li}_2\left (-e^{x}\right ) + \frac{1}{2} \, x{\rm Li}_2\left (e^{x}\right ) + \frac{1}{2} \,{\rm Li}_{3}(-e^{x}) - \frac{1}{2} \,{\rm Li}_{3}(e^{x}) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccoth(exp(x)),x, algorithm="maxima")

[Out]

1/2*x^2*arccoth(e^x) - 1/4*x^2*log(e^x + 1) + 1/4*x^2*log(-e^x + 1) - 1/2*x*dilog(-e^x) + 1/2*x*dilog(e^x) + 1
/2*polylog(3, -e^x) - 1/2*polylog(3, e^x)

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Fricas [C]  time = 1.6609, size = 374, normalized size = 7.33 \begin{align*} \frac{1}{4} \, x^{2} \log \left (\frac{\cosh \left (x\right ) + \sinh \left (x\right ) + 1}{\cosh \left (x\right ) + \sinh \left (x\right ) - 1}\right ) - \frac{1}{4} \, x^{2} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + \frac{1}{4} \, x^{2} \log \left (-\cosh \left (x\right ) - \sinh \left (x\right ) + 1\right ) + \frac{1}{2} \, x{\rm Li}_2\left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - \frac{1}{2} \, x{\rm Li}_2\left (-\cosh \left (x\right ) - \sinh \left (x\right )\right ) - \frac{1}{2} \,{\rm polylog}\left (3, \cosh \left (x\right ) + \sinh \left (x\right )\right ) + \frac{1}{2} \,{\rm polylog}\left (3, -\cosh \left (x\right ) - \sinh \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccoth(exp(x)),x, algorithm="fricas")

[Out]

1/4*x^2*log((cosh(x) + sinh(x) + 1)/(cosh(x) + sinh(x) - 1)) - 1/4*x^2*log(cosh(x) + sinh(x) + 1) + 1/4*x^2*lo
g(-cosh(x) - sinh(x) + 1) + 1/2*x*dilog(cosh(x) + sinh(x)) - 1/2*x*dilog(-cosh(x) - sinh(x)) - 1/2*polylog(3,
cosh(x) + sinh(x)) + 1/2*polylog(3, -cosh(x) - sinh(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*acoth(exp(x)),x)

[Out]

Integral(x*acoth(exp(x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccoth(exp(x)),x, algorithm="giac")

[Out]

integrate(x*arccoth(e^x), x)