3.201 \(\int x \coth ^{-1}(\cosh (x)) \, dx\)

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

[Out]

(x^2*ArcCoth[Cosh[x]])/2 - x^2*ArcTanh[E^x] - x*PolyLog[2, -E^x] + x*PolyLog[2, E^x] + PolyLog[3, -E^x] - Poly
Log[3, E^x]

________________________________________________________________________________________

Rubi [A]  time = 0.0664171, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 5, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {6274, 4182, 2531, 2282, 6589} \[ -x \text{PolyLog}\left (2,-e^x\right )+x \text{PolyLog}\left (2,e^x\right )+\text{PolyLog}\left (3,-e^x\right )-\text{PolyLog}\left (3,e^x\right )-x^2 \tanh ^{-1}\left (e^x\right )+\frac{1}{2} x^2 \coth ^{-1}(\cosh (x)) \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCoth[Cosh[x]],x]

[Out]

(x^2*ArcCoth[Cosh[x]])/2 - x^2*ArcTanh[E^x] - x*PolyLog[2, -E^x] + x*PolyLog[2, E^x] + PolyLog[3, -E^x] - Poly
Log[3, E^x]

Rule 6274

Int[((a_.) + ArcCoth[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcCot
h[u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/(1 - u^2), x],
x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(m
+ 1), u, x] && FalseQ[PowerVariableExpn[u, m + 1, x]]

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

Mathematica [A]  time = 0.0135662, size = 81, normalized size = 1.59 \[ \frac{1}{2} \left (2 x \text{PolyLog}\left (2,-e^{-x}\right )-2 x \text{PolyLog}\left (2,e^{-x}\right )+2 \text{PolyLog}\left (3,-e^{-x}\right )-2 \text{PolyLog}\left (3,e^{-x}\right )+x^2 \log \left (1-e^{-x}\right )-x^2 \log \left (e^{-x}+1\right )+x^2 \coth ^{-1}(\cosh (x))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCoth[Cosh[x]],x]

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.154, size = 449, normalized size = 8.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccoth(cosh(x)),x)

[Out]

x*polylog(2,exp(x))-x*polylog(2,-exp(x))+1/8*I*Pi*csgn(I*(exp(x)-1)^2)^3*x^2-1/8*I*Pi*csgn(I*exp(-x))*csgn(I*e
xp(-x)*(exp(x)-1)^2)^2*x^2+1/8*I*Pi*csgn(I*(exp(x)-1)^2)*csgn(I*exp(-x))*csgn(I*exp(-x)*(exp(x)-1)^2)*x^2-1/8*
I*Pi*csgn(I*(exp(x)-1)^2)*csgn(I*exp(-x)*(exp(x)-1)^2)^2*x^2-1/8*I*Pi*csgn(I*(exp(x)+1)^2)*csgn(I*exp(-x))*csg
n(I*exp(-x)*(exp(x)+1)^2)*x^2+1/8*I*Pi*csgn(I*exp(-x)*(exp(x)-1)^2)^3*x^2+1/8*I*Pi*csgn(I*(exp(x)+1)^2)*csgn(I
*exp(-x)*(exp(x)+1)^2)^2*x^2+1/4*I*Pi*csgn(I*(exp(x)+1))*csgn(I*(exp(x)+1)^2)^2*x^2-1/8*I*Pi*csgn(I*(exp(x)+1)
)^2*csgn(I*(exp(x)+1)^2)*x^2-1/8*I*Pi*csgn(I*exp(-x)*(exp(x)+1)^2)^3*x^2+1/8*I*Pi*csgn(I*exp(-x))*csgn(I*exp(-
x)*(exp(x)+1)^2)^2*x^2+polylog(3,-exp(x))-polylog(3,exp(x))-1/8*I*Pi*csgn(I*(exp(x)+1)^2)^3*x^2+1/2*x^2*ln(1-e
xp(x))-1/2*x^2*ln(exp(x)-1)-1/4*I*Pi*csgn(I*(exp(x)-1))*csgn(I*(exp(x)-1)^2)^2*x^2+1/8*I*Pi*csgn(I*(exp(x)-1))
^2*csgn(I*(exp(x)-1)^2)*x^2

________________________________________________________________________________________

Maxima [A]  time = 1.18363, size = 76, normalized size = 1.49 \begin{align*} \frac{1}{2} \, x^{2} \operatorname{arcoth}\left (\cosh \left (x\right )\right ) - \frac{1}{2} \, x^{2} \log \left (e^{x} + 1\right ) + \frac{1}{2} \, x^{2} \log \left (-e^{x} + 1\right ) - x{\rm Li}_2\left (-e^{x}\right ) + x{\rm Li}_2\left (e^{x}\right ) +{\rm Li}_{3}(-e^{x}) -{\rm Li}_{3}(e^{x}) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [C]  time = 1.61809, size = 325, normalized size = 6.37 \begin{align*} \frac{1}{4} \, x^{2} \log \left (\frac{\cosh \left (x\right ) + 1}{\cosh \left (x\right ) - 1}\right ) - \frac{1}{2} \, x^{2} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + \frac{1}{2} \, x^{2} \log \left (-\cosh \left (x\right ) - \sinh \left (x\right ) + 1\right ) + x{\rm Li}_2\left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - x{\rm Li}_2\left (-\cosh \left (x\right ) - \sinh \left (x\right )\right ) -{\rm polylog}\left (3, \cosh \left (x\right ) + \sinh \left (x\right )\right ) +{\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(cosh(x)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*acoth(cosh(x)),x)

[Out]

Integral(x*acoth(cosh(x)), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arccoth(cosh(x)), x)