\(\int x \coth ^{-1}(e^x) \, dx\) [284]

Optimal result

Integrand size = 6, antiderivative size = 51 \[ \int x \coth ^{-1}\left (e^x\right ) \, dx=\frac {1}{2} x \operatorname {PolyLog}\left (2,-e^{-x}\right )-\frac {1}{2} x \operatorname {PolyLog}\left (2,e^{-x}\right )+\frac {\operatorname {PolyLog}\left (3,-e^{-x}\right )}{2}-\frac {\operatorname {PolyLog}\left (3,e^{-x}\right )}{2} \]

[Out]

1/2*x*polylog(2,-1/exp(x))-1/2*x*polylog(2,exp(-x))+1/2*polylog(3,-1/exp(x))-1/2*polylog(3,exp(-x))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {6349, 2611, 2320, 6724} \[ \int x \coth ^{-1}\left (e^x\right ) \, dx=\frac {1}{2} x \operatorname {PolyLog}\left (2,-e^{-x}\right )-\frac {1}{2} x \operatorname {PolyLog}\left (2,e^{-x}\right )+\frac {\operatorname {PolyLog}\left (3,-e^{-x}\right )}{2}-\frac {\operatorname {PolyLog}\left (3,e^{-x}\right )}{2} \]

[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 2320

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 2611

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 6349

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 6724

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*} \text {integral}& = -\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 \operatorname {PolyLog}\left (2,-e^{-x}\right )-\frac {1}{2} x \operatorname {PolyLog}\left (2,e^{-x}\right )-\frac {1}{2} \int \operatorname {PolyLog}\left (2,-e^{-x}\right ) \, dx+\frac {1}{2} \int \operatorname {PolyLog}\left (2,e^{-x}\right ) \, dx \\ & = \frac {1}{2} x \operatorname {PolyLog}\left (2,-e^{-x}\right )-\frac {1}{2} x \operatorname {PolyLog}\left (2,e^{-x}\right )+\frac {1}{2} \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-x)}{x} \, dx,x,e^{-x}\right )-\frac {1}{2} \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,x)}{x} \, dx,x,e^{-x}\right ) \\ & = \frac {1}{2} x \operatorname {PolyLog}\left (2,-e^{-x}\right )-\frac {1}{2} x \operatorname {PolyLog}\left (2,e^{-x}\right )+\frac {\operatorname {PolyLog}\left (3,-e^{-x}\right )}{2}-\frac {\operatorname {PolyLog}\left (3,e^{-x}\right )}{2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.39 \[ \int x \coth ^{-1}\left (e^x\right ) \, dx=\frac {1}{4} \left (2 x^2 \coth ^{-1}\left (e^x\right )+x^2 \log \left (1-e^x\right )-x^2 \log \left (1+e^x\right )-2 x \operatorname {PolyLog}\left (2,-e^x\right )+2 x \operatorname {PolyLog}\left (2,e^x\right )+2 \operatorname {PolyLog}\left (3,-e^x\right )-2 \operatorname {PolyLog}\left (3,e^x\right )\right ) \]

[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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.06

method	result	size
risch	\(-\frac {x^{2} \ln \left ({\mathrm e}^{x}-1\right )}{4}-\frac {x \operatorname {polylog}\left (2, -{\mathrm e}^{x}\right )}{2}+\frac {\operatorname {polylog}\left (3, -{\mathrm e}^{x}\right )}{2}+\frac {x^{2} \ln \left (1-{\mathrm e}^{x}\right )}{4}+\frac {x \operatorname {polylog}\left (2, {\mathrm e}^{x}\right )}{2}-\frac {\operatorname {polylog}\left (3, {\mathrm e}^{x}\right )}{2}\)	\(54\)
default	\(\frac {x^{2} \operatorname {arccoth}\left ({\mathrm e}^{x}\right )}{2}+\frac {x^{2} \ln \left (1-{\mathrm e}^{x}\right )}{4}+\frac {x \operatorname {polylog}\left (2, {\mathrm e}^{x}\right )}{2}-\frac {\operatorname {polylog}\left (3, {\mathrm e}^{x}\right )}{2}-\frac {x^{2} \ln \left (1+{\mathrm e}^{x}\right )}{4}-\frac {x \operatorname {polylog}\left (2, -{\mathrm e}^{x}\right )}{2}+\frac {\operatorname {polylog}\left (3, -{\mathrm e}^{x}\right )}{2}\)	\(62\)
parts	\(\frac {x^{2} \operatorname {arccoth}\left ({\mathrm e}^{x}\right )}{2}+\frac {x^{2} \ln \left (1-{\mathrm e}^{x}\right )}{4}+\frac {x \operatorname {polylog}\left (2, {\mathrm e}^{x}\right )}{2}-\frac {\operatorname {polylog}\left (3, {\mathrm e}^{x}\right )}{2}-\frac {x^{2} \ln \left (1+{\mathrm e}^{x}\right )}{4}-\frac {x \operatorname {polylog}\left (2, -{\mathrm e}^{x}\right )}{2}+\frac {\operatorname {polylog}\left (3, -{\mathrm e}^{x}\right )}{2}\)	\(62\)

[In]

int(x*arccoth(exp(x)),x,method=_RETURNVERBOSE)

[Out]

-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))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (37) = 74\).

Time = 0.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.84 \[ \int x \coth ^{-1}\left (e^x\right ) \, dx=\frac {1}{4} \, x^{2} \log \left (\frac {\cosh \left (x\right ) + \sinh \left (x\right ) + 1}{\cosh \left (x\right ) + \sinh \left (x\right ) - 1}\right ) - \frac {1}{4} \, x^{2} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + \frac {1}{4} \, x^{2} \log \left (-\cosh \left (x\right ) - \sinh \left (x\right ) + 1\right ) + \frac {1}{2} \, x {\rm Li}_2\left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - \frac {1}{2} \, x {\rm Li}_2\left (-\cosh \left (x\right ) - \sinh \left (x\right )\right ) - \frac {1}{2} \, {\rm polylog}\left (3, \cosh \left (x\right ) + \sinh \left (x\right )\right ) + \frac {1}{2} \, {\rm polylog}\left (3, -\cosh \left (x\right ) - \sinh \left (x\right )\right ) \]

[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))

Sympy [F]

\[ \int x \coth ^{-1}\left (e^x\right ) \, dx=\int x \operatorname {acoth}{\left (e^{x} \right )}\, dx \]

[In]

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

[Out]

Integral(x*acoth(exp(x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.16 \[ \int x \coth ^{-1}\left (e^x\right ) \, dx=\frac {1}{2} \, x^{2} \operatorname {arcoth}\left (e^{x}\right ) - \frac {1}{4} \, x^{2} \log \left (e^{x} + 1\right ) + \frac {1}{4} \, x^{2} \log \left (-e^{x} + 1\right ) - \frac {1}{2} \, x {\rm Li}_2\left (-e^{x}\right ) + \frac {1}{2} \, x {\rm Li}_2\left (e^{x}\right ) + \frac {1}{2} \, {\rm Li}_{3}(-e^{x}) - \frac {1}{2} \, {\rm Li}_{3}(e^{x}) \]

[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)

Giac [F]

\[ \int x \coth ^{-1}\left (e^x\right ) \, dx=\int { x \operatorname {arcoth}\left (e^{x}\right ) \,d x } \]

[In]

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

[Out]

integrate(x*arccoth(e^x), x)

Mupad [F(-1)]

Timed out. \[ \int x \coth ^{-1}\left (e^x\right ) \, dx=\int x\,\mathrm {acoth}\left ({\mathrm {e}}^x\right ) \,d x \]

[In]

int(x*acoth(exp(x)),x)

[Out]

int(x*acoth(exp(x)), x)