Integrand size = 6, antiderivative size = 51 \[ \int x \coth ^{-1}\left (e^x\right ) \, dx=\frac {1}{2} x \operatorname {PolyLog}\left (2,-e^{-x}\right )-\frac {1}{2} x \operatorname {PolyLog}\left (2,e^{-x}\right )+\frac {\operatorname {PolyLog}\left (3,-e^{-x}\right )}{2}-\frac {\operatorname {PolyLog}\left (3,e^{-x}\right )}{2} \]
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))
Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.39 \[ \int x \coth ^{-1}\left (e^x\right ) \, dx=\frac {1}{4} \left (2 x^2 \coth ^{-1}\left (e^x\right )+x^2 \log \left (1-e^x\right )-x^2 \log \left (1+e^x\right )-2 x \operatorname {PolyLog}\left (2,-e^x\right )+2 x \operatorname {PolyLog}\left (2,e^x\right )+2 \operatorname {PolyLog}\left (3,-e^x\right )-2 \operatorname {PolyLog}\left (3,e^x\right )\right ) \]
(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*PolyLog[3, -E^x] - 2*PolyLog[3, E^x])/4
Time = 0.35 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {6768, 3011, 2720, 7143}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x \coth ^{-1}\left (e^x\right ) \, dx\) |
\(\Big \downarrow \) 6768 |
\(\displaystyle \frac {1}{2} \int x \log \left (1+e^{-x}\right )dx-\frac {1}{2} \int x \log \left (1-e^{-x}\right )dx\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {1}{2} \left (x \operatorname {PolyLog}\left (2,-e^{-x}\right )-\int \operatorname {PolyLog}\left (2,-e^{-x}\right )dx\right )+\frac {1}{2} \left (\int \operatorname {PolyLog}\left (2,e^{-x}\right )dx-x \operatorname {PolyLog}\left (2,e^{-x}\right )\right )\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {1}{2} \left (\int e^x \operatorname {PolyLog}\left (2,-e^{-x}\right )de^{-x}+x \operatorname {PolyLog}\left (2,-e^{-x}\right )\right )+\frac {1}{2} \left (-\int e^x \operatorname {PolyLog}\left (2,e^{-x}\right )de^{-x}-x \operatorname {PolyLog}\left (2,e^{-x}\right )\right )\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {1}{2} \left (x \operatorname {PolyLog}\left (2,-e^{-x}\right )+\operatorname {PolyLog}\left (3,-e^{-x}\right )\right )+\frac {1}{2} \left (-x \operatorname {PolyLog}\left (2,e^{-x}\right )-\operatorname {PolyLog}\left (3,e^{-x}\right )\right )\) |
(x*PolyLog[2, -E^(-x)] + PolyLog[3, -E^(-x)])/2 + (-(x*PolyLog[2, E^(-x)]) - PolyLog[3, E^(-x)])/2
3.3.84.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
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] + Simp[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]
Int[ArcCoth[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] :> Simp[1/2 Int[x^m*Log[1 + 1/(a + b*f^(c + d*x))], x], x] - Simp[1/2 I nt[x^m*Log[1 - 1/(a + b*f^(c + d*x))], x], x] /; FreeQ[{a, b, c, d, f}, x] && IGtQ[m, 0]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> 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]
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 | \(\frac {x^{2} \operatorname {arccoth}\left ({\mathrm e}^{x}\right )}{2}+\frac {x^{2} \ln \left (1-{\mathrm e}^{x}\right )}{4}+\frac {x \operatorname {polylog}\left (2, {\mathrm e}^{x}\right )}{2}-\frac {\operatorname {polylog}\left (3, {\mathrm e}^{x}\right )}{2}-\frac {x^{2} \ln \left (1+{\mathrm e}^{x}\right )}{4}-\frac {x \operatorname {polylog}\left (2, -{\mathrm e}^{x}\right )}{2}+\frac {\operatorname {polylog}\left (3, -{\mathrm e}^{x}\right )}{2}\) | \(62\) |
-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))
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 \[ \int x \coth ^{-1}\left (e^x\right ) \, dx=\frac {1}{4} \, x^{2} \log \left (\frac {\cosh \left (x\right ) + \sinh \left (x\right ) + 1}{\cosh \left (x\right ) + \sinh \left (x\right ) - 1}\right ) - \frac {1}{4} \, x^{2} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + \frac {1}{4} \, x^{2} \log \left (-\cosh \left (x\right ) - \sinh \left (x\right ) + 1\right ) + \frac {1}{2} \, x {\rm Li}_2\left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - \frac {1}{2} \, x {\rm Li}_2\left (-\cosh \left (x\right ) - \sinh \left (x\right )\right ) - \frac {1}{2} \, {\rm polylog}\left (3, \cosh \left (x\right ) + \sinh \left (x\right )\right ) + \frac {1}{2} \, {\rm polylog}\left (3, -\cosh \left (x\right ) - \sinh \left (x\right )\right ) \]
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*log(-cosh(x) - sinh(x) + 1) + 1/2*x*dilo g(cosh(x) + sinh(x)) - 1/2*x*dilog(-cosh(x) - sinh(x)) - 1/2*polylog(3, co sh(x) + sinh(x)) + 1/2*polylog(3, -cosh(x) - sinh(x))
\[ \int x \coth ^{-1}\left (e^x\right ) \, dx=\int x \operatorname {acoth}{\left (e^{x} \right )}\, dx \]
Time = 0.21 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.16 \[ \int x \coth ^{-1}\left (e^x\right ) \, dx=\frac {1}{2} \, x^{2} \operatorname {arcoth}\left (e^{x}\right ) - \frac {1}{4} \, x^{2} \log \left (e^{x} + 1\right ) + \frac {1}{4} \, x^{2} \log \left (-e^{x} + 1\right ) - \frac {1}{2} \, x {\rm Li}_2\left (-e^{x}\right ) + \frac {1}{2} \, x {\rm Li}_2\left (e^{x}\right ) + \frac {1}{2} \, {\rm Li}_{3}(-e^{x}) - \frac {1}{2} \, {\rm Li}_{3}(e^{x}) \]
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)
\[ \int x \coth ^{-1}\left (e^x\right ) \, dx=\int { x \operatorname {arcoth}\left (e^{x}\right ) \,d x } \]
Timed out. \[ \int x \coth ^{-1}\left (e^x\right ) \, dx=\int x\,\mathrm {acoth}\left ({\mathrm {e}}^x\right ) \,d x \]