3.40 \(\int \frac{e^x x}{1-e^{2 x}} \, dx\)

Optimal. Leaf size=27 \[ \frac{1}{2} \text{PolyLog}\left (2,-e^x\right )-\frac{1}{2} \text{PolyLog}\left (2,e^x\right )+x \tanh ^{-1}\left (e^x\right ) \]

[Out]

x*ArcTanh[E^x] + PolyLog[2, -E^x]/2 - PolyLog[2, E^x]/2

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Rubi [A]  time = 0.0581555, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {2249, 206, 2245, 2282, 5912} \[ \frac{1}{2} \text{PolyLog}\left (2,-e^x\right )-\frac{1}{2} \text{PolyLog}\left (2,e^x\right )+x \tanh ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

Int[(E^x*x)/(1 - E^(2*x)),x]

[Out]

x*ArcTanh[E^x] + PolyLog[2, -E^x]/2 - PolyLog[2, E^x]/2

Rule 2249

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[(d*e*Log[F])/(g*h*Log[G])]}, Dist[Denominator[m]/(g*h*Log[G]), Subst[Int[x^(Denominator[m]
- 1)*(a + b*F^(c*e - (d*e*f)/g)*x^Numerator[m])^p, x], x, G^((h*(f + g*x))/Denominator[m])], x] /; LtQ[m, -1]
|| GtQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2245

Int[(F_)^((e_.)*((c_.) + (d_.)*(x_)))*((a_.) + (b_.)*(F_)^(v_))^(p_)*(x_)^(m_.), x_Symbol] :> With[{u = IntHid
e[F^(e*(c + d*x))*(a + b*F^v)^p, x]}, Dist[x^m, u, x] - Dist[m, Int[x^(m - 1)*u, x], x]] /; FreeQ[{F, a, b, c,
d, e}, x] && EqQ[v, 2*e*(c + d*x)] && GtQ[m, 0] && ILtQ[p, 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 5912

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (-Simp[(b*PolyLog[2, -(c*x)])/2
, x] + Simp[(b*PolyLog[2, c*x])/2, x]) /; FreeQ[{a, b, c}, x]

Rubi steps

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

Mathematica [A]  time = 0.0323428, size = 45, normalized size = 1.67 \[ \frac{1}{2} \text{PolyLog}\left (2,-e^x\right )-\frac{1}{2} \text{PolyLog}\left (2,e^x\right )-\frac{1}{2} x \log \left (1-e^x\right )+\frac{1}{2} x \log \left (e^x+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(E^x*x)/(1 - E^(2*x)),x]

[Out]

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

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Maple [A]  time = 0.007, size = 34, normalized size = 1.3 \begin{align*}{\frac{x\ln \left ( 1+{{\rm e}^{x}} \right ) }{2}}+{\frac{{\it polylog} \left ( 2,-{{\rm e}^{x}} \right ) }{2}}-{\frac{x\ln \left ( 1-{{\rm e}^{x}} \right ) }{2}}-{\frac{{\it polylog} \left ( 2,{{\rm e}^{x}} \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*x/(1-exp(2*x)),x)

[Out]

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

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Maxima [A]  time = 1.21072, size = 42, normalized size = 1.56 \begin{align*} \frac{1}{2} \, x \log \left (e^{x} + 1\right ) - \frac{1}{2} \, x \log \left (-e^{x} + 1\right ) + \frac{1}{2} \,{\rm Li}_2\left (-e^{x}\right ) - \frac{1}{2} \,{\rm Li}_2\left (e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*x/(1-exp(2*x)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.52436, size = 104, normalized size = 3.85 \begin{align*} \frac{1}{2} \, x \log \left (e^{x} + 1\right ) - \frac{1}{2} \, x \log \left (-e^{x} + 1\right ) + \frac{1}{2} \,{\rm Li}_2\left (-e^{x}\right ) - \frac{1}{2} \,{\rm Li}_2\left (e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*x/(1-exp(2*x)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x e^{x}}{e^{2 x} - 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*x/(1-exp(2*x)),x)

[Out]

-Integral(x*exp(x)/(exp(2*x) - 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{x e^{x}}{e^{\left (2 \, x\right )} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*x/(1-exp(2*x)),x, algorithm="giac")

[Out]

integrate(-x*e^x/(e^(2*x) - 1), x)