\(\int \frac {e^x x}{1-e^{2 x}} \, dx\) [40]

Optimal result

Integrand size = 16, antiderivative size = 27 \[ \int \frac {e^x x}{1-e^{2 x}} \, dx=x \text {arctanh}\left (e^x\right )+\frac {\operatorname {PolyLog}\left (2,-e^x\right )}{2}-\frac {\operatorname {PolyLog}\left (2,e^x\right )}{2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {2281, 212, 2277, 2320, 6031} \[ \int \frac {e^x x}{1-e^{2 x}} \, dx=x \text {arctanh}\left (e^x\right )+\frac {\operatorname {PolyLog}\left (2,-e^x\right )}{2}-\frac {\operatorname {PolyLog}\left (2,e^x\right )}{2} \]

[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 212

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

Rule 2277

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 2281

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 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 6031

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

Rubi steps \begin{align*} \text {integral}& = x \tanh ^{-1}\left (e^x\right )-\int \tanh ^{-1}\left (e^x\right ) \, dx \\ & = x \tanh ^{-1}\left (e^x\right )-\text {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] (verified)

Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.67 \[ \int \frac {e^x x}{1-e^{2 x}} \, dx=-\frac {1}{2} x \log \left (1-e^x\right )+\frac {1}{2} x \log \left (1+e^x\right )+\frac {\operatorname {PolyLog}\left (2,-e^x\right )}{2}-\frac {\operatorname {PolyLog}\left (2,e^x\right )}{2} \]

[In]

Integrate[(E^x*x)/(1 - E^(2*x)),x]

[Out]

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

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26

method	result	size
default	\(\frac {x \ln \left (1+{\mathrm e}^{x}\right )}{2}+\frac {\operatorname {Li}_{2}\left (-{\mathrm e}^{x}\right )}{2}-\frac {x \ln \left (1-{\mathrm e}^{x}\right )}{2}-\frac {\operatorname {Li}_{2}\left ({\mathrm e}^{x}\right )}{2}\)	\(34\)
risch	\(\frac {x \ln \left (1+{\mathrm e}^{x}\right )}{2}+\frac {\operatorname {Li}_{2}\left (-{\mathrm e}^{x}\right )}{2}-\frac {x \ln \left (1-{\mathrm e}^{x}\right )}{2}-\frac {\operatorname {Li}_{2}\left ({\mathrm e}^{x}\right )}{2}\)	\(34\)

[In]

int(exp(x)*x/(1-exp(2*x)),x,method=_RETURNVERBOSE)

[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))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {e^x x}{1-e^{2 x}} \, dx=\frac {1}{2} \, x \log \left (e^{x} + 1\right ) - \frac {1}{2} \, x \log \left (-e^{x} + 1\right ) + \frac {1}{2} \, {\rm Li}_2\left (-e^{x}\right ) - \frac {1}{2} \, {\rm Li}_2\left (e^{x}\right ) \]

[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)

Sympy [F]

\[ \int \frac {e^x x}{1-e^{2 x}} \, dx=- \int \frac {x e^{x}}{e^{2 x} - 1}\, dx \]

[In]

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

[Out]

-Integral(x*exp(x)/(exp(2*x) - 1), x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {e^x x}{1-e^{2 x}} \, dx=\frac {1}{2} \, x \log \left (e^{x} + 1\right ) - \frac {1}{2} \, x \log \left (-e^{x} + 1\right ) + \frac {1}{2} \, {\rm Li}_2\left (-e^{x}\right ) - \frac {1}{2} \, {\rm Li}_2\left (e^{x}\right ) \]

[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)

Giac [F]

\[ \int \frac {e^x x}{1-e^{2 x}} \, dx=\int { -\frac {x e^{x}}{e^{\left (2 \, x\right )} - 1} \,d x } \]

[In]

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

[Out]

integrate(-x*e^x/(e^(2*x) - 1), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {e^x x}{1-e^{2 x}} \, dx=-\int \frac {x\,{\mathrm {e}}^x}{{\mathrm {e}}^{2\,x}-1} \,d x \]

[In]

int(-(x*exp(x))/(exp(2*x) - 1),x)

[Out]

-int((x*exp(x))/(exp(2*x) - 1), x)