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

Optimal. Leaf size=180 \[ -\frac{2 \text{PolyLog}\left (2,-\frac{2 e^x}{1-\sqrt{5}}\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{2 \text{PolyLog}\left (2,-\frac{2 e^x}{1+\sqrt{5}}\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}-\frac{x^2}{\sqrt{5} \left (1+\sqrt{5}\right )}+\frac{x^2}{\sqrt{5} \left (1-\sqrt{5}\right )}-\frac{2 x \log \left (\frac{2 e^x}{1-\sqrt{5}}+1\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{2 x \log \left (\frac{2 e^x}{1+\sqrt{5}}+1\right )}{\sqrt{5} \left (1+\sqrt{5}\right )} \]

[Out]

x^2/(Sqrt[5]*(1 - Sqrt[5])) - x^2/(Sqrt[5]*(1 + Sqrt[5])) - (2*x*Log[1 + (2*E^x)/(1 - Sqrt[5])])/(Sqrt[5]*(1 -
Sqrt[5])) + (2*x*Log[1 + (2*E^x)/(1 + Sqrt[5])])/(Sqrt[5]*(1 + Sqrt[5])) - (2*PolyLog[2, (-2*E^x)/(1 - Sqrt[5
])])/(Sqrt[5]*(1 - Sqrt[5])) + (2*PolyLog[2, (-2*E^x)/(1 + Sqrt[5])])/(Sqrt[5]*(1 + Sqrt[5]))

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Rubi [A]  time = 0.188071, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {2263, 2184, 2190, 2279, 2391} \[ -\frac{2 \text{PolyLog}\left (2,-\frac{2 e^x}{1-\sqrt{5}}\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{2 \text{PolyLog}\left (2,-\frac{2 e^x}{1+\sqrt{5}}\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}-\frac{x^2}{\sqrt{5} \left (1+\sqrt{5}\right )}+\frac{x^2}{\sqrt{5} \left (1-\sqrt{5}\right )}-\frac{2 x \log \left (\frac{2 e^x}{1-\sqrt{5}}+1\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{2 x \log \left (\frac{2 e^x}{1+\sqrt{5}}+1\right )}{\sqrt{5} \left (1+\sqrt{5}\right )} \]

Antiderivative was successfully verified.

[In]

Int[x/(-1 + E^x + E^(2*x)),x]

[Out]

x^2/(Sqrt[5]*(1 - Sqrt[5])) - x^2/(Sqrt[5]*(1 + Sqrt[5])) - (2*x*Log[1 + (2*E^x)/(1 - Sqrt[5])])/(Sqrt[5]*(1 -
Sqrt[5])) + (2*x*Log[1 + (2*E^x)/(1 + Sqrt[5])])/(Sqrt[5]*(1 + Sqrt[5])) - (2*PolyLog[2, (-2*E^x)/(1 - Sqrt[5
])])/(Sqrt[5]*(1 - Sqrt[5])) + (2*PolyLog[2, (-2*E^x)/(1 + Sqrt[5])])/(Sqrt[5]*(1 + Sqrt[5]))

Rule 2263

Int[((f_.) + (g_.)*(x_))^(m_.)/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q = Rt[b^2 - 4*
a*c, 2]}, Dist[(2*c)/q, Int[(f + g*x)^m/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[(f + g*x)^m/(b + q + 2*c
*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[
m, 0]

Rule 2184

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[((c + d*x)^m*(F^(g*(e + f*x)))^n)/(a + b*(F^(g*(e + f*x)))^n)
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{x}{-1+e^x+e^{2 x}} \, dx &=\frac{2 \int \frac{x}{1-\sqrt{5}+2 e^x} \, dx}{\sqrt{5}}-\frac{2 \int \frac{x}{1+\sqrt{5}+2 e^x} \, dx}{\sqrt{5}}\\ &=\frac{x^2}{\sqrt{5} \left (1-\sqrt{5}\right )}-\frac{x^2}{\sqrt{5} \left (1+\sqrt{5}\right )}-\frac{4 \int \frac{e^x x}{1-\sqrt{5}+2 e^x} \, dx}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{4 \int \frac{e^x x}{1+\sqrt{5}+2 e^x} \, dx}{\sqrt{5} \left (1+\sqrt{5}\right )}\\ &=\frac{x^2}{\sqrt{5} \left (1-\sqrt{5}\right )}-\frac{x^2}{\sqrt{5} \left (1+\sqrt{5}\right )}-\frac{2 x \log \left (1+\frac{2 e^x}{1-\sqrt{5}}\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{2 x \log \left (1+\frac{2 e^x}{1+\sqrt{5}}\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}+\frac{2 \int \log \left (1+\frac{2 e^x}{1-\sqrt{5}}\right ) \, dx}{\sqrt{5} \left (1-\sqrt{5}\right )}-\frac{2 \int \log \left (1+\frac{2 e^x}{1+\sqrt{5}}\right ) \, dx}{\sqrt{5} \left (1+\sqrt{5}\right )}\\ &=\frac{x^2}{\sqrt{5} \left (1-\sqrt{5}\right )}-\frac{x^2}{\sqrt{5} \left (1+\sqrt{5}\right )}-\frac{2 x \log \left (1+\frac{2 e^x}{1-\sqrt{5}}\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{2 x \log \left (1+\frac{2 e^x}{1+\sqrt{5}}\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}+\frac{2 \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{2 x}{1-\sqrt{5}}\right )}{x} \, dx,x,e^x\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}-\frac{2 \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{2 x}{1+\sqrt{5}}\right )}{x} \, dx,x,e^x\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}\\ &=\frac{x^2}{\sqrt{5} \left (1-\sqrt{5}\right )}-\frac{x^2}{\sqrt{5} \left (1+\sqrt{5}\right )}-\frac{2 x \log \left (1+\frac{2 e^x}{1-\sqrt{5}}\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{2 x \log \left (1+\frac{2 e^x}{1+\sqrt{5}}\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}-\frac{2 \text{Li}_2\left (-\frac{2 e^x}{1-\sqrt{5}}\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{2 \text{Li}_2\left (-\frac{2 e^x}{1+\sqrt{5}}\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}\\ \end{align*}

Mathematica [A]  time = 0.0878204, size = 120, normalized size = 0.67 \[ \frac{-\left (1+\sqrt{5}\right ) \text{PolyLog}\left (2,\frac{1}{2} \left (\sqrt{5}-1\right ) e^{-x}\right )-\left (\sqrt{5}-1\right ) \text{PolyLog}\left (2,-\frac{1}{2} \left (1+\sqrt{5}\right ) e^{-x}\right )+\left (1+\sqrt{5}\right ) x \log \left (1-\frac{1}{2} \left (\sqrt{5}-1\right ) e^{-x}\right )+\left (\sqrt{5}-1\right ) x \log \left (\frac{1}{2} \left (1+\sqrt{5}\right ) e^{-x}+1\right )}{2 \sqrt{5}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(-1 + E^x + E^(2*x)),x]

[Out]

((1 + Sqrt[5])*x*Log[1 - (-1 + Sqrt[5])/(2*E^x)] + (-1 + Sqrt[5])*x*Log[1 + (1 + Sqrt[5])/(2*E^x)] - (1 + Sqrt
[5])*PolyLog[2, (-1 + Sqrt[5])/(2*E^x)] - (-1 + Sqrt[5])*PolyLog[2, -(1 + Sqrt[5])/(2*E^x)])/(2*Sqrt[5])

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Maple [A]  time = 0.01, size = 183, normalized size = 1. \begin{align*} -{\frac{{x}^{2}}{2}}+{\frac{x}{2}\ln \left ({\frac{\sqrt{5}-1-2\,{{\rm e}^{x}}}{\sqrt{5}-1}} \right ) }+{\frac{\sqrt{5}x}{10}\ln \left ({\frac{\sqrt{5}-1-2\,{{\rm e}^{x}}}{\sqrt{5}-1}} \right ) }-{\frac{\sqrt{5}x}{10}\ln \left ({\frac{1+2\,{{\rm e}^{x}}+\sqrt{5}}{\sqrt{5}+1}} \right ) }+{\frac{x}{2}\ln \left ({\frac{1+2\,{{\rm e}^{x}}+\sqrt{5}}{\sqrt{5}+1}} \right ) }+{\frac{1}{2}{\it dilog} \left ({\frac{\sqrt{5}-1-2\,{{\rm e}^{x}}}{\sqrt{5}-1}} \right ) }+{\frac{\sqrt{5}}{10}{\it dilog} \left ({\frac{\sqrt{5}-1-2\,{{\rm e}^{x}}}{\sqrt{5}-1}} \right ) }-{\frac{\sqrt{5}}{10}{\it dilog} \left ({\frac{1+2\,{{\rm e}^{x}}+\sqrt{5}}{\sqrt{5}+1}} \right ) }+{\frac{1}{2}{\it dilog} \left ({\frac{1+2\,{{\rm e}^{x}}+\sqrt{5}}{\sqrt{5}+1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^2+1/2*x*ln((5^(1/2)-1-2*exp(x))/(5^(1/2)-1))+1/10*5^(1/2)*x*ln((5^(1/2)-1-2*exp(x))/(5^(1/2)-1))-1/10*5
^(1/2)*x*ln((1+2*exp(x)+5^(1/2))/(5^(1/2)+1))+1/2*x*ln((1+2*exp(x)+5^(1/2))/(5^(1/2)+1))+1/2*dilog((5^(1/2)-1-
2*exp(x))/(5^(1/2)-1))+1/10*5^(1/2)*dilog((5^(1/2)-1-2*exp(x))/(5^(1/2)-1))-1/10*5^(1/2)*dilog((1+2*exp(x)+5^(
1/2))/(5^(1/2)+1))+1/2*dilog((1+2*exp(x)+5^(1/2))/(5^(1/2)+1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(e^(2*x) + e^x - 1), x)

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Fricas [A]  time = 1.57597, size = 302, normalized size = 1.68 \begin{align*} -\frac{1}{2} \, x^{2} + \frac{1}{10} \,{\left (\sqrt{5} + 5\right )}{\rm Li}_2\left (\frac{1}{2} \,{\left (\sqrt{5} + 1\right )} e^{x}\right ) - \frac{1}{10} \,{\left (\sqrt{5} - 5\right )}{\rm Li}_2\left (-\frac{1}{2} \,{\left (\sqrt{5} - 1\right )} e^{x}\right ) + \frac{1}{10} \,{\left (\sqrt{5} x + 5 \, x\right )} \log \left (-\frac{1}{2} \,{\left (\sqrt{5} + 1\right )} e^{x} + 1\right ) - \frac{1}{10} \,{\left (\sqrt{5} x - 5 \, x\right )} \log \left (\frac{1}{2} \,{\left (\sqrt{5} - 1\right )} e^{x} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^2 + 1/10*(sqrt(5) + 5)*dilog(1/2*(sqrt(5) + 1)*e^x) - 1/10*(sqrt(5) - 5)*dilog(-1/2*(sqrt(5) - 1)*e^x)
+ 1/10*(sqrt(5)*x + 5*x)*log(-1/2*(sqrt(5) + 1)*e^x + 1) - 1/10*(sqrt(5)*x - 5*x)*log(1/2*(sqrt(5) - 1)*e^x +
1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/(exp(2*x) + exp(x) - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(e^(2*x) + e^x - 1), x)