\(\int \frac {x}{-1+e^x+e^{2 x}} \, dx\) [512]

Optimal result

Integrand size = 14, antiderivative size = 180 \[ \int \frac {x}{-1+e^x+e^{2 x}} \, dx=\frac {x^2}{\sqrt {5} \left (1-\sqrt {5}\right )}-\frac {x^2}{\sqrt {5} \left (1+\sqrt {5}\right )}-\frac {2 x \log \left (1+\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 x \log \left (1+\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}-\frac {2 \operatorname {PolyLog}\left (2,-\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 \operatorname {PolyLog}\left (2,-\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )} \]

[Out]

1/5*x^2/(-5^(1/2)+1)*5^(1/2)-2/5*x*ln(1+2*exp(x)/(-5^(1/2)+1))/(-5^(1/2)+1)*5^(1/2)-2/5*polylog(2,-2*exp(x)/(-
5^(1/2)+1))/(-5^(1/2)+1)*5^(1/2)-1/5*x^2*5^(1/2)/(5^(1/2)+1)+2/5*x*ln(1+2*exp(x)/(5^(1/2)+1))*5^(1/2)/(5^(1/2)
+1)+2/5*polylog(2,-2*exp(x)/(5^(1/2)+1))*5^(1/2)/(5^(1/2)+1)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {2295, 2215, 2221, 2317, 2438} \[ \int \frac {x}{-1+e^x+e^{2 x}} \, dx=-\frac {2 \operatorname {PolyLog}\left (2,-\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 \operatorname {PolyLog}\left (2,-\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}-\frac {x^2}{\sqrt {5} \left (1+\sqrt {5}\right )}+\frac {x^2}{\sqrt {5} \left (1-\sqrt {5}\right )}-\frac {2 x \log \left (\frac {2 e^x}{1-\sqrt {5}}+1\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 x \log \left (\frac {2 e^x}{1+\sqrt {5}}+1\right )}{\sqrt {5} \left (1+\sqrt {5}\right )} \]

[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 2215

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2295

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 2317

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 2438

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*} \text {integral}& = \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 \text {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 \text {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] (verified)

Time = 0.08 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.67 \[ \int \frac {x}{-1+e^x+e^{2 x}} \, dx=\frac {\left (1+\sqrt {5}\right ) x \log \left (1-\frac {1}{2} \left (-1+\sqrt {5}\right ) e^{-x}\right )+\left (-1+\sqrt {5}\right ) x \log \left (1+\frac {1}{2} \left (1+\sqrt {5}\right ) e^{-x}\right )-\left (1+\sqrt {5}\right ) \operatorname {PolyLog}\left (2,\frac {1}{2} \left (-1+\sqrt {5}\right ) e^{-x}\right )-\left (-1+\sqrt {5}\right ) \operatorname {PolyLog}\left (2,-\frac {1}{2} \left (1+\sqrt {5}\right ) e^{-x}\right )}{2 \sqrt {5}} \]

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

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.02

method	result	size
default	\(\frac {\sqrt {5}\, x \ln \left (\frac {\sqrt {5}-1-2 \,{\mathrm e}^{x}}{\sqrt {5}-1}\right )}{10}+\frac {x \ln \left (\frac {\sqrt {5}-1-2 \,{\mathrm e}^{x}}{\sqrt {5}-1}\right )}{2}-\frac {\sqrt {5}\, x \ln \left (\frac {1+2 \,{\mathrm e}^{x}+\sqrt {5}}{\sqrt {5}+1}\right )}{10}+\frac {x \ln \left (\frac {1+2 \,{\mathrm e}^{x}+\sqrt {5}}{\sqrt {5}+1}\right )}{2}+\frac {\sqrt {5}\, \operatorname {dilog}\left (\frac {\sqrt {5}-1-2 \,{\mathrm e}^{x}}{\sqrt {5}-1}\right )}{10}+\frac {\operatorname {dilog}\left (\frac {\sqrt {5}-1-2 \,{\mathrm e}^{x}}{\sqrt {5}-1}\right )}{2}-\frac {\sqrt {5}\, \operatorname {dilog}\left (\frac {1+2 \,{\mathrm e}^{x}+\sqrt {5}}{\sqrt {5}+1}\right )}{10}+\frac {\operatorname {dilog}\left (\frac {1+2 \,{\mathrm e}^{x}+\sqrt {5}}{\sqrt {5}+1}\right )}{2}-\frac {x^{2}}{2}\)	\(183\)
risch	\(\frac {\sqrt {5}\, x \ln \left (\frac {\sqrt {5}-1-2 \,{\mathrm e}^{x}}{\sqrt {5}-1}\right )}{10}+\frac {x \ln \left (\frac {\sqrt {5}-1-2 \,{\mathrm e}^{x}}{\sqrt {5}-1}\right )}{2}-\frac {\sqrt {5}\, x \ln \left (\frac {1+2 \,{\mathrm e}^{x}+\sqrt {5}}{\sqrt {5}+1}\right )}{10}+\frac {x \ln \left (\frac {1+2 \,{\mathrm e}^{x}+\sqrt {5}}{\sqrt {5}+1}\right )}{2}+\frac {\sqrt {5}\, \operatorname {dilog}\left (\frac {\sqrt {5}-1-2 \,{\mathrm e}^{x}}{\sqrt {5}-1}\right )}{10}+\frac {\operatorname {dilog}\left (\frac {\sqrt {5}-1-2 \,{\mathrm e}^{x}}{\sqrt {5}-1}\right )}{2}-\frac {\sqrt {5}\, \operatorname {dilog}\left (\frac {1+2 \,{\mathrm e}^{x}+\sqrt {5}}{\sqrt {5}+1}\right )}{10}+\frac {\operatorname {dilog}\left (\frac {1+2 \,{\mathrm e}^{x}+\sqrt {5}}{\sqrt {5}+1}\right )}{2}-\frac {x^{2}}{2}\)	\(183\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.48 \[ \int \frac {x}{-1+e^x+e^{2 x}} \, dx=-\frac {1}{2} \, x^{2} + \frac {1}{10} \, {\left (\sqrt {5} + 5\right )} {\rm Li}_2\left (\frac {1}{2} \, {\left (\sqrt {5} + 1\right )} e^{x}\right ) - \frac {1}{10} \, {\left (\sqrt {5} - 5\right )} {\rm Li}_2\left (-\frac {1}{2} \, {\left (\sqrt {5} - 1\right )} e^{x}\right ) + \frac {1}{10} \, {\left (\sqrt {5} x + 5 \, x\right )} \log \left (-\frac {1}{2} \, {\left (\sqrt {5} + 1\right )} e^{x} + 1\right ) - \frac {1}{10} \, {\left (\sqrt {5} x - 5 \, x\right )} \log \left (\frac {1}{2} \, {\left (\sqrt {5} - 1\right )} e^{x} + 1\right ) \]

[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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate(x/(e^(2*x) + e^x - 1), x)

Giac [F]

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

[In]

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

[Out]

integrate(x/(e^(2*x) + e^x - 1), x)

Mupad [F(-1)]

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

[In]

int(x/(exp(2*x) + exp(x) - 1),x)

[Out]

int(x/(exp(2*x) + exp(x) - 1), x)