Integrand size = 14, antiderivative size = 180 \[ \int \frac {x}{-1+e^x+e^{2 x}} \, dx=\frac {x^2}{\sqrt {5} \left (1-\sqrt {5}\right )}-\frac {x^2}{\sqrt {5} \left (1+\sqrt {5}\right )}-\frac {2 x \log \left (1+\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 x \log \left (1+\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}-\frac {2 \operatorname {PolyLog}\left (2,-\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 \operatorname {PolyLog}\left (2,-\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )} \]
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)
Time = 0.08 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.67 \[ \int \frac {x}{-1+e^x+e^{2 x}} \, dx=\frac {\left (1+\sqrt {5}\right ) x \log \left (1-\frac {1}{2} \left (-1+\sqrt {5}\right ) e^{-x}\right )+\left (-1+\sqrt {5}\right ) x \log \left (1+\frac {1}{2} \left (1+\sqrt {5}\right ) e^{-x}\right )-\left (1+\sqrt {5}\right ) \operatorname {PolyLog}\left (2,\frac {1}{2} \left (-1+\sqrt {5}\right ) e^{-x}\right )-\left (-1+\sqrt {5}\right ) \operatorname {PolyLog}\left (2,-\frac {1}{2} \left (1+\sqrt {5}\right ) e^{-x}\right )}{2 \sqrt {5}} \]
((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])
Time = 0.54 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {2693, 2615, 2620, 2715, 2838}
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.
\(\displaystyle \int \frac {x}{e^x+e^{2 x}-1} \, dx\) |
\(\Big \downarrow \) 2693 |
\(\displaystyle \frac {2 \int \frac {x}{1-\sqrt {5}+2 e^x}dx}{\sqrt {5}}-\frac {2 \int \frac {x}{1+\sqrt {5}+2 e^x}dx}{\sqrt {5}}\) |
\(\Big \downarrow \) 2615 |
\(\displaystyle \frac {2 \left (\frac {x^2}{2 \left (1-\sqrt {5}\right )}-\frac {2 \int \frac {e^x x}{1-\sqrt {5}+2 e^x}dx}{1-\sqrt {5}}\right )}{\sqrt {5}}-\frac {2 \left (\frac {x^2}{2 \left (1+\sqrt {5}\right )}-\frac {2 \int \frac {e^x x}{1+\sqrt {5}+2 e^x}dx}{1+\sqrt {5}}\right )}{\sqrt {5}}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {2 \left (\frac {x^2}{2 \left (1-\sqrt {5}\right )}-\frac {2 \left (\frac {1}{2} x \log \left (\frac {2 e^x}{1-\sqrt {5}}+1\right )-\frac {1}{2} \int \log \left (1+\frac {2 e^x}{1-\sqrt {5}}\right )dx\right )}{1-\sqrt {5}}\right )}{\sqrt {5}}-\frac {2 \left (\frac {x^2}{2 \left (1+\sqrt {5}\right )}-\frac {2 \left (\frac {1}{2} x \log \left (\frac {2 e^x}{1+\sqrt {5}}+1\right )-\frac {1}{2} \int \log \left (1+\frac {2 e^x}{1+\sqrt {5}}\right )dx\right )}{1+\sqrt {5}}\right )}{\sqrt {5}}\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle \frac {2 \left (\frac {x^2}{2 \left (1-\sqrt {5}\right )}-\frac {2 \left (\frac {1}{2} x \log \left (\frac {2 e^x}{1-\sqrt {5}}+1\right )-\frac {1}{2} \int e^{-x} \log \left (1+\frac {2 e^x}{1-\sqrt {5}}\right )de^x\right )}{1-\sqrt {5}}\right )}{\sqrt {5}}-\frac {2 \left (\frac {x^2}{2 \left (1+\sqrt {5}\right )}-\frac {2 \left (\frac {1}{2} x \log \left (\frac {2 e^x}{1+\sqrt {5}}+1\right )-\frac {1}{2} \int e^{-x} \log \left (1+\frac {2 e^x}{1+\sqrt {5}}\right )de^x\right )}{1+\sqrt {5}}\right )}{\sqrt {5}}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {2 \left (\frac {x^2}{2 \left (1-\sqrt {5}\right )}-\frac {2 \left (\frac {1}{2} \operatorname {PolyLog}\left (2,-\frac {2 e^x}{1-\sqrt {5}}\right )+\frac {1}{2} x \log \left (\frac {2 e^x}{1-\sqrt {5}}+1\right )\right )}{1-\sqrt {5}}\right )}{\sqrt {5}}-\frac {2 \left (\frac {x^2}{2 \left (1+\sqrt {5}\right )}-\frac {2 \left (\frac {1}{2} \operatorname {PolyLog}\left (2,-\frac {2 e^x}{1+\sqrt {5}}\right )+\frac {1}{2} x \log \left (\frac {2 e^x}{1+\sqrt {5}}+1\right )\right )}{1+\sqrt {5}}\right )}{\sqrt {5}}\) |
(2*(x^2/(2*(1 - Sqrt[5])) - (2*((x*Log[1 + (2*E^x)/(1 - Sqrt[5])])/2 + Pol yLog[2, (-2*E^x)/(1 - Sqrt[5])]/2))/(1 - Sqrt[5])))/Sqrt[5] - (2*(x^2/(2*( 1 + Sqrt[5])) - (2*((x*Log[1 + (2*E^x)/(1 + Sqrt[5])])/2 + PolyLog[2, (-2* E^x)/(1 + Sqrt[5])]/2))/(1 + Sqrt[5])))/Sqrt[5]
3.6.12.3.1 Defintions of rubi rules used
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] - Simp[ 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]
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] - Si mp[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]
Int[((f_.) + (g_.)*(x_))^(m_.)/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q) Int[(f + g*x)^m /(b - q + 2*c*F^u), x], x] - Simp[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]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[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]
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]
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 | \(\frac {\sqrt {5}\, x \ln \left (\frac {\sqrt {5}-1-2 \,{\mathrm e}^{x}}{\sqrt {5}-1}\right )}{10}+\frac {x \ln \left (\frac {\sqrt {5}-1-2 \,{\mathrm e}^{x}}{\sqrt {5}-1}\right )}{2}-\frac {\sqrt {5}\, x \ln \left (\frac {1+2 \,{\mathrm e}^{x}+\sqrt {5}}{\sqrt {5}+1}\right )}{10}+\frac {x \ln \left (\frac {1+2 \,{\mathrm e}^{x}+\sqrt {5}}{\sqrt {5}+1}\right )}{2}+\frac {\sqrt {5}\, \operatorname {dilog}\left (\frac {\sqrt {5}-1-2 \,{\mathrm e}^{x}}{\sqrt {5}-1}\right )}{10}+\frac {\operatorname {dilog}\left (\frac {\sqrt {5}-1-2 \,{\mathrm e}^{x}}{\sqrt {5}-1}\right )}{2}-\frac {\sqrt {5}\, \operatorname {dilog}\left (\frac {1+2 \,{\mathrm e}^{x}+\sqrt {5}}{\sqrt {5}+1}\right )}{10}+\frac {\operatorname {dilog}\left (\frac {1+2 \,{\mathrm e}^{x}+\sqrt {5}}{\sqrt {5}+1}\right )}{2}-\frac {x^{2}}{2}\) | \(183\) |
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
Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.48 \[ \int \frac {x}{-1+e^x+e^{2 x}} \, dx=-\frac {1}{2} \, x^{2} + \frac {1}{10} \, {\left (\sqrt {5} + 5\right )} {\rm Li}_2\left (\frac {1}{2} \, {\left (\sqrt {5} + 1\right )} e^{x}\right ) - \frac {1}{10} \, {\left (\sqrt {5} - 5\right )} {\rm Li}_2\left (-\frac {1}{2} \, {\left (\sqrt {5} - 1\right )} e^{x}\right ) + \frac {1}{10} \, {\left (\sqrt {5} x + 5 \, x\right )} \log \left (-\frac {1}{2} \, {\left (\sqrt {5} + 1\right )} e^{x} + 1\right ) - \frac {1}{10} \, {\left (\sqrt {5} x - 5 \, x\right )} \log \left (\frac {1}{2} \, {\left (\sqrt {5} - 1\right )} e^{x} + 1\right ) \]
-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*(sqr t(5) + 1)*e^x + 1) - 1/10*(sqrt(5)*x - 5*x)*log(1/2*(sqrt(5) - 1)*e^x + 1)
\[ \int \frac {x}{-1+e^x+e^{2 x}} \, dx=\int \frac {x}{e^{2 x} + e^{x} - 1}\, dx \]
\[ \int \frac {x}{-1+e^x+e^{2 x}} \, dx=\int { \frac {x}{e^{\left (2 \, x\right )} + e^{x} - 1} \,d x } \]
\[ \int \frac {x}{-1+e^x+e^{2 x}} \, dx=\int { \frac {x}{e^{\left (2 \, x\right )} + e^{x} - 1} \,d x } \]
Timed out. \[ \int \frac {x}{-1+e^x+e^{2 x}} \, dx=\int \frac {x}{{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x-1} \,d x \]