\(\int \frac {x}{-1+e^x+e^{2 x}} \, dx\) [435]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.67 \[ \int \frac {x}{-1+e^x+e^{2 x}} \, dx=\frac {\left (1+\sqrt {5}\right ) x \log \left (1-\frac {1}{2} \left (-1+\sqrt {5}\right ) e^{-x}\right )+\left (-1+\sqrt {5}\right ) x \log \left (1+\frac {1}{2} \left (1+\sqrt {5}\right ) e^{-x}\right )-\left (1+\sqrt {5}\right ) \operatorname {PolyLog}\left (2,\frac {1}{2} \left (-1+\sqrt {5}\right ) e^{-x}\right )-\left (-1+\sqrt {5}\right ) \operatorname {PolyLog}\left (2,-\frac {1}{2} \left (1+\sqrt {5}\right ) e^{-x}\right )}{2 \sqrt {5}} \] Input:

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

Output:

((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])
 

Rubi [A] (verified)

Time = 0.87 (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

\(\displaystyle \frac {2 \left (\frac {x^2}{2 \left (1-\sqrt {5}\right )}-\frac {2 \left (\frac {1}{2} \operatorname {PolyLog}\left (2,-\frac {2 e^x}{1-\sqrt {5}}\right )+\frac {1}{2} x \log \left (\frac {2 e^x}{1-\sqrt {5}}+1\right )\right )}{1-\sqrt {5}}\right )}{\sqrt {5}}-\frac {2 \left (\frac {x^2}{2 \left (1+\sqrt {5}\right )}-\frac {2 \left (\frac {1}{2} \operatorname {PolyLog}\left (2,-\frac {2 e^x}{1+\sqrt {5}}\right )+\frac {1}{2} x \log \left (\frac {2 e^x}{1+\sqrt {5}}+1\right )\right )}{1+\sqrt {5}}\right )}{\sqrt {5}}\)

Input:

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

Output:

(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]
 

Defintions of rubi rules used

rule 2615

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]
 

rule 2620

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]
 

rule 2693

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]
 

rule 2715

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]
 

rule 2838

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]
 

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.02

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.48 \[ \int \frac {x}{-1+e^x+e^{2 x}} \, dx=-\frac {1}{2} \, x^{2} + \frac {1}{10} \, {\left (\sqrt {5} + 5\right )} {\rm Li}_2\left (\frac {1}{2} \, {\left (\sqrt {5} + 1\right )} e^{x}\right ) - \frac {1}{10} \, {\left (\sqrt {5} - 5\right )} {\rm Li}_2\left (-\frac {1}{2} \, {\left (\sqrt {5} - 1\right )} e^{x}\right ) + \frac {1}{10} \, {\left (\sqrt {5} x + 5 \, x\right )} \log \left (-\frac {1}{2} \, {\left (\sqrt {5} + 1\right )} e^{x} + 1\right ) - \frac {1}{10} \, {\left (\sqrt {5} x - 5 \, x\right )} \log \left (\frac {1}{2} \, {\left (\sqrt {5} - 1\right )} e^{x} + 1\right ) \] Input:

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

Output:

-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)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate(x/(e^(2*x) + e^x - 1), x)
 

Giac [F]

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

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

Output:

integrate(x/(e^(2*x) + e^x - 1), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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