\(\int \frac {x}{2+3 e^x+e^{2 x}} \, dx\) [511]

Optimal result

Integrand size = 16, antiderivative size = 54 \[ \int \frac {x}{2+3 e^x+e^{2 x}} \, dx=\frac {x^2}{4}+\frac {1}{2} x \log \left (1+\frac {e^x}{2}\right )-x \log \left (1+e^x\right )-\operatorname {PolyLog}\left (2,-e^x\right )+\frac {1}{2} \operatorname {PolyLog}\left (2,-\frac {e^x}{2}\right ) \]

[Out]

1/4*x^2+1/2*x*ln(1+1/2*exp(x))-x*ln(1+exp(x))-polylog(2,-exp(x))+1/2*polylog(2,-1/2*exp(x))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {2295, 2215, 2221, 2317, 2438} \[ \int \frac {x}{2+3 e^x+e^{2 x}} \, dx=-\operatorname {PolyLog}\left (2,-e^x\right )+\frac {1}{2} \operatorname {PolyLog}\left (2,-\frac {e^x}{2}\right )+\frac {x^2}{4}+\frac {1}{2} x \log \left (\frac {e^x}{2}+1\right )-x \log \left (e^x+1\right ) \]

[In]

Int[x/(2 + 3*E^x + E^(2*x)),x]

[Out]

x^2/4 + (x*Log[1 + E^x/2])/2 - x*Log[1 + E^x] - PolyLog[2, -E^x] + PolyLog[2, -1/2*E^x]/2

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.91 \[ \int \frac {x}{2+3 e^x+e^{2 x}} \, dx=-x \log \left (1+e^{-x}\right )+\frac {1}{2} x \log \left (1+2 e^{-x}\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,-2 e^{-x}\right )+\operatorname {PolyLog}\left (2,-e^{-x}\right ) \]

[In]

Integrate[x/(2 + 3*E^x + E^(2*x)),x]

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.76

method	result	size
default	\(\frac {x^{2}}{4}+\frac {x \ln \left (1+\frac {{\mathrm e}^{x}}{2}\right )}{2}-x \ln \left (1+{\mathrm e}^{x}\right )-\operatorname {Li}_{2}\left (-{\mathrm e}^{x}\right )+\frac {\operatorname {Li}_{2}\left (-\frac {{\mathrm e}^{x}}{2}\right )}{2}\)	\(41\)
risch	\(\frac {x^{2}}{4}+\frac {x \ln \left (1+\frac {{\mathrm e}^{x}}{2}\right )}{2}-x \ln \left (1+{\mathrm e}^{x}\right )-\operatorname {Li}_{2}\left (-{\mathrm e}^{x}\right )+\frac {\operatorname {Li}_{2}\left (-\frac {{\mathrm e}^{x}}{2}\right )}{2}\)	\(41\)

[In]

int(x/(2+3*exp(x)+exp(2*x)),x,method=_RETURNVERBOSE)

[Out]

1/4*x^2+1/2*x*ln(1+1/2*exp(x))-x*ln(1+exp(x))-polylog(2,-exp(x))+1/2*polylog(2,-1/2*exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.70 \[ \int \frac {x}{2+3 e^x+e^{2 x}} \, dx=\frac {1}{4} \, x^{2} - x \log \left (e^{x} + 1\right ) + \frac {1}{2} \, x \log \left (\frac {1}{2} \, e^{x} + 1\right ) + \frac {1}{2} \, {\rm Li}_2\left (-\frac {1}{2} \, e^{x}\right ) - {\rm Li}_2\left (-e^{x}\right ) \]

[In]

integrate(x/(2+3*exp(x)+exp(2*x)),x, algorithm="fricas")

[Out]

1/4*x^2 - x*log(e^x + 1) + 1/2*x*log(1/2*e^x + 1) + 1/2*dilog(-1/2*e^x) - dilog(-e^x)

Sympy [F]

\[ \int \frac {x}{2+3 e^x+e^{2 x}} \, dx=\int \frac {x}{\left (e^{x} + 1\right ) \left (e^{x} + 2\right )}\, dx \]

[In]

integrate(x/(2+3*exp(x)+exp(2*x)),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.70 \[ \int \frac {x}{2+3 e^x+e^{2 x}} \, dx=\frac {1}{4} \, x^{2} - x \log \left (e^{x} + 1\right ) + \frac {1}{2} \, x \log \left (\frac {1}{2} \, e^{x} + 1\right ) + \frac {1}{2} \, {\rm Li}_2\left (-\frac {1}{2} \, e^{x}\right ) - {\rm Li}_2\left (-e^{x}\right ) \]

[In]

integrate(x/(2+3*exp(x)+exp(2*x)),x, algorithm="maxima")

[Out]

1/4*x^2 - x*log(e^x + 1) + 1/2*x*log(1/2*e^x + 1) + 1/2*dilog(-1/2*e^x) - dilog(-e^x)

Giac [F]

\[ \int \frac {x}{2+3 e^x+e^{2 x}} \, dx=\int { \frac {x}{e^{\left (2 \, x\right )} + 3 \, e^{x} + 2} \,d x } \]

[In]

integrate(x/(2+3*exp(x)+exp(2*x)),x, algorithm="giac")

[Out]

integrate(x/(e^(2*x) + 3*e^x + 2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{2+3 e^x+e^{2 x}} \, dx=\int \frac {x}{{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^x+2} \,d x \]

[In]

int(x/(exp(2*x) + 3*exp(x) + 2),x)

[Out]

int(x/(exp(2*x) + 3*exp(x) + 2), x)