3.511 \(\int \frac{x}{2+3 e^x+e^{2 x}} \, dx\)

Optimal. Leaf size=54 \[ -\text{PolyLog}\left (2,-e^x\right )+\frac{1}{2} \text{PolyLog}\left (2,-\frac{e^x}{2}\right )+\frac{x^2}{4}+\frac{1}{2} x \log \left (\frac{e^x}{2}+1\right )-x \log \left (e^x+1\right ) \]

[Out]

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

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Rubi [A]  time = 0.124028, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {2263, 2184, 2190, 2279, 2391} \[ -\text{PolyLog}\left (2,-e^x\right )+\frac{1}{2} \text{PolyLog}\left (2,-\frac{e^x}{2}\right )+\frac{x^2}{4}+\frac{1}{2} x \log \left (\frac{e^x}{2}+1\right )-x \log \left (e^x+1\right ) \]

Antiderivative was successfully verified.

[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, -E^x/2]/2

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}{2+3 e^x+e^{2 x}} \, dx &=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} \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{x}{2}\right )}{x} \, dx,x,e^x\right )+\operatorname{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]  time = 0.0034946, size = 49, normalized size = 0.91 \[ -\frac{1}{2} \text{PolyLog}\left (2,-2 e^{-x}\right )+\text{PolyLog}\left (2,-e^{-x}\right )-x \log \left (e^{-x}+1\right )+\frac{1}{2} x \log \left (2 e^{-x}+1\right ) \]

Antiderivative was successfully verified.

[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)]

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Maple [A]  time = 0.006, size = 41, normalized size = 0.8 \begin{align*}{\frac{{x}^{2}}{4}}+{\frac{x}{2}\ln \left ( 1+{\frac{{{\rm e}^{x}}}{2}} \right ) }-x\ln \left ( 1+{{\rm e}^{x}} \right ) -{\it polylog} \left ( 2,-{{\rm e}^{x}} \right ) +{\frac{1}{2}{\it polylog} \left ( 2,-{\frac{{{\rm e}^{x}}}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))

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Maxima [A]  time = 0.981413, size = 51, normalized size = 0.94 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.56201, size = 117, normalized size = 2.17 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{e^{\left (2 \, x\right )} + 3 \, e^{x} + 2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(e^(2*x) + 3*e^x + 2), x)