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

Optimal. Leaf size=77 \[ -2 x \text{PolyLog}\left (2,-e^x\right )+x \text{PolyLog}\left (2,-\frac{e^x}{2}\right )+2 \text{PolyLog}\left (3,-e^x\right )-\text{PolyLog}\left (3,-\frac{e^x}{2}\right )+\frac{x^3}{6}+\frac{1}{2} x^2 \log \left (\frac{e^x}{2}+1\right )-x^2 \log \left (e^x+1\right ) \]

[Out]

x^3/6 + (x^2*Log[1 + E^x/2])/2 - x^2*Log[1 + E^x] - 2*x*PolyLog[2, -E^x] + x*PolyLog[2, -E^x/2] + 2*PolyLog[3,
-E^x] - PolyLog[3, -E^x/2]

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Rubi [A]  time = 0.222342, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2263, 2184, 2190, 2531, 2282, 6589} \[ -2 x \text{PolyLog}\left (2,-e^x\right )+x \text{PolyLog}\left (2,-\frac{e^x}{2}\right )+2 \text{PolyLog}\left (3,-e^x\right )-\text{PolyLog}\left (3,-\frac{e^x}{2}\right )+\frac{x^3}{6}+\frac{1}{2} x^2 \log \left (\frac{e^x}{2}+1\right )-x^2 \log \left (e^x+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2/(2 + 3*E^x + E^(2*x)),x]

[Out]

x^3/6 + (x^2*Log[1 + E^x/2])/2 - x^2*Log[1 + E^x] - 2*x*PolyLog[2, -E^x] + x*PolyLog[2, -E^x/2] + 2*PolyLog[3,
-E^x] - PolyLog[3, -E^x/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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int \frac{x^2}{2+3 e^x+e^{2 x}} \, dx &=2 \int \frac{x^2}{2+2 e^x} \, dx-2 \int \frac{x^2}{4+2 e^x} \, dx\\ &=\frac{x^3}{6}-2 \int \frac{e^x x^2}{2+2 e^x} \, dx+\int \frac{e^x x^2}{4+2 e^x} \, dx\\ &=\frac{x^3}{6}+\frac{1}{2} x^2 \log \left (1+\frac{e^x}{2}\right )-x^2 \log \left (1+e^x\right )+2 \int x \log \left (1+e^x\right ) \, dx-\int x \log \left (1+\frac{e^x}{2}\right ) \, dx\\ &=\frac{x^3}{6}+\frac{1}{2} x^2 \log \left (1+\frac{e^x}{2}\right )-x^2 \log \left (1+e^x\right )-2 x \text{Li}_2\left (-e^x\right )+x \text{Li}_2\left (-\frac{e^x}{2}\right )+2 \int \text{Li}_2\left (-e^x\right ) \, dx-\int \text{Li}_2\left (-\frac{e^x}{2}\right ) \, dx\\ &=\frac{x^3}{6}+\frac{1}{2} x^2 \log \left (1+\frac{e^x}{2}\right )-x^2 \log \left (1+e^x\right )-2 x \text{Li}_2\left (-e^x\right )+x \text{Li}_2\left (-\frac{e^x}{2}\right )+2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^x\right )-\operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{x}{2}\right )}{x} \, dx,x,e^x\right )\\ &=\frac{x^3}{6}+\frac{1}{2} x^2 \log \left (1+\frac{e^x}{2}\right )-x^2 \log \left (1+e^x\right )-2 x \text{Li}_2\left (-e^x\right )+x \text{Li}_2\left (-\frac{e^x}{2}\right )+2 \text{Li}_3\left (-e^x\right )-\text{Li}_3\left (-\frac{e^x}{2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0098889, size = 77, normalized size = 1. \[ -x \text{PolyLog}\left (2,-2 e^{-x}\right )+2 x \text{PolyLog}\left (2,-e^{-x}\right )-\text{PolyLog}\left (3,-2 e^{-x}\right )+2 \text{PolyLog}\left (3,-e^{-x}\right )+x^2 \left (-\log \left (e^{-x}+1\right )\right )+\frac{1}{2} x^2 \log \left (2 e^{-x}+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(2 + 3*E^x + E^(2*x)),x]

[Out]

-(x^2*Log[1 + E^(-x)]) + (x^2*Log[1 + 2/E^x])/2 - x*PolyLog[2, -2/E^x] + 2*x*PolyLog[2, -E^(-x)] - PolyLog[3,
-2/E^x] + 2*PolyLog[3, -E^(-x)]

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Maple [A]  time = 0.006, size = 62, normalized size = 0.8 \begin{align*}{\frac{{x}^{3}}{6}}+{\frac{{x}^{2}}{2}\ln \left ( 1+{\frac{{{\rm e}^{x}}}{2}} \right ) }-{x}^{2}\ln \left ( 1+{{\rm e}^{x}} \right ) -2\,x{\it polylog} \left ( 2,-{{\rm e}^{x}} \right ) +x{\it polylog} \left ( 2,-{\frac{{{\rm e}^{x}}}{2}} \right ) +2\,{\it polylog} \left ( 3,-{{\rm e}^{x}} \right ) -{\it polylog} \left ( 3,-{\frac{{{\rm e}^{x}}}{2}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3+1/2*x^2*ln(1+1/2*exp(x))-x^2*ln(1+exp(x))-2*x*polylog(2,-exp(x))+x*polylog(2,-1/2*exp(x))+2*polylog(3,
-exp(x))-polylog(3,-1/2*exp(x))

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Maxima [A]  time = 0.989142, size = 80, normalized size = 1.04 \begin{align*} \frac{1}{6} \, x^{3} - x^{2} \log \left (e^{x} + 1\right ) + \frac{1}{2} \, x^{2} \log \left (\frac{1}{2} \, e^{x} + 1\right ) + x{\rm Li}_2\left (-\frac{1}{2} \, e^{x}\right ) - 2 \, x{\rm Li}_2\left (-e^{x}\right ) -{\rm Li}_{3}(-\frac{1}{2} \, e^{x}) + 2 \,{\rm Li}_{3}(-e^{x}) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3 - x^2*log(e^x + 1) + 1/2*x^2*log(1/2*e^x + 1) + x*dilog(-1/2*e^x) - 2*x*dilog(-e^x) - polylog(3, -1/2*
e^x) + 2*polylog(3, -e^x)

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Fricas [C]  time = 1.48589, size = 185, normalized size = 2.4 \begin{align*} \frac{1}{6} \, x^{3} - x^{2} \log \left (e^{x} + 1\right ) + \frac{1}{2} \, x^{2} \log \left (\frac{1}{2} \, e^{x} + 1\right ) + x{\rm Li}_2\left (-\frac{1}{2} \, e^{x}\right ) - 2 \, x{\rm Li}_2\left (-e^{x}\right ) -{\rm polylog}\left (3, -\frac{1}{2} \, e^{x}\right ) + 2 \,{\rm polylog}\left (3, -e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3 - x^2*log(e^x + 1) + 1/2*x^2*log(1/2*e^x + 1) + x*dilog(-1/2*e^x) - 2*x*dilog(-e^x) - polylog(3, -1/2*
e^x) + 2*polylog(3, -e^x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\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/(2+3*exp(x)+exp(2*x)),x)

[Out]

Integral(x**2/((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^{2}}{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/(2+3*exp(x)+exp(2*x)),x, algorithm="giac")

[Out]

integrate(x^2/(e^(2*x) + 3*e^x + 2), x)