3.531 \(\int \frac{x^2}{2+e^{-x}+e^x} \, dx\)

Optimal. Leaf size=34 \[ -2 \text{PolyLog}\left (2,-e^x\right )-\frac{x^2}{e^x+1}+x^2-2 x \log \left (e^x+1\right ) \]

[Out]

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

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Rubi [A]  time = 0.249505, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {2267, 6688, 2191, 2184, 2190, 2279, 2391} \[ -2 \text{PolyLog}\left (2,-e^x\right )-\frac{x^2}{e^x+1}+x^2-2 x \log \left (e^x+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2/(2 + E^(-x) + E^x),x]

[Out]

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

Rule 2267

Int[(u_)/((a_) + (b_.)*(F_)^(v_) + (c_.)*(F_)^(w_)), x_Symbol] :> Int[(u*F^v)/(c + a*F^v + b*F^(2*v)), x] /; F
reeQ[{F, a, b, c}, x] && EqQ[w, -v] && LinearQ[v, x] && If[RationalQ[Coefficient[v, x, 1]], GtQ[Coefficient[v,
x, 1], 0], LtQ[LeafCount[v], LeafCount[w]]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 2191

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

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

Mathematica [A]  time = 0.0469111, size = 33, normalized size = 0.97 \[ x \left (\frac{e^x x}{e^x+1}-2 \log \left (e^x+1\right )\right )-2 \text{PolyLog}\left (2,-e^x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(2 + E^(-x) + E^x),x]

[Out]

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

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Maple [A]  time = 0.02, size = 32, normalized size = 0.9 \begin{align*}{x}^{2}-{\frac{{x}^{2}}{1+{{\rm e}^{x}}}}-2\,x\ln \left ( 1+{{\rm e}^{x}} \right ) -2\,{\it polylog} \left ( 2,-{{\rm e}^{x}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(2+exp(-x)+exp(x)),x)

[Out]

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

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Maxima [A]  time = 0.980707, size = 41, normalized size = 1.21 \begin{align*} x^{2} - 2 \, x \log \left (e^{x} + 1\right ) - \frac{x^{2}}{e^{x} + 1} - 2 \,{\rm Li}_2\left (-e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.31795, size = 103, normalized size = 3.03 \begin{align*} \frac{x^{2} e^{x} - 2 \,{\left (e^{x} + 1\right )}{\rm Li}_2\left (-e^{x}\right ) - 2 \,{\left (x e^{x} + x\right )} \log \left (e^{x} + 1\right )}{e^{x} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(e^(-x) + e^x + 2), x)