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

Optimal. Leaf size=259 \[ -\frac{4 x \text{PolyLog}\left (2,-\frac{2 e^x}{1-\sqrt{5}}\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{4 x \text{PolyLog}\left (2,-\frac{2 e^x}{1+\sqrt{5}}\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}+\frac{4 \text{PolyLog}\left (3,-\frac{2 e^x}{1-\sqrt{5}}\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}-\frac{4 \text{PolyLog}\left (3,-\frac{2 e^x}{1+\sqrt{5}}\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}-\frac{2 x^3}{3 \sqrt{5} \left (1+\sqrt{5}\right )}+\frac{2 x^3}{3 \sqrt{5} \left (1-\sqrt{5}\right )}-\frac{2 x^2 \log \left (\frac{2 e^x}{1-\sqrt{5}}+1\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{2 x^2 \log \left (\frac{2 e^x}{1+\sqrt{5}}+1\right )}{\sqrt{5} \left (1+\sqrt{5}\right )} \]

[Out]

(2*x^3)/(3*Sqrt[5]*(1 - Sqrt[5])) - (2*x^3)/(3*Sqrt[5]*(1 + Sqrt[5])) - (2*x^2*Log[1 + (2*E^x)/(1 - Sqrt[5])])
/(Sqrt[5]*(1 - Sqrt[5])) + (2*x^2*Log[1 + (2*E^x)/(1 + Sqrt[5])])/(Sqrt[5]*(1 + Sqrt[5])) - (4*x*PolyLog[2, (-
2*E^x)/(1 - Sqrt[5])])/(Sqrt[5]*(1 - Sqrt[5])) + (4*x*PolyLog[2, (-2*E^x)/(1 + Sqrt[5])])/(Sqrt[5]*(1 + Sqrt[5
])) + (4*PolyLog[3, (-2*E^x)/(1 - Sqrt[5])])/(Sqrt[5]*(1 - Sqrt[5])) - (4*PolyLog[3, (-2*E^x)/(1 + Sqrt[5])])/
(Sqrt[5]*(1 + Sqrt[5]))

________________________________________________________________________________________

Rubi [A]  time = 0.296272, antiderivative size = 259, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {2263, 2184, 2190, 2531, 2282, 6589} \[ -\frac{4 x \text{PolyLog}\left (2,-\frac{2 e^x}{1-\sqrt{5}}\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{4 x \text{PolyLog}\left (2,-\frac{2 e^x}{1+\sqrt{5}}\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}+\frac{4 \text{PolyLog}\left (3,-\frac{2 e^x}{1-\sqrt{5}}\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}-\frac{4 \text{PolyLog}\left (3,-\frac{2 e^x}{1+\sqrt{5}}\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}-\frac{2 x^3}{3 \sqrt{5} \left (1+\sqrt{5}\right )}+\frac{2 x^3}{3 \sqrt{5} \left (1-\sqrt{5}\right )}-\frac{2 x^2 \log \left (\frac{2 e^x}{1-\sqrt{5}}+1\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{2 x^2 \log \left (\frac{2 e^x}{1+\sqrt{5}}+1\right )}{\sqrt{5} \left (1+\sqrt{5}\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(-1 + E^x + E^(2*x)),x]

[Out]

(2*x^3)/(3*Sqrt[5]*(1 - Sqrt[5])) - (2*x^3)/(3*Sqrt[5]*(1 + Sqrt[5])) - (2*x^2*Log[1 + (2*E^x)/(1 - Sqrt[5])])
/(Sqrt[5]*(1 - Sqrt[5])) + (2*x^2*Log[1 + (2*E^x)/(1 + Sqrt[5])])/(Sqrt[5]*(1 + Sqrt[5])) - (4*x*PolyLog[2, (-
2*E^x)/(1 - Sqrt[5])])/(Sqrt[5]*(1 - Sqrt[5])) + (4*x*PolyLog[2, (-2*E^x)/(1 + Sqrt[5])])/(Sqrt[5]*(1 + Sqrt[5
])) + (4*PolyLog[3, (-2*E^x)/(1 - Sqrt[5])])/(Sqrt[5]*(1 - Sqrt[5])) - (4*PolyLog[3, (-2*E^x)/(1 + Sqrt[5])])/
(Sqrt[5]*(1 + Sqrt[5]))

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}{-1+e^x+e^{2 x}} \, dx &=\frac{2 \int \frac{x^2}{1-\sqrt{5}+2 e^x} \, dx}{\sqrt{5}}-\frac{2 \int \frac{x^2}{1+\sqrt{5}+2 e^x} \, dx}{\sqrt{5}}\\ &=\frac{2 x^3}{3 \sqrt{5} \left (1-\sqrt{5}\right )}-\frac{2 x^3}{3 \sqrt{5} \left (1+\sqrt{5}\right )}-\frac{4 \int \frac{e^x x^2}{1-\sqrt{5}+2 e^x} \, dx}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{4 \int \frac{e^x x^2}{1+\sqrt{5}+2 e^x} \, dx}{\sqrt{5} \left (1+\sqrt{5}\right )}\\ &=\frac{2 x^3}{3 \sqrt{5} \left (1-\sqrt{5}\right )}-\frac{2 x^3}{3 \sqrt{5} \left (1+\sqrt{5}\right )}-\frac{2 x^2 \log \left (1+\frac{2 e^x}{1-\sqrt{5}}\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{2 x^2 \log \left (1+\frac{2 e^x}{1+\sqrt{5}}\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}+\frac{4 \int x \log \left (1+\frac{2 e^x}{1-\sqrt{5}}\right ) \, dx}{\sqrt{5} \left (1-\sqrt{5}\right )}-\frac{4 \int x \log \left (1+\frac{2 e^x}{1+\sqrt{5}}\right ) \, dx}{\sqrt{5} \left (1+\sqrt{5}\right )}\\ &=\frac{2 x^3}{3 \sqrt{5} \left (1-\sqrt{5}\right )}-\frac{2 x^3}{3 \sqrt{5} \left (1+\sqrt{5}\right )}-\frac{2 x^2 \log \left (1+\frac{2 e^x}{1-\sqrt{5}}\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{2 x^2 \log \left (1+\frac{2 e^x}{1+\sqrt{5}}\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}-\frac{4 x \text{Li}_2\left (-\frac{2 e^x}{1-\sqrt{5}}\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{4 x \text{Li}_2\left (-\frac{2 e^x}{1+\sqrt{5}}\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}+\frac{4 \int \text{Li}_2\left (-\frac{2 e^x}{1-\sqrt{5}}\right ) \, dx}{\sqrt{5} \left (1-\sqrt{5}\right )}-\frac{4 \int \text{Li}_2\left (-\frac{2 e^x}{1+\sqrt{5}}\right ) \, dx}{\sqrt{5} \left (1+\sqrt{5}\right )}\\ &=\frac{2 x^3}{3 \sqrt{5} \left (1-\sqrt{5}\right )}-\frac{2 x^3}{3 \sqrt{5} \left (1+\sqrt{5}\right )}-\frac{2 x^2 \log \left (1+\frac{2 e^x}{1-\sqrt{5}}\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{2 x^2 \log \left (1+\frac{2 e^x}{1+\sqrt{5}}\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}-\frac{4 x \text{Li}_2\left (-\frac{2 e^x}{1-\sqrt{5}}\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{4 x \text{Li}_2\left (-\frac{2 e^x}{1+\sqrt{5}}\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}+\frac{4 \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{2 x}{-1+\sqrt{5}}\right )}{x} \, dx,x,e^x\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}-\frac{4 \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{2 x}{1+\sqrt{5}}\right )}{x} \, dx,x,e^x\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}\\ &=\frac{2 x^3}{3 \sqrt{5} \left (1-\sqrt{5}\right )}-\frac{2 x^3}{3 \sqrt{5} \left (1+\sqrt{5}\right )}-\frac{2 x^2 \log \left (1+\frac{2 e^x}{1-\sqrt{5}}\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{2 x^2 \log \left (1+\frac{2 e^x}{1+\sqrt{5}}\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}-\frac{4 x \text{Li}_2\left (-\frac{2 e^x}{1-\sqrt{5}}\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}+\frac{4 x \text{Li}_2\left (-\frac{2 e^x}{1+\sqrt{5}}\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}+\frac{4 \text{Li}_3\left (-\frac{2 e^x}{1-\sqrt{5}}\right )}{\sqrt{5} \left (1-\sqrt{5}\right )}-\frac{4 \text{Li}_3\left (-\frac{2 e^x}{1+\sqrt{5}}\right )}{\sqrt{5} \left (1+\sqrt{5}\right )}\\ \end{align*}

Mathematica [A]  time = 0.150859, size = 172, normalized size = 0.66 \[ \frac{2 \left (-\frac{2 \left (x \text{PolyLog}\left (2,\frac{1}{2} \left (\sqrt{5}-1\right ) e^{-x}\right )+\text{PolyLog}\left (3,\frac{1}{2} \left (\sqrt{5}-1\right ) e^{-x}\right )\right )}{\sqrt{5}-1}-\frac{2 \left (x \text{PolyLog}\left (2,-\frac{1}{2} \left (1+\sqrt{5}\right ) e^{-x}\right )+\text{PolyLog}\left (3,-\frac{1}{2} \left (1+\sqrt{5}\right ) e^{-x}\right )\right )}{1+\sqrt{5}}+\frac{x^2 \log \left (1-\frac{1}{2} \left (\sqrt{5}-1\right ) e^{-x}\right )}{\sqrt{5}-1}+\frac{x^2 \log \left (\frac{1}{2} \left (1+\sqrt{5}\right ) e^{-x}+1\right )}{1+\sqrt{5}}\right )}{\sqrt{5}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(-1 + E^x + E^(2*x)),x]

[Out]

(2*((x^2*Log[1 - (-1 + Sqrt[5])/(2*E^x)])/(-1 + Sqrt[5]) + (x^2*Log[1 + (1 + Sqrt[5])/(2*E^x)])/(1 + Sqrt[5])
- (2*(x*PolyLog[2, (-1 + Sqrt[5])/(2*E^x)] + PolyLog[3, (-1 + Sqrt[5])/(2*E^x)]))/(-1 + Sqrt[5]) - (2*(x*PolyL
og[2, -(1 + Sqrt[5])/(2*E^x)] + PolyLog[3, -(1 + Sqrt[5])/(2*E^x)]))/(1 + Sqrt[5])))/Sqrt[5]

________________________________________________________________________________________

Maple [F]  time = 0.028, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{2}}{-1+{{\rm e}^{x}}+{{\rm e}^{2\,x}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(e^(2*x) + e^x - 1), x)

________________________________________________________________________________________

Fricas [C]  time = 1.49855, size = 468, normalized size = 1.81 \begin{align*} -\frac{1}{3} \, x^{3} + \frac{1}{5} \,{\left (\sqrt{5} x + 5 \, x\right )}{\rm Li}_2\left (\frac{1}{2} \,{\left (\sqrt{5} + 1\right )} e^{x}\right ) - \frac{1}{5} \,{\left (\sqrt{5} x - 5 \, x\right )}{\rm Li}_2\left (-\frac{1}{2} \,{\left (\sqrt{5} - 1\right )} e^{x}\right ) + \frac{1}{10} \,{\left (\sqrt{5} x^{2} + 5 \, x^{2}\right )} \log \left (-\frac{1}{2} \,{\left (\sqrt{5} + 1\right )} e^{x} + 1\right ) - \frac{1}{10} \,{\left (\sqrt{5} x^{2} - 5 \, x^{2}\right )} \log \left (\frac{1}{2} \,{\left (\sqrt{5} - 1\right )} e^{x} + 1\right ) - \frac{1}{5} \,{\left (\sqrt{5} + 5\right )}{\rm polylog}\left (3, \frac{1}{2} \,{\left (\sqrt{5} + 1\right )} e^{x}\right ) + \frac{1}{5} \,{\left (\sqrt{5} - 5\right )}{\rm polylog}\left (3, -\frac{1}{2} \,{\left (\sqrt{5} - 1\right )} e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*x^3 + 1/5*(sqrt(5)*x + 5*x)*dilog(1/2*(sqrt(5) + 1)*e^x) - 1/5*(sqrt(5)*x - 5*x)*dilog(-1/2*(sqrt(5) - 1)
*e^x) + 1/10*(sqrt(5)*x^2 + 5*x^2)*log(-1/2*(sqrt(5) + 1)*e^x + 1) - 1/10*(sqrt(5)*x^2 - 5*x^2)*log(1/2*(sqrt(
5) - 1)*e^x + 1) - 1/5*(sqrt(5) + 5)*polylog(3, 1/2*(sqrt(5) + 1)*e^x) + 1/5*(sqrt(5) - 5)*polylog(3, -1/2*(sq
rt(5) - 1)*e^x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(e^(2*x) + e^x - 1), x)