3.6.17 \(\int \frac {x^2}{-1+e^x+e^{2 x}} \, dx\) [517]

Optimal. Leaf size=259 \[ \frac {2 x^3}{3 \sqrt {5} \left (1-\sqrt {5}\right )}-\frac {2 x^3}{3 \sqrt {5} \left (1+\sqrt {5}\right )}-\frac {2 x^2 \log \left (1+\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {2 x^2 \log \left (1+\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}-\frac {4 x \text {Li}_2\left (-\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}+\frac {4 x \text {Li}_2\left (-\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )}+\frac {4 \text {Li}_3\left (-\frac {2 e^x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right )}-\frac {4 \text {Li}_3\left (-\frac {2 e^x}{1+\sqrt {5}}\right )}{\sqrt {5} \left (1+\sqrt {5}\right )} \]

[Out]

2/15*x^3/(-5^(1/2)+1)*5^(1/2)-2/5*x^2*ln(1+2*exp(x)/(-5^(1/2)+1))/(-5^(1/2)+1)*5^(1/2)-4/5*x*polylog(2,-2*exp(
x)/(-5^(1/2)+1))/(-5^(1/2)+1)*5^(1/2)+4/5*polylog(3,-2*exp(x)/(-5^(1/2)+1))/(-5^(1/2)+1)*5^(1/2)-2/15*x^3*5^(1
/2)/(5^(1/2)+1)+2/5*x^2*ln(1+2*exp(x)/(5^(1/2)+1))*5^(1/2)/(5^(1/2)+1)+4/5*x*polylog(2,-2*exp(x)/(5^(1/2)+1))*
5^(1/2)/(5^(1/2)+1)-4/5*polylog(3,-2*exp(x)/(5^(1/2)+1))*5^(1/2)/(5^(1/2)+1)

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Rubi [A]
time = 0.19, antiderivative size = 259, normalized size of antiderivative = 1.00, 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 = {2295, 2215, 2221, 2611, 2320, 6724} \begin {gather*} -\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 )} \end {gather*}

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 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 2320

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 2611

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 6724

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 \text {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 \text {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*}

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Mathematica [A]
time = 0.11, size = 172, normalized size = 0.66 \begin {gather*} \frac {2 \left (\frac {x^2 \log \left (1-\frac {1}{2} \left (-1+\sqrt {5}\right ) e^{-x}\right )}{-1+\sqrt {5}}+\frac {x^2 \log \left (1+\frac {1}{2} \left (1+\sqrt {5}\right ) e^{-x}\right )}{1+\sqrt {5}}-\frac {2 \left (x \text {Li}_2\left (\frac {1}{2} \left (-1+\sqrt {5}\right ) e^{-x}\right )+\text {Li}_3\left (\frac {1}{2} \left (-1+\sqrt {5}\right ) e^{-x}\right )\right )}{-1+\sqrt {5}}-\frac {2 \left (x \text {Li}_2\left (-\frac {1}{2} \left (1+\sqrt {5}\right ) e^{-x}\right )+\text {Li}_3\left (-\frac {1}{2} \left (1+\sqrt {5}\right ) e^{-x}\right )\right )}{1+\sqrt {5}}\right )}{\sqrt {5}} \end {gather*}

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {x^{2}}{-1+{\mathrm e}^{x}+{\mathrm e}^{2 x}}\, dx\]

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)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

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Fricas [A]
time = 0.40, size = 138, normalized size = 0.53 \begin {gather*} -\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 {gather*}

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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{e^{2 x} + e^{x} - 1}\, dx \end {gather*}

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)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^2}{{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x-1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(exp(2*x) + exp(x) - 1),x)

[Out]

int(x^2/(exp(2*x) + exp(x) - 1), x)

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