3.539 \(\int \frac{x^2}{a+b e^{-x}+c e^x} \, dx\)

Optimal. Leaf size=244 \[ \frac{2 x \text{PolyLog}\left (2,-\frac{2 c e^x}{a-\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}-\frac{2 x \text{PolyLog}\left (2,-\frac{2 c e^x}{\sqrt{a^2-4 b c}+a}\right )}{\sqrt{a^2-4 b c}}-\frac{2 \text{PolyLog}\left (3,-\frac{2 c e^x}{a-\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}+\frac{2 \text{PolyLog}\left (3,-\frac{2 c e^x}{\sqrt{a^2-4 b c}+a}\right )}{\sqrt{a^2-4 b c}}+\frac{x^2 \log \left (\frac{2 c e^x}{a-\sqrt{a^2-4 b c}}+1\right )}{\sqrt{a^2-4 b c}}-\frac{x^2 \log \left (\frac{2 c e^x}{\sqrt{a^2-4 b c}+a}+1\right )}{\sqrt{a^2-4 b c}} \]

[Out]

(x^2*Log[1 + (2*c*E^x)/(a - Sqrt[a^2 - 4*b*c])])/Sqrt[a^2 - 4*b*c] - (x^2*Log[1 + (2*c*E^x)/(a + Sqrt[a^2 - 4*
b*c])])/Sqrt[a^2 - 4*b*c] + (2*x*PolyLog[2, (-2*c*E^x)/(a - Sqrt[a^2 - 4*b*c])])/Sqrt[a^2 - 4*b*c] - (2*x*Poly
Log[2, (-2*c*E^x)/(a + Sqrt[a^2 - 4*b*c])])/Sqrt[a^2 - 4*b*c] - (2*PolyLog[3, (-2*c*E^x)/(a - Sqrt[a^2 - 4*b*c
])])/Sqrt[a^2 - 4*b*c] + (2*PolyLog[3, (-2*c*E^x)/(a + Sqrt[a^2 - 4*b*c])])/Sqrt[a^2 - 4*b*c]

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Rubi [A]  time = 0.498767, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {2267, 2264, 2190, 2531, 2282, 6589} \[ \frac{2 x \text{PolyLog}\left (2,-\frac{2 c e^x}{a-\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}-\frac{2 x \text{PolyLog}\left (2,-\frac{2 c e^x}{\sqrt{a^2-4 b c}+a}\right )}{\sqrt{a^2-4 b c}}-\frac{2 \text{PolyLog}\left (3,-\frac{2 c e^x}{a-\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}+\frac{2 \text{PolyLog}\left (3,-\frac{2 c e^x}{\sqrt{a^2-4 b c}+a}\right )}{\sqrt{a^2-4 b c}}+\frac{x^2 \log \left (\frac{2 c e^x}{a-\sqrt{a^2-4 b c}}+1\right )}{\sqrt{a^2-4 b c}}-\frac{x^2 \log \left (\frac{2 c e^x}{\sqrt{a^2-4 b c}+a}+1\right )}{\sqrt{a^2-4 b c}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b/E^x + c*E^x),x]

[Out]

(x^2*Log[1 + (2*c*E^x)/(a - Sqrt[a^2 - 4*b*c])])/Sqrt[a^2 - 4*b*c] - (x^2*Log[1 + (2*c*E^x)/(a + Sqrt[a^2 - 4*
b*c])])/Sqrt[a^2 - 4*b*c] + (2*x*PolyLog[2, (-2*c*E^x)/(a - Sqrt[a^2 - 4*b*c])])/Sqrt[a^2 - 4*b*c] - (2*x*Poly
Log[2, (-2*c*E^x)/(a + Sqrt[a^2 - 4*b*c])])/Sqrt[a^2 - 4*b*c] - (2*PolyLog[3, (-2*c*E^x)/(a - Sqrt[a^2 - 4*b*c
])])/Sqrt[a^2 - 4*b*c] + (2*PolyLog[3, (-2*c*E^x)/(a + Sqrt[a^2 - 4*b*c])])/Sqrt[a^2 - 4*b*c]

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 2264

Int[((F_)^(u_)*((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*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(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 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}{a+b e^{-x}+c e^x} \, dx &=\int \frac{e^x x^2}{b+a e^x+c e^{2 x}} \, dx\\ &=\frac{(2 c) \int \frac{e^x x^2}{a-\sqrt{a^2-4 b c}+2 c e^x} \, dx}{\sqrt{a^2-4 b c}}-\frac{(2 c) \int \frac{e^x x^2}{a+\sqrt{a^2-4 b c}+2 c e^x} \, dx}{\sqrt{a^2-4 b c}}\\ &=\frac{x^2 \log \left (1+\frac{2 c e^x}{a-\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}-\frac{x^2 \log \left (1+\frac{2 c e^x}{a+\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}-\frac{2 \int x \log \left (1+\frac{2 c e^x}{a-\sqrt{a^2-4 b c}}\right ) \, dx}{\sqrt{a^2-4 b c}}+\frac{2 \int x \log \left (1+\frac{2 c e^x}{a+\sqrt{a^2-4 b c}}\right ) \, dx}{\sqrt{a^2-4 b c}}\\ &=\frac{x^2 \log \left (1+\frac{2 c e^x}{a-\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}-\frac{x^2 \log \left (1+\frac{2 c e^x}{a+\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}+\frac{2 x \text{Li}_2\left (-\frac{2 c e^x}{a-\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}-\frac{2 x \text{Li}_2\left (-\frac{2 c e^x}{a+\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}-\frac{2 \int \text{Li}_2\left (-\frac{2 c e^x}{a-\sqrt{a^2-4 b c}}\right ) \, dx}{\sqrt{a^2-4 b c}}+\frac{2 \int \text{Li}_2\left (-\frac{2 c e^x}{a+\sqrt{a^2-4 b c}}\right ) \, dx}{\sqrt{a^2-4 b c}}\\ &=\frac{x^2 \log \left (1+\frac{2 c e^x}{a-\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}-\frac{x^2 \log \left (1+\frac{2 c e^x}{a+\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}+\frac{2 x \text{Li}_2\left (-\frac{2 c e^x}{a-\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}-\frac{2 x \text{Li}_2\left (-\frac{2 c e^x}{a+\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}-\frac{2 \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{2 c x}{-a+\sqrt{a^2-4 b c}}\right )}{x} \, dx,x,e^x\right )}{\sqrt{a^2-4 b c}}+\frac{2 \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{2 c x}{a+\sqrt{a^2-4 b c}}\right )}{x} \, dx,x,e^x\right )}{\sqrt{a^2-4 b c}}\\ &=\frac{x^2 \log \left (1+\frac{2 c e^x}{a-\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}-\frac{x^2 \log \left (1+\frac{2 c e^x}{a+\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}+\frac{2 x \text{Li}_2\left (-\frac{2 c e^x}{a-\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}-\frac{2 x \text{Li}_2\left (-\frac{2 c e^x}{a+\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}-\frac{2 \text{Li}_3\left (-\frac{2 c e^x}{a-\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}+\frac{2 \text{Li}_3\left (-\frac{2 c e^x}{a+\sqrt{a^2-4 b c}}\right )}{\sqrt{a^2-4 b c}}\\ \end{align*}

Mathematica [A]  time = 0.0374542, size = 185, normalized size = 0.76 \[ \frac{2 x \text{PolyLog}\left (2,\frac{2 c e^x}{\sqrt{a^2-4 b c}-a}\right )-2 x \text{PolyLog}\left (2,-\frac{2 c e^x}{\sqrt{a^2-4 b c}+a}\right )-2 \text{PolyLog}\left (3,\frac{2 c e^x}{\sqrt{a^2-4 b c}-a}\right )+2 \text{PolyLog}\left (3,-\frac{2 c e^x}{\sqrt{a^2-4 b c}+a}\right )+x^2 \log \left (\frac{2 c e^x}{a-\sqrt{a^2-4 b c}}+1\right )-x^2 \log \left (\frac{2 c e^x}{\sqrt{a^2-4 b c}+a}+1\right )}{\sqrt{a^2-4 b c}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b/E^x + c*E^x),x]

[Out]

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

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Maple [F]  time = 0.017, size = 0, normalized size = 0. \begin{align*} \int{{x}^{2} \left ( a+{\frac{b}{{{\rm e}^{x}}}}+c{{\rm e}^{x}} \right ) ^{-1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a+b/exp(x)+c*exp(x)),x)

[Out]

int(x^2/(a+b/exp(x)+c*exp(x)),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+b/exp(x)+c*exp(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [C]  time = 1.38465, size = 759, normalized size = 3.11 \begin{align*} \frac{b x^{2} \sqrt{\frac{a^{2} - 4 \, b c}{b^{2}}} \log \left (\frac{b \sqrt{\frac{a^{2} - 4 \, b c}{b^{2}}} e^{x} + a e^{x} + 2 \, b}{2 \, b}\right ) - b x^{2} \sqrt{\frac{a^{2} - 4 \, b c}{b^{2}}} \log \left (-\frac{b \sqrt{\frac{a^{2} - 4 \, b c}{b^{2}}} e^{x} - a e^{x} - 2 \, b}{2 \, b}\right ) + 2 \, b x \sqrt{\frac{a^{2} - 4 \, b c}{b^{2}}}{\rm Li}_2\left (-\frac{b \sqrt{\frac{a^{2} - 4 \, b c}{b^{2}}} e^{x} + a e^{x} + 2 \, b}{2 \, b} + 1\right ) - 2 \, b x \sqrt{\frac{a^{2} - 4 \, b c}{b^{2}}}{\rm Li}_2\left (\frac{b \sqrt{\frac{a^{2} - 4 \, b c}{b^{2}}} e^{x} - a e^{x} - 2 \, b}{2 \, b} + 1\right ) - 2 \, b \sqrt{\frac{a^{2} - 4 \, b c}{b^{2}}}{\rm polylog}\left (3, -\frac{b \sqrt{\frac{a^{2} - 4 \, b c}{b^{2}}} e^{x} + a e^{x}}{2 \, b}\right ) + 2 \, b \sqrt{\frac{a^{2} - 4 \, b c}{b^{2}}}{\rm polylog}\left (3, \frac{b \sqrt{\frac{a^{2} - 4 \, b c}{b^{2}}} e^{x} - a e^{x}}{2 \, b}\right )}{a^{2} - 4 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+b/exp(x)+c*exp(x)),x, algorithm="fricas")

[Out]

(b*x^2*sqrt((a^2 - 4*b*c)/b^2)*log(1/2*(b*sqrt((a^2 - 4*b*c)/b^2)*e^x + a*e^x + 2*b)/b) - b*x^2*sqrt((a^2 - 4*
b*c)/b^2)*log(-1/2*(b*sqrt((a^2 - 4*b*c)/b^2)*e^x - a*e^x - 2*b)/b) + 2*b*x*sqrt((a^2 - 4*b*c)/b^2)*dilog(-1/2
*(b*sqrt((a^2 - 4*b*c)/b^2)*e^x + a*e^x + 2*b)/b + 1) - 2*b*x*sqrt((a^2 - 4*b*c)/b^2)*dilog(1/2*(b*sqrt((a^2 -
4*b*c)/b^2)*e^x - a*e^x - 2*b)/b + 1) - 2*b*sqrt((a^2 - 4*b*c)/b^2)*polylog(3, -1/2*(b*sqrt((a^2 - 4*b*c)/b^2
)*e^x + a*e^x)/b) + 2*b*sqrt((a^2 - 4*b*c)/b^2)*polylog(3, 1/2*(b*sqrt((a^2 - 4*b*c)/b^2)*e^x - a*e^x)/b))/(a^
2 - 4*b*c)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(a+b/exp(x)+c*exp(x)),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+b/exp(x)+c*exp(x)),x, algorithm="giac")

[Out]

integrate(x^2/(b*e^(-x) + c*e^x + a), x)