3.6.19 \(\int \frac {x^2}{a+b e^x+c e^{2 x}} \, dx\) [519]

Optimal. Leaf size=391 \[ -\frac {2 c x^3}{3 \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right )}-\frac {2 c x^3}{3 \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right )}+\frac {2 c x^2 \log \left (1+\frac {2 c e^x}{b-\sqrt {b^2-4 a c}}\right )}{b^2-4 a c-b \sqrt {b^2-4 a c}}+\frac {2 c x^2 \log \left (1+\frac {2 c e^x}{b+\sqrt {b^2-4 a c}}\right )}{b^2-4 a c+b \sqrt {b^2-4 a c}}+\frac {4 c x \text {Li}_2\left (-\frac {2 c e^x}{b-\sqrt {b^2-4 a c}}\right )}{b^2-4 a c-b \sqrt {b^2-4 a c}}+\frac {4 c x \text {Li}_2\left (-\frac {2 c e^x}{b+\sqrt {b^2-4 a c}}\right )}{b^2-4 a c+b \sqrt {b^2-4 a c}}-\frac {4 c \text {Li}_3\left (-\frac {2 c e^x}{b-\sqrt {b^2-4 a c}}\right )}{b^2-4 a c-b \sqrt {b^2-4 a c}}-\frac {4 c \text {Li}_3\left (-\frac {2 c e^x}{b+\sqrt {b^2-4 a c}}\right )}{b^2-4 a c+b \sqrt {b^2-4 a c}} \]

[Out]

-2/3*c*x^3/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))+2*c*x^2*ln(1+2*c*exp(x)/(b-(-4*a*c+b^2)^(1/2)))/(b^2-4*a*c-b*(-4*a
*c+b^2)^(1/2))+4*c*x*polylog(2,-2*c*exp(x)/(b-(-4*a*c+b^2)^(1/2)))/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))-4*c*polylo
g(3,-2*c*exp(x)/(b-(-4*a*c+b^2)^(1/2)))/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))-2/3*c*x^3/(b^2-4*a*c+b*(-4*a*c+b^2)^(
1/2))+2*c*x^2*ln(1+2*c*exp(x)/(b+(-4*a*c+b^2)^(1/2)))/(b^2-4*a*c+b*(-4*a*c+b^2)^(1/2))+4*c*x*polylog(2,-2*c*ex
p(x)/(b+(-4*a*c+b^2)^(1/2)))/(b^2-4*a*c+b*(-4*a*c+b^2)^(1/2))-4*c*polylog(3,-2*c*exp(x)/(b+(-4*a*c+b^2)^(1/2))
)/(b^2-4*a*c+b*(-4*a*c+b^2)^(1/2))

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Rubi [A]
time = 0.42, antiderivative size = 391, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2295, 2215, 2221, 2611, 2320, 6724} \begin {gather*} \frac {4 c x \text {PolyLog}\left (2,-\frac {2 c e^x}{b-\sqrt {b^2-4 a c}}\right )}{-b \sqrt {b^2-4 a c}-4 a c+b^2}+\frac {4 c x \text {PolyLog}\left (2,-\frac {2 c e^x}{\sqrt {b^2-4 a c}+b}\right )}{b \sqrt {b^2-4 a c}-4 a c+b^2}-\frac {4 c \text {PolyLog}\left (3,-\frac {2 c e^x}{b-\sqrt {b^2-4 a c}}\right )}{-b \sqrt {b^2-4 a c}-4 a c+b^2}-\frac {4 c \text {PolyLog}\left (3,-\frac {2 c e^x}{\sqrt {b^2-4 a c}+b}\right )}{b \sqrt {b^2-4 a c}-4 a c+b^2}-\frac {2 c x^3}{3 \left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right )}-\frac {2 c x^3}{3 \left (b \sqrt {b^2-4 a c}-4 a c+b^2\right )}+\frac {2 c x^2 \log \left (\frac {2 c e^x}{b-\sqrt {b^2-4 a c}}+1\right )}{-b \sqrt {b^2-4 a c}-4 a c+b^2}+\frac {2 c x^2 \log \left (\frac {2 c e^x}{\sqrt {b^2-4 a c}+b}+1\right )}{b \sqrt {b^2-4 a c}-4 a c+b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*E^x + c*E^(2*x)),x]

[Out]

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

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

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Mathematica [A]
time = 0.16, size = 407, normalized size = 1.04 \begin {gather*} \frac {2 \sqrt {b^2-4 a c} x^3-3 b x^2 \log \left (1+\frac {2 c e^x}{b-\sqrt {b^2-4 a c}}\right )-3 \sqrt {b^2-4 a c} x^2 \log \left (1+\frac {2 c e^x}{b-\sqrt {b^2-4 a c}}\right )+3 b x^2 \log \left (1+\frac {2 c e^x}{b+\sqrt {b^2-4 a c}}\right )-3 \sqrt {b^2-4 a c} x^2 \log \left (1+\frac {2 c e^x}{b+\sqrt {b^2-4 a c}}\right )-6 \left (b+\sqrt {b^2-4 a c}\right ) x \text {Li}_2\left (\frac {2 c e^x}{-b+\sqrt {b^2-4 a c}}\right )+6 \left (b-\sqrt {b^2-4 a c}\right ) x \text {Li}_2\left (-\frac {2 c e^x}{b+\sqrt {b^2-4 a c}}\right )+6 b \text {Li}_3\left (\frac {2 c e^x}{-b+\sqrt {b^2-4 a c}}\right )+6 \sqrt {b^2-4 a c} \text {Li}_3\left (\frac {2 c e^x}{-b+\sqrt {b^2-4 a c}}\right )-6 b \text {Li}_3\left (-\frac {2 c e^x}{b+\sqrt {b^2-4 a c}}\right )+6 \sqrt {b^2-4 a c} \text {Li}_3\left (-\frac {2 c e^x}{b+\sqrt {b^2-4 a c}}\right )}{6 a \sqrt {b^2-4 a c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*E^x + c*E^(2*x)),x]

[Out]

(2*Sqrt[b^2 - 4*a*c]*x^3 - 3*b*x^2*Log[1 + (2*c*E^x)/(b - Sqrt[b^2 - 4*a*c])] - 3*Sqrt[b^2 - 4*a*c]*x^2*Log[1
+ (2*c*E^x)/(b - Sqrt[b^2 - 4*a*c])] + 3*b*x^2*Log[1 + (2*c*E^x)/(b + Sqrt[b^2 - 4*a*c])] - 3*Sqrt[b^2 - 4*a*c
]*x^2*Log[1 + (2*c*E^x)/(b + Sqrt[b^2 - 4*a*c])] - 6*(b + Sqrt[b^2 - 4*a*c])*x*PolyLog[2, (2*c*E^x)/(-b + Sqrt
[b^2 - 4*a*c])] + 6*(b - Sqrt[b^2 - 4*a*c])*x*PolyLog[2, (-2*c*E^x)/(b + Sqrt[b^2 - 4*a*c])] + 6*b*PolyLog[3,
(2*c*E^x)/(-b + Sqrt[b^2 - 4*a*c])] + 6*Sqrt[b^2 - 4*a*c]*PolyLog[3, (2*c*E^x)/(-b + Sqrt[b^2 - 4*a*c])] - 6*b
*PolyLog[3, (-2*c*E^x)/(b + Sqrt[b^2 - 4*a*c])] + 6*Sqrt[b^2 - 4*a*c]*PolyLog[3, (-2*c*E^x)/(b + Sqrt[b^2 - 4*
a*c])])/(6*a*Sqrt[b^2 - 4*a*c])

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

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Fricas [A]
time = 0.37, size = 415, normalized size = 1.06 \begin {gather*} \frac {2 \, {\left (b^{2} - 4 \, a c\right )} x^{3} - 6 \, {\left (a b x \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} + {\left (b^{2} - 4 \, a c\right )} x\right )} {\rm Li}_2\left (-\frac {a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} e^{x} + b e^{x} + 2 \, a}{2 \, a} + 1\right ) + 6 \, {\left (a b x \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} - {\left (b^{2} - 4 \, a c\right )} x\right )} {\rm Li}_2\left (\frac {a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} e^{x} - b e^{x} - 2 \, a}{2 \, a} + 1\right ) - 3 \, {\left (a b x^{2} \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} + {\left (b^{2} - 4 \, a c\right )} x^{2}\right )} \log \left (\frac {a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} e^{x} + b e^{x} + 2 \, a}{2 \, a}\right ) + 3 \, {\left (a b x^{2} \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} - {\left (b^{2} - 4 \, a c\right )} x^{2}\right )} \log \left (-\frac {a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} e^{x} - b e^{x} - 2 \, a}{2 \, a}\right ) + 6 \, {\left (a b \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} + b^{2} - 4 \, a c\right )} {\rm polylog}\left (3, -\frac {a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} e^{x} + b e^{x}}{2 \, a}\right ) - 6 \, {\left (a b \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} - b^{2} + 4 \, a c\right )} {\rm polylog}\left (3, \frac {a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} e^{x} - b e^{x}}{2 \, a}\right )}{6 \, {\left (a b^{2} - 4 \, a^{2} c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*(b^2 - 4*a*c)*x^3 - 6*(a*b*x*sqrt((b^2 - 4*a*c)/a^2) + (b^2 - 4*a*c)*x)*dilog(-1/2*(a*sqrt((b^2 - 4*a*c
)/a^2)*e^x + b*e^x + 2*a)/a + 1) + 6*(a*b*x*sqrt((b^2 - 4*a*c)/a^2) - (b^2 - 4*a*c)*x)*dilog(1/2*(a*sqrt((b^2
- 4*a*c)/a^2)*e^x - b*e^x - 2*a)/a + 1) - 3*(a*b*x^2*sqrt((b^2 - 4*a*c)/a^2) + (b^2 - 4*a*c)*x^2)*log(1/2*(a*s
qrt((b^2 - 4*a*c)/a^2)*e^x + b*e^x + 2*a)/a) + 3*(a*b*x^2*sqrt((b^2 - 4*a*c)/a^2) - (b^2 - 4*a*c)*x^2)*log(-1/
2*(a*sqrt((b^2 - 4*a*c)/a^2)*e^x - b*e^x - 2*a)/a) + 6*(a*b*sqrt((b^2 - 4*a*c)/a^2) + b^2 - 4*a*c)*polylog(3,
-1/2*(a*sqrt((b^2 - 4*a*c)/a^2)*e^x + b*e^x)/a) - 6*(a*b*sqrt((b^2 - 4*a*c)/a^2) - b^2 + 4*a*c)*polylog(3, 1/2
*(a*sqrt((b^2 - 4*a*c)/a^2)*e^x - b*e^x)/a))/(a*b^2 - 4*a^2*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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