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

Optimal. Leaf size=84 \[ -\frac{2 x \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a d^2}+\frac{2 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a}\right )}{a d^3}-\frac{x^2 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a d}+\frac{x^3}{3 a} \]

[Out]

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

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Rubi [A]  time = 0.167754, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {2184, 2190, 2531, 2282, 6589} \[ -\frac{2 x \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a d^2}+\frac{2 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a}\right )}{a d^3}-\frac{x^2 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a d}+\frac{x^3}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*E^(c + d*x)),x]

[Out]

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

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

Mathematica [A]  time = 0.0060089, size = 83, normalized size = 0.99 \[ \frac{2 x \text{PolyLog}\left (2,-\frac{a e^{-c-d x}}{b}\right )}{a d^2}+\frac{2 \text{PolyLog}\left (3,-\frac{a e^{-c-d x}}{b}\right )}{a d^3}-\frac{x^2 \log \left (\frac{a e^{-c-d x}}{b}+1\right )}{a d} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*E^(c + d*x)),x]

[Out]

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

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Maple [B]  time = 0.003, size = 166, normalized size = 2. \begin{align*}{\frac{{x}^{3}}{3\,a}}-{\frac{{c}^{2}x}{a{d}^{2}}}-{\frac{2\,{c}^{3}}{3\,{d}^{3}a}}-{\frac{{x}^{2}}{ad}\ln \left ( 1+{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }+{\frac{{c}^{2}}{{d}^{3}a}\ln \left ( 1+{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }-2\,{\frac{x}{a{d}^{2}}{\it polylog} \left ( 2,-{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }+2\,{\frac{1}{{d}^{3}a}{\it polylog} \left ( 3,-{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }+{\frac{{c}^{2}\ln \left ({{\rm e}^{dx+c}} \right ) }{{d}^{3}a}}-{\frac{{c}^{2}\ln \left ( a+b{{\rm e}^{dx+c}} \right ) }{{d}^{3}a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.07884, size = 97, normalized size = 1.15 \begin{align*} \frac{x^{3}}{3 \, a} - \frac{d^{2} x^{2} \log \left (\frac{b e^{\left (d x + c\right )}}{a} + 1\right ) + 2 \, d x{\rm Li}_2\left (-\frac{b e^{\left (d x + c\right )}}{a}\right ) - 2 \,{\rm Li}_{3}(-\frac{b e^{\left (d x + c\right )}}{a})}{a d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 1.52682, size = 238, normalized size = 2.83 \begin{align*} \frac{d^{3} x^{3} - 6 \, d x{\rm Li}_2\left (-\frac{b e^{\left (d x + c\right )} + a}{a} + 1\right ) - 3 \, c^{2} \log \left (b e^{\left (d x + c\right )} + a\right ) - 3 \,{\left (d^{2} x^{2} - c^{2}\right )} \log \left (\frac{b e^{\left (d x + c\right )} + a}{a}\right ) + 6 \,{\rm polylog}\left (3, -\frac{b e^{\left (d x + c\right )}}{a}\right )}{3 \, a d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(b*e^(d*x + c) + a), x)