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

Optimal. Leaf size=165 \[ -\frac{2 x \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^2 d^2}+\frac{2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^2 d^3}+\frac{2 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a}\right )}{a^2 d^3}+\frac{2 x \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a^2 d^2}-\frac{x^2 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a^2 d}-\frac{x^2}{a^2 d}+\frac{x^3}{3 a^2}+\frac{x^2}{a d \left (a+b e^{c+d x}\right )} \]

[Out]

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

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Rubi [A]  time = 0.399357, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.529, Rules used = {2185, 2184, 2190, 2531, 2282, 6589, 2191, 2279, 2391} \[ -\frac{2 x \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^2 d^2}+\frac{2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^2 d^3}+\frac{2 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a}\right )}{a^2 d^3}+\frac{2 x \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a^2 d^2}-\frac{x^2 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a^2 d}-\frac{x^2}{a^2 d}+\frac{x^3}{3 a^2}+\frac{x^2}{a d \left (a+b e^{c+d x}\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*E^(c + d*x))^2,x]

[Out]

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

Rule 2185

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dis
t[1/a, Int[(c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1), x], x] - Dist[b/a, Int[(c + d*x)^m*(F^(g*(e + f*x)
))^n*(a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && ILtQ[p, 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]

Rule 2191

Int[((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((a_.) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*
((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1))/(b*f*g*n*(p +
1)*Log[F]), x] - Dist[(d*m)/(b*f*g*n*(p + 1)*Log[F]), Int[(c + d*x)^(m - 1)*(a + b*(F^(g*(e + f*x)))^n)^(p + 1
), x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n, p}, x] && NeQ[p, -1]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

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

Mathematica [A]  time = 0.232652, size = 113, normalized size = 0.68 \[ \frac{(6-6 d x) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )+6 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a}\right )+\frac{d^2 x^2 \left (a d x+b (d x-3) e^{c+d x}\right )}{a+b e^{c+d x}}-3 d x (d x-2) \log \left (\frac{b e^{c+d x}}{a}+1\right )}{3 a^2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.063, size = 324, normalized size = 2. \begin{align*}{\frac{{x}^{2}}{da \left ( a+b{{\rm e}^{dx+c}} \right ) }}+{\frac{{c}^{2}\ln \left ({{\rm e}^{dx+c}} \right ) }{{d}^{3}{a}^{2}}}-{\frac{{c}^{2}\ln \left ( a+b{{\rm e}^{dx+c}} \right ) }{{d}^{3}{a}^{2}}}+{\frac{{x}^{3}}{3\,{a}^{2}}}-{\frac{{c}^{2}x}{{a}^{2}{d}^{2}}}-{\frac{2\,{c}^{3}}{3\,{d}^{3}{a}^{2}}}-{\frac{{x}^{2}}{{a}^{2}d}\ln \left ( 1+{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }+{\frac{{c}^{2}}{{d}^{3}{a}^{2}}\ln \left ( 1+{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }-2\,{\frac{x}{{a}^{2}{d}^{2}}{\it polylog} \left ( 2,-{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }+2\,{\frac{1}{{d}^{3}{a}^{2}}{\it polylog} \left ( 3,-{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }+2\,{\frac{c\ln \left ({{\rm e}^{dx+c}} \right ) }{{d}^{3}{a}^{2}}}-2\,{\frac{c\ln \left ( a+b{{\rm e}^{dx+c}} \right ) }{{d}^{3}{a}^{2}}}-{\frac{{x}^{2}}{{a}^{2}d}}-2\,{\frac{cx}{{a}^{2}{d}^{2}}}-{\frac{{c}^{2}}{{d}^{3}{a}^{2}}}+2\,{\frac{x}{{a}^{2}{d}^{2}}\ln \left ( 1+{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }+2\,{\frac{c}{{d}^{3}{a}^{2}}\ln \left ( 1+{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }+2\,{\frac{1}{{d}^{3}{a}^{2}}{\it polylog} \left ( 2,-{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2/a/d/(a+b*exp(d*x+c))+1/d^3/a^2*c^2*ln(exp(d*x+c))-1/d^3/a^2*c^2*ln(a+b*exp(d*x+c))+1/3*x^3/a^2-1/d^2/a^2*c
^2*x-2/3/d^3/a^2*c^3-x^2*ln(1+b*exp(d*x+c)/a)/a^2/d+1/d^3/a^2*ln(1+b*exp(d*x+c)/a)*c^2-2*x*polylog(2,-b*exp(d*
x+c)/a)/a^2/d^2+2*polylog(3,-b*exp(d*x+c)/a)/a^2/d^3+2/d^3/a^2*c*ln(exp(d*x+c))-2/d^3/a^2*c*ln(a+b*exp(d*x+c))
-x^2/a^2/d-2/d^2/a^2*c*x-1/d^3/a^2*c^2+2*x*ln(1+b*exp(d*x+c)/a)/a^2/d^2+2/d^3/a^2*ln(1+b*exp(d*x+c)/a)*c+2*pol
ylog(2,-b*exp(d*x+c)/a)/a^2/d^3

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Maxima [A]  time = 1.21669, size = 201, normalized size = 1.22 \begin{align*} \frac{x^{2}}{a b d e^{\left (d x + c\right )} + a^{2} d} + \frac{d^{3} x^{3} - 3 \, d^{2} x^{2}}{3 \, a^{2} d^{3}} - \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^{2} d^{3}} + \frac{2 \,{\left (d x \log \left (\frac{b e^{\left (d x + c\right )}}{a} + 1\right ) +{\rm Li}_2\left (-\frac{b e^{\left (d x + c\right )}}{a}\right )\right )}}{a^{2} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 1.48576, size = 608, normalized size = 3.68 \begin{align*} \frac{a d^{3} x^{3} + a c^{3} + 3 \, a c^{2} - 6 \,{\left (a d x +{\left (b d x - b\right )} e^{\left (d x + c\right )} - a\right )}{\rm Li}_2\left (-\frac{b e^{\left (d x + c\right )} + a}{a} + 1\right ) +{\left (b d^{3} x^{3} - 3 \, b d^{2} x^{2} + b c^{3} + 3 \, b c^{2}\right )} e^{\left (d x + c\right )} - 3 \,{\left (a c^{2} + 2 \, a c +{\left (b c^{2} + 2 \, b c\right )} e^{\left (d x + c\right )}\right )} \log \left (b e^{\left (d x + c\right )} + a\right ) - 3 \,{\left (a d^{2} x^{2} - a c^{2} - 2 \, a d x - 2 \, a c +{\left (b d^{2} x^{2} - b c^{2} - 2 \, b d x - 2 \, b c\right )} e^{\left (d x + c\right )}\right )} \log \left (\frac{b e^{\left (d x + c\right )} + a}{a}\right ) + 6 \,{\left (b e^{\left (d x + c\right )} + a\right )}{\rm polylog}\left (3, -\frac{b e^{\left (d x + c\right )}}{a}\right )}{3 \,{\left (a^{2} b d^{3} e^{\left (d x + c\right )} + a^{3} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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