3.18 \(\int \frac{x^2}{(a+b e^{c+d x})^3} \, dx\)

Optimal. Leaf size=243 \[ -\frac{2 x \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^3 d^2}+\frac{3 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^3 d^3}+\frac{2 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a}\right )}{a^3 d^3}-\frac{x}{a^2 d^2 \left (a+b e^{c+d x}\right )}+\frac{3 x \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a^3 d^2}-\frac{\log \left (a+b e^{c+d x}\right )}{a^3 d^3}+\frac{x^2}{a^2 d \left (a+b e^{c+d x}\right )}-\frac{x^2 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a^3 d}+\frac{x}{a^3 d^2}-\frac{3 x^2}{2 a^3 d}+\frac{x^3}{3 a^3}+\frac{x^2}{2 a d \left (a+b e^{c+d x}\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.764287, antiderivative size = 243, normalized size of antiderivative = 1., number of steps used = 23, number of rules used = 12, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.706, Rules used = {2185, 2184, 2190, 2531, 2282, 6589, 2191, 2279, 2391, 36, 29, 31} \[ -\frac{2 x \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^3 d^2}+\frac{3 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^3 d^3}+\frac{2 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a}\right )}{a^3 d^3}-\frac{x}{a^2 d^2 \left (a+b e^{c+d x}\right )}+\frac{3 x \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a^3 d^2}-\frac{\log \left (a+b e^{c+d x}\right )}{a^3 d^3}+\frac{x^2}{a^2 d \left (a+b e^{c+d x}\right )}-\frac{x^2 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a^3 d}+\frac{x}{a^3 d^2}-\frac{3 x^2}{2 a^3 d}+\frac{x^3}{3 a^3}+\frac{x^2}{2 a d \left (a+b e^{c+d x}\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*E^(c + d*x))^3,x]

[Out]

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.168021, size = 203, normalized size = 0.84 \[ \frac{-\frac{6 (2 d x-3) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{d^3}+\frac{12 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a}\right )}{d^3}+\frac{3 a^2 x^2}{d \left (a+b e^{c+d x}\right )^2}-\frac{6 a x}{d^2 \left (a+b e^{c+d x}\right )}+\frac{18 x \log \left (\frac{b e^{c+d x}}{a}+1\right )}{d^2}-\frac{6 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{d^3}+\frac{6 a x^2}{a d+b d e^{c+d x}}-\frac{6 x^2 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{d}+\frac{6 x}{d^2}-\frac{9 x^2}{d}+2 x^3}{6 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*E^(c + d*x))^3,x]

[Out]

((6*x)/d^2 - (6*a*x)/(d^2*(a + b*E^(c + d*x))) - (9*x^2)/d + (3*a^2*x^2)/(d*(a + b*E^(c + d*x))^2) + (6*a*x^2)
/(a*d + b*d*E^(c + d*x)) + 2*x^3 - (6*Log[1 + (b*E^(c + d*x))/a])/d^3 + (18*x*Log[1 + (b*E^(c + d*x))/a])/d^2
- (6*x^2*Log[1 + (b*E^(c + d*x))/a])/d - (6*(-3 + 2*d*x)*PolyLog[2, -((b*E^(c + d*x))/a)])/d^3 + (12*PolyLog[3
, -((b*E^(c + d*x))/a)])/d^3)/(6*a^3)

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Maple [A]  time = 0.077, size = 385, normalized size = 1.6 \begin{align*}{\frac{x \left ( 2\,xbd{{\rm e}^{dx+c}}+3\,axd-2\,b{{\rm e}^{dx+c}}-2\,a \right ) }{2\,{d}^{2}{a}^{2} \left ( a+b{{\rm e}^{dx+c}} \right ) ^{2}}}+{\frac{\ln \left ({{\rm e}^{dx+c}} \right ) }{{a}^{3}{d}^{3}}}-{\frac{\ln \left ( a+b{{\rm e}^{dx+c}} \right ) }{{a}^{3}{d}^{3}}}+{\frac{{c}^{2}\ln \left ({{\rm e}^{dx+c}} \right ) }{{a}^{3}{d}^{3}}}-{\frac{{c}^{2}\ln \left ( a+b{{\rm e}^{dx+c}} \right ) }{{a}^{3}{d}^{3}}}+{\frac{{x}^{3}}{3\,{a}^{3}}}-{\frac{{c}^{2}x}{{d}^{2}{a}^{3}}}-{\frac{2\,{c}^{3}}{3\,{a}^{3}{d}^{3}}}-{\frac{{x}^{2}}{{a}^{3}d}\ln \left ( 1+{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }+{\frac{{c}^{2}}{{a}^{3}{d}^{3}}\ln \left ( 1+{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }-2\,{\frac{x}{{d}^{2}{a}^{3}}{\it polylog} \left ( 2,-{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }+2\,{\frac{1}{{a}^{3}{d}^{3}}{\it polylog} \left ( 3,-{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }+3\,{\frac{c\ln \left ({{\rm e}^{dx+c}} \right ) }{{a}^{3}{d}^{3}}}-3\,{\frac{c\ln \left ( a+b{{\rm e}^{dx+c}} \right ) }{{a}^{3}{d}^{3}}}-{\frac{3\,{x}^{2}}{2\,{a}^{3}d}}-3\,{\frac{cx}{{d}^{2}{a}^{3}}}-{\frac{3\,{c}^{2}}{2\,{a}^{3}{d}^{3}}}+3\,{\frac{x}{{d}^{2}{a}^{3}}\ln \left ( 1+{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }+3\,{\frac{c}{{a}^{3}{d}^{3}}\ln \left ( 1+{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }+3\,{\frac{1}{{a}^{3}{d}^{3}}{\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))^3,x)

[Out]

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

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Maxima [A]  time = 1.05542, size = 316, normalized size = 1.3 \begin{align*} \frac{3 \, a d x^{2} - 2 \, a x + 2 \,{\left (b d x^{2} e^{c} - b x e^{c}\right )} e^{\left (d x\right )}}{2 \,{\left (a^{2} b^{2} d^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{3} b d^{2} e^{\left (d x + c\right )} + a^{4} d^{2}\right )}} + \frac{x}{a^{3} d^{2}} + \frac{2 \, d^{3} x^{3} - 9 \, d^{2} x^{2}}{6 \, a^{3} 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^{3} d^{3}} + \frac{3 \,{\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^{3} d^{3}} - \frac{\log \left (b e^{\left (d x + c\right )} + a\right )}{a^{3} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+b*exp(d*x+c))^3,x, algorithm="maxima")

[Out]

1/2*(3*a*d*x^2 - 2*a*x + 2*(b*d*x^2*e^c - b*x*e^c)*e^(d*x))/(a^2*b^2*d^2*e^(2*d*x + 2*c) + 2*a^3*b*d^2*e^(d*x
+ c) + a^4*d^2) + x/(a^3*d^2) + 1/6*(2*d^3*x^3 - 9*d^2*x^2)/(a^3*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^3*d^3) + 3*(d*x*log(b*e^(d*x + c)/a + 1) + di
log(-b*e^(d*x + c)/a))/(a^3*d^3) - log(b*e^(d*x + c) + a)/(a^3*d^3)

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Fricas [C]  time = 1.5318, size = 1170, normalized size = 4.81 \begin{align*} \frac{2 \, a^{2} d^{3} x^{3} + 2 \, a^{2} c^{3} + 9 \, a^{2} c^{2} + 6 \, a^{2} c - 6 \,{\left (2 \, a^{2} d x - 3 \, a^{2} +{\left (2 \, b^{2} d x - 3 \, b^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \,{\left (2 \, a b d x - 3 \, a b\right )} e^{\left (d x + c\right )}\right )}{\rm Li}_2\left (-\frac{b e^{\left (d x + c\right )} + a}{a} + 1\right ) +{\left (2 \, b^{2} d^{3} x^{3} - 9 \, b^{2} d^{2} x^{2} + 2 \, b^{2} c^{3} + 9 \, b^{2} c^{2} + 6 \, b^{2} d x + 6 \, b^{2} c\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \,{\left (2 \, a b d^{3} x^{3} - 6 \, a b d^{2} x^{2} + 2 \, a b c^{3} + 9 \, a b c^{2} + 3 \, a b d x + 6 \, a b c\right )} e^{\left (d x + c\right )} - 6 \,{\left (a^{2} c^{2} + 3 \, a^{2} c + a^{2} +{\left (b^{2} c^{2} + 3 \, b^{2} c + b^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \,{\left (a b c^{2} + 3 \, a b c + a b\right )} e^{\left (d x + c\right )}\right )} \log \left (b e^{\left (d x + c\right )} + a\right ) - 6 \,{\left (a^{2} d^{2} x^{2} - a^{2} c^{2} - 3 \, a^{2} d x - 3 \, a^{2} c +{\left (b^{2} d^{2} x^{2} - b^{2} c^{2} - 3 \, b^{2} d x - 3 \, b^{2} c\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \,{\left (a b d^{2} x^{2} - a b c^{2} - 3 \, a b d x - 3 \, a b c\right )} e^{\left (d x + c\right )}\right )} \log \left (\frac{b e^{\left (d x + c\right )} + a}{a}\right ) + 12 \,{\left (b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b e^{\left (d x + c\right )} + a^{2}\right )}{\rm polylog}\left (3, -\frac{b e^{\left (d x + c\right )}}{a}\right )}{6 \,{\left (a^{3} b^{2} d^{3} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{4} b d^{3} e^{\left (d x + c\right )} + a^{5} 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))^3,x, algorithm="fricas")

[Out]

1/6*(2*a^2*d^3*x^3 + 2*a^2*c^3 + 9*a^2*c^2 + 6*a^2*c - 6*(2*a^2*d*x - 3*a^2 + (2*b^2*d*x - 3*b^2)*e^(2*d*x + 2
*c) + 2*(2*a*b*d*x - 3*a*b)*e^(d*x + c))*dilog(-(b*e^(d*x + c) + a)/a + 1) + (2*b^2*d^3*x^3 - 9*b^2*d^2*x^2 +
2*b^2*c^3 + 9*b^2*c^2 + 6*b^2*d*x + 6*b^2*c)*e^(2*d*x + 2*c) + 2*(2*a*b*d^3*x^3 - 6*a*b*d^2*x^2 + 2*a*b*c^3 +
9*a*b*c^2 + 3*a*b*d*x + 6*a*b*c)*e^(d*x + c) - 6*(a^2*c^2 + 3*a^2*c + a^2 + (b^2*c^2 + 3*b^2*c + b^2)*e^(2*d*x
+ 2*c) + 2*(a*b*c^2 + 3*a*b*c + a*b)*e^(d*x + c))*log(b*e^(d*x + c) + a) - 6*(a^2*d^2*x^2 - a^2*c^2 - 3*a^2*d
*x - 3*a^2*c + (b^2*d^2*x^2 - b^2*c^2 - 3*b^2*d*x - 3*b^2*c)*e^(2*d*x + 2*c) + 2*(a*b*d^2*x^2 - a*b*c^2 - 3*a*
b*d*x - 3*a*b*c)*e^(d*x + c))*log((b*e^(d*x + c) + a)/a) + 12*(b^2*e^(2*d*x + 2*c) + 2*a*b*e^(d*x + c) + a^2)*
polylog(3, -b*e^(d*x + c)/a))/(a^3*b^2*d^3*e^(2*d*x + 2*c) + 2*a^4*b*d^3*e^(d*x + c) + a^5*d^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{3 a d x^{2} - 2 a x + \left (2 b d x^{2} - 2 b x\right ) e^{c + d x}}{2 a^{4} d^{2} + 4 a^{3} b d^{2} e^{c + d x} + 2 a^{2} b^{2} d^{2} e^{2 c + 2 d x}} + \frac{\int - \frac{3 d x}{a + b e^{c} e^{d x}}\, dx + \int \frac{d^{2} x^{2}}{a + b e^{c} e^{d x}}\, dx + \int \frac{1}{a + b e^{c} e^{d x}}\, dx}{a^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(a+b*exp(d*x+c))**3,x)

[Out]

(3*a*d*x**2 - 2*a*x + (2*b*d*x**2 - 2*b*x)*exp(c + d*x))/(2*a**4*d**2 + 4*a**3*b*d**2*exp(c + d*x) + 2*a**2*b*
*2*d**2*exp(2*c + 2*d*x)) + (Integral(-3*d*x/(a + b*exp(c)*exp(d*x)), x) + Integral(d**2*x**2/(a + b*exp(c)*ex
p(d*x)), x) + Integral(1/(a + b*exp(c)*exp(d*x)), x))/(a**2*d**2)

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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 )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+b*exp(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(x^2/(b*e^(d*x + c) + a)^3, x)