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

Optimal. Leaf size=333 \[ -\frac{3 x^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^3 d^2}+\frac{9 x \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^3 d^3}+\frac{6 x \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a}\right )}{a^3 d^3}-\frac{3 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^3 d^4}-\frac{9 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a}\right )}{a^3 d^4}-\frac{6 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{a}\right )}{a^3 d^4}-\frac{3 x^2}{2 a^2 d^2 \left (a+b e^{c+d x}\right )}+\frac{9 x^2 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{2 a^3 d^2}-\frac{3 x \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a^3 d^3}+\frac{x^3}{a^2 d \left (a+b e^{c+d x}\right )}-\frac{x^3 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a^3 d}+\frac{3 x^2}{2 a^3 d^2}-\frac{3 x^3}{2 a^3 d}+\frac{x^4}{4 a^3}+\frac{x^3}{2 a d \left (a+b e^{c+d x}\right )^2} \]

[Out]

(3*x^2)/(2*a^3*d^2) - (3*x^2)/(2*a^2*d^2*(a + b*E^(c + d*x))) - (3*x^3)/(2*a^3*d) + x^3/(2*a*d*(a + b*E^(c + d
*x))^2) + x^3/(a^2*d*(a + b*E^(c + d*x))) + x^4/(4*a^3) - (3*x*Log[1 + (b*E^(c + d*x))/a])/(a^3*d^3) + (9*x^2*
Log[1 + (b*E^(c + d*x))/a])/(2*a^3*d^2) - (x^3*Log[1 + (b*E^(c + d*x))/a])/(a^3*d) - (3*PolyLog[2, -((b*E^(c +
d*x))/a)])/(a^3*d^4) + (9*x*PolyLog[2, -((b*E^(c + d*x))/a)])/(a^3*d^3) - (3*x^2*PolyLog[2, -((b*E^(c + d*x))
/a)])/(a^3*d^2) - (9*PolyLog[3, -((b*E^(c + d*x))/a)])/(a^3*d^4) + (6*x*PolyLog[3, -((b*E^(c + d*x))/a)])/(a^3
*d^3) - (6*PolyLog[4, -((b*E^(c + d*x))/a)])/(a^3*d^4)

________________________________________________________________________________________

Rubi [A]  time = 1.0778, antiderivative size = 333, normalized size of antiderivative = 1., number of steps used = 26, number of rules used = 10, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.588, Rules used = {2185, 2184, 2190, 2531, 6609, 2282, 6589, 2191, 2279, 2391} \[ -\frac{3 x^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^3 d^2}+\frac{9 x \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^3 d^3}+\frac{6 x \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a}\right )}{a^3 d^3}-\frac{3 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^3 d^4}-\frac{9 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a}\right )}{a^3 d^4}-\frac{6 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{a}\right )}{a^3 d^4}-\frac{3 x^2}{2 a^2 d^2 \left (a+b e^{c+d x}\right )}+\frac{9 x^2 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{2 a^3 d^2}-\frac{3 x \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a^3 d^3}+\frac{x^3}{a^2 d \left (a+b e^{c+d x}\right )}-\frac{x^3 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a^3 d}+\frac{3 x^2}{2 a^3 d^2}-\frac{3 x^3}{2 a^3 d}+\frac{x^4}{4 a^3}+\frac{x^3}{2 a d \left (a+b e^{c+d x}\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b*E^(c + d*x))^3,x]

[Out]

(3*x^2)/(2*a^3*d^2) - (3*x^2)/(2*a^2*d^2*(a + b*E^(c + d*x))) - (3*x^3)/(2*a^3*d) + x^3/(2*a*d*(a + b*E^(c + d
*x))^2) + x^3/(a^2*d*(a + b*E^(c + d*x))) + x^4/(4*a^3) - (3*x*Log[1 + (b*E^(c + d*x))/a])/(a^3*d^3) + (9*x^2*
Log[1 + (b*E^(c + d*x))/a])/(2*a^3*d^2) - (x^3*Log[1 + (b*E^(c + d*x))/a])/(a^3*d) - (3*PolyLog[2, -((b*E^(c +
d*x))/a)])/(a^3*d^4) + (9*x*PolyLog[2, -((b*E^(c + d*x))/a)])/(a^3*d^3) - (3*x^2*PolyLog[2, -((b*E^(c + d*x))
/a)])/(a^3*d^2) - (9*PolyLog[3, -((b*E^(c + d*x))/a)])/(a^3*d^4) + (6*x*PolyLog[3, -((b*E^(c + d*x))/a)])/(a^3
*d^3) - (6*PolyLog[4, -((b*E^(c + d*x))/a)])/(a^3*d^4)

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 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, 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^3}{\left (a+b e^{c+d x}\right )^3} \, dx &=\frac{\int \frac{x^3}{\left (a+b e^{c+d x}\right )^2} \, dx}{a}-\frac{b \int \frac{e^{c+d x} x^3}{\left (a+b e^{c+d x}\right )^3} \, dx}{a}\\ &=\frac{x^3}{2 a d \left (a+b e^{c+d x}\right )^2}+\frac{\int \frac{x^3}{a+b e^{c+d x}} \, dx}{a^2}-\frac{b \int \frac{e^{c+d x} x^3}{\left (a+b e^{c+d x}\right )^2} \, dx}{a^2}-\frac{3 \int \frac{x^2}{\left (a+b e^{c+d x}\right )^2} \, dx}{2 a d}\\ &=\frac{x^3}{2 a d \left (a+b e^{c+d x}\right )^2}+\frac{x^3}{a^2 d \left (a+b e^{c+d x}\right )}+\frac{x^4}{4 a^3}-\frac{b \int \frac{e^{c+d x} x^3}{a+b e^{c+d x}} \, dx}{a^3}-\frac{3 \int \frac{x^2}{a+b e^{c+d x}} \, dx}{2 a^2 d}-\frac{3 \int \frac{x^2}{a+b e^{c+d x}} \, dx}{a^2 d}+\frac{(3 b) \int \frac{e^{c+d x} x^2}{\left (a+b e^{c+d x}\right )^2} \, dx}{2 a^2 d}\\ &=-\frac{3 x^2}{2 a^2 d^2 \left (a+b e^{c+d x}\right )}-\frac{3 x^3}{2 a^3 d}+\frac{x^3}{2 a d \left (a+b e^{c+d x}\right )^2}+\frac{x^3}{a^2 d \left (a+b e^{c+d x}\right )}+\frac{x^4}{4 a^3}-\frac{x^3 \log \left (1+\frac{b e^{c+d x}}{a}\right )}{a^3 d}+\frac{3 \int \frac{x}{a+b e^{c+d x}} \, dx}{a^2 d^2}+\frac{3 \int x^2 \log \left (1+\frac{b e^{c+d x}}{a}\right ) \, dx}{a^3 d}+\frac{(3 b) \int \frac{e^{c+d x} x^2}{a+b e^{c+d x}} \, dx}{2 a^3 d}+\frac{(3 b) \int \frac{e^{c+d x} x^2}{a+b e^{c+d x}} \, dx}{a^3 d}\\ &=\frac{3 x^2}{2 a^3 d^2}-\frac{3 x^2}{2 a^2 d^2 \left (a+b e^{c+d x}\right )}-\frac{3 x^3}{2 a^3 d}+\frac{x^3}{2 a d \left (a+b e^{c+d x}\right )^2}+\frac{x^3}{a^2 d \left (a+b e^{c+d x}\right )}+\frac{x^4}{4 a^3}+\frac{9 x^2 \log \left (1+\frac{b e^{c+d x}}{a}\right )}{2 a^3 d^2}-\frac{x^3 \log \left (1+\frac{b e^{c+d x}}{a}\right )}{a^3 d}-\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a}\right )}{a^3 d^2}-\frac{3 \int x \log \left (1+\frac{b e^{c+d x}}{a}\right ) \, dx}{a^3 d^2}-\frac{6 \int x \log \left (1+\frac{b e^{c+d x}}{a}\right ) \, dx}{a^3 d^2}+\frac{6 \int x \text{Li}_2\left (-\frac{b e^{c+d x}}{a}\right ) \, dx}{a^3 d^2}-\frac{(3 b) \int \frac{e^{c+d x} x}{a+b e^{c+d x}} \, dx}{a^3 d^2}\\ &=\frac{3 x^2}{2 a^3 d^2}-\frac{3 x^2}{2 a^2 d^2 \left (a+b e^{c+d x}\right )}-\frac{3 x^3}{2 a^3 d}+\frac{x^3}{2 a d \left (a+b e^{c+d x}\right )^2}+\frac{x^3}{a^2 d \left (a+b e^{c+d x}\right )}+\frac{x^4}{4 a^3}-\frac{3 x \log \left (1+\frac{b e^{c+d x}}{a}\right )}{a^3 d^3}+\frac{9 x^2 \log \left (1+\frac{b e^{c+d x}}{a}\right )}{2 a^3 d^2}-\frac{x^3 \log \left (1+\frac{b e^{c+d x}}{a}\right )}{a^3 d}+\frac{9 x \text{Li}_2\left (-\frac{b e^{c+d x}}{a}\right )}{a^3 d^3}-\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a}\right )}{a^3 d^2}+\frac{6 x \text{Li}_3\left (-\frac{b e^{c+d x}}{a}\right )}{a^3 d^3}+\frac{3 \int \log \left (1+\frac{b e^{c+d x}}{a}\right ) \, dx}{a^3 d^3}-\frac{3 \int \text{Li}_2\left (-\frac{b e^{c+d x}}{a}\right ) \, dx}{a^3 d^3}-\frac{6 \int \text{Li}_2\left (-\frac{b e^{c+d x}}{a}\right ) \, dx}{a^3 d^3}-\frac{6 \int \text{Li}_3\left (-\frac{b e^{c+d x}}{a}\right ) \, dx}{a^3 d^3}\\ &=\frac{3 x^2}{2 a^3 d^2}-\frac{3 x^2}{2 a^2 d^2 \left (a+b e^{c+d x}\right )}-\frac{3 x^3}{2 a^3 d}+\frac{x^3}{2 a d \left (a+b e^{c+d x}\right )^2}+\frac{x^3}{a^2 d \left (a+b e^{c+d x}\right )}+\frac{x^4}{4 a^3}-\frac{3 x \log \left (1+\frac{b e^{c+d x}}{a}\right )}{a^3 d^3}+\frac{9 x^2 \log \left (1+\frac{b e^{c+d x}}{a}\right )}{2 a^3 d^2}-\frac{x^3 \log \left (1+\frac{b e^{c+d x}}{a}\right )}{a^3 d}+\frac{9 x \text{Li}_2\left (-\frac{b e^{c+d x}}{a}\right )}{a^3 d^3}-\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a}\right )}{a^3 d^2}+\frac{6 x \text{Li}_3\left (-\frac{b e^{c+d x}}{a}\right )}{a^3 d^3}+\frac{3 \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{b x}{a}\right )}{x} \, dx,x,e^{c+d x}\right )}{a^3 d^4}-\frac{3 \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^4}-\frac{6 \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^4}-\frac{6 \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (-\frac{b x}{a}\right )}{x} \, dx,x,e^{c+d x}\right )}{a^3 d^4}\\ &=\frac{3 x^2}{2 a^3 d^2}-\frac{3 x^2}{2 a^2 d^2 \left (a+b e^{c+d x}\right )}-\frac{3 x^3}{2 a^3 d}+\frac{x^3}{2 a d \left (a+b e^{c+d x}\right )^2}+\frac{x^3}{a^2 d \left (a+b e^{c+d x}\right )}+\frac{x^4}{4 a^3}-\frac{3 x \log \left (1+\frac{b e^{c+d x}}{a}\right )}{a^3 d^3}+\frac{9 x^2 \log \left (1+\frac{b e^{c+d x}}{a}\right )}{2 a^3 d^2}-\frac{x^3 \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^4}+\frac{9 x \text{Li}_2\left (-\frac{b e^{c+d x}}{a}\right )}{a^3 d^3}-\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a}\right )}{a^3 d^2}-\frac{9 \text{Li}_3\left (-\frac{b e^{c+d x}}{a}\right )}{a^3 d^4}+\frac{6 x \text{Li}_3\left (-\frac{b e^{c+d x}}{a}\right )}{a^3 d^3}-\frac{6 \text{Li}_4\left (-\frac{b e^{c+d x}}{a}\right )}{a^3 d^4}\\ \end{align*}

Mathematica [A]  time = 0.226188, size = 241, normalized size = 0.72 \[ \frac{-\frac{12 \left (d^2 x^2-3 d x+1\right ) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{d^4}+\frac{12 (2 d x-3) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a}\right )}{d^4}-\frac{24 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{a}\right )}{d^4}+\frac{2 a^2 x^3}{d \left (a+b e^{c+d x}\right )^2}-\frac{6 a x^2}{d^2 \left (a+b e^{c+d x}\right )}+\frac{18 x^2 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{d^2}-\frac{12 x \log \left (\frac{b e^{c+d x}}{a}+1\right )}{d^3}+\frac{4 a x^3}{a d+b d e^{c+d x}}-\frac{4 x^3 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{d}+\frac{6 x^2}{d^2}-\frac{6 x^3}{d}+x^4}{4 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*x^2)/d^2 - (6*a*x^2)/(d^2*(a + b*E^(c + d*x))) - (6*x^3)/d + (2*a^2*x^3)/(d*(a + b*E^(c + d*x))^2) + (4*a*
x^3)/(a*d + b*d*E^(c + d*x)) + x^4 - (12*x*Log[1 + (b*E^(c + d*x))/a])/d^3 + (18*x^2*Log[1 + (b*E^(c + d*x))/a
])/d^2 - (4*x^3*Log[1 + (b*E^(c + d*x))/a])/d - (12*(1 - 3*d*x + d^2*x^2)*PolyLog[2, -((b*E^(c + d*x))/a)])/d^
4 + (12*(-3 + 2*d*x)*PolyLog[3, -((b*E^(c + d*x))/a)])/d^4 - (24*PolyLog[4, -((b*E^(c + d*x))/a)])/d^4)/(4*a^3
)

________________________________________________________________________________________

Maple [A]  time = 0.092, size = 548, normalized size = 1.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a+b*exp(d*x+c))^3,x)

[Out]

1/4*x^4/a^3-3*x*ln(1+b*exp(d*x+c)/a)/a^3/d^3+9/2*x^2*ln(1+b*exp(d*x+c)/a)/a^3/d^2-x^3*ln(1+b*exp(d*x+c)/a)/a^3
/d+9*x*polylog(2,-b*exp(d*x+c)/a)/a^3/d^3-3*x^2*polylog(2,-b*exp(d*x+c)/a)/a^3/d^2+6*x*polylog(3,-b*exp(d*x+c)
/a)/a^3/d^3+1/2*x^2*(2*x*b*d*exp(d*x+c)+3*a*x*d-3*b*exp(d*x+c)-3*a)/d^2/a^2/(a+b*exp(d*x+c))^2+3/4/d^4/a^3*c^4
+3/2/d^4/a^3*c^2+3/d^4/a^3*c^3+3/2*x^2/a^3/d^2-9/2/d^4/a^3*c^2*ln(exp(d*x+c))+9/2/d^4/a^3*c^2*ln(a+b*exp(d*x+c
))-1/d^4/a^3*c^3*ln(exp(d*x+c))+1/d^4/a^3*c^3*ln(a+b*exp(d*x+c))-3/d^4/a^3*c*ln(exp(d*x+c))+3/d^4/a^3*c*ln(a+b
*exp(d*x+c))-1/d^4/a^3*ln(1+b*exp(d*x+c)/a)*c^3-9/2/d^4/a^3*ln(1+b*exp(d*x+c)/a)*c^2-3/d^4/a^3*c*ln(1+b*exp(d*
x+c)/a)+1/d^3/a^3*c^3*x+9/2/d^3/a^3*c^2*x+3/d^3/a^3*c*x-3/2*x^3/a^3/d-3*polylog(2,-b*exp(d*x+c)/a)/a^3/d^4-9*p
olylog(3,-b*exp(d*x+c)/a)/a^3/d^4-6*polylog(4,-b*exp(d*x+c)/a)/a^3/d^4

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Maxima [A]  time = 1.11851, size = 409, normalized size = 1.23 \begin{align*} \frac{3 \, a d x^{3} - 3 \, a x^{2} +{\left (2 \, b d x^{3} e^{c} - 3 \, b x^{2} 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{d^{4} x^{4} - 6 \, d^{3} x^{3} + 6 \, d^{2} x^{2}}{4 \, a^{3} d^{4}} - \frac{d^{3} x^{3} \log \left (\frac{b e^{\left (d x + c\right )}}{a} + 1\right ) + 3 \, d^{2} x^{2}{\rm Li}_2\left (-\frac{b e^{\left (d x + c\right )}}{a}\right ) - 6 \, d x{\rm Li}_{3}(-\frac{b e^{\left (d x + c\right )}}{a}) + 6 \,{\rm Li}_{4}(-\frac{b e^{\left (d x + c\right )}}{a})}{a^{3} d^{4}} + \frac{9 \,{\left (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})\right )}}{2 \, a^{3} d^{4}} - \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^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 1.57116, size = 1558, normalized size = 4.68 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a^2*d^4*x^4 - a^2*c^4 - 6*a^2*c^3 - 6*a^2*c^2 - 12*(a^2*d^2*x^2 - 3*a^2*d*x + a^2 + (b^2*d^2*x^2 - 3*b^2*
d*x + b^2)*e^(2*d*x + 2*c) + 2*(a*b*d^2*x^2 - 3*a*b*d*x + a*b)*e^(d*x + c))*dilog(-(b*e^(d*x + c) + a)/a + 1)
+ (b^2*d^4*x^4 - 6*b^2*d^3*x^3 - b^2*c^4 + 6*b^2*d^2*x^2 - 6*b^2*c^3 - 6*b^2*c^2)*e^(2*d*x + 2*c) + 2*(a*b*d^4
*x^4 - 4*a*b*d^3*x^3 - a*b*c^4 + 3*a*b*d^2*x^2 - 6*a*b*c^3 - 6*a*b*c^2)*e^(d*x + c) + 2*(2*a^2*c^3 + 9*a^2*c^2
+ 6*a^2*c + (2*b^2*c^3 + 9*b^2*c^2 + 6*b^2*c)*e^(2*d*x + 2*c) + 2*(2*a*b*c^3 + 9*a*b*c^2 + 6*a*b*c)*e^(d*x +
c))*log(b*e^(d*x + c) + a) - 2*(2*a^2*d^3*x^3 - 9*a^2*d^2*x^2 + 2*a^2*c^3 + 9*a^2*c^2 + 6*a^2*d*x + 6*a^2*c +
(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 - 9*a*b*d^2*x^2 + 2*a*b*c^3 + 9*a*b*c^2 + 6*a*b*d*x + 6*a*b*c)*e^(d*x + c))*log((b*e^(d*x + c) + a)/a) - 24
*(b^2*e^(2*d*x + 2*c) + 2*a*b*e^(d*x + c) + a^2)*polylog(4, -b*e^(d*x + c)/a) + 12*(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))*polylog(3, -b*e^(d*x + c)/a))/(a^3*b^2*d^4*
e^(2*d*x + 2*c) + 2*a^4*b*d^4*e^(d*x + c) + a^5*d^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{3 a d x^{3} - 3 a x^{2} + \left (2 b d x^{3} - 3 b x^{2}\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{6 x}{a + b e^{c} e^{d x}}\, dx + \int - \frac{9 d x^{2}}{a + b e^{c} e^{d x}}\, dx + \int \frac{2 d^{2} x^{3}}{a + b e^{c} e^{d x}}\, dx}{2 a^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*a*d*x**3 - 3*a*x**2 + (2*b*d*x**3 - 3*b*x**2)*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(6*x/(a + b*exp(c)*exp(d*x)), x) + Integral(-9*d*x**2/(a + b*exp(c)
*exp(d*x)), x) + Integral(2*d**2*x**3/(a + b*exp(c)*exp(d*x)), x))/(2*a**2*d**2)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{{\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^3/(a+b*exp(d*x+c))^3,x, algorithm="giac")

[Out]

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