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

Optimal. Leaf size=217 \[ -\frac{3 x^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^2 d^2}+\frac{6 x \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^2 d^3}+\frac{6 x \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a}\right )}{a^2 d^3}-\frac{6 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a}\right )}{a^2 d^4}-\frac{6 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{a}\right )}{a^2 d^4}+\frac{3 x^2 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a^2 d^2}-\frac{x^3 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a^2 d}-\frac{x^3}{a^2 d}+\frac{x^4}{4 a^2}+\frac{x^3}{a d \left (a+b e^{c+d x}\right )} \]

[Out]

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

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Rubi [A]  time = 0.506581, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.471, Rules used = {2185, 2184, 2190, 2531, 6609, 2282, 6589, 2191} \[ -\frac{3 x^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^2 d^2}+\frac{6 x \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^2 d^3}+\frac{6 x \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a}\right )}{a^2 d^3}-\frac{6 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a}\right )}{a^2 d^4}-\frac{6 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{a}\right )}{a^2 d^4}+\frac{3 x^2 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a^2 d^2}-\frac{x^3 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a^2 d}-\frac{x^3}{a^2 d}+\frac{x^4}{4 a^2}+\frac{x^3}{a d \left (a+b e^{c+d x}\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b*E^(c + d*x))^2,x]

[Out]

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

Rubi steps

\begin{align*} \int \frac{x^3}{\left (a+b e^{c+d x}\right )^2} \, dx &=\frac{\int \frac{x^3}{a+b e^{c+d x}} \, dx}{a}-\frac{b \int \frac{e^{c+d x} x^3}{\left (a+b e^{c+d x}\right )^2} \, dx}{a}\\ &=\frac{x^3}{a d \left (a+b e^{c+d x}\right )}+\frac{x^4}{4 a^2}-\frac{b \int \frac{e^{c+d x} x^3}{a+b e^{c+d x}} \, dx}{a^2}-\frac{3 \int \frac{x^2}{a+b e^{c+d x}} \, dx}{a d}\\ &=-\frac{x^3}{a^2 d}+\frac{x^3}{a d \left (a+b e^{c+d x}\right )}+\frac{x^4}{4 a^2}-\frac{x^3 \log \left (1+\frac{b e^{c+d x}}{a}\right )}{a^2 d}+\frac{3 \int x^2 \log \left (1+\frac{b e^{c+d x}}{a}\right ) \, dx}{a^2 d}+\frac{(3 b) \int \frac{e^{c+d x} x^2}{a+b e^{c+d x}} \, dx}{a^2 d}\\ &=-\frac{x^3}{a^2 d}+\frac{x^3}{a d \left (a+b e^{c+d x}\right )}+\frac{x^4}{4 a^2}+\frac{3 x^2 \log \left (1+\frac{b e^{c+d x}}{a}\right )}{a^2 d^2}-\frac{x^3 \log \left (1+\frac{b e^{c+d x}}{a}\right )}{a^2 d}-\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a}\right )}{a^2 d^2}-\frac{6 \int x \log \left (1+\frac{b e^{c+d x}}{a}\right ) \, dx}{a^2 d^2}+\frac{6 \int x \text{Li}_2\left (-\frac{b e^{c+d x}}{a}\right ) \, dx}{a^2 d^2}\\ &=-\frac{x^3}{a^2 d}+\frac{x^3}{a d \left (a+b e^{c+d x}\right )}+\frac{x^4}{4 a^2}+\frac{3 x^2 \log \left (1+\frac{b e^{c+d x}}{a}\right )}{a^2 d^2}-\frac{x^3 \log \left (1+\frac{b e^{c+d x}}{a}\right )}{a^2 d}+\frac{6 x \text{Li}_2\left (-\frac{b e^{c+d x}}{a}\right )}{a^2 d^3}-\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a}\right )}{a^2 d^2}+\frac{6 x \text{Li}_3\left (-\frac{b e^{c+d x}}{a}\right )}{a^2 d^3}-\frac{6 \int \text{Li}_2\left (-\frac{b e^{c+d x}}{a}\right ) \, dx}{a^2 d^3}-\frac{6 \int \text{Li}_3\left (-\frac{b e^{c+d x}}{a}\right ) \, dx}{a^2 d^3}\\ &=-\frac{x^3}{a^2 d}+\frac{x^3}{a d \left (a+b e^{c+d x}\right )}+\frac{x^4}{4 a^2}+\frac{3 x^2 \log \left (1+\frac{b e^{c+d x}}{a}\right )}{a^2 d^2}-\frac{x^3 \log \left (1+\frac{b e^{c+d x}}{a}\right )}{a^2 d}+\frac{6 x \text{Li}_2\left (-\frac{b e^{c+d x}}{a}\right )}{a^2 d^3}-\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a}\right )}{a^2 d^2}+\frac{6 x \text{Li}_3\left (-\frac{b e^{c+d x}}{a}\right )}{a^2 d^3}-\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^2 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^2 d^4}\\ &=-\frac{x^3}{a^2 d}+\frac{x^3}{a d \left (a+b e^{c+d x}\right )}+\frac{x^4}{4 a^2}+\frac{3 x^2 \log \left (1+\frac{b e^{c+d x}}{a}\right )}{a^2 d^2}-\frac{x^3 \log \left (1+\frac{b e^{c+d x}}{a}\right )}{a^2 d}+\frac{6 x \text{Li}_2\left (-\frac{b e^{c+d x}}{a}\right )}{a^2 d^3}-\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a}\right )}{a^2 d^2}-\frac{6 \text{Li}_3\left (-\frac{b e^{c+d x}}{a}\right )}{a^2 d^4}+\frac{6 x \text{Li}_3\left (-\frac{b e^{c+d x}}{a}\right )}{a^2 d^3}-\frac{6 \text{Li}_4\left (-\frac{b e^{c+d x}}{a}\right )}{a^2 d^4}\\ \end{align*}

Mathematica [A]  time = 0.201282, size = 158, normalized size = 0.73 \[ \frac{-\frac{12 x (d x-2) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{d^3}+\frac{24 (d x-1) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a}\right )}{d^4}-\frac{24 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{a}\right )}{d^4}+\frac{12 x^2 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{d^2}+\frac{4 a x^3}{a d+b d e^{c+d x}}-\frac{4 x^3 \log \left (\frac{b e^{c+d x}}{a}+1\right )}{d}-\frac{4 x^3}{d}+x^4}{4 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.101, size = 382, normalized size = 1.8 \begin{align*}{\frac{{x}^{3}}{da \left ( a+b{{\rm e}^{dx+c}} \right ) }}-6\,{\frac{1}{{a}^{2}{d}^{4}}{\it polylog} \left ( 3,-{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }+{\frac{3\,{c}^{4}}{4\,{a}^{2}{d}^{4}}}-6\,{\frac{1}{{a}^{2}{d}^{4}}{\it polylog} \left ( 4,-{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }+2\,{\frac{{c}^{3}}{{a}^{2}{d}^{4}}}-3\,{\frac{{c}^{2}}{{a}^{2}{d}^{4}}\ln \left ( 1+{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }-{\frac{{x}^{3}}{{a}^{2}d}\ln \left ( 1+{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }-3\,{\frac{{x}^{2}}{{a}^{2}{d}^{2}}{\it polylog} \left ( 2,-{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }+6\,{\frac{x}{{a}^{2}{d}^{3}}{\it polylog} \left ( 3,-{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }+3\,{\frac{{x}^{2}}{{a}^{2}{d}^{2}}\ln \left ( 1+{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }+6\,{\frac{x}{{a}^{2}{d}^{3}}{\it polylog} \left ( 2,-{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }+3\,{\frac{{c}^{2}x}{{a}^{2}{d}^{3}}}+{\frac{{c}^{3}x}{{a}^{2}{d}^{3}}}+{\frac{{x}^{4}}{4\,{a}^{2}}}-3\,{\frac{{c}^{2}\ln \left ({{\rm e}^{dx+c}} \right ) }{{a}^{2}{d}^{4}}}+3\,{\frac{{c}^{2}\ln \left ( a+b{{\rm e}^{dx+c}} \right ) }{{a}^{2}{d}^{4}}}-{\frac{{c}^{3}\ln \left ({{\rm e}^{dx+c}} \right ) }{{a}^{2}{d}^{4}}}+{\frac{{c}^{3}\ln \left ( a+b{{\rm e}^{dx+c}} \right ) }{{a}^{2}{d}^{4}}}-{\frac{{x}^{3}}{{a}^{2}d}}-{\frac{{c}^{3}}{{a}^{2}{d}^{4}}\ln \left ( 1+{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.19674, size = 263, normalized size = 1.21 \begin{align*} \frac{x^{3}}{a b d e^{\left (d x + c\right )} + a^{2} d} + \frac{d^{4} x^{4} - 4 \, d^{3} x^{3}}{4 \, a^{2} 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^{2} d^{4}} + \frac{3 \,{\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 )}}{a^{2} d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 1.52375, size = 756, normalized size = 3.48 \begin{align*} \frac{a d^{4} x^{4} - a c^{4} - 4 \, a c^{3} - 12 \,{\left (a d^{2} x^{2} - 2 \, a d x +{\left (b d^{2} x^{2} - 2 \, b d x\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 (b d^{4} x^{4} - 4 \, b d^{3} x^{3} - b c^{4} - 4 \, b c^{3}\right )} e^{\left (d x + c\right )} + 4 \,{\left (a c^{3} + 3 \, a c^{2} +{\left (b c^{3} + 3 \, b c^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (b e^{\left (d x + c\right )} + a\right ) - 4 \,{\left (a d^{3} x^{3} - 3 \, a d^{2} x^{2} + a c^{3} + 3 \, a c^{2} +{\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 )}\right )} \log \left (\frac{b e^{\left (d x + c\right )} + a}{a}\right ) - 24 \,{\left (b e^{\left (d x + c\right )} + a\right )}{\rm polylog}\left (4, -\frac{b e^{\left (d x + c\right )}}{a}\right ) + 24 \,{\left (a d x +{\left (b d x - b\right )} e^{\left (d x + c\right )} - a\right )}{\rm polylog}\left (3, -\frac{b e^{\left (d x + c\right )}}{a}\right )}{4 \,{\left (a^{2} b d^{4} e^{\left (d x + c\right )} + a^{3} d^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

x**3/(a**2*d + a*b*d*exp(c + d*x)) + (Integral(-3*x**2/(a + b*exp(c)*exp(d*x)), x) + Integral(d*x**3/(a + b*ex
p(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^{3}}{{\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^3/(a+b*exp(d*x+c))^2,x, algorithm="giac")

[Out]

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