\(\int \frac {x^3}{(a+b e^{c+d x})^2} \, dx\) [9]

Optimal result

Integrand size = 17, antiderivative size = 217 \[ \int \frac {x^3}{\left (a+b e^{c+d x}\right )^2} \, dx=-\frac {x^3}{a^2 d}+\frac {x^3}{a d \left (a+b e^{c+d x}\right )}+\frac {x^4}{4 a^2}+\frac {3 x^2 \log \left (1+\frac {b e^{c+d x}}{a}\right )}{a^2 d^2}-\frac {x^3 \log \left (1+\frac {b e^{c+d x}}{a}\right )}{a^2 d}+\frac {6 x \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{a^2 d^3}-\frac {3 x^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{a^2 d^2}-\frac {6 \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a}\right )}{a^2 d^4}+\frac {6 x \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a}\right )}{a^2 d^3}-\frac {6 \operatorname {PolyLog}\left (4,-\frac {b e^{c+d x}}{a}\right )}{a^2 d^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.73 \[ \int \frac {x^3}{\left (a+b e^{c+d x}\right )^2} \, dx=\frac {-\frac {4 x^3}{d}+\frac {4 a x^3}{a d+b d e^{c+d x}}+x^4+\frac {12 x^2 \log \left (1+\frac {b e^{c+d x}}{a}\right )}{d^2}-\frac {4 x^3 \log \left (1+\frac {b e^{c+d x}}{a}\right )}{d}-\frac {12 x (-2+d x) \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{d^3}+\frac {24 (-1+d x) \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a}\right )}{d^4}-\frac {24 \operatorname {PolyLog}\left (4,-\frac {b e^{c+d x}}{a}\right )}{d^4}}{4 a^2} \] Input:

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

Output:

((-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)
 

Rubi [A] (verified)

Time = 2.76 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.18, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.706, Rules used = {2616, 2615, 2620, 2621, 2615, 2620, 3011, 2720, 7143, 7163, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2616

\(\displaystyle \frac {\int \frac {x^3}{a+b e^{c+d x}}dx}{a}-\frac {b \int \frac {e^{c+d x} x^3}{\left (a+b e^{c+d x}\right )^2}dx}{a}\)

\(\Big \downarrow \) 2615

\(\displaystyle \frac {\frac {x^4}{4 a}-\frac {b \int \frac {e^{c+d x} x^3}{a+b e^{c+d x}}dx}{a}}{a}-\frac {b \int \frac {e^{c+d x} x^3}{\left (a+b e^{c+d x}\right )^2}dx}{a}\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {\frac {x^4}{4 a}-\frac {b \left (\frac {x^3 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {3 \int x^2 \log \left (\frac {e^{c+d x} b}{a}+1\right )dx}{b d}\right )}{a}}{a}-\frac {b \int \frac {e^{c+d x} x^3}{\left (a+b e^{c+d x}\right )^2}dx}{a}\)

\(\Big \downarrow \) 2621

\(\displaystyle \frac {\frac {x^4}{4 a}-\frac {b \left (\frac {x^3 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {3 \int x^2 \log \left (\frac {e^{c+d x} b}{a}+1\right )dx}{b d}\right )}{a}}{a}-\frac {b \left (\frac {3 \int \frac {x^2}{a+b e^{c+d x}}dx}{b d}-\frac {x^3}{b d \left (a+b e^{c+d x}\right )}\right )}{a}\)

\(\Big \downarrow \) 2615

\(\displaystyle \frac {\frac {x^4}{4 a}-\frac {b \left (\frac {x^3 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {3 \int x^2 \log \left (\frac {e^{c+d x} b}{a}+1\right )dx}{b d}\right )}{a}}{a}-\frac {b \left (\frac {3 \left (\frac {x^3}{3 a}-\frac {b \int \frac {e^{c+d x} x^2}{a+b e^{c+d x}}dx}{a}\right )}{b d}-\frac {x^3}{b d \left (a+b e^{c+d x}\right )}\right )}{a}\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {\frac {x^4}{4 a}-\frac {b \left (\frac {x^3 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {3 \int x^2 \log \left (\frac {e^{c+d x} b}{a}+1\right )dx}{b d}\right )}{a}}{a}-\frac {b \left (\frac {3 \left (\frac {x^3}{3 a}-\frac {b \left (\frac {x^2 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {2 \int x \log \left (\frac {e^{c+d x} b}{a}+1\right )dx}{b d}\right )}{a}\right )}{b d}-\frac {x^3}{b d \left (a+b e^{c+d x}\right )}\right )}{a}\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {\frac {x^4}{4 a}-\frac {b \left (\frac {x^3 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {3 \left (\frac {2 \int x \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )dx}{d}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{d}\right )}{b d}\right )}{a}}{a}-\frac {b \left (\frac {3 \left (\frac {x^3}{3 a}-\frac {b \left (\frac {x^2 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {2 \left (\frac {\int \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )dx}{d}-\frac {x \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{d}\right )}{b d}\right )}{a}\right )}{b d}-\frac {x^3}{b d \left (a+b e^{c+d x}\right )}\right )}{a}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\frac {x^4}{4 a}-\frac {b \left (\frac {x^3 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {3 \left (\frac {2 \int x \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )dx}{d}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{d}\right )}{b d}\right )}{a}}{a}-\frac {b \left (\frac {3 \left (\frac {x^3}{3 a}-\frac {b \left (\frac {x^2 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {2 \left (\frac {\int e^{-c-d x} \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )de^{c+d x}}{d^2}-\frac {x \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{d}\right )}{b d}\right )}{a}\right )}{b d}-\frac {x^3}{b d \left (a+b e^{c+d x}\right )}\right )}{a}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {\frac {x^4}{4 a}-\frac {b \left (\frac {x^3 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {3 \left (\frac {2 \int x \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )dx}{d}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{d}\right )}{b d}\right )}{a}}{a}-\frac {b \left (\frac {3 \left (\frac {x^3}{3 a}-\frac {b \left (\frac {x^2 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {2 \left (\frac {\operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a}\right )}{d^2}-\frac {x \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{d}\right )}{b d}\right )}{a}\right )}{b d}-\frac {x^3}{b d \left (a+b e^{c+d x}\right )}\right )}{a}\)

\(\Big \downarrow \) 7163

\(\displaystyle \frac {\frac {x^4}{4 a}-\frac {b \left (\frac {x^3 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {3 \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a}\right )}{d}-\frac {\int \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a}\right )dx}{d}\right )}{d}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{d}\right )}{b d}\right )}{a}}{a}-\frac {b \left (\frac {3 \left (\frac {x^3}{3 a}-\frac {b \left (\frac {x^2 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {2 \left (\frac {\operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a}\right )}{d^2}-\frac {x \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{d}\right )}{b d}\right )}{a}\right )}{b d}-\frac {x^3}{b d \left (a+b e^{c+d x}\right )}\right )}{a}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\frac {x^4}{4 a}-\frac {b \left (\frac {x^3 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {3 \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a}\right )}{d}-\frac {\int e^{-c-d x} \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a}\right )de^{c+d x}}{d^2}\right )}{d}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{d}\right )}{b d}\right )}{a}}{a}-\frac {b \left (\frac {3 \left (\frac {x^3}{3 a}-\frac {b \left (\frac {x^2 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {2 \left (\frac {\operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a}\right )}{d^2}-\frac {x \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{d}\right )}{b d}\right )}{a}\right )}{b d}-\frac {x^3}{b d \left (a+b e^{c+d x}\right )}\right )}{a}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {\frac {x^4}{4 a}-\frac {b \left (\frac {x^3 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {3 \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a}\right )}{d}-\frac {\operatorname {PolyLog}\left (4,-\frac {b e^{c+d x}}{a}\right )}{d^2}\right )}{d}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{d}\right )}{b d}\right )}{a}}{a}-\frac {b \left (\frac {3 \left (\frac {x^3}{3 a}-\frac {b \left (\frac {x^2 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{b d}-\frac {2 \left (\frac {\operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a}\right )}{d^2}-\frac {x \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{d}\right )}{b d}\right )}{a}\right )}{b d}-\frac {x^3}{b d \left (a+b e^{c+d x}\right )}\right )}{a}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2615

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] - Simp[ 
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 2616

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_)*((c_.) + 
(d_.)*(x_))^(m_.), x_Symbol] :> Simp[1/a   Int[(c + d*x)^m*(a + b*(F^(g*(e 
+ f*x)))^n)^(p + 1), x], x] - Simp[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 2620

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[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 2621

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] - Simp[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 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{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 3011

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] + Simp[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 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> 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 7163

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] - Simp[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]
 

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.76

method	result	size
risch	\(\frac {x^{3}}{a d \left (a +b \,{\mathrm e}^{d x +c}\right )}+\frac {3 c^{2} \ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a^{2} d^{4}}+\frac {3 x^{2} \ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{2} d^{2}}+\frac {6 x \operatorname {polylog}\left (2, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{2} d^{3}}+\frac {c^{3} x}{a^{2} d^{3}}-\frac {x^{3} \ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{2} d}-\frac {3 x^{2} \operatorname {polylog}\left (2, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{2} d^{2}}+\frac {6 x \operatorname {polylog}\left (3, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{2} d^{3}}+\frac {3 c^{2} x}{a^{2} d^{3}}+\frac {c^{3} \ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a^{2} d^{4}}-\frac {c^{3} \ln \left ({\mathrm e}^{d x +c}\right )}{a^{2} d^{4}}+\frac {x^{4}}{4 a^{2}}-\frac {\ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right ) c^{3}}{a^{2} d^{4}}-\frac {x^{3}}{a^{2} d}-\frac {3 \ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right ) c^{2}}{a^{2} d^{4}}-\frac {3 c^{2} \ln \left ({\mathrm e}^{d x +c}\right )}{a^{2} d^{4}}+\frac {3 c^{4}}{4 a^{2} d^{4}}-\frac {6 \operatorname {polylog}\left (4, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{2} d^{4}}+\frac {2 c^{3}}{a^{2} d^{4}}-\frac {6 \operatorname {polylog}\left (3, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{2} d^{4}}\)	\(382\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.52 \[ \int \frac {x^3}{\left (a+b e^{c+d x}\right )^2} \, dx=\frac {a d^{4} x^{4} - a c^{4} - 4 \, a c^{3} - 12 \, {\left (a d^{2} x^{2} - 2 \, a d x + {\left (b d^{2} x^{2} - 2 \, b d x\right )} e^{\left (d x + c\right )}\right )} {\rm Li}_2\left (-\frac {b e^{\left (d x + c\right )} + a}{a} + 1\right ) + {\left (b d^{4} x^{4} - 4 \, b d^{3} x^{3} - b c^{4} - 4 \, b c^{3}\right )} e^{\left (d x + c\right )} + 4 \, {\left (a c^{3} + 3 \, a c^{2} + {\left (b c^{3} + 3 \, b c^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (b e^{\left (d x + c\right )} + a\right ) - 4 \, {\left (a d^{3} x^{3} - 3 \, a d^{2} x^{2} + a c^{3} + 3 \, a c^{2} + {\left (b d^{3} x^{3} - 3 \, b d^{2} x^{2} + b c^{3} + 3 \, b c^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (\frac {b e^{\left (d x + c\right )} + a}{a}\right ) - 24 \, {\left (b e^{\left (d x + c\right )} + a\right )} {\rm polylog}\left (4, -\frac {b e^{\left (d x + c\right )}}{a}\right ) + 24 \, {\left (a d x + {\left (b d x - b\right )} e^{\left (d x + c\right )} - a\right )} {\rm polylog}\left (3, -\frac {b e^{\left (d x + c\right )}}{a}\right )}{4 \, {\left (a^{2} b d^{4} e^{\left (d x + c\right )} + a^{3} d^{4}\right )}} \] Input:

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

Output:

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)
 

Sympy [F]

\[ \int \frac {x^3}{\left (a+b e^{c+d x}\right )^2} \, dx=\frac {x^{3}}{a^{2} d + a b d e^{c + d x}} + \frac {\int \left (- \frac {3 x^{2}}{a + b e^{c} e^{d x}}\right )\, dx + \int \frac {d x^{3}}{a + b e^{c} e^{d x}}\, dx}{a d} \] Input:

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

Output:

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*exp(c)*exp(d*x)), x))/(a*d)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.90 \[ \int \frac {x^3}{\left (a+b e^{c+d x}\right )^2} \, dx=\frac {x^{3}}{a b d e^{\left (d x + c\right )} + a^{2} d} + \frac {d^{4} x^{4} - 4 \, d^{3} x^{3}}{4 \, a^{2} d^{4}} - \frac {d^{3} x^{3} \log \left (\frac {b e^{\left (d x + c\right )}}{a} + 1\right ) + 3 \, d^{2} x^{2} {\rm Li}_2\left (-\frac {b e^{\left (d x + c\right )}}{a}\right ) - 6 \, d x {\rm Li}_{3}(-\frac {b e^{\left (d x + c\right )}}{a}) + 6 \, {\rm Li}_{4}(-\frac {b e^{\left (d x + c\right )}}{a})}{a^{2} d^{4}} + \frac {3 \, {\left (d^{2} x^{2} \log \left (\frac {b e^{\left (d x + c\right )}}{a} + 1\right ) + 2 \, d x {\rm Li}_2\left (-\frac {b e^{\left (d x + c\right )}}{a}\right ) - 2 \, {\rm Li}_{3}(-\frac {b e^{\left (d x + c\right )}}{a})\right )}}{a^{2} d^{4}} \] Input:

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

Output:

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)
 

Giac [F]

\[ \int \frac {x^3}{\left (a+b e^{c+d x}\right )^2} \, dx=\int { \frac {x^{3}}{{\left (b e^{\left (d x + c\right )} + a\right )}^{2}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\left (a+b e^{c+d x}\right )^2} \, dx=\int \frac {x^3}{{\left (a+b\,{\mathrm {e}}^{c+d\,x}\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {x^3}{\left (a+b e^{c+d x}\right )^2} \, dx=\int \frac {x^{3}}{e^{2 d x +2 c} b^{2}+2 e^{d x +c} a b +a^{2}}d x \] Input:

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

Output:

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