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

Optimal. Leaf size=243 \[ \frac {3 \text {Li}_2\left (-\frac {b e^{c+d x}}{a}\right )}{a^3 d^3}+\frac {2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a}\right )}{a^3 d^3}-\frac {\log \left (a+b e^{c+d x}\right )}{a^3 d^3}-\frac {2 x \text {Li}_2\left (-\frac {b e^{c+d x}}{a}\right )}{a^3 d^2}+\frac {3 x \log \left (\frac {b e^{c+d x}}{a}+1\right )}{a^3 d^2}-\frac {x^2 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{a^3 d}+\frac {x}{a^3 d^2}-\frac {3 x^2}{2 a^3 d}+\frac {x^3}{3 a^3}-\frac {x}{a^2 d^2 \left (a+b e^{c+d x}\right )}+\frac {x^2}{a^2 d \left (a+b e^{c+d x}\right )}+\frac {x^2}{2 a d \left (a+b e^{c+d x}\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.76, antiderivative size = 243, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.19, size = 203, normalized size = 0.84 \[ \frac {\frac {3 a^2 x^2}{d \left (a+b e^{c+d x}\right )^2}-\frac {6 (2 d x-3) \text {Li}_2\left (-\frac {b e^{c+d x}}{a}\right )}{d^3}+\frac {12 \text {Li}_3\left (-\frac {b e^{c+d x}}{a}\right )}{d^3}-\frac {6 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{d^3}-\frac {6 a x}{d^2 \left (a+b e^{c+d x}\right )}+\frac {18 x \log \left (\frac {b e^{c+d x}}{a}+1\right )}{d^2}+\frac {6 a x^2}{a d+b d e^{c+d x}}-\frac {6 x^2 \log \left (\frac {b e^{c+d x}}{a}+1\right )}{d}+\frac {6 x}{d^2}-\frac {9 x^2}{d}+2 x^3}{6 a^3} \]

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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fricas [C]  time = 0.41, size = 521, normalized size = 2.14 \[ \frac {2 \, a^{2} d^{3} x^{3} + 2 \, a^{2} c^{3} + 9 \, a^{2} c^{2} + 6 \, a^{2} c - 6 \, {\left (2 \, a^{2} d x - 3 \, a^{2} + {\left (2 \, b^{2} d x - 3 \, b^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (2 \, a b d x - 3 \, a b\right )} e^{\left (d x + c\right )}\right )} {\rm Li}_2\left (-\frac {b e^{\left (d x + c\right )} + a}{a} + 1\right ) + {\left (2 \, b^{2} d^{3} x^{3} - 9 \, b^{2} d^{2} x^{2} + 2 \, b^{2} c^{3} + 9 \, b^{2} c^{2} + 6 \, b^{2} d x + 6 \, b^{2} c\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (2 \, a b d^{3} x^{3} - 6 \, a b d^{2} x^{2} + 2 \, a b c^{3} + 9 \, a b c^{2} + 3 \, a b d x + 6 \, a b c\right )} e^{\left (d x + c\right )} - 6 \, {\left (a^{2} c^{2} + 3 \, a^{2} c + a^{2} + {\left (b^{2} c^{2} + 3 \, b^{2} c + b^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (a b c^{2} + 3 \, a b c + a b\right )} e^{\left (d x + c\right )}\right )} \log \left (b e^{\left (d x + c\right )} + a\right ) - 6 \, {\left (a^{2} d^{2} x^{2} - a^{2} c^{2} - 3 \, a^{2} d x - 3 \, a^{2} c + {\left (b^{2} d^{2} x^{2} - b^{2} c^{2} - 3 \, b^{2} d x - 3 \, b^{2} c\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (a b d^{2} x^{2} - a b c^{2} - 3 \, a b d x - 3 \, a b c\right )} e^{\left (d x + c\right )}\right )} \log \left (\frac {b e^{\left (d x + c\right )} + a}{a}\right ) + 12 \, {\left (b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b e^{\left (d x + c\right )} + a^{2}\right )} {\rm polylog}\left (3, -\frac {b e^{\left (d x + c\right )}}{a}\right )}{6 \, {\left (a^{3} b^{2} d^{3} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{4} b d^{3} e^{\left (d x + c\right )} + a^{5} d^{3}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{{\left (b e^{\left (d x + c\right )} + a\right )}^{3}}\,{d x} \]

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)

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maple [A]  time = 0.14, size = 385, normalized size = 1.58 \[ \frac {x^{3}}{3 a^{3}}-\frac {x^{2} \ln \left (\frac {b \,{\mathrm e}^{d x +c}}{a}+1\right )}{a^{3} d}-\frac {c^{2} x}{a^{3} d^{2}}-\frac {3 x^{2}}{2 a^{3} d}+\frac {\left (2 b d x \,{\mathrm e}^{d x +c}+3 a d x -2 b \,{\mathrm e}^{d x +c}-2 a \right ) x}{2 \left (b \,{\mathrm e}^{d x +c}+a \right )^{2} a^{2} d^{2}}-\frac {2 c^{3}}{3 a^{3} d^{3}}+\frac {c^{2} \ln \left (\frac {b \,{\mathrm e}^{d x +c}}{a}+1\right )}{a^{3} d^{3}}-\frac {c^{2} \ln \left (b \,{\mathrm e}^{d x +c}+a \right )}{a^{3} d^{3}}+\frac {c^{2} \ln \left ({\mathrm e}^{d x +c}\right )}{a^{3} d^{3}}-\frac {3 c x}{a^{3} d^{2}}-\frac {2 x \polylog \left (2, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{3} d^{2}}+\frac {3 x \ln \left (\frac {b \,{\mathrm e}^{d x +c}}{a}+1\right )}{a^{3} d^{2}}-\frac {3 c^{2}}{2 a^{3} d^{3}}+\frac {3 c \ln \left (\frac {b \,{\mathrm e}^{d x +c}}{a}+1\right )}{a^{3} d^{3}}-\frac {3 c \ln \left (b \,{\mathrm e}^{d x +c}+a \right )}{a^{3} d^{3}}+\frac {3 c \ln \left ({\mathrm e}^{d x +c}\right )}{a^{3} d^{3}}+\frac {3 \polylog \left (2, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{3} d^{3}}+\frac {2 \polylog \left (3, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a^{3} d^{3}}-\frac {\ln \left (b \,{\mathrm e}^{d x +c}+a \right )}{a^{3} d^{3}}+\frac {\ln \left ({\mathrm e}^{d x +c}\right )}{a^{3} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.06, size = 234, normalized size = 0.96 \[ \frac {3 \, a d x^{2} - 2 \, a x + 2 \, {\left (b d x^{2} e^{c} - b x e^{c}\right )} e^{\left (d x\right )}}{2 \, {\left (a^{2} b^{2} d^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{3} b d^{2} e^{\left (d x + c\right )} + a^{4} d^{2}\right )}} + \frac {x}{a^{3} d^{2}} + \frac {2 \, d^{3} x^{3} - 9 \, d^{2} x^{2}}{6 \, a^{3} d^{3}} - \frac {d^{2} x^{2} \log \left (\frac {b e^{\left (d x + c\right )}}{a} + 1\right ) + 2 \, d x {\rm Li}_2\left (-\frac {b e^{\left (d x + c\right )}}{a}\right ) - 2 \, {\rm Li}_{3}(-\frac {b e^{\left (d x + c\right )}}{a})}{a^{3} d^{3}} + \frac {3 \, {\left (d x \log \left (\frac {b e^{\left (d x + c\right )}}{a} + 1\right ) + {\rm Li}_2\left (-\frac {b e^{\left (d x + c\right )}}{a}\right )\right )}}{a^{3} d^{3}} - \frac {\log \left (b e^{\left (d x + c\right )} + a\right )}{a^{3} d^{3}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2}{{\left (a+b\,{\mathrm {e}}^{c+d\,x}\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*exp(c + d*x))^3,x)

[Out]

int(x^2/(a + b*exp(c + d*x))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {3 a d x^{2} - 2 a x + \left (2 b d x^{2} - 2 b x\right ) e^{c + d x}}{2 a^{4} d^{2} + 4 a^{3} b d^{2} e^{c + d x} + 2 a^{2} b^{2} d^{2} e^{2 c + 2 d x}} + \frac {\int \left (- \frac {3 d x}{a + b e^{c} e^{d x}}\right )\, dx + \int \frac {d^{2} x^{2}}{a + b e^{c} e^{d x}}\, dx + \int \frac {1}{a + b e^{c} e^{d x}}\, dx}{a^{2} d^{2}} \]

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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