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

Optimal. Leaf size=159 \[ -\frac {\text {Li}_2\left (-\frac {b e^{c+d x}}{a}\right )}{a^3 d^2}+\frac {3 \log \left (a+b e^{c+d x}\right )}{2 a^3 d^2}-\frac {x \log \left (\frac {b e^{c+d x}}{a}+1\right )}{a^3 d}-\frac {3 x}{2 a^3 d}+\frac {x^2}{2 a^3}-\frac {1}{2 a^2 d^2 \left (a+b e^{c+d x}\right )}+\frac {x}{a^2 d \left (a+b e^{c+d x}\right )}+\frac {x}{2 a d \left (a+b e^{c+d x}\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.33, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.733, Rules used = {2185, 2184, 2190, 2279, 2391, 2191, 2282, 36, 29, 31, 44} \[ -\frac {\text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{a^3 d^2}-\frac {1}{2 a^2 d^2 \left (a+b e^{c+d x}\right )}+\frac {3 \log \left (a+b e^{c+d x}\right )}{2 a^3 d^2}+\frac {x}{a^2 d \left (a+b e^{c+d x}\right )}-\frac {x \log \left (\frac {b e^{c+d x}}{a}+1\right )}{a^3 d}-\frac {3 x}{2 a^3 d}+\frac {x^2}{2 a^3}+\frac {x}{2 a d \left (a+b e^{c+d x}\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*E^(c + d*x))^3,x]

[Out]

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

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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 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]

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 120, normalized size = 0.75 \[ \frac {\frac {d x \left (d x-2 \log \left (\frac {b e^{c+d x}}{a}+1\right )\right )-2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a}\right )}{a}+\frac {a d x}{\left (a+b e^{c+d x}\right )^2}+\frac {2 d x-1}{a+b e^{c+d x}}+\frac {3 \log \left (\frac {b e^{c+d x}}{a}+1\right )-3 d x}{a}}{2 a^2 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*E^(c + d*x))^3,x]

[Out]

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

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fricas [B]  time = 0.47, size = 338, normalized size = 2.13 \[ \frac {a^{2} d^{2} x^{2} - a^{2} c^{2} - 3 \, a^{2} c - a^{2} - 2 \, {\left (b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b e^{\left (d x + c\right )} + a^{2}\right )} {\rm Li}_2\left (-\frac {b e^{\left (d x + c\right )} + a}{a} + 1\right ) + {\left (b^{2} d^{2} x^{2} - b^{2} c^{2} - 3 \, b^{2} d x - 3 \, b^{2} c\right )} e^{\left (2 \, d x + 2 \, c\right )} + {\left (2 \, a b d^{2} x^{2} - 2 \, a b c^{2} - 4 \, a b d x - 6 \, a b c - a b\right )} e^{\left (d x + c\right )} + {\left (2 \, a^{2} c + 3 \, a^{2} + {\left (2 \, b^{2} c + 3 \, b^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (2 \, a b c + 3 \, a b\right )} e^{\left (d x + c\right )}\right )} \log \left (b e^{\left (d x + c\right )} + a\right ) - 2 \, {\left (a^{2} d x + a^{2} c + {\left (b^{2} d x + b^{2} c\right )} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, {\left (a b d x + a b c\right )} e^{\left (d x + c\right )}\right )} \log \left (\frac {b e^{\left (d x + c\right )} + a}{a}\right )}{2 \, {\left (a^{3} b^{2} d^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{4} b d^{2} e^{\left (d x + c\right )} + a^{5} d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(b*e^(d*x + c) + a)^3, x)

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maple [B]  time = 0.04, size = 393, normalized size = 2.47 \[ -\frac {b x \,{\mathrm e}^{d x +c}}{\left (b \,{\mathrm e}^{d x +c}+a \right )^{2} a^{2} d}-\frac {b^{2} x \,{\mathrm e}^{2 d x +2 c}}{2 \left (b \,{\mathrm e}^{d x +c}+a \right )^{2} a^{3} d}-\frac {b c \,{\mathrm e}^{d x +c}}{\left (b \,{\mathrm e}^{d x +c}+a \right )^{2} a^{2} d^{2}}-\frac {b^{2} c \,{\mathrm e}^{2 d x +2 c}}{2 \left (b \,{\mathrm e}^{d x +c}+a \right )^{2} a^{3} d^{2}}-\frac {b x \,{\mathrm e}^{d x +c}}{\left (b \,{\mathrm e}^{d x +c}+a \right ) a^{3} d}+\frac {x^{2}}{2 a^{3}}-\frac {c}{2 \left (b \,{\mathrm e}^{d x +c}+a \right )^{2} a \,d^{2}}-\frac {b c \,{\mathrm e}^{d x +c}}{\left (b \,{\mathrm e}^{d x +c}+a \right ) a^{3} d^{2}}+\frac {c x}{a^{3} d}-\frac {x \ln \left (\frac {b \,{\mathrm e}^{d x +c}+a}{a}\right )}{a^{3} d}-\frac {c}{\left (b \,{\mathrm e}^{d x +c}+a \right ) a^{2} d^{2}}+\frac {c^{2}}{2 a^{3} d^{2}}-\frac {c \ln \left (\frac {b \,{\mathrm e}^{d x +c}+a}{a}\right )}{a^{3} d^{2}}+\frac {c \ln \left (b \,{\mathrm e}^{d x +c}+a \right )}{a^{3} d^{2}}-\frac {c \ln \left ({\mathrm e}^{d x +c}\right )}{a^{3} d^{2}}-\frac {1}{2 \left (b \,{\mathrm e}^{d x +c}+a \right ) a^{2} d^{2}}-\frac {\dilog \left (\frac {b \,{\mathrm e}^{d x +c}+a}{a}\right )}{a^{3} d^{2}}+\frac {3 \ln \left (b \,{\mathrm e}^{d x +c}+a \right )}{2 a^{3} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.05, size = 149, normalized size = 0.94 \[ \frac {3 \, a d x + {\left (2 \, b d x e^{c} - b e^{c}\right )} e^{\left (d x\right )} - a}{2 \, {\left (a^{2} b^{2} d^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{3} b d^{2} e^{\left (d x + c\right )} + a^{4} d^{2}\right )}} + \frac {x^{2}}{2 \, a^{3}} - \frac {3 \, x}{2 \, a^{3} d} - \frac {d x \log \left (\frac {b e^{\left (d x + c\right )}}{a} + 1\right ) + {\rm Li}_2\left (-\frac {b e^{\left (d x + c\right )}}{a}\right )}{a^{3} d^{2}} + \frac {3 \, \log \left (b e^{\left (d x + c\right )} + a\right )}{2 \, a^{3} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{{\left (a+b\,{\mathrm {e}}^{c+d\,x}\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*exp(c + d*x))^3,x)

[Out]

int(x/(a + b*exp(c + d*x))^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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