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

Optimal. Leaf size=107 \[ -\frac{\text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^2 d^2}+\frac{\log \left (a+b e^{c+d x}\right )}{a^2 d^2}-\frac{x \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a^2 d}-\frac{x}{a^2 d}+\frac{x^2}{2 a^2}+\frac{x}{a d \left (a+b e^{c+d x}\right )} \]

[Out]

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

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Rubi [A]  time = 0.2014, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {2185, 2184, 2190, 2279, 2391, 2191, 2282, 36, 29, 31} \[ -\frac{\text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a^2 d^2}+\frac{\log \left (a+b e^{c+d x}\right )}{a^2 d^2}-\frac{x \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a^2 d}-\frac{x}{a^2 d}+\frac{x^2}{2 a^2}+\frac{x}{a d \left (a+b e^{c+d x}\right )} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*E^(c + d*x))^2,x]

[Out]

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

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 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]

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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.118005, size = 85, normalized size = 0.79 \[ \frac{-2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )+\frac{d x \left (a d x+b (d x-2) e^{c+d x}\right )}{a+b e^{c+d x}}-2 (d x-1) \log \left (\frac{b e^{c+d x}}{a}+1\right )}{2 a^2 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*E^(c + d*x))^2,x]

[Out]

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

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Maple [B]  time = 0.023, size = 231, normalized size = 2.2 \begin{align*}{\frac{{x}^{2}}{2\,{a}^{2}}}+{\frac{cx}{{a}^{2}d}}+{\frac{{c}^{2}}{2\,{d}^{2}{a}^{2}}}-{\frac{1}{{d}^{2}{a}^{2}}{\it dilog} \left ({\frac{a+b{{\rm e}^{dx+c}}}{a}} \right ) }-{\frac{x}{{a}^{2}d}\ln \left ({\frac{a+b{{\rm e}^{dx+c}}}{a}} \right ) }-{\frac{c}{{d}^{2}{a}^{2}}\ln \left ({\frac{a+b{{\rm e}^{dx+c}}}{a}} \right ) }+{\frac{\ln \left ( a+b{{\rm e}^{dx+c}} \right ) }{{d}^{2}{a}^{2}}}-{\frac{b{{\rm e}^{dx+c}}x}{{a}^{2}d \left ( a+b{{\rm e}^{dx+c}} \right ) }}-{\frac{b{{\rm e}^{dx+c}}c}{{d}^{2}{a}^{2} \left ( a+b{{\rm e}^{dx+c}} \right ) }}-{\frac{c\ln \left ({{\rm e}^{dx+c}} \right ) }{{d}^{2}{a}^{2}}}+{\frac{c\ln \left ( a+b{{\rm e}^{dx+c}} \right ) }{{d}^{2}{a}^{2}}}-{\frac{c}{a{d}^{2} \left ( a+b{{\rm e}^{dx+c}} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.11172, size = 128, normalized size = 1.2 \begin{align*} \frac{x}{a b d e^{\left (d x + c\right )} + a^{2} d} + \frac{x^{2}}{2 \, a^{2}} - \frac{x}{a^{2} 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^{2} d^{2}} + \frac{\log \left (b e^{\left (d x + c\right )} + a\right )}{a^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.53214, size = 420, normalized size = 3.93 \begin{align*} \frac{a d^{2} x^{2} - a c^{2} - 2 \, a c - 2 \,{\left (b e^{\left (d x + c\right )} + a\right )}{\rm Li}_2\left (-\frac{b e^{\left (d x + c\right )} + a}{a} + 1\right ) +{\left (b d^{2} x^{2} - b c^{2} - 2 \, b d x - 2 \, b c\right )} e^{\left (d x + c\right )} + 2 \,{\left (a c +{\left (b c + b\right )} e^{\left (d x + c\right )} + a\right )} \log \left (b e^{\left (d x + c\right )} + a\right ) - 2 \,{\left (a d x + a c +{\left (b d x + 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^{2} b d^{2} e^{\left (d x + c\right )} + a^{3} d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{x}{a^{2} d + a b d e^{c + d x}} + \frac{\int \frac{d x}{a + b e^{c} e^{d x}}\, dx + \int - \frac{1}{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/(a+b*exp(d*x+c))**2,x)

[Out]

x/(a**2*d + a*b*d*exp(c + d*x)) + (Integral(d*x/(a + b*exp(c)*exp(d*x)), x) + Integral(-1/(a + b*exp(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}{{\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/(a+b*exp(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(x/(b*e^(d*x + c) + a)^2, x)