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

Optimal. Leaf size=58 \[ -\frac{\text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a d^2}-\frac{x \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a d}+\frac{x^2}{2 a} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0991882, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {2184, 2190, 2279, 2391} \[ -\frac{\text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a}\right )}{a d^2}-\frac{x \log \left (\frac{b e^{c+d x}}{a}+1\right )}{a d}+\frac{x^2}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*E^(c + d*x)),x]

[Out]

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

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]

Rubi steps

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

Mathematica [A]  time = 0.0030927, size = 53, normalized size = 0.91 \[ \frac{\text{PolyLog}\left (2,-\frac{a e^{-c-d x}}{b}\right )}{a d^2}-\frac{x \log \left (\frac{a e^{-c-d x}}{b}+1\right )}{a d} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*E^(c + d*x)),x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.005, size = 133, normalized size = 2.3 \begin{align*}{\frac{{x}^{2}}{2\,a}}+{\frac{cx}{ad}}+{\frac{{c}^{2}}{2\,a{d}^{2}}}-{\frac{x}{ad}\ln \left ( 1+{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }-{\frac{c}{a{d}^{2}}\ln \left ( 1+{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }-{\frac{1}{a{d}^{2}}{\it polylog} \left ( 2,-{\frac{b{{\rm e}^{dx+c}}}{a}} \right ) }-{\frac{c\ln \left ({{\rm e}^{dx+c}} \right ) }{a{d}^{2}}}+{\frac{c\ln \left ( a+b{{\rm e}^{dx+c}} \right ) }{a{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.08588, size = 65, normalized size = 1.12 \begin{align*} \frac{x^{2}}{2 \, a} - \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 d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.49365, size = 177, normalized size = 3.05 \begin{align*} \frac{d^{2} x^{2} + 2 \, c \log \left (b e^{\left (d x + c\right )} + a\right ) - 2 \,{\left (d x + c\right )} \log \left (\frac{b e^{\left (d x + c\right )} + a}{a}\right ) - 2 \,{\rm Li}_2\left (-\frac{b e^{\left (d x + c\right )} + a}{a} + 1\right )}{2 \, a d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{a + b e^{c} e^{d x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{b e^{\left (d x + c\right )} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(b*e^(d*x + c) + a), x)