3.115 \(\int x \log (a+b e^x) \, dx\)

Optimal. Leaf size=59 \[ -x \text{PolyLog}\left (2,-\frac{b e^x}{a}\right )+\text{PolyLog}\left (3,-\frac{b e^x}{a}\right )+\frac{1}{2} x^2 \log \left (a+b e^x\right )-\frac{1}{2} x^2 \log \left (\frac{b e^x}{a}+1\right ) \]

[Out]

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

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Rubi [A]  time = 0.036611, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {2532, 2531, 2282, 6589} \[ -x \text{PolyLog}\left (2,-\frac{b e^x}{a}\right )+\text{PolyLog}\left (3,-\frac{b e^x}{a}\right )+\frac{1}{2} x^2 \log \left (a+b e^x\right )-\frac{1}{2} x^2 \log \left (\frac{b e^x}{a}+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*Log[a + b*E^x],x]

[Out]

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

Rule 2532

Int[Log[(d_) + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[
((f + g*x)^(m + 1)*Log[d + e*(F^(c*(a + b*x)))^n])/(g*(m + 1)), x] + (Int[(f + g*x)^m*Log[1 + (e*(F^(c*(a + b*
x)))^n)/d], x] - Simp[((f + g*x)^(m + 1)*Log[1 + (e*(F^(c*(a + b*x)))^n)/d])/(g*(m + 1)), x]) /; FreeQ[{F, a,
b, c, d, e, f, g, n}, x] && GtQ[m, 0] && NeQ[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 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 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 x \log \left (a+b e^x\right ) \, dx &=\frac{1}{2} x^2 \log \left (a+b e^x\right )-\frac{1}{2} x^2 \log \left (1+\frac{b e^x}{a}\right )+\int x \log \left (1+\frac{b e^x}{a}\right ) \, dx\\ &=\frac{1}{2} x^2 \log \left (a+b e^x\right )-\frac{1}{2} x^2 \log \left (1+\frac{b e^x}{a}\right )-x \text{Li}_2\left (-\frac{b e^x}{a}\right )+\int \text{Li}_2\left (-\frac{b e^x}{a}\right ) \, dx\\ &=\frac{1}{2} x^2 \log \left (a+b e^x\right )-\frac{1}{2} x^2 \log \left (1+\frac{b e^x}{a}\right )-x \text{Li}_2\left (-\frac{b e^x}{a}\right )+\operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{b x}{a}\right )}{x} \, dx,x,e^x\right )\\ &=\frac{1}{2} x^2 \log \left (a+b e^x\right )-\frac{1}{2} x^2 \log \left (1+\frac{b e^x}{a}\right )-x \text{Li}_2\left (-\frac{b e^x}{a}\right )+\text{Li}_3\left (-\frac{b e^x}{a}\right )\\ \end{align*}

Mathematica [A]  time = 0.0045427, size = 59, normalized size = 1. \[ -x \text{PolyLog}\left (2,-\frac{b e^x}{a}\right )+\text{PolyLog}\left (3,-\frac{b e^x}{a}\right )+\frac{1}{2} x^2 \log \left (a+b e^x\right )-\frac{1}{2} x^2 \log \left (\frac{b e^x}{a}+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Log[a + b*E^x],x]

[Out]

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

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Maple [A]  time = 0.007, size = 52, normalized size = 0.9 \begin{align*}{\frac{{x}^{2}\ln \left ( a+b{{\rm e}^{x}} \right ) }{2}}-{\frac{{x}^{2}}{2}\ln \left ( 1+{\frac{b{{\rm e}^{x}}}{a}} \right ) }-x{\it polylog} \left ( 2,-{\frac{b{{\rm e}^{x}}}{a}} \right ) +{\it polylog} \left ( 3,-{\frac{b{{\rm e}^{x}}}{a}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*ln(a+b*exp(x)),x)

[Out]

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

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Maxima [A]  time = 1.12416, size = 68, normalized size = 1.15 \begin{align*} \frac{1}{2} \, x^{2} \log \left (b e^{x} + a\right ) - \frac{1}{2} \, x^{2} \log \left (\frac{b e^{x}}{a} + 1\right ) - x{\rm Li}_2\left (-\frac{b e^{x}}{a}\right ) +{\rm Li}_{3}(-\frac{b e^{x}}{a}) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(a+b*exp(x)),x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 2.13465, size = 143, normalized size = 2.42 \begin{align*} \frac{1}{2} \, x^{2} \log \left (b e^{x} + a\right ) - \frac{1}{2} \, x^{2} \log \left (\frac{b e^{x} + a}{a}\right ) - x{\rm Li}_2\left (-\frac{b e^{x} + a}{a} + 1\right ) +{\rm polylog}\left (3, -\frac{b e^{x}}{a}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(a+b*exp(x)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{b \int \frac{x^{2} e^{x}}{a + b e^{x}}\, dx}{2} + \frac{x^{2} \log{\left (a + b e^{x} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*ln(a+b*exp(x)),x)

[Out]

-b*Integral(x**2*exp(x)/(a + b*exp(x)), x)/2 + x**2*log(a + b*exp(x))/2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(a+b*exp(x)),x, algorithm="giac")

[Out]

integrate(x*log(b*e^x + a), x)