3.113 \(\int x^3 \log (a+b e^x) \, dx\)

Optimal. Leaf size=93 \[ -x^3 \text{PolyLog}\left (2,-\frac{b e^x}{a}\right )+3 x^2 \text{PolyLog}\left (3,-\frac{b e^x}{a}\right )-6 x \text{PolyLog}\left (4,-\frac{b e^x}{a}\right )+6 \text{PolyLog}\left (5,-\frac{b e^x}{a}\right )+\frac{1}{4} x^4 \log \left (a+b e^x\right )-\frac{1}{4} x^4 \log \left (\frac{b e^x}{a}+1\right ) \]

[Out]

(x^4*Log[a + b*E^x])/4 - (x^4*Log[1 + (b*E^x)/a])/4 - x^3*PolyLog[2, -((b*E^x)/a)] + 3*x^2*PolyLog[3, -((b*E^x
)/a)] - 6*x*PolyLog[4, -((b*E^x)/a)] + 6*PolyLog[5, -((b*E^x)/a)]

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Rubi [A]  time = 0.0730538, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {2532, 2531, 6609, 2282, 6589} \[ -x^3 \text{PolyLog}\left (2,-\frac{b e^x}{a}\right )+3 x^2 \text{PolyLog}\left (3,-\frac{b e^x}{a}\right )-6 x \text{PolyLog}\left (4,-\frac{b e^x}{a}\right )+6 \text{PolyLog}\left (5,-\frac{b e^x}{a}\right )+\frac{1}{4} x^4 \log \left (a+b e^x\right )-\frac{1}{4} x^4 \log \left (\frac{b e^x}{a}+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^3*Log[a + b*E^x],x]

[Out]

(x^4*Log[a + b*E^x])/4 - (x^4*Log[1 + (b*E^x)/a])/4 - x^3*PolyLog[2, -((b*E^x)/a)] + 3*x^2*PolyLog[3, -((b*E^x
)/a)] - 6*x*PolyLog[4, -((b*E^x)/a)] + 6*PolyLog[5, -((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 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, 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^3 \log \left (a+b e^x\right ) \, dx &=\frac{1}{4} x^4 \log \left (a+b e^x\right )-\frac{1}{4} x^4 \log \left (1+\frac{b e^x}{a}\right )+\int x^3 \log \left (1+\frac{b e^x}{a}\right ) \, dx\\ &=\frac{1}{4} x^4 \log \left (a+b e^x\right )-\frac{1}{4} x^4 \log \left (1+\frac{b e^x}{a}\right )-x^3 \text{Li}_2\left (-\frac{b e^x}{a}\right )+3 \int x^2 \text{Li}_2\left (-\frac{b e^x}{a}\right ) \, dx\\ &=\frac{1}{4} x^4 \log \left (a+b e^x\right )-\frac{1}{4} x^4 \log \left (1+\frac{b e^x}{a}\right )-x^3 \text{Li}_2\left (-\frac{b e^x}{a}\right )+3 x^2 \text{Li}_3\left (-\frac{b e^x}{a}\right )-6 \int x \text{Li}_3\left (-\frac{b e^x}{a}\right ) \, dx\\ &=\frac{1}{4} x^4 \log \left (a+b e^x\right )-\frac{1}{4} x^4 \log \left (1+\frac{b e^x}{a}\right )-x^3 \text{Li}_2\left (-\frac{b e^x}{a}\right )+3 x^2 \text{Li}_3\left (-\frac{b e^x}{a}\right )-6 x \text{Li}_4\left (-\frac{b e^x}{a}\right )+6 \int \text{Li}_4\left (-\frac{b e^x}{a}\right ) \, dx\\ &=\frac{1}{4} x^4 \log \left (a+b e^x\right )-\frac{1}{4} x^4 \log \left (1+\frac{b e^x}{a}\right )-x^3 \text{Li}_2\left (-\frac{b e^x}{a}\right )+3 x^2 \text{Li}_3\left (-\frac{b e^x}{a}\right )-6 x \text{Li}_4\left (-\frac{b e^x}{a}\right )+6 \operatorname{Subst}\left (\int \frac{\text{Li}_4\left (-\frac{b x}{a}\right )}{x} \, dx,x,e^x\right )\\ &=\frac{1}{4} x^4 \log \left (a+b e^x\right )-\frac{1}{4} x^4 \log \left (1+\frac{b e^x}{a}\right )-x^3 \text{Li}_2\left (-\frac{b e^x}{a}\right )+3 x^2 \text{Li}_3\left (-\frac{b e^x}{a}\right )-6 x \text{Li}_4\left (-\frac{b e^x}{a}\right )+6 \text{Li}_5\left (-\frac{b e^x}{a}\right )\\ \end{align*}

Mathematica [A]  time = 0.0071216, size = 93, normalized size = 1. \[ -x^3 \text{PolyLog}\left (2,-\frac{b e^x}{a}\right )+3 x^2 \text{PolyLog}\left (3,-\frac{b e^x}{a}\right )-6 x \text{PolyLog}\left (4,-\frac{b e^x}{a}\right )+6 \text{PolyLog}\left (5,-\frac{b e^x}{a}\right )+\frac{1}{4} x^4 \log \left (a+b e^x\right )-\frac{1}{4} x^4 \log \left (\frac{b e^x}{a}+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Log[a + b*E^x],x]

[Out]

(x^4*Log[a + b*E^x])/4 - (x^4*Log[1 + (b*E^x)/a])/4 - x^3*PolyLog[2, -((b*E^x)/a)] + 3*x^2*PolyLog[3, -((b*E^x
)/a)] - 6*x*PolyLog[4, -((b*E^x)/a)] + 6*PolyLog[5, -((b*E^x)/a)]

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Maple [A]  time = 0.007, size = 84, normalized size = 0.9 \begin{align*}{\frac{{x}^{4}\ln \left ( a+b{{\rm e}^{x}} \right ) }{4}}-{\frac{{x}^{4}}{4}\ln \left ( 1+{\frac{b{{\rm e}^{x}}}{a}} \right ) }-{x}^{3}{\it polylog} \left ( 2,-{\frac{b{{\rm e}^{x}}}{a}} \right ) +3\,{x}^{2}{\it polylog} \left ( 3,-{\frac{b{{\rm e}^{x}}}{a}} \right ) -6\,x{\it polylog} \left ( 4,-{\frac{b{{\rm e}^{x}}}{a}} \right ) +6\,{\it polylog} \left ( 5,-{\frac{b{{\rm e}^{x}}}{a}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*ln(a+b*exp(x)),x)

[Out]

1/4*x^4*ln(a+b*exp(x))-1/4*x^4*ln(1+b*exp(x)/a)-x^3*polylog(2,-b*exp(x)/a)+3*x^2*polylog(3,-b*exp(x)/a)-6*x*po
lylog(4,-b*exp(x)/a)+6*polylog(5,-b*exp(x)/a)

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Maxima [A]  time = 1.20719, size = 111, normalized size = 1.19 \begin{align*} \frac{1}{4} \, x^{4} \log \left (b e^{x} + a\right ) - \frac{1}{4} \, x^{4} \log \left (\frac{b e^{x}}{a} + 1\right ) - x^{3}{\rm Li}_2\left (-\frac{b e^{x}}{a}\right ) + 3 \, x^{2}{\rm Li}_{3}(-\frac{b e^{x}}{a}) - 6 \, x{\rm Li}_{4}(-\frac{b e^{x}}{a}) + 6 \,{\rm Li}_{5}(-\frac{b e^{x}}{a}) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*log(b*e^x + a) - 1/4*x^4*log(b*e^x/a + 1) - x^3*dilog(-b*e^x/a) + 3*x^2*polylog(3, -b*e^x/a) - 6*x*pol
ylog(4, -b*e^x/a) + 6*polylog(5, -b*e^x/a)

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Fricas [C]  time = 2.04759, size = 224, normalized size = 2.41 \begin{align*} \frac{1}{4} \, x^{4} \log \left (b e^{x} + a\right ) - \frac{1}{4} \, x^{4} \log \left (\frac{b e^{x} + a}{a}\right ) - x^{3}{\rm Li}_2\left (-\frac{b e^{x} + a}{a} + 1\right ) + 3 \, x^{2}{\rm polylog}\left (3, -\frac{b e^{x}}{a}\right ) - 6 \, x{\rm polylog}\left (4, -\frac{b e^{x}}{a}\right ) + 6 \,{\rm polylog}\left (5, -\frac{b e^{x}}{a}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{b \int \frac{x^{4} e^{x}}{a + b e^{x}}\, dx}{4} + \frac{x^{4} \log{\left (a + b e^{x} \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*ln(a+b*exp(x)),x)

[Out]

-b*Integral(x**4*exp(x)/(a + b*exp(x)), x)/4 + x**4*log(a + b*exp(x))/4

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \log \left (b e^{x} + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*log(b*e^x + a), x)