\(\int x^2 \log (a+b e^x) \, dx\) [114]

Optimal result

Integrand size = 12, antiderivative size = 77 \[ \int x^2 \log \left (a+b e^x\right ) \, dx=\frac {1}{3} x^3 \log \left (a+b e^x\right )-\frac {1}{3} x^3 \log \left (1+\frac {b e^x}{a}\right )-x^2 \operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )+2 x \operatorname {PolyLog}\left (3,-\frac {b e^x}{a}\right )-2 \operatorname {PolyLog}\left (4,-\frac {b e^x}{a}\right ) \]

[Out]

1/3*x^3*ln(a+b*exp(x))-1/3*x^3*ln(1+b*exp(x)/a)-x^2*polylog(2,-b*exp(x)/a)+2*x*polylog(3,-b*exp(x)/a)-2*polylo
g(4,-b*exp(x)/a)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {2612, 2611, 6744, 2320, 6724} \[ \int x^2 \log \left (a+b e^x\right ) \, dx=-x^2 \operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )+2 x \operatorname {PolyLog}\left (3,-\frac {b e^x}{a}\right )-2 \operatorname {PolyLog}\left (4,-\frac {b e^x}{a}\right )+\frac {1}{3} x^3 \log \left (a+b e^x\right )-\frac {1}{3} x^3 \log \left (\frac {b e^x}{a}+1\right ) \]

[In]

Int[x^2*Log[a + b*E^x],x]

[Out]

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

Rule 2320

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 2611

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 2612

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/d)*(F^(c*(a +
b*x)))^n], x] - Simp[(f + g*x)^(m + 1)*(Log[1 + (e/d)*(F^(c*(a + b*x)))^n]/(g*(m + 1))), x]) /; FreeQ[{F, a,
b, c, d, e, f, g, n}, x] && GtQ[m, 0] && NeQ[d, 1]

Rule 6724

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]

Rule 6744

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} x^3 \log \left (a+b e^x\right )-\frac {1}{3} x^3 \log \left (1+\frac {b e^x}{a}\right )+\int x^2 \log \left (1+\frac {b e^x}{a}\right ) \, dx \\ & = \frac {1}{3} x^3 \log \left (a+b e^x\right )-\frac {1}{3} x^3 \log \left (1+\frac {b e^x}{a}\right )-x^2 \text {Li}_2\left (-\frac {b e^x}{a}\right )+2 \int x \text {Li}_2\left (-\frac {b e^x}{a}\right ) \, dx \\ & = \frac {1}{3} x^3 \log \left (a+b e^x\right )-\frac {1}{3} x^3 \log \left (1+\frac {b e^x}{a}\right )-x^2 \text {Li}_2\left (-\frac {b e^x}{a}\right )+2 x \text {Li}_3\left (-\frac {b e^x}{a}\right )-2 \int \text {Li}_3\left (-\frac {b e^x}{a}\right ) \, dx \\ & = \frac {1}{3} x^3 \log \left (a+b e^x\right )-\frac {1}{3} x^3 \log \left (1+\frac {b e^x}{a}\right )-x^2 \text {Li}_2\left (-\frac {b e^x}{a}\right )+2 x \text {Li}_3\left (-\frac {b e^x}{a}\right )-2 \text {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {b x}{a}\right )}{x} \, dx,x,e^x\right ) \\ & = \frac {1}{3} x^3 \log \left (a+b e^x\right )-\frac {1}{3} x^3 \log \left (1+\frac {b e^x}{a}\right )-x^2 \text {Li}_2\left (-\frac {b e^x}{a}\right )+2 x \text {Li}_3\left (-\frac {b e^x}{a}\right )-2 \text {Li}_4\left (-\frac {b e^x}{a}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00 \[ \int x^2 \log \left (a+b e^x\right ) \, dx=\frac {1}{3} x^3 \log \left (a+b e^x\right )-\frac {1}{3} x^3 \log \left (1+\frac {b e^x}{a}\right )-x^2 \operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )+2 x \operatorname {PolyLog}\left (3,-\frac {b e^x}{a}\right )-2 \operatorname {PolyLog}\left (4,-\frac {b e^x}{a}\right ) \]

[In]

Integrate[x^2*Log[a + b*E^x],x]

[Out]

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

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.90

method	result	size
default	\(\frac {x^{3} \ln \left (a +b \,{\mathrm e}^{x}\right )}{3}-\frac {x^{3} \ln \left (1+\frac {b \,{\mathrm e}^{x}}{a}\right )}{3}-x^{2} \operatorname {Li}_{2}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )+2 x \,\operatorname {Li}_{3}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )-2 \,\operatorname {Li}_{4}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )\)	\(69\)
risch	\(\frac {x^{3} \ln \left (a +b \,{\mathrm e}^{x}\right )}{3}-\frac {x^{3} \ln \left (1+\frac {b \,{\mathrm e}^{x}}{a}\right )}{3}-x^{2} \operatorname {Li}_{2}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )+2 x \,\operatorname {Li}_{3}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )-2 \,\operatorname {Li}_{4}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )\)	\(69\)
parts	\(\frac {x^{3} \ln \left (a +b \,{\mathrm e}^{x}\right )}{3}-\frac {x^{3} \ln \left (1+\frac {b \,{\mathrm e}^{x}}{a}\right )}{3}-x^{2} \operatorname {Li}_{2}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )+2 x \,\operatorname {Li}_{3}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )-2 \,\operatorname {Li}_{4}\left (-\frac {b \,{\mathrm e}^{x}}{a}\right )\)	\(69\)

[In]

int(x^2*ln(a+b*exp(x)),x,method=_RETURNVERBOSE)

[Out]

1/3*x^3*ln(a+b*exp(x))-1/3*x^3*ln(1+b*exp(x)/a)-x^2*polylog(2,-b*exp(x)/a)+2*x*polylog(3,-b*exp(x)/a)-2*polylo
g(4,-b*exp(x)/a)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.95 \[ \int x^2 \log \left (a+b e^x\right ) \, dx=\frac {1}{3} \, x^{3} \log \left (b e^{x} + a\right ) - \frac {1}{3} \, x^{3} \log \left (\frac {b e^{x} + a}{a}\right ) - x^{2} {\rm Li}_2\left (-\frac {b e^{x} + a}{a} + 1\right ) + 2 \, x {\rm polylog}\left (3, -\frac {b e^{x}}{a}\right ) - 2 \, {\rm polylog}\left (4, -\frac {b e^{x}}{a}\right ) \]

[In]

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

[Out]

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

Sympy [F]

\[ \int x^2 \log \left (a+b e^x\right ) \, dx=- \frac {b \int \frac {x^{3} e^{x}}{a + b e^{x}}\, dx}{3} + \frac {x^{3} \log {\left (a + b e^{x} \right )}}{3} \]

[In]

integrate(x**2*ln(a+b*exp(x)),x)

[Out]

-b*Integral(x**3*exp(x)/(a + b*exp(x)), x)/3 + x**3*log(a + b*exp(x))/3

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.87 \[ \int x^2 \log \left (a+b e^x\right ) \, dx=\frac {1}{3} \, x^{3} \log \left (b e^{x} + a\right ) - \frac {1}{3} \, x^{3} \log \left (\frac {b e^{x}}{a} + 1\right ) - x^{2} {\rm Li}_2\left (-\frac {b e^{x}}{a}\right ) + 2 \, x {\rm Li}_{3}(-\frac {b e^{x}}{a}) - 2 \, {\rm Li}_{4}(-\frac {b e^{x}}{a}) \]

[In]

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

[Out]

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

Giac [F]

\[ \int x^2 \log \left (a+b e^x\right ) \, dx=\int { x^{2} \log \left (b e^{x} + a\right ) \,d x } \]

[In]

integrate(x^2*log(a+b*exp(x)),x, algorithm="giac")

[Out]

integrate(x^2*log(b*e^x + a), x)

Mupad [F(-1)]

Timed out. \[ \int x^2 \log \left (a+b e^x\right ) \, dx=\int x^2\,\ln \left (a+b\,{\mathrm {e}}^x\right ) \,d x \]

[In]

int(x^2*log(a + b*exp(x)),x)

[Out]

int(x^2*log(a + b*exp(x)), x)