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

Optimal result

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

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*polylog(4,-b*exp(x)/a)
 

Mathematica [A] (verified)

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

Integrate[x^2*Log[a + b*E^x],x]
 

Output:

(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)]
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3012, 3011, 7163, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^2 \log \left (a+b e^x\right ) \, dx\)

\(\Big \downarrow \) 3012

\(\displaystyle \int x^2 \log \left (\frac {e^x b}{a}+1\right )dx+\frac {1}{3} x^3 \log \left (a+b e^x\right )-\frac {1}{3} x^3 \log \left (\frac {b e^x}{a}+1\right )\)

\(\Big \downarrow \) 3011

\(\displaystyle 2 \int x \operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )dx-x^2 \operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )+\frac {1}{3} x^3 \log \left (a+b e^x\right )-\frac {1}{3} x^3 \log \left (\frac {b e^x}{a}+1\right )\)

\(\Big \downarrow \) 7163

\(\displaystyle 2 \left (x \operatorname {PolyLog}\left (3,-\frac {b e^x}{a}\right )-\int \operatorname {PolyLog}\left (3,-\frac {b e^x}{a}\right )dx\right )-x^2 \operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )+\frac {1}{3} x^3 \log \left (a+b e^x\right )-\frac {1}{3} x^3 \log \left (\frac {b e^x}{a}+1\right )\)

\(\Big \downarrow \) 2720

\(\displaystyle 2 \left (x \operatorname {PolyLog}\left (3,-\frac {b e^x}{a}\right )-\int e^{-x} \operatorname {PolyLog}\left (3,-\frac {b e^x}{a}\right )de^x\right )-x^2 \operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )+\frac {1}{3} x^3 \log \left (a+b e^x\right )-\frac {1}{3} x^3 \log \left (\frac {b e^x}{a}+1\right )\)

\(\Big \downarrow \) 7143

\(\displaystyle -x^2 \operatorname {PolyLog}\left (2,-\frac {b e^x}{a}\right )+2 \left (x \operatorname {PolyLog}\left (3,-\frac {b e^x}{a}\right )-\operatorname {PolyLog}\left (4,-\frac {b e^x}{a}\right )\right )+\frac {1}{3} x^3 \log \left (a+b e^x\right )-\frac {1}{3} x^3 \log \left (\frac {b e^x}{a}+1\right )\)

Input:

Int[x^2*Log[a + b*E^x],x]
 

Output:

(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)] - PolyLog[4, -((b*E^x)/a)])
 

Defintions of rubi rules used

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{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 3011

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] + Simp[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 3012

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 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> 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 7163

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] - Simp[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]
 

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.90

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

Input:

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

Output:

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*polylog(4,-b*exp(x)/a)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.95 \[ \int x^2 \log \left (a+b e^x\right ) \, dx=\frac {1}{3} \, x^{3} \log \left (b e^{x} + a\right ) - \frac {1}{3} \, x^{3} \log \left (\frac {b e^{x} + a}{a}\right ) - x^{2} {\rm Li}_2\left (-\frac {b e^{x} + a}{a} + 1\right ) + 2 \, x {\rm polylog}\left (3, -\frac {b e^{x}}{a}\right ) - 2 \, {\rm polylog}\left (4, -\frac {b e^{x}}{a}\right ) \] Input:

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

Output:

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]

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

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

Output:

-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)

Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.87 \[ \int x^2 \log \left (a+b e^x\right ) \, dx=\frac {1}{3} \, x^{3} \log \left (b e^{x} + a\right ) - \frac {1}{3} \, x^{3} \log \left (\frac {b e^{x}}{a} + 1\right ) - x^{2} {\rm Li}_2\left (-\frac {b e^{x}}{a}\right ) + 2 \, x {\rm Li}_{3}(-\frac {b e^{x}}{a}) - 2 \, {\rm Li}_{4}(-\frac {b e^{x}}{a}) \] Input:

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

Output:

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]

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

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

Output:

integrate(x^2*log(b*e^x + a), x)
 

Mupad [F(-1)]

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

int(x^2*log(a + b*exp(x)),x)
 

Output:

int(x^2*log(a + b*exp(x)), x)
 

Reduce [F]

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

int(x^2*log(a+b*exp(x)),x)
 

Output:

int(log(e**x*b + a)*x**2,x)