\(\int \frac {x^2 \log (1+x+x^2)}{2+3 x+x^2} \, dx\) [98]

Optimal result

Integrand size = 21, antiderivative size = 311 \[ \int \frac {x^2 \log \left (1+x+x^2\right )}{2+3 x+x^2} \, dx=-2 x+\sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-\log (2+2 x) \log \left (-\frac {1-i \sqrt {3}+2 x}{1+i \sqrt {3}}\right )+4 \log (4+2 x) \log \left (-\frac {1-i \sqrt {3}+2 x}{3+i \sqrt {3}}\right )-\log (2+2 x) \log \left (-\frac {1+i \sqrt {3}+2 x}{1-i \sqrt {3}}\right )+4 \log (4+2 x) \log \left (-\frac {1+i \sqrt {3}+2 x}{3-i \sqrt {3}}\right )+\frac {1}{2} \log \left (1+x+x^2\right )+x \log \left (1+x+x^2\right )+\log (2+2 x) \log \left (1+x+x^2\right )-4 \log (4+2 x) \log \left (1+x+x^2\right )-\operatorname {PolyLog}\left (2,\frac {2 (1+x)}{1-i \sqrt {3}}\right )-\operatorname {PolyLog}\left (2,\frac {2 (1+x)}{1+i \sqrt {3}}\right )+4 \operatorname {PolyLog}\left (2,\frac {2 (2+x)}{3-i \sqrt {3}}\right )+4 \operatorname {PolyLog}\left (2,\frac {2 (2+x)}{3+i \sqrt {3}}\right ) \] Output:

-2*x+3^(1/2)*arctan(1/3*(1+2*x)*3^(1/2))-ln(2+2*x)*ln(-(1-I*3^(1/2)+2*x)/( 
1+I*3^(1/2)))+4*ln(4+2*x)*ln(-(1-I*3^(1/2)+2*x)/(3+I*3^(1/2)))-ln(2+2*x)*l 
n(-(1+I*3^(1/2)+2*x)/(1-I*3^(1/2)))+4*ln(4+2*x)*ln(-(1+I*3^(1/2)+2*x)/(3-I 
*3^(1/2)))+1/2*ln(x^2+x+1)+x*ln(x^2+x+1)+ln(2+2*x)*ln(x^2+x+1)-4*ln(4+2*x) 
*ln(x^2+x+1)-polylog(2,2*(1+x)/(1-I*3^(1/2)))-polylog(2,2*(1+x)/(1+I*3^(1/ 
2)))+4*polylog(2,2*(2+x)/(3-I*3^(1/2)))+4*polylog(2,2*(2+x)/(3+I*3^(1/2)))
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.93 \[ \int \frac {x^2 \log \left (1+x+x^2\right )}{2+3 x+x^2} \, dx=-2 x+\sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-\log \left (\frac {-i+\sqrt {3}-2 i x}{i+\sqrt {3}}\right ) \log (2 (1+x))-\log \left (\frac {i+\sqrt {3}+2 i x}{-i+\sqrt {3}}\right ) \log (2 (1+x))+\frac {1}{2} \log \left (1+x+x^2\right )+x \log \left (1+x+x^2\right )+\log (2 (1+x)) \log \left (1+x+x^2\right )-4 \log (2 (2+x)) \log \left (1+x+x^2\right )-\operatorname {PolyLog}\left (2,\frac {2 (1+x)}{1+i \sqrt {3}}\right )-\operatorname {PolyLog}\left (2,\frac {2 i (1+x)}{i+\sqrt {3}}\right )+4 \left (\left (\log \left (\frac {-i+\sqrt {3}-2 i x}{3 i+\sqrt {3}}\right )+\log \left (\frac {i+\sqrt {3}+2 i x}{-3 i+\sqrt {3}}\right )\right ) \log (2 (2+x))+\operatorname {PolyLog}\left (2,\frac {2 (2+x)}{3+i \sqrt {3}}\right )+\operatorname {PolyLog}\left (2,\frac {2 i (2+x)}{3 i+\sqrt {3}}\right )\right ) \] Input:

Integrate[(x^2*Log[1 + x + x^2])/(2 + 3*x + x^2),x]
 

Output:

-2*x + Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] - Log[(-I + Sqrt[3] - (2*I)*x)/(I 
 + Sqrt[3])]*Log[2*(1 + x)] - Log[(I + Sqrt[3] + (2*I)*x)/(-I + Sqrt[3])]* 
Log[2*(1 + x)] + Log[1 + x + x^2]/2 + x*Log[1 + x + x^2] + Log[2*(1 + x)]* 
Log[1 + x + x^2] - 4*Log[2*(2 + x)]*Log[1 + x + x^2] - PolyLog[2, (2*(1 + 
x))/(1 + I*Sqrt[3])] - PolyLog[2, ((2*I)*(1 + x))/(I + Sqrt[3])] + 4*((Log 
[(-I + Sqrt[3] - (2*I)*x)/(3*I + Sqrt[3])] + Log[(I + Sqrt[3] + (2*I)*x)/( 
-3*I + Sqrt[3])])*Log[2*(2 + x)] + PolyLog[2, (2*(2 + x))/(3 + I*Sqrt[3])] 
 + PolyLog[2, ((2*I)*(2 + x))/(3*I + Sqrt[3])])
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3008, 2009}

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.

\(\displaystyle \int \frac {x^2 \log \left (x^2+x+1\right )}{x^2+3 x+2} \, dx\)

\(\Big \downarrow \) 3008

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

\(\Big \downarrow \) 2009

\(\displaystyle \sqrt {3} \arctan \left (\frac {2 x+1}{\sqrt {3}}\right )-\operatorname {PolyLog}\left (2,\frac {2 (x+1)}{1-i \sqrt {3}}\right )-\operatorname {PolyLog}\left (2,\frac {2 (x+1)}{1+i \sqrt {3}}\right )+4 \operatorname {PolyLog}\left (2,\frac {2 (x+2)}{3-i \sqrt {3}}\right )+4 \operatorname {PolyLog}\left (2,\frac {2 (x+2)}{3+i \sqrt {3}}\right )+x \log \left (x^2+x+1\right )+\log (2 x+2) \log \left (x^2+x+1\right )-4 \log (2 x+4) \log \left (x^2+x+1\right )+\frac {1}{2} \log \left (x^2+x+1\right )-2 x-\log (2 x+2) \log \left (-\frac {2 x-i \sqrt {3}+1}{1+i \sqrt {3}}\right )+4 \log (2 x+4) \log \left (-\frac {2 x-i \sqrt {3}+1}{3+i \sqrt {3}}\right )-\log (2 x+2) \log \left (-\frac {2 x+i \sqrt {3}+1}{1-i \sqrt {3}}\right )+4 \log (2 x+4) \log \left (-\frac {2 x+i \sqrt {3}+1}{3-i \sqrt {3}}\right )\)

Input:

Int[(x^2*Log[1 + x + x^2])/(2 + 3*x + x^2),x]
 

Output:

-2*x + Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] - Log[2 + 2*x]*Log[-((1 - I*Sqrt[ 
3] + 2*x)/(1 + I*Sqrt[3]))] + 4*Log[4 + 2*x]*Log[-((1 - I*Sqrt[3] + 2*x)/( 
3 + I*Sqrt[3]))] - Log[2 + 2*x]*Log[-((1 + I*Sqrt[3] + 2*x)/(1 - I*Sqrt[3] 
))] + 4*Log[4 + 2*x]*Log[-((1 + I*Sqrt[3] + 2*x)/(3 - I*Sqrt[3]))] + Log[1 
 + x + x^2]/2 + x*Log[1 + x + x^2] + Log[2 + 2*x]*Log[1 + x + x^2] - 4*Log 
[4 + 2*x]*Log[1 + x + x^2] - PolyLog[2, (2*(1 + x))/(1 - I*Sqrt[3])] - Pol 
yLog[2, (2*(1 + x))/(1 + I*Sqrt[3])] + 4*PolyLog[2, (2*(2 + x))/(3 - I*Sqr 
t[3])] + 4*PolyLog[2, (2*(2 + x))/(3 + I*Sqrt[3])]
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3008

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*(RGx_), x_Symbol] :> With 
[{u = ExpandIntegrand[(a + b*Log[c*RFx^p])^n, RGx, x]}, Int[u, x] /; SumQ[u 
]] /; FreeQ[{a, b, c, p}, x] && RationalFunctionQ[RFx, x] && RationalFuncti 
onQ[RGx, x] && IGtQ[n, 0]
 

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 279, normalized size of antiderivative = 0.90

method	result	size
default	\(-2 x +\sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )+\frac {\ln \left (x^{2}+x +1\right )}{2}+x \ln \left (x^{2}+x +1\right )+\ln \left (1+x \right ) \ln \left (x^{2}+x +1\right )-\ln \left (1+x \right ) \ln \left (\frac {i \sqrt {3}-1-2 x}{1+i \sqrt {3}}\right )-\ln \left (1+x \right ) \ln \left (\frac {1+i \sqrt {3}+2 x}{i \sqrt {3}-1}\right )-\operatorname {dilog}\left (\frac {i \sqrt {3}-1-2 x}{1+i \sqrt {3}}\right )-\operatorname {dilog}\left (\frac {1+i \sqrt {3}+2 x}{i \sqrt {3}-1}\right )-4 \ln \left (x +2\right ) \ln \left (x^{2}+x +1\right )+4 \ln \left (x +2\right ) \ln \left (\frac {i \sqrt {3}-1-2 x}{3+i \sqrt {3}}\right )+4 \ln \left (x +2\right ) \ln \left (\frac {1+i \sqrt {3}+2 x}{i \sqrt {3}-3}\right )+4 \operatorname {dilog}\left (\frac {i \sqrt {3}-1-2 x}{3+i \sqrt {3}}\right )+4 \operatorname {dilog}\left (\frac {1+i \sqrt {3}+2 x}{i \sqrt {3}-3}\right )\)	\(279\)
risch	\(-2 x +\sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )+\frac {\ln \left (x^{2}+x +1\right )}{2}+x \ln \left (x^{2}+x +1\right )+\ln \left (1+x \right ) \ln \left (x^{2}+x +1\right )-\ln \left (1+x \right ) \ln \left (\frac {i \sqrt {3}-1-2 x}{1+i \sqrt {3}}\right )-\ln \left (1+x \right ) \ln \left (\frac {1+i \sqrt {3}+2 x}{i \sqrt {3}-1}\right )-\operatorname {dilog}\left (\frac {i \sqrt {3}-1-2 x}{1+i \sqrt {3}}\right )-\operatorname {dilog}\left (\frac {1+i \sqrt {3}+2 x}{i \sqrt {3}-1}\right )-4 \ln \left (x +2\right ) \ln \left (x^{2}+x +1\right )+4 \ln \left (x +2\right ) \ln \left (\frac {i \sqrt {3}-1-2 x}{3+i \sqrt {3}}\right )+4 \ln \left (x +2\right ) \ln \left (\frac {1+i \sqrt {3}+2 x}{i \sqrt {3}-3}\right )+4 \operatorname {dilog}\left (\frac {i \sqrt {3}-1-2 x}{3+i \sqrt {3}}\right )+4 \operatorname {dilog}\left (\frac {1+i \sqrt {3}+2 x}{i \sqrt {3}-3}\right )\)	\(279\)
parts	\(-2 x +\sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )+\frac {\ln \left (x^{2}+x +1\right )}{2}+x \ln \left (x^{2}+x +1\right )+\ln \left (1+x \right ) \ln \left (x^{2}+x +1\right )-\ln \left (1+x \right ) \ln \left (\frac {i \sqrt {3}-1-2 x}{1+i \sqrt {3}}\right )-\ln \left (1+x \right ) \ln \left (\frac {1+i \sqrt {3}+2 x}{i \sqrt {3}-1}\right )-\operatorname {dilog}\left (\frac {i \sqrt {3}-1-2 x}{1+i \sqrt {3}}\right )-\operatorname {dilog}\left (\frac {1+i \sqrt {3}+2 x}{i \sqrt {3}-1}\right )-4 \ln \left (x +2\right ) \ln \left (x^{2}+x +1\right )+4 \ln \left (x +2\right ) \ln \left (\frac {i \sqrt {3}-1-2 x}{3+i \sqrt {3}}\right )+4 \ln \left (x +2\right ) \ln \left (\frac {1+i \sqrt {3}+2 x}{i \sqrt {3}-3}\right )+4 \operatorname {dilog}\left (\frac {i \sqrt {3}-1-2 x}{3+i \sqrt {3}}\right )+4 \operatorname {dilog}\left (\frac {1+i \sqrt {3}+2 x}{i \sqrt {3}-3}\right )\)	\(279\)

Input:

int(x^2*ln(x^2+x+1)/(x^2+3*x+2),x,method=_RETURNVERBOSE)
 

Output:

-2*x+3^(1/2)*arctan(1/3*(2*x+1)*3^(1/2))+1/2*ln(x^2+x+1)+x*ln(x^2+x+1)+ln( 
1+x)*ln(x^2+x+1)-ln(1+x)*ln((I*3^(1/2)-1-2*x)/(1+I*3^(1/2)))-ln(1+x)*ln((1 
+I*3^(1/2)+2*x)/(I*3^(1/2)-1))-dilog((I*3^(1/2)-1-2*x)/(1+I*3^(1/2)))-dilo 
g((1+I*3^(1/2)+2*x)/(I*3^(1/2)-1))-4*ln(x+2)*ln(x^2+x+1)+4*ln(x+2)*ln((I*3 
^(1/2)-1-2*x)/(3+I*3^(1/2)))+4*ln(x+2)*ln((1+I*3^(1/2)+2*x)/(I*3^(1/2)-3)) 
+4*dilog((I*3^(1/2)-1-2*x)/(3+I*3^(1/2)))+4*dilog((1+I*3^(1/2)+2*x)/(I*3^( 
1/2)-3))
 

Fricas [F]

\[ \int \frac {x^2 \log \left (1+x+x^2\right )}{2+3 x+x^2} \, dx=\int { \frac {x^{2} \log \left (x^{2} + x + 1\right )}{x^{2} + 3 \, x + 2} \,d x } \] Input:

integrate(x^2*log(x^2+x+1)/(x^2+3*x+2),x, algorithm="fricas")
 

Output:

integral(x^2*log(x^2 + x + 1)/(x^2 + 3*x + 2), x)
 

Sympy [F]

\[ \int \frac {x^2 \log \left (1+x+x^2\right )}{2+3 x+x^2} \, dx=\int \frac {x^{2} \log {\left (x^{2} + x + 1 \right )}}{\left (x + 1\right ) \left (x + 2\right )}\, dx \] Input:

integrate(x**2*ln(x**2+x+1)/(x**2+3*x+2),x)
 

Output:

Integral(x**2*log(x**2 + x + 1)/((x + 1)*(x + 2)), x)
 

Maxima [F]

\[ \int \frac {x^2 \log \left (1+x+x^2\right )}{2+3 x+x^2} \, dx=\int { \frac {x^{2} \log \left (x^{2} + x + 1\right )}{x^{2} + 3 \, x + 2} \,d x } \] Input:

integrate(x^2*log(x^2+x+1)/(x^2+3*x+2),x, algorithm="maxima")
 

Output:

integrate(x^2*log(x^2 + x + 1)/(x^2 + 3*x + 2), x)
 

Giac [F]

\[ \int \frac {x^2 \log \left (1+x+x^2\right )}{2+3 x+x^2} \, dx=\int { \frac {x^{2} \log \left (x^{2} + x + 1\right )}{x^{2} + 3 \, x + 2} \,d x } \] Input:

integrate(x^2*log(x^2+x+1)/(x^2+3*x+2),x, algorithm="giac")
 

Output:

integrate(x^2*log(x^2 + x + 1)/(x^2 + 3*x + 2), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \log \left (1+x+x^2\right )}{2+3 x+x^2} \, dx=\int \frac {x^2\,\ln \left (x^2+x+1\right )}{x^2+3\,x+2} \,d x \] Input:

int((x^2*log(x + x^2 + 1))/(3*x + x^2 + 2),x)
 

Output:

int((x^2*log(x + x^2 + 1))/(3*x + x^2 + 2), x)
 

Reduce [F]

\[ \int \frac {x^2 \log \left (1+x+x^2\right )}{2+3 x+x^2} \, dx=\sqrt {3}\, \mathit {atan} \left (\frac {2 x +1}{\sqrt {3}}\right )-\frac {4 \left (\int \frac {\mathrm {log}\left (x^{2}+x +1\right )}{x^{4}+4 x^{3}+6 x^{2}+5 x +2}d x \right )}{7}-\frac {11 \left (\int \frac {\mathrm {log}\left (x^{2}+x +1\right ) x^{3}}{x^{4}+4 x^{3}+6 x^{2}+5 x +2}d x \right )}{7}-\frac {5 \mathrm {log}\left (x^{2}+x +1\right )^{2}}{14}+\mathrm {log}\left (x^{2}+x +1\right ) x +\frac {\mathrm {log}\left (x^{2}+x +1\right )}{2}-2 x \] Input:

int(x^2*log(x^2+x+1)/(x^2+3*x+2),x)
 

Output:

(14*sqrt(3)*atan((2*x + 1)/sqrt(3)) - 8*int(log(x**2 + x + 1)/(x**4 + 4*x* 
*3 + 6*x**2 + 5*x + 2),x) - 22*int((log(x**2 + x + 1)*x**3)/(x**4 + 4*x**3 
 + 6*x**2 + 5*x + 2),x) - 5*log(x**2 + x + 1)**2 + 14*log(x**2 + x + 1)*x 
+ 7*log(x**2 + x + 1) - 28*x)/14