Optimal. Leaf size=311 \[ -\text {Li}_2\left (\frac {2 (x+1)}{1-i \sqrt {3}}\right )-\text {Li}_2\left (\frac {2 (x+1)}{1+i \sqrt {3}}\right )+4 \text {Li}_2\left (\frac {2 (x+2)}{3-i \sqrt {3}}\right )+4 \text {Li}_2\left (\frac {2 (x+2)}{3+i \sqrt {3}}\right )+x \log \left (x^2+x+1\right )+\log (2 x+2) \log \left (x^2+x+1\right )-4 \log (2 x+4) \log \left (x^2+x+1\right )+\frac {1}{2} \log \left (x^2+x+1\right )-2 x-\log (2 x+2) \log \left (-\frac {2 x-i \sqrt {3}+1}{1+i \sqrt {3}}\right )+4 \log (2 x+4) \log \left (-\frac {2 x-i \sqrt {3}+1}{3+i \sqrt {3}}\right )-\log (2 x+2) \log \left (-\frac {2 x+i \sqrt {3}+1}{1-i \sqrt {3}}\right )+4 \log (2 x+4) \log \left (-\frac {2 x+i \sqrt {3}+1}{3-i \sqrt {3}}\right )+\sqrt {3} \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right ) \]
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Rubi [A] time = 0.48, antiderivative size = 311, normalized size of antiderivative = 1.00, number of steps used = 28, number of rules used = 12, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {2528, 2523, 773, 634, 618, 204, 628, 2524, 2418, 2394, 2393, 2391} \[ -\text {PolyLog}\left (2,\frac {2 (x+1)}{1-i \sqrt {3}}\right )-\text {PolyLog}\left (2,\frac {2 (x+1)}{1+i \sqrt {3}}\right )+4 \text {PolyLog}\left (2,\frac {2 (x+2)}{3-i \sqrt {3}}\right )+4 \text {PolyLog}\left (2,\frac {2 (x+2)}{3+i \sqrt {3}}\right )+x \log \left (x^2+x+1\right )+\log (2 x+2) \log \left (x^2+x+1\right )-4 \log (2 x+4) \log \left (x^2+x+1\right )+\frac {1}{2} \log \left (x^2+x+1\right )-2 x-\log (2 x+2) \log \left (-\frac {2 x-i \sqrt {3}+1}{1+i \sqrt {3}}\right )+4 \log (2 x+4) \log \left (-\frac {2 x-i \sqrt {3}+1}{3+i \sqrt {3}}\right )-\log (2 x+2) \log \left (-\frac {2 x+i \sqrt {3}+1}{1-i \sqrt {3}}\right )+4 \log (2 x+4) \log \left (-\frac {2 x+i \sqrt {3}+1}{3-i \sqrt {3}}\right )+\sqrt {3} \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right ) \]
Antiderivative was successfully verified.
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Rule 204
Rule 618
Rule 628
Rule 634
Rule 773
Rule 2391
Rule 2393
Rule 2394
Rule 2418
Rule 2523
Rule 2524
Rule 2528
Rubi steps
\begin {align*} \int \frac {x^2 \log \left (1+x+x^2\right )}{2+3 x+x^2} \, dx &=\int \left (\log \left (1+x+x^2\right )-\frac {(2+3 x) \log \left (1+x+x^2\right )}{2+3 x+x^2}\right ) \, dx\\ &=\int \log \left (1+x+x^2\right ) \, dx-\int \frac {(2+3 x) \log \left (1+x+x^2\right )}{2+3 x+x^2} \, dx\\ &=x \log \left (1+x+x^2\right )-\int \frac {x (1+2 x)}{1+x+x^2} \, dx-\int \left (-\frac {2 \log \left (1+x+x^2\right )}{2+2 x}+\frac {8 \log \left (1+x+x^2\right )}{4+2 x}\right ) \, dx\\ &=-2 x+x \log \left (1+x+x^2\right )+2 \int \frac {\log \left (1+x+x^2\right )}{2+2 x} \, dx-8 \int \frac {\log \left (1+x+x^2\right )}{4+2 x} \, dx-\int \frac {-2-x}{1+x+x^2} \, dx\\ &=-2 x+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 )+\frac {1}{2} \int \frac {1+2 x}{1+x+x^2} \, dx+\frac {3}{2} \int \frac {1}{1+x+x^2} \, dx+4 \int \frac {(1+2 x) \log (4+2 x)}{1+x+x^2} \, dx-\int \frac {(1+2 x) \log (2+2 x)}{1+x+x^2} \, dx\\ &=-2 x+\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 )-3 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right )+4 \int \left (\frac {2 \log (4+2 x)}{1-i \sqrt {3}+2 x}+\frac {2 \log (4+2 x)}{1+i \sqrt {3}+2 x}\right ) \, dx-\int \left (\frac {2 \log (2+2 x)}{1-i \sqrt {3}+2 x}+\frac {2 \log (2+2 x)}{1+i \sqrt {3}+2 x}\right ) \, dx\\ &=-2 x+\sqrt {3} \tan ^{-1}\left (\frac {1+2 x}{\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 )-2 \int \frac {\log (2+2 x)}{1-i \sqrt {3}+2 x} \, dx-2 \int \frac {\log (2+2 x)}{1+i \sqrt {3}+2 x} \, dx+8 \int \frac {\log (4+2 x)}{1-i \sqrt {3}+2 x} \, dx+8 \int \frac {\log (4+2 x)}{1+i \sqrt {3}+2 x} \, dx\\ &=-2 x+\sqrt {3} \tan ^{-1}\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 )+2 \int \frac {\log \left (\frac {2 \left (1-i \sqrt {3}+2 x\right )}{-4+2 \left (1-i \sqrt {3}\right )}\right )}{2+2 x} \, dx+2 \int \frac {\log \left (\frac {2 \left (1+i \sqrt {3}+2 x\right )}{-4+2 \left (1+i \sqrt {3}\right )}\right )}{2+2 x} \, dx-8 \int \frac {\log \left (\frac {2 \left (1-i \sqrt {3}+2 x\right )}{-8+2 \left (1-i \sqrt {3}\right )}\right )}{4+2 x} \, dx-8 \int \frac {\log \left (\frac {2 \left (1+i \sqrt {3}+2 x\right )}{-8+2 \left (1+i \sqrt {3}\right )}\right )}{4+2 x} \, dx\\ &=-2 x+\sqrt {3} \tan ^{-1}\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 )-4 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 x}{-8+2 \left (1-i \sqrt {3}\right )}\right )}{x} \, dx,x,4+2 x\right )-4 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 x}{-8+2 \left (1+i \sqrt {3}\right )}\right )}{x} \, dx,x,4+2 x\right )+\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 x}{-4+2 \left (1-i \sqrt {3}\right )}\right )}{x} \, dx,x,2+2 x\right )+\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 x}{-4+2 \left (1+i \sqrt {3}\right )}\right )}{x} \, dx,x,2+2 x\right )\\ &=-2 x+\sqrt {3} \tan ^{-1}\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 )-\text {Li}_2\left (\frac {2 (1+x)}{1+i \sqrt {3}}\right )-\text {Li}_2\left (\frac {2 i (1+x)}{i+\sqrt {3}}\right )+4 \text {Li}_2\left (\frac {2 (2+x)}{3-i \sqrt {3}}\right )+4 \text {Li}_2\left (\frac {2 (2+x)}{3+i \sqrt {3}}\right )\\ \end {align*}
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Mathematica [A] time = 0.20, size = 290, normalized size = 0.93 \[ -\text {Li}_2\left (\frac {2 (x+1)}{1+i \sqrt {3}}\right )-\text {Li}_2\left (\frac {2 i (x+1)}{i+\sqrt {3}}\right )+4 \left (\text {Li}_2\left (\frac {2 (x+2)}{3+i \sqrt {3}}\right )+\text {Li}_2\left (\frac {2 i (x+2)}{3 i+\sqrt {3}}\right )+\left (\log \left (\frac {-2 i x+\sqrt {3}-i}{\sqrt {3}+3 i}\right )+\log \left (\frac {2 i x+\sqrt {3}+i}{\sqrt {3}-3 i}\right )\right ) \log (2 (x+2))\right )+x \log \left (x^2+x+1\right )+\log (2 (x+1)) \log \left (x^2+x+1\right )-4 \log (2 (x+2)) \log \left (x^2+x+1\right )+\frac {1}{2} \log \left (x^2+x+1\right )-2 x-\log \left (\frac {-2 i x+\sqrt {3}-i}{\sqrt {3}+i}\right ) \log (2 (x+1))-\log \left (\frac {2 i x+\sqrt {3}+i}{\sqrt {3}-i}\right ) \log (2 (x+1))+\sqrt {3} \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right ) \]
Antiderivative was successfully verified.
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fricas [F] time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2} \log \left (x^{2} + x + 1\right )}{x^{2} + 3 \, x + 2}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \log \left (x^{2} + x + 1\right )}{x^{2} + 3 \, x + 2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.09, size = 279, normalized size = 0.90 \[ x \ln \left (x^{2}+x +1\right )-\ln \left (\frac {-2 x -1+i \sqrt {3}}{1+i \sqrt {3}}\right ) \ln \left (x +1\right )+4 \ln \left (\frac {-2 x -1+i \sqrt {3}}{3+i \sqrt {3}}\right ) \ln \left (x +2\right )+4 \ln \left (\frac {2 x +1+i \sqrt {3}}{-3+i \sqrt {3}}\right ) \ln \left (x +2\right )-\ln \left (\frac {2 x +1+i \sqrt {3}}{i \sqrt {3}-1}\right ) \ln \left (x +1\right )+\ln \left (x +1\right ) \ln \left (x^{2}+x +1\right )-4 \ln \left (x +2\right ) \ln \left (x^{2}+x +1\right )-2 x -\dilog \left (\frac {-2 x -1+i \sqrt {3}}{1+i \sqrt {3}}\right )+4 \dilog \left (\frac {-2 x -1+i \sqrt {3}}{3+i \sqrt {3}}\right )+4 \dilog \left (\frac {2 x +1+i \sqrt {3}}{-3+i \sqrt {3}}\right )-\dilog \left (\frac {2 x +1+i \sqrt {3}}{i \sqrt {3}-1}\right )+\sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )+\frac {\ln \left (x^{2}+x +1\right )}{2} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \log \left (x^{2} + x + 1\right )}{x^{2} + 3 \, x + 2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2\,\ln \left (x^2+x+1\right )}{x^2+3\,x+2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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