Optimal. Leaf size=103 \[ -\frac {1}{3} \text {Ei}(3 \log (x))+\frac {1}{2} i x^2 \text {Li}_2\left (e^{2 i x}\right )-\frac {1}{2} x \text {Li}_3\left (e^{2 i x}\right )-\frac {1}{4} i \text {Li}_4\left (e^{2 i x}\right )+\left (-\frac {1}{12}+\frac {i}{12}\right ) x^4-\frac {1}{3} x^3 \log \left (1-e^{2 i x}\right )+\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right ) \]
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Rubi [A] time = 0.20, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.923, Rules used = {30, 2555, 12, 14, 3717, 2190, 2531, 6609, 2282, 6589, 2309, 2178} \[ \frac {1}{2} i x^2 \text {PolyLog}\left (2,e^{2 i x}\right )-\frac {1}{2} x \text {PolyLog}\left (3,e^{2 i x}\right )-\frac {1}{4} i \text {PolyLog}\left (4,e^{2 i x}\right )-\frac {1}{3} \text {Ei}(3 \log (x))+\left (-\frac {1}{12}+\frac {i}{12}\right ) x^4-\frac {1}{3} x^3 \log \left (1-e^{2 i x}\right )+\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right ) \]
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 30
Rule 2178
Rule 2190
Rule 2282
Rule 2309
Rule 2531
Rule 2555
Rule 3717
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x^2 \log \left (e^x \log (x) \sin (x)\right ) \, dx &=\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )-\int \frac {1}{3} x^3 \left (1+\cot (x)+\frac {1}{x \log (x)}\right ) \, dx\\ &=\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )-\frac {1}{3} \int x^3 \left (1+\cot (x)+\frac {1}{x \log (x)}\right ) \, dx\\ &=\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )-\frac {1}{3} \int \left (x^3 (1+\cot (x))+\frac {x^2}{\log (x)}\right ) \, dx\\ &=\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )-\frac {1}{3} \int x^3 (1+\cot (x)) \, dx-\frac {1}{3} \int \frac {x^2}{\log (x)} \, dx\\ &=\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )-\frac {1}{3} \int \left (x^3+x^3 \cot (x)\right ) \, dx-\frac {1}{3} \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )\\ &=-\frac {x^4}{12}-\frac {1}{3} \text {Ei}(3 \log (x))+\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )-\frac {1}{3} \int x^3 \cot (x) \, dx\\ &=\left (-\frac {1}{12}+\frac {i}{12}\right ) x^4-\frac {1}{3} \text {Ei}(3 \log (x))+\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )+\frac {2}{3} i \int \frac {e^{2 i x} x^3}{1-e^{2 i x}} \, dx\\ &=\left (-\frac {1}{12}+\frac {i}{12}\right ) x^4-\frac {1}{3} \text {Ei}(3 \log (x))-\frac {1}{3} x^3 \log \left (1-e^{2 i x}\right )+\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )+\int x^2 \log \left (1-e^{2 i x}\right ) \, dx\\ &=\left (-\frac {1}{12}+\frac {i}{12}\right ) x^4-\frac {1}{3} \text {Ei}(3 \log (x))-\frac {1}{3} x^3 \log \left (1-e^{2 i x}\right )+\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )+\frac {1}{2} i x^2 \text {Li}_2\left (e^{2 i x}\right )-i \int x \text {Li}_2\left (e^{2 i x}\right ) \, dx\\ &=\left (-\frac {1}{12}+\frac {i}{12}\right ) x^4-\frac {1}{3} \text {Ei}(3 \log (x))-\frac {1}{3} x^3 \log \left (1-e^{2 i x}\right )+\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )+\frac {1}{2} i x^2 \text {Li}_2\left (e^{2 i x}\right )-\frac {1}{2} x \text {Li}_3\left (e^{2 i x}\right )+\frac {1}{2} \int \text {Li}_3\left (e^{2 i x}\right ) \, dx\\ &=\left (-\frac {1}{12}+\frac {i}{12}\right ) x^4-\frac {1}{3} \text {Ei}(3 \log (x))-\frac {1}{3} x^3 \log \left (1-e^{2 i x}\right )+\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )+\frac {1}{2} i x^2 \text {Li}_2\left (e^{2 i x}\right )-\frac {1}{2} x \text {Li}_3\left (e^{2 i x}\right )-\frac {1}{4} i \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{2 i x}\right )\\ &=\left (-\frac {1}{12}+\frac {i}{12}\right ) x^4-\frac {1}{3} \text {Ei}(3 \log (x))-\frac {1}{3} x^3 \log \left (1-e^{2 i x}\right )+\frac {1}{3} x^3 \log \left (e^x \log (x) \sin (x)\right )+\frac {1}{2} i x^2 \text {Li}_2\left (e^{2 i x}\right )-\frac {1}{2} x \text {Li}_3\left (e^{2 i x}\right )-\frac {1}{4} i \text {Li}_4\left (e^{2 i x}\right )\\ \end {align*}
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Mathematica [A] time = 0.07, size = 100, normalized size = 0.97 \[ \frac {1}{192} i \left (64 i \text {Ei}(3 \log (x))-96 x^2 \text {Li}_2\left (e^{-2 i x}\right )+96 i x \text {Li}_3\left (e^{-2 i x}\right )+48 \text {Li}_4\left (e^{-2 i x}\right )+(-16+16 i) x^4+64 i x^3 \log \left (1-e^{-2 i x}\right )-64 i x^3 \log \left (e^x \log (x) \sin (x)\right )+\pi ^4\right ) \]
Antiderivative was successfully verified.
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fricas [C] time = 0.49, size = 241, normalized size = 2.34 \[ -\frac {1}{12} \, x^{4} + \frac {1}{3} \, x^{3} \log \left (e^{x} \log \relax (x) \sin \relax (x)\right ) - \frac {1}{6} \, x^{3} \log \left (\cos \relax (x) + i \, \sin \relax (x) + 1\right ) - \frac {1}{6} \, x^{3} \log \left (\cos \relax (x) - i \, \sin \relax (x) + 1\right ) - \frac {1}{6} \, x^{3} \log \left (-\cos \relax (x) + i \, \sin \relax (x) + 1\right ) - \frac {1}{6} \, x^{3} \log \left (-\cos \relax (x) - i \, \sin \relax (x) + 1\right ) + \frac {1}{2} i \, x^{2} {\rm Li}_2\left (\cos \relax (x) + i \, \sin \relax (x)\right ) - \frac {1}{2} i \, x^{2} {\rm Li}_2\left (\cos \relax (x) - i \, \sin \relax (x)\right ) - \frac {1}{2} i \, x^{2} {\rm Li}_2\left (-\cos \relax (x) + i \, \sin \relax (x)\right ) + \frac {1}{2} i \, x^{2} {\rm Li}_2\left (-\cos \relax (x) - i \, \sin \relax (x)\right ) - x {\rm polylog}\left (3, \cos \relax (x) + i \, \sin \relax (x)\right ) - x {\rm polylog}\left (3, \cos \relax (x) - i \, \sin \relax (x)\right ) - x {\rm polylog}\left (3, -\cos \relax (x) + i \, \sin \relax (x)\right ) - x {\rm polylog}\left (3, -\cos \relax (x) - i \, \sin \relax (x)\right ) - \frac {1}{3} \, \operatorname {log\_integral}\left (x^{3}\right ) - i \, {\rm polylog}\left (4, \cos \relax (x) + i \, \sin \relax (x)\right ) + i \, {\rm polylog}\left (4, \cos \relax (x) - i \, \sin \relax (x)\right ) + i \, {\rm polylog}\left (4, -\cos \relax (x) + i \, \sin \relax (x)\right ) - i \, {\rm polylog}\left (4, -\cos \relax (x) - i \, \sin \relax (x)\right ) \]
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 \[ \int x^{2} \log \left (e^{x} \log \relax (x) \sin \relax (x)\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 2.47, size = 0, normalized size = 0.00 \[ \int x^{2} \ln \left ({\mathrm e}^{x} \ln \relax (x ) \sin \relax (x )\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 2.42, size = 94, normalized size = 0.91 \[ \frac {1}{12} \, {\left (2 i \, \pi - 4 \, \log \relax (2)\right )} x^{3} - \left (\frac {1}{4} i - \frac {1}{4}\right ) \, x^{4} + \frac {1}{3} \, x^{3} \log \left (\log \relax (x)\right ) + i \, x^{2} {\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) + i \, x^{2} {\rm Li}_2\left (e^{\left (i \, x\right )}\right ) - 2 \, x {\rm Li}_{3}(-e^{\left (i \, x\right )}) - 2 \, x {\rm Li}_{3}(e^{\left (i \, x\right )}) - \frac {1}{3} \, {\rm Ei}\left (3 \, \log \relax (x)\right ) - 2 i \, {\rm Li}_{4}(-e^{\left (i \, x\right )}) - 2 i \, {\rm Li}_{4}(e^{\left (i \, x\right )}) \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\ln \left ({\mathrm {e}}^x\,\ln \relax (x)\,\sin \relax (x)\right ) \,d x \]
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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