Optimal. Leaf size=20 \[ e^{\frac {4 \log ^2(4)}{5}} x^2 \log \left (\frac {x}{e^4}\right ) \]
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Rubi [A] time = 0.01, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {2304} \begin {gather*} x^2 e^{\frac {4 \log ^2(4)}{5}} \log \left (\frac {x}{e^4}\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 2304
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} e^{\frac {4 \log ^2(4)}{5}} x^2+\left (2 e^{\frac {4 \log ^2(4)}{5}}\right ) \int x \log \left (\frac {x}{e^4}\right ) \, dx\\ &=e^{\frac {4 \log ^2(4)}{5}} x^2 \log \left (\frac {x}{e^4}\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.00, size = 23, normalized size = 1.15 \begin {gather*} e^{\frac {4 \log ^2(4)}{5}} \left (-4 x^2+x^2 \log (x)\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 16, normalized size = 0.80 \begin {gather*} x^{2} e^{\left (\frac {16}{5} \, \log \relax (2)^{2}\right )} \log \left (x e^{\left (-4\right )}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.21, size = 38, normalized size = 1.90 \begin {gather*} \frac {1}{2} \, x^{2} e^{\left (\frac {16}{5} \, \log \relax (2)^{2}\right )} + \frac {1}{2} \, {\left (2 \, x^{2} \log \left (x e^{\left (-4\right )}\right ) - x^{2}\right )} e^{\left (\frac {16}{5} \, \log \relax (2)^{2}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 17, normalized size = 0.85
method | result | size |
risch | \({\mathrm e}^{\frac {16 \ln \relax (2)^{2}}{5}} x^{2} \ln \left ({\mathrm e}^{-4} x \right )\) | \(17\) |
derivativedivides | \({\mathrm e}^{\frac {16 \ln \relax (2)^{2}}{5}} x^{2} \ln \left ({\mathrm e}^{-4} x \right )\) | \(19\) |
default | \({\mathrm e}^{\frac {16 \ln \relax (2)^{2}}{5}} x^{2} \ln \left ({\mathrm e}^{-4} x \right )\) | \(19\) |
norman | \({\mathrm e}^{\frac {16 \ln \relax (2)^{2}}{5}} x^{2} \ln \left ({\mathrm e}^{-4} x \right )\) | \(19\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.53, size = 38, normalized size = 1.90 \begin {gather*} \frac {1}{2} \, x^{2} e^{\left (\frac {16}{5} \, \log \relax (2)^{2}\right )} + \frac {1}{2} \, {\left (2 \, x^{2} \log \left (x e^{\left (-4\right )}\right ) - x^{2}\right )} e^{\left (\frac {16}{5} \, \log \relax (2)^{2}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.81, size = 15, normalized size = 0.75 \begin {gather*} x^2\,{\mathrm {e}}^{\frac {16\,{\ln \relax (2)}^2}{5}}\,\left (\ln \relax (x)-4\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.11, size = 19, normalized size = 0.95 \begin {gather*} x^{2} e^{\frac {16 \log {\relax (2 )}^{2}}{5}} \log {\left (\frac {x}{e^{4}} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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