Optimal. Leaf size=22 \[ \frac {1}{5} x \left (2 x+e^{e^4} x\right ) (1+\log (2)) \log (x) \]
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Rubi [B] time = 0.05, antiderivative size = 66, normalized size of antiderivative = 3.00, number of steps used = 7, number of rules used = 4, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {6, 12, 2332, 2304} \begin {gather*} \frac {1}{5} \left (2+e^{e^4}\right ) x^2 (1+\log (2)) \log (x)-\frac {1}{10} \left (2+e^{e^4}\right ) x^2 (1+\log (2))+\frac {1}{10} e^{e^4} x^2 (1+\log (2))+\frac {1}{5} x^2 (1+\log (2)) \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 12
Rule 2304
Rule 2332
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1}{5} \left (x (2+2 \log (2))+e^{e^4} (x+x \log (2))+\left (4 x+4 x \log (2)+e^{e^4} (2 x+2 x \log (2))\right ) \log (x)\right ) \, dx\\ &=\frac {1}{5} \int \left (x (2+2 \log (2))+e^{e^4} (x+x \log (2))+\left (4 x+4 x \log (2)+e^{e^4} (2 x+2 x \log (2))\right ) \log (x)\right ) \, dx\\ &=\frac {1}{5} x^2 (1+\log (2))+\frac {1}{10} e^{e^4} x^2 (1+\log (2))+\frac {1}{5} \int \left (4 x+4 x \log (2)+e^{e^4} (2 x+2 x \log (2))\right ) \log (x) \, dx\\ &=\frac {1}{5} x^2 (1+\log (2))+\frac {1}{10} e^{e^4} x^2 (1+\log (2))+\frac {1}{5} \int \left (x (4+4 \log (2))+e^{e^4} (2 x+2 x \log (2))\right ) \log (x) \, dx\\ &=\frac {1}{5} x^2 (1+\log (2))+\frac {1}{10} e^{e^4} x^2 (1+\log (2))+\frac {1}{5} \int 2 \left (2+e^{e^4}\right ) x (1+\log (2)) \log (x) \, dx\\ &=\frac {1}{5} x^2 (1+\log (2))+\frac {1}{10} e^{e^4} x^2 (1+\log (2))+\frac {1}{5} \left (2 \left (2+e^{e^4}\right ) (1+\log (2))\right ) \int x \log (x) \, dx\\ &=\frac {1}{5} x^2 (1+\log (2))+\frac {1}{10} e^{e^4} x^2 (1+\log (2))-\frac {1}{10} \left (2+e^{e^4}\right ) x^2 (1+\log (2))+\frac {1}{5} \left (2+e^{e^4}\right ) x^2 (1+\log (2)) \log (x)\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.00, size = 20, normalized size = 0.91 \begin {gather*} \frac {1}{5} \left (2+e^{e^4}\right ) x^2 (1+\log (2)) \log (x) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.63, size = 31, normalized size = 1.41 \begin {gather*} \frac {1}{5} \, {\left (2 \, x^{2} \log \relax (2) + 2 \, x^{2} + {\left (x^{2} \log \relax (2) + x^{2}\right )} e^{\left (e^{4}\right )}\right )} \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.19, size = 72, normalized size = 3.27 \begin {gather*} \frac {1}{5} \, x^{2} e^{\left (e^{4}\right )} \log \relax (2) \log \relax (x) - \frac {1}{10} \, x^{2} e^{\left (e^{4}\right )} \log \relax (2) + \frac {1}{5} \, x^{2} e^{\left (e^{4}\right )} \log \relax (x) + \frac {2}{5} \, x^{2} \log \relax (2) \log \relax (x) - \frac {1}{10} \, x^{2} e^{\left (e^{4}\right )} + \frac {2}{5} \, x^{2} \log \relax (x) + \frac {1}{10} \, {\left (x^{2} \log \relax (2) + x^{2}\right )} e^{\left (e^{4}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 25, normalized size = 1.14
method | result | size |
norman | \(\left (\frac {{\mathrm e}^{{\mathrm e}^{4}} \ln \relax (2)}{5}+\frac {2 \ln \relax (2)}{5}+\frac {{\mathrm e}^{{\mathrm e}^{4}}}{5}+\frac {2}{5}\right ) x^{2} \ln \relax (x )\) | \(25\) |
default | \(\frac {{\mathrm e}^{{\mathrm e}^{4}} \ln \relax (2) x^{2} \ln \relax (x )}{5}+\frac {2 x^{2} \ln \relax (2) \ln \relax (x )}{5}+\frac {{\mathrm e}^{{\mathrm e}^{4}} x^{2} \ln \relax (x )}{5}+\frac {2 x^{2} \ln \relax (x )}{5}\) | \(40\) |
risch | \(\frac {{\mathrm e}^{{\mathrm e}^{4}} \ln \relax (2) x^{2} \ln \relax (x )}{5}+\frac {2 x^{2} \ln \relax (2) \ln \relax (x )}{5}+\frac {{\mathrm e}^{{\mathrm e}^{4}} x^{2} \ln \relax (x )}{5}+\frac {2 x^{2} \ln \relax (x )}{5}\) | \(40\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.40, size = 77, normalized size = 3.50 \begin {gather*} -\frac {1}{10} \, {\left ({\left (e^{\left (e^{4}\right )} + 2\right )} \log \relax (2) + e^{\left (e^{4}\right )} + 2\right )} x^{2} + \frac {1}{5} \, x^{2} \log \relax (2) + \frac {1}{5} \, x^{2} + \frac {1}{10} \, {\left (x^{2} \log \relax (2) + x^{2}\right )} e^{\left (e^{4}\right )} + \frac {1}{5} \, {\left (2 \, x^{2} \log \relax (2) + 2 \, x^{2} + {\left (x^{2} \log \relax (2) + x^{2}\right )} e^{\left (e^{4}\right )}\right )} \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.31, size = 16, normalized size = 0.73 \begin {gather*} \frac {x^2\,\ln \relax (x)\,\left ({\mathrm {e}}^{{\mathrm {e}}^4}+2\right )\,\left (\ln \relax (2)+1\right )}{5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.16, size = 42, normalized size = 1.91 \begin {gather*} \left (\frac {2 x^{2} \log {\relax (2 )}}{5} + \frac {2 x^{2}}{5} + \frac {x^{2} e^{e^{4}} \log {\relax (2 )}}{5} + \frac {x^{2} e^{e^{4}}}{5}\right ) \log {\relax (x )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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