Optimal. Leaf size=27 \[ 1-e^{12}-\log ^2\left (x+\frac {\log \left (16 x \log ^2(\log (5))\right )}{x}\right ) \]
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Rubi [A] time = 0.72, antiderivative size = 22, normalized size of antiderivative = 0.81, number of steps used = 2, number of rules used = 6, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2561, 6741, 12, 6742, 6684, 6686} \begin {gather*} -\log ^2\left (\frac {x^2+\log \left (16 x \log ^2(\log (5))\right )}{x}\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2561
Rule 6684
Rule 6686
Rule 6741
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (-2-2 x^2+2 \log \left (16 x \log ^2(\log (5))\right )\right ) \log \left (\frac {x^2+\log \left (16 x \log ^2(\log (5))\right )}{x}\right )}{x \left (x^2+\log \left (16 x \log ^2(\log (5))\right )\right )} \, dx\\ &=-\log ^2\left (\frac {x^2+\log \left (16 x \log ^2(\log (5))\right )}{x}\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.02, size = 20, normalized size = 0.74 \begin {gather*} -\log ^2\left (x+\frac {\log \left (16 x \log ^2(\log (5))\right )}{x}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.87, size = 22, normalized size = 0.81 \begin {gather*} -\log \left (\frac {x^{2} + \log \left (16 \, x \log \left (\log \relax (5)\right )^{2}\right )}{x}\right )^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.20, size = 60, normalized size = 2.22 \begin {gather*} -2 \, {\left (\log \left (x^{2} + \log \left (16 \, \log \left (\log \relax (5)\right )^{2}\right ) + \log \relax (x)\right ) - \log \relax (x)\right )} \log \left (x^{2} + \log \left (16 \, x \log \left (\log \relax (5)\right )^{2}\right )\right ) + \log \left (x^{2} + \log \left (16 \, \log \left (\log \relax (5)\right )^{2}\right ) + \log \relax (x)\right )^{2} - \log \relax (x)^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (2 \ln \left (16 x \ln \left (\ln \relax (5)\right )^{2}\right )-2 x^{2}-2\right ) \ln \left (\frac {\ln \left (16 x \ln \left (\ln \relax (5)\right )^{2}\right )+x^{2}}{x}\right )}{x \ln \left (16 x \ln \left (\ln \relax (5)\right )^{2}\right )+x^{3}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.48, size = 87, normalized size = 3.22 \begin {gather*} \log \left (x^{2} + 4 \, \log \relax (2) + \log \relax (x) + 2 \, \log \left (\log \left (\log \relax (5)\right )\right )\right )^{2} - 2 \, \log \left (x^{2} + 4 \, \log \relax (2) + \log \relax (x) + 2 \, \log \left (\log \left (\log \relax (5)\right )\right )\right ) \log \relax (x) + \log \relax (x)^{2} - 2 \, {\left (\log \left (x^{2} + 4 \, \log \relax (2) + \log \relax (x) + 2 \, \log \left (\log \left (\log \relax (5)\right )\right )\right ) - \log \relax (x)\right )} \log \left (\frac {x^{2} + \log \left (16 \, x \log \left (\log \relax (5)\right )^{2}\right )}{x}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.34, size = 22, normalized size = 0.81 \begin {gather*} -{\ln \left (\frac {\ln \left (16\,x\,{\ln \left (\ln \relax (5)\right )}^2\right )+x^2}{x}\right )}^2 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.33, size = 20, normalized size = 0.74 \begin {gather*} - \log {\left (\frac {x^{2} + \log {\left (16 x \log {\left (\log {\relax (5 )} \right )}^{2} \right )}}{x} \right )}^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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