Optimal. Leaf size=21 \[ \frac {5}{(5+2 x+\log (5))^2-\log ^2(\log (4))} \]
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Rubi [A] time = 0.07, antiderivative size = 28, normalized size of antiderivative = 1.33, number of steps used = 3, number of rules used = 3, integrand size = 114, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {1680, 12, 261} \begin {gather*} \frac {5}{4 \left (x+\frac {1}{64} (160+32 \log (5))\right )^2-\log ^2(\log (4))} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 261
Rule 1680
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int -\frac {40 x}{\left (4 x^2-\log ^2(\log (4))\right )^2} \, dx,x,x+\frac {1}{64} (160+32 \log (5))\right )\\ &=-\left (40 \operatorname {Subst}\left (\int \frac {x}{\left (4 x^2-\log ^2(\log (4))\right )^2} \, dx,x,x+\frac {1}{64} (160+32 \log (5))\right )\right )\\ &=\frac {5}{(5+2 x+\log (5))^2-\log ^2(\log (4))}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.02, size = 34, normalized size = 1.62 \begin {gather*} \frac {5}{25+20 x+4 x^2+10 \log (5)+4 x \log (5)+\log ^2(5)-\log ^2(\log (4))} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 3.23, size = 36, normalized size = 1.71 \begin {gather*} \frac {5}{4 \, x^{2} + 2 \, {\left (2 \, x + 5\right )} \log \relax (5) + \log \relax (5)^{2} - \log \left (2 \, \log \relax (2)\right )^{2} + 20 \, x + 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.68, size = 36, normalized size = 1.71 \begin {gather*} \frac {5}{4 \, x^{2} + 4 \, x \log \relax (5) + \log \relax (5)^{2} - \log \left (2 \, \log \relax (2)\right )^{2} + 20 \, x + 10 \, \log \relax (5) + 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.35, size = 37, normalized size = 1.76
method | result | size |
gosper | \(-\frac {5}{\ln \left (2 \ln \relax (2)\right )^{2}-\ln \relax (5)^{2}-4 x \ln \relax (5)-4 x^{2}-10 \ln \relax (5)-20 x -25}\) | \(37\) |
norman | \(-\frac {5}{\ln \left (2 \ln \relax (2)\right )^{2}-\ln \relax (5)^{2}-4 x \ln \relax (5)-4 x^{2}-10 \ln \relax (5)-20 x -25}\) | \(37\) |
risch | \(-\frac {5}{\ln \relax (2)^{2}+2 \ln \relax (2) \ln \left (\ln \relax (2)\right )-\ln \relax (5)^{2}-4 x \ln \relax (5)+\ln \left (\ln \relax (2)\right )^{2}-4 x^{2}-10 \ln \relax (5)-20 x -25}\) | \(46\) |
default | \(-\frac {5}{2 \ln \left (2 \ln \relax (2)\right ) \left (2 x +5+\ln \left (2 \ln \relax (2)\right )+\ln \relax (5)\right )}+\frac {5}{2 \ln \left (2 \ln \relax (2)\right ) \left (2 x +5-\ln \left (2 \ln \relax (2)\right )+\ln \relax (5)\right )}\) | \(50\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.34, size = 35, normalized size = 1.67 \begin {gather*} \frac {5}{4 \, x^{2} + 4 \, x {\left (\log \relax (5) + 5\right )} + \log \relax (5)^{2} - \log \left (2 \, \log \relax (2)\right )^{2} + 10 \, \log \relax (5) + 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.32, size = 34, normalized size = 1.62 \begin {gather*} \frac {5}{4\,x^2+\left (4\,\ln \relax (5)+20\right )\,x+10\,\ln \relax (5)-{\ln \left (\ln \relax (4)\right )}^2+{\ln \relax (5)}^2+25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 1.39, size = 48, normalized size = 2.29 \begin {gather*} \frac {5}{4 x^{2} + x \left (4 \log {\relax (5 )} + 20\right ) - \log {\relax (2 )}^{2} - \log {\left (\log {\relax (2 )} \right )}^{2} - 2 \log {\relax (2 )} \log {\left (\log {\relax (2 )} \right )} + \log {\relax (5 )}^{2} + 10 \log {\relax (5 )} + 25} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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