Optimal. Leaf size=14 \[ x \log \left (-4+(x+4 \log (\log (50)))^2\right ) \]
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Rubi [A] time = 0.19, antiderivative size = 24, normalized size of antiderivative = 1.71, number of steps used = 10, number of rules used = 7, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.099, Rules used = {6728, 142, 2523, 12, 773, 632, 31} \begin {gather*} x \log \left (x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 31
Rule 142
Rule 632
Rule 773
Rule 2523
Rule 6728
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 x (x+4 \log (\log (50)))}{(-2+x+4 \log (\log (50))) (2+x+4 \log (\log (50)))}+\log \left (x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )\right )\right ) \, dx\\ &=2 \int \frac {x (x+4 \log (\log (50)))}{(-2+x+4 \log (\log (50))) (2+x+4 \log (\log (50)))} \, dx+\int \log \left (x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )\right ) \, dx\\ &=x \log \left (x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )\right )+2 \int \left (1+\frac {1-2 \log (\log (50))}{-2+x+4 \log (\log (50))}+\frac {-1-2 \log (\log (50))}{2+x+4 \log (\log (50))}\right ) \, dx-\int \frac {2 x (x+4 \log (\log (50)))}{x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )} \, dx\\ &=2 x+2 (1-2 \log (\log (50))) \log (-x+2 (1-2 \log (\log (50))))-2 (1+2 \log (\log (50))) \log (x+2 (1+2 \log (\log (50))))+x \log \left (x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )\right )-2 \int \frac {x (x+4 \log (\log (50)))}{x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )} \, dx\\ &=2 (1-2 \log (\log (50))) \log (-x+2 (1-2 \log (\log (50))))-2 (1+2 \log (\log (50))) \log (x+2 (1+2 \log (\log (50))))+x \log \left (x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )\right )-2 \int \frac {-4 x \log (\log (50))+4 \left (1-4 \log ^2(\log (50))\right )}{x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )} \, dx\\ &=2 (1-2 \log (\log (50))) \log (-x+2 (1-2 \log (\log (50))))-2 (1+2 \log (\log (50))) \log (x+2 (1+2 \log (\log (50))))+x \log \left (x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )\right )-(2 (1-2 \log (\log (50)))) \int \frac {1}{-2+x+4 \log (\log (50))} \, dx+(2 (1+2 \log (\log (50)))) \int \frac {1}{2+x+4 \log (\log (50))} \, dx\\ &=x \log \left (x^2+8 x \log (\log (50))-4 \left (1-4 \log ^2(\log (50))\right )\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.06, size = 21, normalized size = 1.50 \begin {gather*} x \log \left (-4+x^2+8 x \log (\log (50))+16 \log ^2(\log (50))\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 21, normalized size = 1.50 \begin {gather*} x \log \left (x^{2} + 8 \, x \log \left (\log \left (50\right )\right ) + 16 \, \log \left (\log \left (50\right )\right )^{2} - 4\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.21, size = 21, normalized size = 1.50 \begin {gather*} x \log \left (x^{2} + 8 \, x \log \left (\log \left (50\right )\right ) + 16 \, \log \left (\log \left (50\right )\right )^{2} - 4\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.53, size = 22, normalized size = 1.57
method | result | size |
default | \(x \ln \left (16 \ln \left (\ln \left (50\right )\right )^{2}+8 x \ln \left (\ln \left (50\right )\right )+x^{2}-4\right )\) | \(22\) |
norman | \(x \ln \left (16 \ln \left (\ln \left (50\right )\right )^{2}+8 x \ln \left (\ln \left (50\right )\right )+x^{2}-4\right )\) | \(22\) |
risch | \(x \ln \left (16 \ln \left (\ln \relax (2)+2 \ln \relax (5)\right )^{2}+8 x \ln \left (\ln \relax (2)+2 \ln \relax (5)\right )+x^{2}-4\right )\) | \(32\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.58, size = 148, normalized size = 10.57 \begin {gather*} {\left (x + 4 \, \log \left (2 \, \log \relax (5) + \log \relax (2)\right ) + 2\right )} \log \left (x + 4 \, \log \left (2 \, \log \relax (5) + \log \relax (2)\right ) + 2\right ) + {\left (x + 4 \, \log \left (2 \, \log \relax (5) + \log \relax (2)\right ) - 2\right )} \log \left (x + 4 \, \log \left (2 \, \log \relax (5) + \log \relax (2)\right ) - 2\right ) - 2 \, {\left (4 \, \log \left (\log \left (50\right )\right )^{2} + 4 \, \log \left (\log \left (50\right )\right ) + 1\right )} \log \left (x + 4 \, \log \left (\log \left (50\right )\right ) + 2\right ) + 2 \, {\left (4 \, \log \left (\log \left (50\right )\right )^{2} - 4 \, \log \left (\log \left (50\right )\right ) + 1\right )} \log \left (x + 4 \, \log \left (\log \left (50\right )\right ) - 2\right ) + 4 \, {\left ({\left (2 \, \log \left (\log \left (50\right )\right ) + 1\right )} \log \left (x + 4 \, \log \left (\log \left (50\right )\right ) + 2\right ) - {\left (2 \, \log \left (\log \left (50\right )\right ) - 1\right )} \log \left (x + 4 \, \log \left (\log \left (50\right )\right ) - 2\right )\right )} \log \left (\log \left (50\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.62, size = 21, normalized size = 1.50 \begin {gather*} x\,\ln \left (x^2+8\,\ln \left (\ln \left (50\right )\right )\,x+16\,{\ln \left (\ln \left (50\right )\right )}^2-4\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.16, size = 24, normalized size = 1.71 \begin {gather*} x \log {\left (x^{2} + 8 x \log {\left (\log {\left (50 \right )} \right )} - 4 + 16 \log {\left (\log {\left (50 \right )} \right )}^{2} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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