Optimal. Leaf size=21 \[ -x+\left (\log (\log (5))+\log \left (\log \left (-2 x+5 x^2\right )\right )\right )^2 \]
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Rubi [A] time = 0.61, antiderivative size = 19, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {1593, 6688, 6742, 6686} \begin {gather*} \log ^2(\log (5) \log (-((2-5 x) x)))-x \end {gather*}
Antiderivative was successfully verified.
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Rule 1593
Rule 6686
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (2 x-5 x^2\right ) \log \left (-2 x+5 x^2\right )+(-4+20 x) \log (\log (5))+(-4+20 x) \log \left (\log \left (-2 x+5 x^2\right )\right )}{x (-2+5 x) \log \left (-2 x+5 x^2\right )} \, dx\\ &=\int \frac {-2+5 x-\frac {4 (-1+5 x) \log (\log (5) \log (x (-2+5 x)))}{x \log (x (-2+5 x))}}{2-5 x} \, dx\\ &=\int \left (-1+\frac {4 (-1+5 x) \log (\log (5) \log (x (-2+5 x)))}{x (-2+5 x) \log (x (-2+5 x))}\right ) \, dx\\ &=-x+4 \int \frac {(-1+5 x) \log (\log (5) \log (x (-2+5 x)))}{x (-2+5 x) \log (x (-2+5 x))} \, dx\\ &=-x+\log ^2(\log (5) \log (-((2-5 x) x)))\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.10, size = 18, normalized size = 0.86 \begin {gather*} -x+\log ^2(\log (5) \log (x (-2+5 x))) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 33, normalized size = 1.57 \begin {gather*} 2 \, \log \left (\log \relax (5)\right ) \log \left (\log \left (5 \, x^{2} - 2 \, x\right )\right ) + \log \left (\log \left (5 \, x^{2} - 2 \, x\right )\right )^{2} - x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.24, size = 33, normalized size = 1.57 \begin {gather*} 2 \, \log \left (\log \relax (5)\right ) \log \left (\log \left (5 \, x^{2} - 2 \, x\right )\right ) + \log \left (\log \left (5 \, x^{2} - 2 \, x\right )\right )^{2} - x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.11, size = 34, normalized size = 1.62
method | result | size |
default | \(-x +\ln \left (\ln \left (5 x^{2}-2 x \right )\right )^{2}+2 \ln \left (\ln \relax (5)\right ) \ln \left (\ln \left (5 x^{2}-2 x \right )\right )\) | \(34\) |
norman | \(-x +\ln \left (\ln \left (5 x^{2}-2 x \right )\right )^{2}+2 \ln \left (\ln \relax (5)\right ) \ln \left (\ln \left (5 x^{2}-2 x \right )\right )\) | \(34\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.51, size = 31, normalized size = 1.48 \begin {gather*} \log \left (\log \left (5 \, x - 2\right ) + \log \relax (x)\right )^{2} + 2 \, \log \left (\log \left (5 \, x - 2\right ) + \log \relax (x)\right ) \log \left (\log \relax (5)\right ) - x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.59, size = 33, normalized size = 1.57 \begin {gather*} {\ln \left (\ln \left (5\,x^2-2\,x\right )\right )}^2+2\,\ln \left (\ln \relax (5)\right )\,\ln \left (\ln \left (5\,x^2-2\,x\right )\right )-x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.45, size = 32, normalized size = 1.52 \begin {gather*} - x + \log {\left (\log {\left (5 x^{2} - 2 x \right )} \right )}^{2} + 2 \log {\left (\log {\relax (5 )} \right )} \log {\left (\log {\left (5 x^{2} - 2 x \right )} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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