Optimal. Leaf size=24 \[ -3+\log \left (-4+x-\frac {5+x}{x+\frac {25 x}{3 \log (5)}}\right ) \]
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Rubi [A] time = 0.11, antiderivative size = 27, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 2, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {2074, 628} \begin {gather*} \log \left (-\left (x^2 (25+\log (125))\right )+5 x (20+\log (125))+15 \log (5)\right )-\log (x) \end {gather*}
Antiderivative was successfully verified.
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Rule 628
Rule 2074
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {1}{x}+\frac {5 (20+\log (125))-2 x (25+\log (125))}{15 \log (5)+5 x (20+\log (125))-x^2 (25+\log (125))}\right ) \, dx\\ &=-\log (x)+\int \frac {5 (20+\log (125))-2 x (25+\log (125))}{15 \log (5)+x^2 (-25-\log (125))+5 x (20+\log (125))} \, dx\\ &=-\log (x)+\log \left (15 \log (5)+5 x (20+\log (125))-x^2 (25+\log (125))\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.02, size = 30, normalized size = 1.25 \begin {gather*} -\log (x)+\log \left (-100 x+25 x^2-15 \log (5)-5 x \log (125)+x^2 \log (125)\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.44, size = 27, normalized size = 1.12 \begin {gather*} \log \left (25 \, x^{2} + 3 \, {\left (x^{2} - 5 \, x - 5\right )} \log \relax (5) - 100 \, x\right ) - \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.25, size = 33, normalized size = 1.38 \begin {gather*} \log \left ({\left | 3 \, x^{2} \log \relax (5) + 25 \, x^{2} - 15 \, x \log \relax (5) - 100 \, x - 15 \, \log \relax (5) \right |}\right ) - \log \left ({\left | x \right |}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 30, normalized size = 1.25
method | result | size |
risch | \(-\ln \relax (x )+\ln \left (\left (-3 \ln \relax (5)-25\right ) x^{2}+\left (15 \ln \relax (5)+100\right ) x +15 \ln \relax (5)\right )\) | \(30\) |
default | \(-\ln \relax (x )+\ln \left (3 x^{2} \ln \relax (5)-15 x \ln \relax (5)+25 x^{2}-15 \ln \relax (5)-100 x \right )\) | \(32\) |
norman | \(-\ln \relax (x )+\ln \left (3 x^{2} \ln \relax (5)-15 x \ln \relax (5)+25 x^{2}-15 \ln \relax (5)-100 x \right )\) | \(32\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.44, size = 30, normalized size = 1.25 \begin {gather*} \log \left (x^{2} {\left (3 \, \log \relax (5) + 25\right )} - 5 \, x {\left (3 \, \log \relax (5) + 20\right )} - 15 \, \log \relax (5)\right ) - \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.00, size = 31, normalized size = 1.29 \begin {gather*} \ln \left (200\,x+30\,\ln \relax (5)+30\,x\,\ln \relax (5)-6\,x^2\,\ln \relax (5)-50\,x^2\right )-\ln \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.64, size = 36, normalized size = 1.50 \begin {gather*} - \log {\relax (x )} + \log {\left (x^{2} + \frac {x \left (-100 - 15 \log {\relax (5 )}\right )}{3 \log {\relax (5 )} + 25} - \frac {15 \log {\relax (5 )}}{3 \log {\relax (5 )} + 25} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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