Optimal. Leaf size=30 \[ -4-x-x^2-\log (x)+\frac {(-2+x-\log (2)) \log (x)}{5 x} \]
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Rubi [A] time = 0.04, antiderivative size = 28, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 14, 2304} \begin {gather*} -x^2-x-\frac {4 \log (x)}{5}-\frac {(2+\log (2)) \log (x)}{5 x} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 2304
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {-2-4 x-5 x^2-10 x^3-\log (2)+(2+\log (2)) \log (x)}{x^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {-2-4 x-5 x^2-10 x^3-\log (2)}{x^2}+\frac {(2+\log (2)) \log (x)}{x^2}\right ) \, dx\\ &=\frac {1}{5} \int \frac {-2-4 x-5 x^2-10 x^3-\log (2)}{x^2} \, dx+\frac {1}{5} (2+\log (2)) \int \frac {\log (x)}{x^2} \, dx\\ &=-\frac {2+\log (2)}{5 x}-\frac {(2+\log (2)) \log (x)}{5 x}+\frac {1}{5} \int \left (-5-\frac {4}{x}-10 x+\frac {-2-\log (2)}{x^2}\right ) \, dx\\ &=-x-x^2-\frac {4 \log (x)}{5}-\frac {(2+\log (2)) \log (x)}{5 x}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.01, size = 35, normalized size = 1.17 \begin {gather*} -x-x^2-\frac {4 \log (x)}{5}-\frac {2 \log (x)}{5 x}-\frac {\log (2) \log (x)}{5 x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.55, size = 26, normalized size = 0.87 \begin {gather*} -\frac {5 \, x^{3} + 5 \, x^{2} + {\left (4 \, x + \log \relax (2) + 2\right )} \log \relax (x)}{5 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.24, size = 24, normalized size = 0.80 \begin {gather*} -x^{2} - x - \frac {{\left (\log \relax (2) + 2\right )} \log \relax (x)}{5 \, x} - \frac {4}{5} \, \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 25, normalized size = 0.83
method | result | size |
risch | \(-\frac {\left (\ln \relax (2)+2\right ) \ln \relax (x )}{5 x}-x^{2}-x -\frac {4 \ln \relax (x )}{5}\) | \(25\) |
norman | \(\frac {\left (-\frac {2}{5}-\frac {\ln \relax (2)}{5}\right ) \ln \relax (x )-\frac {4 x \ln \relax (x )}{5}-x^{2}-x^{3}}{x}\) | \(30\) |
default | \(-x^{2}+\frac {\ln \relax (2) \left (-\frac {\ln \relax (x )}{x}-\frac {1}{x}\right )}{5}-x -\frac {2 \ln \relax (x )}{5 x}+\frac {\ln \relax (2)}{5 x}-\frac {4 \ln \relax (x )}{5}\) | \(45\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 41, normalized size = 1.37 \begin {gather*} -x^{2} - \frac {1}{5} \, {\left (\frac {\log \relax (x)}{x} + \frac {1}{x}\right )} \log \relax (2) - x + \frac {\log \relax (2)}{5 \, x} - \frac {2 \, \log \relax (x)}{5 \, x} - \frac {4}{5} \, \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.10, size = 24, normalized size = 0.80 \begin {gather*} -x-\frac {4\,\ln \relax (x)}{5}-x^2-\frac {\ln \relax (x)\,\left (\ln \relax (2)+2\right )}{5\,x} \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 = 0.80 \begin {gather*} - x^{2} - x - \frac {4 \log {\relax (x )}}{5} + \frac {\left (-2 - \log {\relax (2 )}\right ) \log {\relax (x )}}{5 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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