Optimal. Leaf size=25 \[ \frac {1}{5}+\frac {\left (4-x+\frac {1}{4} x \log (4)\right ) \log (x)}{5 x^2} \]
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Rubi [B] time = 0.09, antiderivative size = 77, normalized size of antiderivative = 3.08, number of steps used = 9, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {6, 12, 14, 37, 2334, 43} \begin {gather*} \frac {2}{5 x^2}-\frac {(16-x (4-\log (4)))^2}{640 x^2}+\frac {(32-x (4-\log (4)))^2 \log (x)}{1280 x^2}-\frac {(4-\log (4))^2 \log (x)}{1280}-\frac {4-\log (4)}{20 x} \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 12
Rule 14
Rule 37
Rule 43
Rule 2334
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {16+x (-4+\log (4))+(-32+4 x-x \log (4)) \log (x)}{20 x^3} \, dx\\ &=\frac {1}{20} \int \frac {16+x (-4+\log (4))+(-32+4 x-x \log (4)) \log (x)}{x^3} \, dx\\ &=\frac {1}{20} \int \left (\frac {16-x (4-\log (4))}{x^3}+\frac {(-32+x (4-\log (4))) \log (x)}{x^3}\right ) \, dx\\ &=\frac {1}{20} \int \frac {16-x (4-\log (4))}{x^3} \, dx+\frac {1}{20} \int \frac {(-32+x (4-\log (4))) \log (x)}{x^3} \, dx\\ &=-\frac {(16-x (4-\log (4)))^2}{640 x^2}+\frac {(32-x (4-\log (4)))^2 \log (x)}{1280 x^2}-\frac {1}{20} \int \frac {(32+x (-4+\log (4)))^2}{64 x^3} \, dx\\ &=-\frac {(16-x (4-\log (4)))^2}{640 x^2}+\frac {(32-x (4-\log (4)))^2 \log (x)}{1280 x^2}-\frac {\int \frac {(32+x (-4+\log (4)))^2}{x^3} \, dx}{1280}\\ &=-\frac {(16-x (4-\log (4)))^2}{640 x^2}+\frac {(32-x (4-\log (4)))^2 \log (x)}{1280 x^2}-\frac {\int \left (\frac {1024}{x^3}+\frac {64 (-4+\log (4))}{x^2}+\frac {(-4+\log (4))^2}{x}\right ) \, dx}{1280}\\ &=\frac {2}{5 x^2}-\frac {(16-x (4-\log (4)))^2}{640 x^2}-\frac {4-\log (4)}{20 x}+\frac {(32-x (4-\log (4)))^2 \log (x)}{1280 x^2}-\frac {(4-\log (4))^2 \log (x)}{1280}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.02, size = 27, normalized size = 1.08 \begin {gather*} \frac {1}{20} \left (\frac {16 \log (x)}{x^2}-\frac {4 \log (x)}{x}+\frac {\log (4) \log (x)}{x}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.69, size = 16, normalized size = 0.64 \begin {gather*} \frac {{\left (x \log \relax (2) - 2 \, x + 8\right )} \log \relax (x)}{10 \, x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.33, size = 16, normalized size = 0.64 \begin {gather*} \frac {{\left (x \log \relax (2) - 2 \, x + 8\right )} \log \relax (x)}{10 \, x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 17, normalized size = 0.68
method | result | size |
risch | \(\frac {\left (x \ln \relax (2)-2 x +8\right ) \ln \relax (x )}{10 x^{2}}\) | \(17\) |
norman | \(\frac {\left (\frac {\ln \relax (2)}{10}-\frac {1}{5}\right ) x \ln \relax (x )+\frac {4 \ln \relax (x )}{5}}{x^{2}}\) | \(20\) |
default | \(-\frac {\ln \relax (2) \left (-\frac {\ln \relax (x )}{x}-\frac {1}{x}\right )}{10}-\frac {\ln \relax (x )}{5 x}-\frac {\ln \relax (2)}{10 x}+\frac {4 \ln \relax (x )}{5 x^{2}}\) | \(40\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.40, size = 36, normalized size = 1.44 \begin {gather*} \frac {1}{10} \, {\left (\frac {\log \relax (x)}{x} + \frac {1}{x}\right )} \log \relax (2) - \frac {\log \relax (2)}{10 \, x} - \frac {\log \relax (x)}{5 \, x} + \frac {4 \, \log \relax (x)}{5 \, x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.89, size = 16, normalized size = 0.64 \begin {gather*} \frac {\ln \relax (x)\,\left (x\,\ln \relax (2)-2\,x+8\right )}{10\,x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.15, size = 17, normalized size = 0.68 \begin {gather*} \frac {\left (- 2 x + x \log {\relax (2 )} + 8\right ) \log {\relax (x )}}{10 x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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