Optimal. Leaf size=27 \[ \frac {1+\frac {\log \left (e^5+x\right )}{20 x}}{1-x+\log (4 x)} \]
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Rubi [F] time = 2.79, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {x-21 x^2+20 x^3+e^5 \left (-20 x+20 x^2\right )+x \log (4 x)+\left (-2 x+2 x^2+e^5 (-2+2 x)+\left (-e^5-x\right ) \log (4 x)\right ) \log \left (e^5+x\right )}{20 x^3-40 x^4+20 x^5+e^5 \left (20 x^2-40 x^3+20 x^4\right )+\left (40 x^3-40 x^4+e^5 \left (40 x^2-40 x^3\right )\right ) \log (4 x)+\left (20 e^5 x^2+20 x^3\right ) \log ^2(4 x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\log (4 x) \left (x-\left (e^5+x\right ) \log \left (e^5+x\right )\right )+(-1+x) \left (x \left (-1+20 e^5+20 x\right )+2 \left (e^5+x\right ) \log \left (e^5+x\right )\right )}{20 x^2 \left (e^5+x\right ) (1-x+\log (4 x))^2} \, dx\\ &=\frac {1}{20} \int \frac {\log (4 x) \left (x-\left (e^5+x\right ) \log \left (e^5+x\right )\right )+(-1+x) \left (x \left (-1+20 e^5+20 x\right )+2 \left (e^5+x\right ) \log \left (e^5+x\right )\right )}{x^2 \left (e^5+x\right ) (1-x+\log (4 x))^2} \, dx\\ &=\frac {1}{20} \int \left (\frac {1-20 e^5-21 \left (1-\frac {20 e^5}{21}\right ) x+20 x^2+\log (4 x)}{x \left (e^5+x\right ) (1-x+\log (4 x))^2}+\frac {(-2+2 x-\log (4 x)) \log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2}\right ) \, dx\\ &=\frac {1}{20} \int \frac {1-20 e^5-21 \left (1-\frac {20 e^5}{21}\right ) x+20 x^2+\log (4 x)}{x \left (e^5+x\right ) (1-x+\log (4 x))^2} \, dx+\frac {1}{20} \int \frac {(-2+2 x-\log (4 x)) \log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2} \, dx\\ &=\frac {1}{20} \int \left (\frac {20 (-1+x)}{x (-1+x-\log (4 x))^2}-\frac {1}{x \left (e^5+x\right ) (-1+x-\log (4 x))}\right ) \, dx+\frac {1}{20} \int \left (-\frac {2 \log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2}+\frac {2 \log \left (e^5+x\right )}{x (-1+x-\log (4 x))^2}-\frac {\log (4 x) \log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2}\right ) \, dx\\ &=-\left (\frac {1}{20} \int \frac {1}{x \left (e^5+x\right ) (-1+x-\log (4 x))} \, dx\right )-\frac {1}{20} \int \frac {\log (4 x) \log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2} \, dx-\frac {1}{10} \int \frac {\log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2} \, dx+\frac {1}{10} \int \frac {\log \left (e^5+x\right )}{x (-1+x-\log (4 x))^2} \, dx+\int \frac {-1+x}{x (-1+x-\log (4 x))^2} \, dx\\ &=\frac {1}{1-x+\log (4 x)}-\frac {1}{20} \int \left (\frac {1}{e^5 x (-1+x-\log (4 x))}-\frac {1}{e^5 \left (e^5+x\right ) (-1+x-\log (4 x))}\right ) \, dx-\frac {1}{20} \int \frac {\log (4 x) \log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2} \, dx-\frac {1}{10} \int \frac {\log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2} \, dx+\frac {1}{10} \int \frac {\log \left (e^5+x\right )}{x (-1+x-\log (4 x))^2} \, dx\\ &=\frac {1}{1-x+\log (4 x)}-\frac {1}{20} \int \frac {\log (4 x) \log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2} \, dx-\frac {1}{10} \int \frac {\log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2} \, dx+\frac {1}{10} \int \frac {\log \left (e^5+x\right )}{x (-1+x-\log (4 x))^2} \, dx-\frac {\int \frac {1}{x (-1+x-\log (4 x))} \, dx}{20 e^5}+\frac {\int \frac {1}{\left (e^5+x\right ) (-1+x-\log (4 x))} \, dx}{20 e^5}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.80, size = 29, normalized size = 1.07 \begin {gather*} \frac {20 x+\log \left (e^5+x\right )}{20 \left (x-x^2+x \log (4 x)\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.83, size = 27, normalized size = 1.00 \begin {gather*} -\frac {20 \, x + \log \left (x + e^{5}\right )}{20 \, {\left (x^{2} - x \log \left (4 \, x\right ) - x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.34, size = 27, normalized size = 1.00 \begin {gather*} -\frac {20 \, x + \log \left (x + e^{5}\right )}{20 \, {\left (x^{2} - x \log \left (4 \, x\right ) - x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.47, size = 36, normalized size = 1.33
method | result | size |
risch | \(-\frac {\ln \left ({\mathrm e}^{5}+x \right )}{20 x \left (x -\ln \left (4 x \right )-1\right )}-\frac {1}{x -\ln \left (4 x \right )-1}\) | \(36\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.99, size = 31, normalized size = 1.15 \begin {gather*} -\frac {20 \, x + \log \left (x + e^{5}\right )}{20 \, {\left (x^{2} - x {\left (2 \, \log \relax (2) + 1\right )} - x \log \relax (x)\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.37, size = 25, normalized size = 0.93 \begin {gather*} \frac {20\,x+\ln \left (x+{\mathrm {e}}^5\right )}{20\,x\,\left (\ln \left (4\,x\right )-x+1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.55, size = 32, normalized size = 1.19 \begin {gather*} - \frac {\log {\left (x + e^{5} \right )}}{20 x^{2} - 20 x \log {\left (4 x \right )} - 20 x} + \frac {1}{- x + \log {\left (4 x \right )} + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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