Optimal. Leaf size=24 \[ 2+\frac {5 \log \left (-\frac {e^{-2+x} \log (5) \log (4 x)}{x}\right )}{x} \]
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Rubi [A]
time = 0.30, antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps
used = 16, number of rules used = 8, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.178, Rules used = {6874, 6820,
14, 45, 2346, 2209, 30, 2635} \begin {gather*} \frac {5 \log \left (-\frac {e^{x-2} \log (5) \log (4 x)}{x}\right )}{x} \end {gather*}
Antiderivative was successfully verified.
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Rule 14
Rule 30
Rule 45
Rule 2209
Rule 2346
Rule 2635
Rule 6820
Rule 6874
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {5 (1-\log (4 x)+x \log (4 x))}{x^2 \log (4 x)}-\frac {5 \log \left (-\frac {e^{-2+x} \log (5) \log (4 x)}{x}\right )}{x^2}\right ) \, dx\\ &=5 \int \frac {1-\log (4 x)+x \log (4 x)}{x^2 \log (4 x)} \, dx-5 \int \frac {\log \left (-\frac {e^{-2+x} \log (5) \log (4 x)}{x}\right )}{x^2} \, dx\\ &=\frac {5 \log \left (-\frac {e^{-2+x} \log (5) \log (4 x)}{x}\right )}{x}+5 \int \frac {-1+x+\frac {1}{\log (4 x)}}{x^2} \, dx+5 \int \frac {-1-(-1+x) \log (4 x)}{x^2 \log (4 x)} \, dx\\ &=\frac {5 \log \left (-\frac {e^{-2+x} \log (5) \log (4 x)}{x}\right )}{x}+5 \int \left (\frac {1-x}{x^2}-\frac {1}{x^2 \log (4 x)}\right ) \, dx+5 \int \left (\frac {-1+x}{x^2}+\frac {1}{x^2 \log (4 x)}\right ) \, dx\\ &=\frac {5 \log \left (-\frac {e^{-2+x} \log (5) \log (4 x)}{x}\right )}{x}+5 \int \frac {1-x}{x^2} \, dx+5 \int \frac {-1+x}{x^2} \, dx\\ &=\frac {5 \log \left (-\frac {e^{-2+x} \log (5) \log (4 x)}{x}\right )}{x}+5 \int \left (\frac {1}{x^2}-\frac {1}{x}\right ) \, dx+5 \int \left (-\frac {1}{x^2}+\frac {1}{x}\right ) \, dx\\ &=\frac {5 \log \left (-\frac {e^{-2+x} \log (5) \log (4 x)}{x}\right )}{x}\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.08, size = 24, normalized size = 1.00 \begin {gather*} -5+\frac {5 \log \left (-\frac {e^{-2+x} \log (5) \log (4 x)}{x}\right )}{x} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {-5 \ln \left (4 x \right ) \ln \left (-\frac {\ln \left (5\right ) \ln \left (4 x \right ) {\mathrm e}^{x -2}}{x}\right )+\left (5 x -5\right ) \ln \left (4 x \right )+5}{x^{2} \ln \left (4 x \right )}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.40, size = 21, normalized size = 0.88 \begin {gather*} \frac {5 \, \log \left (-\frac {e^{\left (x - 2\right )} \log \left (5\right ) \log \left (4 \, x\right )}{x}\right )}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.17, size = 20, normalized size = 0.83 \begin {gather*} \frac {5 \log {\left (- \frac {e^{x - 2} \log {\left (5 \right )} \log {\left (4 x \right )}}{x} \right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.44, size = 27, normalized size = 1.12 \begin {gather*} \frac {5 \, \log \left (-\log \left (5\right ) \log \left (4 \, x\right )\right )}{x} - \frac {5 \, \log \left (x\right )}{x} - \frac {10}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 5.69, size = 20, normalized size = 0.83 \begin {gather*} \frac {5\,\ln \left (-\frac {\ln \left (4\,x\right )\,\ln \left (5\right )}{x}\right )-10}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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