Optimal. Leaf size=16 \[ e+\frac {4-\log (\log (x))}{25 x} \]
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Rubi [C] time = 0.18, antiderivative size = 53, normalized size of antiderivative = 3.31, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {12, 6742, 2309, 2178, 2366, 6482, 2522} \begin {gather*} \frac {4}{25} \log (x) \text {Ei}(-\log (x))-\frac {1}{25} (4 \log (x)+1) \text {Ei}(-\log (x))+\frac {\text {Ei}(-\log (x))}{25}+\frac {4}{25 x}-\frac {\log (\log (x))}{25 x} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2178
Rule 2309
Rule 2366
Rule 2522
Rule 6482
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{25} \int \frac {-1-4 \log (x)+\log (x) \log (\log (x))}{x^2 \log (x)} \, dx\\ &=\frac {1}{25} \int \left (\frac {-1-4 \log (x)}{x^2 \log (x)}+\frac {\log (\log (x))}{x^2}\right ) \, dx\\ &=\frac {1}{25} \int \frac {-1-4 \log (x)}{x^2 \log (x)} \, dx+\frac {1}{25} \int \frac {\log (\log (x))}{x^2} \, dx\\ &=-\frac {1}{25} \text {Ei}(-\log (x)) (1+4 \log (x))-\frac {\log (\log (x))}{25 x}+\frac {1}{25} \int \frac {1}{x^2 \log (x)} \, dx+\frac {4}{25} \int \frac {\text {Ei}(-\log (x))}{x} \, dx\\ &=-\frac {1}{25} \text {Ei}(-\log (x)) (1+4 \log (x))-\frac {\log (\log (x))}{25 x}+\frac {1}{25} \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right )+\frac {4}{25} \operatorname {Subst}(\int \text {Ei}(-x) \, dx,x,\log (x))\\ &=\frac {4}{25 x}+\frac {\text {Ei}(-\log (x))}{25}+\frac {4}{25} \text {Ei}(-\log (x)) \log (x)-\frac {1}{25} \text {Ei}(-\log (x)) (1+4 \log (x))-\frac {\log (\log (x))}{25 x}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.01, size = 18, normalized size = 1.12 \begin {gather*} \frac {4}{25 x}-\frac {\log (\log (x))}{25 x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.66, size = 10, normalized size = 0.62 \begin {gather*} -\frac {\log \left (\log \relax (x)\right ) - 4}{25 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.21, size = 14, normalized size = 0.88 \begin {gather*} -\frac {\log \left (\log \relax (x)\right )}{25 \, x} + \frac {4}{25 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 12, normalized size = 0.75
method | result | size |
norman | \(\frac {\frac {4}{25}-\frac {\ln \left (\ln \relax (x )\right )}{25}}{x}\) | \(12\) |
risch | \(-\frac {\ln \left (\ln \relax (x )\right )}{25 x}+\frac {4}{25 x}\) | \(15\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.37, size = 14, normalized size = 0.88 \begin {gather*} -\frac {\log \left (\log \relax (x)\right )}{25 \, x} + \frac {4}{25 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.37, size = 10, normalized size = 0.62 \begin {gather*} -\frac {\ln \left (\ln \relax (x)\right )-4}{25\,x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.34, size = 12, normalized size = 0.75 \begin {gather*} - \frac {\log {\left (\log {\relax (x )} \right )}}{25 x} + \frac {4}{25 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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