Optimal. Leaf size=27 \[ \frac {2 \left (-e^{3 \left (1+e^{e^2}\right )}+\frac {5 x}{\log (\log (5))}\right )}{x^3} \]
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Rubi [A]
time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {12, 45}
\begin {gather*} \frac {10}{x^2 \log (\log (5))}-\frac {2 e^{3+3 e^{e^2}}}{x^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 45
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-20 x+6 e^{3+3 e^{e^2}} \log (\log (5))}{x^4} \, dx}{\log (\log (5))}\\ &=\frac {\int \left (-\frac {20}{x^3}+\frac {6 e^{3+3 e^{e^2}} \log (\log (5))}{x^4}\right ) \, dx}{\log (\log (5))}\\ &=-\frac {2 e^{3+3 e^{e^2}}}{x^3}+\frac {10}{x^2 \log (\log (5))}\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.02, size = 29, normalized size = 1.07 \begin {gather*} 2 \left (-\frac {e^{3+3 e^{e^2}}}{x^3}+\frac {5}{x^2 \log (\log (5))}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.11, size = 30, normalized size = 1.11
method | result | size |
norman | \(\frac {\frac {10 x}{\ln \left (\ln \left (5\right )\right )}-2 \,{\mathrm e}^{3 \,{\mathrm e}^{{\mathrm e}^{2}}} {\mathrm e}^{3}}{x^{3}}\) | \(24\) |
gosper | \(-\frac {2 \left (\ln \left (\ln \left (5\right )\right ) {\mathrm e}^{3 \,{\mathrm e}^{{\mathrm e}^{2}}+3}-5 x \right )}{\ln \left (\ln \left (5\right )\right ) x^{3}}\) | \(27\) |
risch | \(\frac {-2 \ln \left (\ln \left (5\right )\right ) {\mathrm e}^{3 \,{\mathrm e}^{{\mathrm e}^{2}}+3}+10 x}{\ln \left (\ln \left (5\right )\right ) x^{3}}\) | \(27\) |
default | \(\frac {-\frac {2 \ln \left (\ln \left (5\right )\right ) {\mathrm e}^{3 \,{\mathrm e}^{{\mathrm e}^{2}}+3}}{x^{3}}+\frac {10}{x^{2}}}{\ln \left (\ln \left (5\right )\right )}\) | \(30\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.26, size = 26, normalized size = 0.96 \begin {gather*} -\frac {2 \, {\left (e^{\left (3 \, e^{\left (e^{2}\right )} + 3\right )} \log \left (\log \left (5\right )\right ) - 5 \, x\right )}}{x^{3} \log \left (\log \left (5\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.33, size = 26, normalized size = 0.96 \begin {gather*} -\frac {2 \, {\left (e^{\left (3 \, e^{\left (e^{2}\right )} + 3\right )} \log \left (\log \left (5\right )\right ) - 5 \, x\right )}}{x^{3} \log \left (\log \left (5\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.07, size = 31, normalized size = 1.15 \begin {gather*} - \frac {- 10 x + 2 e^{3} e^{3 e^{e^{2}}} \log {\left (\log {\left (5 \right )} \right )}}{x^{3} \log {\left (\log {\left (5 \right )} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.42, size = 26, normalized size = 0.96 \begin {gather*} -\frac {2 \, {\left (e^{\left (3 \, e^{\left (e^{2}\right )} + 3\right )} \log \left (\log \left (5\right )\right ) - 5 \, x\right )}}{x^{3} \log \left (\log \left (5\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.34, size = 24, normalized size = 0.89 \begin {gather*} \frac {10}{x^2\,\ln \left (\ln \left (5\right )\right )}-\frac {2\,{\mathrm {e}}^{3\,{\mathrm {e}}^{{\mathrm {e}}^2}+3}}{x^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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