Optimal. Leaf size=25 \[ \frac {x (3+x) \left (5+\frac {1}{\log (5)}\right ) \log (\log (25))}{e^2-\frac {x}{2}} \]
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Rubi [B] time = 0.05, antiderivative size = 53, normalized size of antiderivative = 2.12, number of steps used = 4, number of rules used = 3, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {12, 27, 1850} \begin {gather*} \frac {4 e^2 \left (3+2 e^2\right ) (1+5 \log (5)) \log (\log (25))}{\left (2 e^2-x\right ) \log (5)}-\frac {2 x (1+5 \log (5)) \log (\log (25))}{\log (5)} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 27
Rule 1850
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {\log (\log (25)) \int \frac {-2 x^2+e^2 (12+8 x)+\left (-10 x^2+e^2 (60+40 x)\right ) \log (5)}{4 e^4-4 e^2 x+x^2} \, dx}{\log (5)}\\ &=\frac {\log (\log (25)) \int \frac {-2 x^2+e^2 (12+8 x)+\left (-10 x^2+e^2 (60+40 x)\right ) \log (5)}{\left (-2 e^2+x\right )^2} \, dx}{\log (5)}\\ &=\frac {\log (\log (25)) \int \left (-2 (1+5 \log (5))+\frac {4 e^2 \left (3+2 e^2\right ) (1+5 \log (5))}{\left (2 e^2-x\right )^2}\right ) \, dx}{\log (5)}\\ &=\frac {4 e^2 \left (3+2 e^2\right ) (1+5 \log (5)) \log (\log (25))}{\left (2 e^2-x\right ) \log (5)}-\frac {2 x (1+5 \log (5)) \log (\log (25))}{\log (5)}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.03, size = 40, normalized size = 1.60 \begin {gather*} \frac {2 \left (-x-\frac {2 e^2 \left (3+2 e^2\right )}{-2 e^2+x}\right ) (1+5 \log (5)) \log (\log (25))}{\log (5)} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.81, size = 53, normalized size = 2.12 \begin {gather*} -\frac {2 \, {\left (x^{2} - 2 \, {\left (x - 3\right )} e^{2} + 5 \, {\left (x^{2} - 2 \, {\left (x - 3\right )} e^{2} + 4 \, e^{4}\right )} \log \relax (5) + 4 \, e^{4}\right )} \log \left (2 \, \log \relax (5)\right )}{{\left (x - 2 \, e^{2}\right )} \log \relax (5)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.35, size = 36, normalized size = 1.44
method | result | size |
gosper | \(\frac {2 \left (x^{2}+6 \,{\mathrm e}^{2}\right ) \left (5 \ln \relax (5)+1\right ) \ln \left (2 \ln \relax (5)\right )}{\ln \relax (5) \left (2 \,{\mathrm e}^{2}-x \right )}\) | \(36\) |
norman | \(\frac {\frac {2 \left (5 \ln \relax (2) \ln \relax (5)+5 \ln \relax (5) \ln \left (\ln \relax (5)\right )+\ln \relax (2)+\ln \left (\ln \relax (5)\right )\right ) x^{2}}{\ln \relax (5)}+\frac {12 \,{\mathrm e}^{2} \left (5 \ln \relax (2) \ln \relax (5)+5 \ln \relax (5) \ln \left (\ln \relax (5)\right )+\ln \relax (2)+\ln \left (\ln \relax (5)\right )\right )}{\ln \relax (5)}}{2 \,{\mathrm e}^{2}-x}\) | \(68\) |
meijerg | \(-\frac {8 \,{\mathrm e}^{2} \left (-\frac {5 \ln \relax (5)}{2}-\frac {1}{2}\right ) \ln \left (2 \ln \relax (5)\right ) \left (-\frac {x \,{\mathrm e}^{-2} \left (-\frac {3 x \,{\mathrm e}^{-2}}{2}+6\right )}{6 \left (1-\frac {x \,{\mathrm e}^{-2}}{2}\right )}-2 \ln \left (1-\frac {x \,{\mathrm e}^{-2}}{2}\right )\right )}{\ln \relax (5)}+\frac {4 \left (10 \,{\mathrm e}^{2} \ln \relax (5)+2 \,{\mathrm e}^{2}\right ) \ln \left (2 \ln \relax (5)\right ) \left (\frac {x \,{\mathrm e}^{-2}}{2-x \,{\mathrm e}^{-2}}+\ln \left (1-\frac {x \,{\mathrm e}^{-2}}{2}\right )\right )}{\ln \relax (5)}+\frac {15 \,{\mathrm e}^{-2} \ln \left (2 \ln \relax (5)\right ) x}{1-\frac {x \,{\mathrm e}^{-2}}{2}}+\frac {3 \,{\mathrm e}^{-2} \ln \left (2 \ln \relax (5)\right ) x}{\ln \relax (5) \left (1-\frac {x \,{\mathrm e}^{-2}}{2}\right )}\) | \(140\) |
risch | \(-10 x \ln \relax (2)-10 x \ln \left (\ln \relax (5)\right )-\frac {2 x \ln \relax (2)}{\ln \relax (5)}-\frac {2 x \ln \left (\ln \relax (5)\right )}{\ln \relax (5)}+\frac {30 \,{\mathrm e}^{2} \ln \relax (2)}{{\mathrm e}^{2}-\frac {x}{2}}+\frac {30 \,{\mathrm e}^{2} \ln \left (\ln \relax (5)\right )}{{\mathrm e}^{2}-\frac {x}{2}}+\frac {20 \ln \relax (2) {\mathrm e}^{4}}{{\mathrm e}^{2}-\frac {x}{2}}+\frac {20 \,{\mathrm e}^{4} \ln \left (\ln \relax (5)\right )}{{\mathrm e}^{2}-\frac {x}{2}}+\frac {6 \,{\mathrm e}^{2} \ln \relax (2)}{\ln \relax (5) \left ({\mathrm e}^{2}-\frac {x}{2}\right )}+\frac {6 \,{\mathrm e}^{2} \ln \left (\ln \relax (5)\right )}{\ln \relax (5) \left ({\mathrm e}^{2}-\frac {x}{2}\right )}+\frac {4 \ln \relax (2) {\mathrm e}^{4}}{\ln \relax (5) \left ({\mathrm e}^{2}-\frac {x}{2}\right )}+\frac {4 \,{\mathrm e}^{4} \ln \left (\ln \relax (5)\right )}{\ln \relax (5) \left ({\mathrm e}^{2}-\frac {x}{2}\right )}\) | \(164\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.36, size = 52, normalized size = 2.08 \begin {gather*} -\frac {2 \, {\left (x {\left (5 \, \log \relax (5) + 1\right )} + \frac {2 \, {\left (5 \, {\left (2 \, e^{4} + 3 \, e^{2}\right )} \log \relax (5) + 2 \, e^{4} + 3 \, e^{2}\right )}}{x - 2 \, e^{2}}\right )} \log \left (2 \, \log \relax (5)\right )}{\log \relax (5)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.53, size = 42, normalized size = 1.68 \begin {gather*} -\frac {2\,\ln \left (2\,\ln \relax (5)\right )\,\left (5\,\ln \relax (5)+1\right )\,\left (x^2-2\,{\mathrm {e}}^2\,x+6\,{\mathrm {e}}^2+4\,{\mathrm {e}}^4\right )}{\ln \relax (5)\,\left (x-2\,{\mathrm {e}}^2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.62, size = 134, normalized size = 5.36 \begin {gather*} - x \left (\frac {2 \log {\left (\log {\relax (5 )} \right )}}{\log {\relax (5 )}} + \frac {2 \log {\relax (2 )}}{\log {\relax (5 )}} + 10 \log {\left (\log {\relax (5 )} \right )} + 10 \log {\relax (2 )}\right ) - \frac {12 e^{2} \log {\left (\log {\relax (5 )} \right )} + 12 e^{2} \log {\relax (2 )} + 8 e^{4} \log {\left (\log {\relax (5 )} \right )} + 8 e^{4} \log {\relax (2 )} + 60 e^{2} \log {\relax (5 )} \log {\left (\log {\relax (5 )} \right )} + 60 e^{2} \log {\relax (2 )} \log {\relax (5 )} + 40 e^{4} \log {\relax (5 )} \log {\left (\log {\relax (5 )} \right )} + 40 e^{4} \log {\relax (2 )} \log {\relax (5 )}}{x \log {\relax (5 )} - 2 e^{2} \log {\relax (5 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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