Optimal. Leaf size=26 \[ 2+\frac {9 \log \left (e^x-\log (\log (4))\right )}{x^2 \left (-e^3+x\right )} \]
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Rubi [F] time = 2.75, 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 {e^x \left (9 e^3 x-9 x^2\right )+\left (e^x \left (-18 e^3+27 x\right )+\left (18 e^3-27 x\right ) \log (\log (4))\right ) \log \left (e^x-\log (\log (4))\right )}{e^x \left (-e^6 x^3+2 e^3 x^4-x^5\right )+\left (e^6 x^3-2 e^3 x^4+x^5\right ) \log (\log (4))} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {9 \left (-\frac {e^x \left (e^3-x\right ) x}{e^x-\log (\log (4))}+\left (2 e^3-3 x\right ) \log \left (e^x-\log (\log (4))\right )\right )}{\left (e^3-x\right )^2 x^3} \, dx\\ &=9 \int \frac {-\frac {e^x \left (e^3-x\right ) x}{e^x-\log (\log (4))}+\left (2 e^3-3 x\right ) \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right )^2 x^3} \, dx\\ &=9 \int \left (-\frac {\log (\log (4))}{\left (e^3-x\right ) x^2 \left (e^x-\log (\log (4))\right )}+\frac {-e^3 x+x^2+2 e^3 \log \left (e^x-\log (\log (4))\right )-3 x \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right )^2 x^3}\right ) \, dx\\ &=9 \int \frac {-e^3 x+x^2+2 e^3 \log \left (e^x-\log (\log (4))\right )-3 x \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right )^2 x^3} \, dx-(9 \log (\log (4))) \int \frac {1}{\left (e^3-x\right ) x^2 \left (e^x-\log (\log (4))\right )} \, dx\\ &=9 \int \left (-\frac {1}{\left (e^3-x\right ) x^2}+\frac {\left (2 e^3-3 x\right ) \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right )^2 x^3}\right ) \, dx-(9 \log (\log (4))) \int \left (\frac {1}{e^6 \left (e^3-x\right ) \left (e^x-\log (\log (4))\right )}+\frac {1}{e^3 x^2 \left (e^x-\log (\log (4))\right )}+\frac {1}{e^6 x \left (e^x-\log (\log (4))\right )}\right ) \, dx\\ &=-\left (9 \int \frac {1}{\left (e^3-x\right ) x^2} \, dx\right )+9 \int \frac {\left (2 e^3-3 x\right ) \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right )^2 x^3} \, dx-\frac {(9 \log (\log (4))) \int \frac {1}{\left (e^3-x\right ) \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x^2 \left (e^x-\log (\log (4))\right )} \, dx}{e^3}\\ &=-\frac {9 \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right ) x^2}-9 \int \left (\frac {1}{e^6 \left (e^3-x\right )}+\frac {1}{e^3 x^2}+\frac {1}{e^6 x}\right ) \, dx+9 \int \frac {e^x}{\left (e^3-x\right ) x^2 \left (e^x-\log (\log (4))\right )} \, dx-\frac {(9 \log (\log (4))) \int \frac {1}{\left (e^3-x\right ) \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x^2 \left (e^x-\log (\log (4))\right )} \, dx}{e^3}\\ &=\frac {9}{e^3 x}+\frac {9 \log \left (e^3-x\right )}{e^6}-\frac {9 \log (x)}{e^6}-\frac {9 \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right ) x^2}+9 \int \left (\frac {e^{-6+x}}{\left (e^3-x\right ) \left (e^x-\log (\log (4))\right )}+\frac {e^{-3+x}}{x^2 \left (e^x-\log (\log (4))\right )}+\frac {e^{-6+x}}{x \left (e^x-\log (\log (4))\right )}\right ) \, dx-\frac {(9 \log (\log (4))) \int \frac {1}{\left (e^3-x\right ) \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x^2 \left (e^x-\log (\log (4))\right )} \, dx}{e^3}\\ &=\frac {9}{e^3 x}+\frac {9 \log \left (e^3-x\right )}{e^6}-\frac {9 \log (x)}{e^6}-\frac {9 \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right ) x^2}+9 \int \frac {e^{-6+x}}{\left (e^3-x\right ) \left (e^x-\log (\log (4))\right )} \, dx+9 \int \frac {e^{-3+x}}{x^2 \left (e^x-\log (\log (4))\right )} \, dx+9 \int \frac {e^{-6+x}}{x \left (e^x-\log (\log (4))\right )} \, dx-\frac {(9 \log (\log (4))) \int \frac {1}{\left (e^3-x\right ) \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x^2 \left (e^x-\log (\log (4))\right )} \, dx}{e^3}\\ &=\frac {9}{e^3 x}+\frac {9 \log \left (e^3-x\right )}{e^6}-\frac {9 \log (x)}{e^6}-\frac {9 \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right ) x^2}+\frac {9 \int \frac {e^x}{\left (e^3-x\right ) \left (e^x-\log (\log (4))\right )} \, dx}{e^6}+\frac {9 \int \frac {e^x}{x \left (e^x-\log (\log (4))\right )} \, dx}{e^6}+\frac {9 \int \frac {e^x}{x^2 \left (e^x-\log (\log (4))\right )} \, dx}{e^3}-\frac {(9 \log (\log (4))) \int \frac {1}{\left (e^3-x\right ) \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x^2 \left (e^x-\log (\log (4))\right )} \, dx}{e^3}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.52, size = 24, normalized size = 0.92 \begin {gather*} -\frac {9 \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right ) x^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.06, size = 26, normalized size = 1.00 \begin {gather*} \frac {9 \, \log \left (e^{x} - \log \left (2 \, \log \relax (2)\right )\right )}{x^{3} - x^{2} e^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.09, size = 28, normalized size = 1.08 \begin {gather*} \frac {9 \, \log \left (e^{x} - \log \relax (2) - \log \left (\log \relax (2)\right )\right )}{x^{3} - x^{2} e^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.16, size = 27, normalized size = 1.04
method | result | size |
risch | \(-\frac {9 \ln \left (-\ln \relax (2)-\ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )}{x^{2} \left (-x +{\mathrm e}^{3}\right )}\) | \(27\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.69, size = 28, normalized size = 1.08 \begin {gather*} \frac {9 \, \log \left (e^{x} - \log \relax (2) - \log \left (\log \relax (2)\right )\right )}{x^{3} - x^{2} e^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.65, size = 27, normalized size = 1.04 \begin {gather*} -\frac {9\,\ln \left ({\mathrm {e}}^x-\ln \left (2\,\ln \relax (2)\right )\right )}{x^2\,{\mathrm {e}}^3-x^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.35, size = 22, normalized size = 0.85 \begin {gather*} \frac {9 \log {\left (e^{x} - \log {\left (2 \log {\relax (2 )} \right )} \right )}}{x^{3} - x^{2} e^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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