Optimal. Leaf size=15 \[ x \log \left (\log \left (e^x-x-\log (3)\right )\right ) \]
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Rubi [F] time = 1.50, 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 {-x+e^x x+\left (e^x-x-\log (3)\right ) \log \left (e^x-x-\log (3)\right ) \log \left (\log \left (e^x-x-\log (3)\right )\right )}{\left (e^x-x-\log (3)\right ) \log \left (e^x-x-\log (3)\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {\left (-1+e^x\right ) x}{\left (-e^x+x+\log (3)\right ) \log \left (e^x-x-\log (3)\right )}+\log \left (\log \left (e^x-x-\log (3)\right )\right )\right ) \, dx\\ &=-\int \frac {\left (-1+e^x\right ) x}{\left (-e^x+x+\log (3)\right ) \log \left (e^x-x-\log (3)\right )} \, dx+\int \log \left (\log \left (e^x-x-\log (3)\right )\right ) \, dx\\ &=-\int \left (-\frac {x}{\log \left (e^x-x-\log (3)\right )}+\frac {x (-1+x+\log (3))}{\left (-e^x+x+\log (3)\right ) \log \left (e^x-x-\log (3)\right )}\right ) \, dx+\int \log \left (\log \left (e^x-x-\log (3)\right )\right ) \, dx\\ &=\int \frac {x}{\log \left (e^x-x-\log (3)\right )} \, dx-\int \frac {x (-1+x+\log (3))}{\left (-e^x+x+\log (3)\right ) \log \left (e^x-x-\log (3)\right )} \, dx+\int \log \left (\log \left (e^x-x-\log (3)\right )\right ) \, dx\\ &=-\int \left (\frac {x^2}{\left (-e^x+x+\log (3)\right ) \log \left (e^x-x-\log (3)\right )}+\frac {x (-1+\log (3))}{\left (-e^x+x+\log (3)\right ) \log \left (e^x-x-\log (3)\right )}\right ) \, dx+\int \frac {x}{\log \left (e^x-x-\log (3)\right )} \, dx+\int \log \left (\log \left (e^x-x-\log (3)\right )\right ) \, dx\\ &=-\left ((-1+\log (3)) \int \frac {x}{\left (-e^x+x+\log (3)\right ) \log \left (e^x-x-\log (3)\right )} \, dx\right )+\int \frac {x}{\log \left (e^x-x-\log (3)\right )} \, dx-\int \frac {x^2}{\left (-e^x+x+\log (3)\right ) \log \left (e^x-x-\log (3)\right )} \, dx+\int \log \left (\log \left (e^x-x-\log (3)\right )\right ) \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.24, size = 15, normalized size = 1.00 \begin {gather*} x \log \left (\log \left (e^x-x-\log (3)\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.54, size = 14, normalized size = 0.93 \begin {gather*} x \log \left (\log \left (-x + e^{x} - \log \relax (3)\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x - e^{x} + \log \relax (3)\right )} \log \left (-x + e^{x} - \log \relax (3)\right ) \log \left (\log \left (-x + e^{x} - \log \relax (3)\right )\right ) - x e^{x} + x}{{\left (x - e^{x} + \log \relax (3)\right )} \log \left (-x + e^{x} - \log \relax (3)\right )}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 15, normalized size = 1.00
method | result | size |
risch | \(\ln \left (\ln \left ({\mathrm e}^{x}-\ln \relax (3)-x \right )\right ) x\) | \(15\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.49, size = 14, normalized size = 0.93 \begin {gather*} x \log \left (\log \left (-x + e^{x} - \log \relax (3)\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.33, size = 14, normalized size = 0.93 \begin {gather*} x\,\ln \left (\ln \left ({\mathrm {e}}^x-\ln \relax (3)-x\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.40, size = 12, normalized size = 0.80 \begin {gather*} x \log {\left (\log {\left (- x + e^{x} - \log {\relax (3 )} \right )} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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