Optimal. Leaf size=31 \[ x \left (-3+e^5+x+\frac {e}{\log \left (-2+(1-x) \left (-x+\frac {\log (x)}{e^5}\right )\right )}\right ) \]
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Rubi [F] time = 2.32, 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 (1-x)+e^6 \left (-x+2 x^2\right )-e x \log (x)+\left (e^6 \left (2+x-x^2\right )+e (-1+x) \log (x)\right ) \log \left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )+\left (e^{10} \left (2+x-x^2\right )+e^5 \left (-6+x+5 x^2-2 x^3\right )+\left (3+e^5 (-1+x)-5 x+2 x^2\right ) \log (x)\right ) \log ^2\left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )}{\left (e^5 \left (2+x-x^2\right )+(-1+x) \log (x)\right ) \log ^2\left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e-e x+e^6 x (-1+2 x)-e x \log (x)+\left (e^6 \left (2+x-x^2\right )+e (-1+x) \log (x)\right ) \log \left (-2-x+x^2-\frac {(-1+x) \log (x)}{e^5}\right )-\left (-3+e^5+2 x\right ) \left (e^5 \left (-2-x+x^2\right )-(-1+x) \log (x)\right ) \log ^2\left (-2-x+x^2-\frac {(-1+x) \log (x)}{e^5}\right )}{\left (e^5 \left (2+x-x^2\right )+(-1+x) \log (x)\right ) \log ^2\left (-2-x+x^2-\frac {(-1+x) \log (x)}{e^5}\right )} \, dx\\ &=\int \left (-3 \left (1-\frac {e^5}{3}\right )+2 x+\frac {e \left (1-\left (1+e^5\right ) x+2 e^5 x^2-x \log (x)\right )}{\left (2 e^5+e^5 x-e^5 x^2-\log (x)+x \log (x)\right ) \log ^2\left (-2-x+x^2-\frac {(-1+x) \log (x)}{e^5}\right )}+\frac {e}{\log \left (-2-x+x^2-\frac {(-1+x) \log (x)}{e^5}\right )}\right ) \, dx\\ &=-\left (\left (3-e^5\right ) x\right )+x^2+e \int \frac {1-\left (1+e^5\right ) x+2 e^5 x^2-x \log (x)}{\left (2 e^5+e^5 x-e^5 x^2-\log (x)+x \log (x)\right ) \log ^2\left (-2-x+x^2-\frac {(-1+x) \log (x)}{e^5}\right )} \, dx+e \int \frac {1}{\log \left (-2-x+x^2-\frac {(-1+x) \log (x)}{e^5}\right )} \, dx\\ &=-\left (\left (3-e^5\right ) x\right )+x^2+e \int \left (-\frac {1}{\left (-2 e^5-e^5 x+e^5 x^2+\log (x)-x \log (x)\right ) \log ^2\left (-2-x+x^2-\frac {(-1+x) \log (x)}{e^5}\right )}+\frac {\left (1+e^5\right ) x}{\left (-2 e^5-e^5 x+e^5 x^2+\log (x)-x \log (x)\right ) \log ^2\left (-2-x+x^2-\frac {(-1+x) \log (x)}{e^5}\right )}-\frac {2 e^5 x^2}{\left (-2 e^5-e^5 x+e^5 x^2+\log (x)-x \log (x)\right ) \log ^2\left (-2-x+x^2-\frac {(-1+x) \log (x)}{e^5}\right )}+\frac {x \log (x)}{\left (-2 e^5-e^5 x+e^5 x^2+\log (x)-x \log (x)\right ) \log ^2\left (-2-x+x^2-\frac {(-1+x) \log (x)}{e^5}\right )}\right ) \, dx+e \int \frac {1}{\log \left (-2-x+x^2-\frac {(-1+x) \log (x)}{e^5}\right )} \, dx\\ &=-\left (\left (3-e^5\right ) x\right )+x^2-e \int \frac {1}{\left (-2 e^5-e^5 x+e^5 x^2+\log (x)-x \log (x)\right ) \log ^2\left (-2-x+x^2-\frac {(-1+x) \log (x)}{e^5}\right )} \, dx+e \int \frac {x \log (x)}{\left (-2 e^5-e^5 x+e^5 x^2+\log (x)-x \log (x)\right ) \log ^2\left (-2-x+x^2-\frac {(-1+x) \log (x)}{e^5}\right )} \, dx+e \int \frac {1}{\log \left (-2-x+x^2-\frac {(-1+x) \log (x)}{e^5}\right )} \, dx-\left (2 e^6\right ) \int \frac {x^2}{\left (-2 e^5-e^5 x+e^5 x^2+\log (x)-x \log (x)\right ) \log ^2\left (-2-x+x^2-\frac {(-1+x) \log (x)}{e^5}\right )} \, dx+\left (e \left (1+e^5\right )\right ) \int \frac {x}{\left (-2 e^5-e^5 x+e^5 x^2+\log (x)-x \log (x)\right ) \log ^2\left (-2-x+x^2-\frac {(-1+x) \log (x)}{e^5}\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.12, size = 31, normalized size = 1.00 \begin {gather*} x \left (-3+e^5+x+\frac {e}{\log \left (-2-x+x^2-\frac {(-1+x) \log (x)}{e^5}\right )}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.62, size = 66, normalized size = 2.13 \begin {gather*} \frac {x e + {\left (x^{2} + x e^{5} - 3 \, x\right )} \log \left ({\left ({\left (x^{2} - x - 2\right )} e^{5} - {\left (x - 1\right )} \log \relax (x)\right )} e^{\left (-5\right )}\right )}{\log \left ({\left ({\left (x^{2} - x - 2\right )} e^{5} - {\left (x - 1\right )} \log \relax (x)\right )} e^{\left (-5\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.82, size = 130, normalized size = 4.19 \begin {gather*} \frac {x^{2} \log \left (x^{2} e^{5} - x e^{5} - x \log \relax (x) - 2 \, e^{5} + \log \relax (x)\right ) + x e^{5} \log \left (x^{2} e^{5} - x e^{5} - x \log \relax (x) - 2 \, e^{5} + \log \relax (x)\right ) - 5 \, x^{2} - 5 \, x e^{5} + x e - 3 \, x \log \left (x^{2} e^{5} - x e^{5} - x \log \relax (x) - 2 \, e^{5} + \log \relax (x)\right ) + 15 \, x}{\log \left (x^{2} e^{5} - x e^{5} - x \log \relax (x) - 2 \, e^{5} + \log \relax (x)\right ) - 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 42, normalized size = 1.35
method | result | size |
risch | \(x \,{\mathrm e}^{5}+x^{2}-3 x +\frac {x \,{\mathrm e}}{\ln \left (\left (\left (1-x \right ) \ln \relax (x )+\left (x^{2}-x -2\right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-5}\right )}\) | \(42\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.43, size = 84, normalized size = 2.71 \begin {gather*} -\frac {5 \, x^{2} + x {\left (5 \, e^{5} - e - 15\right )} - {\left (x^{2} + x {\left (e^{5} - 3\right )}\right )} \log \left (x^{2} e^{5} - x e^{5} - {\left (x - 1\right )} \log \relax (x) - 2 \, e^{5}\right )}{\log \left (x^{2} e^{5} - x e^{5} - {\left (x - 1\right )} \log \relax (x) - 2 \, e^{5}\right ) - 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.82, size = 165, normalized size = 5.32 \begin {gather*} \frac {2\,{\mathrm {e}}^6+x\,{\mathrm {e}}^6-\ln \relax (x)\,\left (\mathrm {e}-x\,\mathrm {e}\right )-x^2\,{\mathrm {e}}^6}{\ln \relax (x)+\frac {x+x\,{\mathrm {e}}^5-2\,x^2\,{\mathrm {e}}^5-1}{x}}+\frac {x\,\mathrm {e}-\frac {x\,\mathrm {e}\,\ln \left (-{\mathrm {e}}^{-5}\,\left (\ln \relax (x)\,\left (x-1\right )+{\mathrm {e}}^5\,\left (-x^2+x+2\right )\right )\right )\,\left (2\,{\mathrm {e}}^5-\ln \relax (x)+x\,{\mathrm {e}}^5-x^2\,{\mathrm {e}}^5+x\,\ln \relax (x)\right )}{x+x\,{\mathrm {e}}^5-2\,x^2\,{\mathrm {e}}^5+x\,\ln \relax (x)-1}}{\ln \left (-{\mathrm {e}}^{-5}\,\left (\ln \relax (x)\,\left (x-1\right )+{\mathrm {e}}^5\,\left (-x^2+x+2\right )\right )\right )}+x\,\left ({\mathrm {e}}^5-3\right )+x^2 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.67, size = 36, normalized size = 1.16 \begin {gather*} x^{2} + x \left (-3 + e^{5}\right ) + \frac {e x}{\log {\left (\frac {\left (1 - x\right ) \log {\relax (x )} + \left (x^{2} - x - 2\right ) e^{5}}{e^{5}} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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