Optimal. Leaf size=16 \[ -x+5 x (e+x)^{\frac {1}{e^2}} \log (4) \]
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Rubi [A] time = 0.37, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {12, 1593, 6688, 34} \begin {gather*} 5 x (x+e)^{\frac {1}{e^2}} \log (4)-x \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 34
Rule 1593
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^2 \left (-e x-x^2\right )+e^{\frac {\log (e+x)+e^2 \log (5 x \log (4))}{e^2}} \left (x+e^2 (e+x)\right )}{e x+x^2} \, dx}{e^2}\\ &=\frac {\int \frac {e^2 \left (-e x-x^2\right )+e^{\frac {\log (e+x)+e^2 \log (5 x \log (4))}{e^2}} \left (x+e^2 (e+x)\right )}{x (e+x)} \, dx}{e^2}\\ &=\frac {\int \left (-e^2+5 (e+x)^{-1+\frac {1}{e^2}} \left (e^3+\left (1+e^2\right ) x\right ) \log (4)\right ) \, dx}{e^2}\\ &=-x+\frac {(5 \log (4)) \int (e+x)^{-1+\frac {1}{e^2}} \left (e^3+\left (1+e^2\right ) x\right ) \, dx}{e^2}\\ &=-x+5 x (e+x)^{\frac {1}{e^2}} \log (4)\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.09, size = 15, normalized size = 0.94 \begin {gather*} x \left (-1+5 (e+x)^{\frac {1}{e^2}} \log (4)\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.89, size = 23, normalized size = 1.44 \begin {gather*} -x + e^{\left ({\left (e^{2} \log \left (10 \, x \log \relax (2)\right ) + \log \left (x + e\right )\right )} e^{\left (-2\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.22, size = 23, normalized size = 1.44 \begin {gather*} {\left (10 \, {\left (x + e\right )}^{e^{\left (-2\right )}} x e^{2} \log \relax (2) - x e^{2}\right )} e^{\left (-2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.23, size = 17, normalized size = 1.06
method | result | size |
risch | \(-x +10 x \ln \relax (2) \left (x +{\mathrm e}\right )^{{\mathrm e}^{-2}}\) | \(17\) |
norman | \(-x +{\mathrm e}^{\left ({\mathrm e}^{2} \ln \left (10 x \ln \relax (2)\right )+\ln \left (x +{\mathrm e}\right )\right ) {\mathrm e}^{-2}}\) | \(26\) |
default | \({\mathrm e}^{-2} \left ({\mathrm e}^{2} {\mathrm e}^{\left ({\mathrm e}^{2} \ln \left (10 x \ln \relax (2)\right )+\ln \left (x +{\mathrm e}\right )\right ) {\mathrm e}^{-2}}-{\mathrm e}^{2} x \right )\) | \(36\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.38, size = 173, normalized size = 10.81 \begin {gather*} {\left ({\left (e \log \left (x + e\right ) - x\right )} e^{2} + 10 \, {\left (\frac {{\left (x + e\right )} e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 4\right )}}{e^{2} + 1} - \frac {e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 7\right )}}{e^{2} + 1} - \frac {e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 5\right )}}{e^{2} + 1}\right )} \log \relax (2) + 10 \, {\left (\frac {{\left (x + e\right )} e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 2\right )}}{e^{2} + 1} - \frac {e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 5\right )}}{e^{2} + 1} - \frac {e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 3\right )}}{e^{2} + 1}\right )} \log \relax (2) + 10 \, e^{\left (e^{\left (-2\right )} \log \left (x + e\right ) + 5\right )} \log \relax (2) - e^{3} \log \left (x + e\right )\right )} e^{\left (-2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.95, size = 15, normalized size = 0.94 \begin {gather*} x\,\left (10\,\ln \relax (2)\,{\left (x+\mathrm {e}\right )}^{{\mathrm {e}}^{-2}}-1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \left (- \frac {e^{3}}{x + e}\right )\, dx + \int \left (- \frac {x e^{2}}{x + e}\right )\, dx + \int \frac {10 x \left (x + e\right )^{e^{-2}} \log {\relax (2 )}}{x + e}\, dx + \int \frac {10 \left (x + e\right )^{e^{-2}} e^{3} \log {\relax (2 )}}{x + e}\, dx + \int \frac {10 x \left (x + e\right )^{e^{-2}} e^{2} \log {\relax (2 )}}{x + e}\, dx}{e^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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