Optimal. Leaf size=27 \[ x \left (-5-x+\log \left (2 x \left (2+\log \left (5 \left (2+\frac {4}{e^{25}+x}\right )\right )\right )\right )\right ) \]
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Rubi [F] time = 4.72, 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^{50} (-8-4 x)-18 x-16 x^2-4 x^3+e^{25} \left (-16-24 x-8 x^2\right )+\left (e^{50} (-4-2 x)-8 x-8 x^2-2 x^3+e^{25} \left (-8-12 x-4 x^2\right )\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )+\left (2 e^{50}+4 x+2 x^2+e^{25} (4+4 x)+\left (e^{50}+2 x+x^2+e^{25} (2+2 x)\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )\right ) \log \left (4 x+2 x \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )\right )}{2 e^{50}+4 x+2 x^2+e^{25} (4+4 x)+\left (e^{50}+2 x+x^2+e^{25} (2+2 x)\right ) \log \left (\frac {20+10 e^{25}+10 x}{e^{25}+x}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-18 x-16 x^2-4 x^3-4 e^{50} (2+x)-8 e^{25} \left (2+3 x+x^2\right )-2 \left (e^{50} (2+x)+x (2+x)^2+2 e^{25} \left (2+3 x+x^2\right )\right ) \log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )+\left (e^{50}+2 e^{25} (1+x)+x (2+x)\right ) \left (2+\log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )\right ) \log \left (2 x \left (2+\log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )\right )\right )}{\left (e^{25} \left (2+e^{25}\right )+2 \left (1+e^{25}\right ) x+x^2\right ) \left (2+\log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )\right )} \, dx\\ &=\int \left (-\frac {18 x}{\left (e^{25}+x\right ) \left (2+e^{25}+x\right ) \left (2+\log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )\right )}-\frac {16 x^2}{\left (e^{25}+x\right ) \left (2+e^{25}+x\right ) \left (2+\log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )\right )}-\frac {4 x^3}{\left (e^{25}+x\right ) \left (2+e^{25}+x\right ) \left (2+\log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )\right )}-\frac {4 e^{50} (2+x)}{\left (e^{25}+x\right ) \left (2+e^{25}+x\right ) \left (2+\log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )\right )}-\frac {8 e^{25} (1+x) (2+x)}{\left (e^{25}+x\right ) \left (2+e^{25}+x\right ) \left (2+\log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )\right )}-\frac {2 (2+x) \log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )}{2+\log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )}+\log \left (2 x \left (2+\log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )\right )\right )\right ) \, dx\\ &=-\left (2 \int \frac {(2+x) \log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )}{2+\log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )} \, dx\right )-4 \int \frac {x^3}{\left (e^{25}+x\right ) \left (2+e^{25}+x\right ) \left (2+\log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )\right )} \, dx-16 \int \frac {x^2}{\left (e^{25}+x\right ) \left (2+e^{25}+x\right ) \left (2+\log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )\right )} \, dx-18 \int \frac {x}{\left (e^{25}+x\right ) \left (2+e^{25}+x\right ) \left (2+\log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )\right )} \, dx-\left (8 e^{25}\right ) \int \frac {(1+x) (2+x)}{\left (e^{25}+x\right ) \left (2+e^{25}+x\right ) \left (2+\log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )\right )} \, dx-\left (4 e^{50}\right ) \int \frac {2+x}{\left (e^{25}+x\right ) \left (2+e^{25}+x\right ) \left (2+\log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )\right )} \, dx+\int \log \left (2 x \left (2+\log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )\right )\right ) \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.27, size = 33, normalized size = 1.22 \begin {gather*} -5 x-x^2+x \log \left (2 x \left (2+\log \left (\frac {10 \left (2+e^{25}+x\right )}{e^{25}+x}\right )\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.03, size = 33, normalized size = 1.22 \begin {gather*} -x^{2} + x \log \left (2 \, x \log \left (\frac {10 \, {\left (x + e^{25} + 2\right )}}{x + e^{25}}\right ) + 4 \, x\right ) - 5 \, x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 6.71, size = 254, normalized size = 9.41 \begin {gather*} -e^{50} \log \relax (2) \log \left (x + e^{25} + 2\right ) - e^{25} \log \relax (2) \log \left (x + e^{25} + 2\right ) - e^{50} \log \relax (2) \log \left (x + e^{25}\right ) + e^{25} \log \relax (2) \log \left (x + e^{25}\right ) + e^{50} \log \relax (2) \log \left (-x - e^{25}\right ) - e^{25} \log \relax (2) \log \left (-x - e^{25}\right ) + e^{50} \log \relax (2) \log \left (-x - e^{25} - 2\right ) + e^{25} \log \relax (2) \log \left (-x - e^{25} - 2\right ) - x^{2} + x \log \relax (2) + x \log \left (x \log \left (\frac {10 \, {\left (x + e^{25} + 2\right )}}{x + e^{25}}\right ) + 2 \, x\right ) - 2 \, e^{75} \log \left (x + e^{25} + 2\right ) - 2 \, e^{50} \log \left (x + e^{25} + 2\right ) - 2 \, e^{75} \log \left (x + e^{25}\right ) + 6 \, e^{50} \log \left (x + e^{25}\right ) - 4 \, e^{25} \log \left (x + e^{25}\right ) + 2 \, e^{75} \log \left (-x - e^{25}\right ) - 6 \, e^{50} \log \left (-x - e^{25}\right ) + 4 \, e^{25} \log \left (-x - e^{25}\right ) + 2 \, e^{75} \log \left (-x - e^{25} - 2\right ) + 2 \, e^{50} \log \left (-x - e^{25} - 2\right ) - 5 \, x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.79, size = 0, normalized size = 0.00 \[\int \frac {\left (\left ({\mathrm e}^{50}+\left (2 x +2\right ) {\mathrm e}^{25}+x^{2}+2 x \right ) \ln \left (\frac {10 \,{\mathrm e}^{25}+10 x +20}{{\mathrm e}^{25}+x}\right )+2 \,{\mathrm e}^{50}+\left (4 x +4\right ) {\mathrm e}^{25}+2 x^{2}+4 x \right ) \ln \left (2 x \ln \left (\frac {10 \,{\mathrm e}^{25}+10 x +20}{{\mathrm e}^{25}+x}\right )+4 x \right )+\left (\left (-2 x -4\right ) {\mathrm e}^{50}+\left (-4 x^{2}-12 x -8\right ) {\mathrm e}^{25}-2 x^{3}-8 x^{2}-8 x \right ) \ln \left (\frac {10 \,{\mathrm e}^{25}+10 x +20}{{\mathrm e}^{25}+x}\right )+\left (-4 x -8\right ) {\mathrm e}^{50}+\left (-8 x^{2}-24 x -16\right ) {\mathrm e}^{25}-4 x^{3}-16 x^{2}-18 x}{\left ({\mathrm e}^{50}+\left (2 x +2\right ) {\mathrm e}^{25}+x^{2}+2 x \right ) \ln \left (\frac {10 \,{\mathrm e}^{25}+10 x +20}{{\mathrm e}^{25}+x}\right )+2 \,{\mathrm e}^{50}+\left (4 x +4\right ) {\mathrm e}^{25}+2 x^{2}+4 x}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.70, size = 95, normalized size = 3.52 \begin {gather*} -x^{2} + x {\left (\log \relax (2) - 5\right )} + x \log \relax (x) + {\left (x - 4 \, e^{50} - 8 \, e^{25}\right )} \log \left (\log \relax (5) + \log \relax (2) + \log \left (x + e^{25} + 2\right ) - \log \left (x + e^{25}\right ) + 2\right ) + 4 \, e^{50} \log \left (\log \relax (5) + \log \relax (2) + \log \left (x + e^{25} + 2\right ) - \log \left (x + e^{25}\right ) + 2\right ) + 8 \, e^{25} \log \left (\log \relax (5) + \log \relax (2) + \log \left (x + e^{25} + 2\right ) - \log \left (x + e^{25}\right ) + 2\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.04, size = 33, normalized size = 1.22 \begin {gather*} -x\,\left (x-\ln \left (4\,x+2\,x\,\ln \left (\frac {10\,x+10\,{\mathrm {e}}^{25}+20}{x+{\mathrm {e}}^{25}}\right )\right )+5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: CoercionFailed} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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