Optimal. Leaf size=36 \[ \frac {e^{1-x} x (x-\log (2+x))}{\log \left (\log \left (\frac {x^2}{\left (-x+(4+x)^2\right )^2}\right )\right )} \]
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Rubi [F] time = 20.38, 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} \left (-64 x-32 x^2+4 x^3+2 x^4+\left (64+32 x-4 x^2-2 x^3\right ) \log (2+x)+\left (\left (48 x+21 x^2-13 x^3-7 x^4-x^5\right ) \log \left (\frac {x^2}{256+224 x+81 x^2+14 x^3+x^4}\right )+\left (-32+2 x+21 x^2+8 x^3+x^4\right ) \log (2+x) \log \left (\frac {x^2}{256+224 x+81 x^2+14 x^3+x^4}\right )\right ) \log \left (\log \left (\frac {x^2}{256+224 x+81 x^2+14 x^3+x^4}\right )\right )\right )}{\left (32+30 x+9 x^2+x^3\right ) \log \left (\frac {x^2}{256+224 x+81 x^2+14 x^3+x^4}\right ) \log ^2\left (\log \left (\frac {x^2}{256+224 x+81 x^2+14 x^3+x^4}\right )\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{1-x} \left (-64 x-32 x^2+4 x^3+2 x^4-2 \left (-32-16 x+2 x^2+x^3\right ) \log (2+x)-\left (16+7 x+x^2\right ) \left (x \left (-3+x^2\right )-\left (-2+x+x^2\right ) \log (2+x)\right ) \log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right ) \log \left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )\right )}{\left (32+30 x+9 x^2+x^3\right ) \log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right ) \log ^2\left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )} \, dx\\ &=\int \left (\frac {2 e^{1-x} \left (-16+x^2\right ) (x-\log (2+x))}{\left (16+7 x+x^2\right ) \log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right ) \log ^2\left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )}+\frac {e^{1-x} \left (3 x-x^3-2 \log (2+x)+x \log (2+x)+x^2 \log (2+x)\right )}{(2+x) \log \left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )}\right ) \, dx\\ &=2 \int \frac {e^{1-x} \left (-16+x^2\right ) (x-\log (2+x))}{\left (16+7 x+x^2\right ) \log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right ) \log ^2\left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )} \, dx+\int \frac {e^{1-x} \left (3 x-x^3-2 \log (2+x)+x \log (2+x)+x^2 \log (2+x)\right )}{(2+x) \log \left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )} \, dx\\ &=2 \int \left (\frac {e^{1-x} (x-\log (2+x))}{\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right ) \log ^2\left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )}-\frac {e^{1-x} (32+7 x) (x-\log (2+x))}{\left (16+7 x+x^2\right ) \log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right ) \log ^2\left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )}\right ) \, dx+\int \frac {e^{1-x} \left (-x \left (-3+x^2\right )+\left (-2+x+x^2\right ) \log (2+x)\right )}{(2+x) \log \left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )} \, dx\\ &=2 \int \frac {e^{1-x} (x-\log (2+x))}{\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right ) \log ^2\left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )} \, dx-2 \int \frac {e^{1-x} (32+7 x) (x-\log (2+x))}{\left (16+7 x+x^2\right ) \log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right ) \log ^2\left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )} \, dx+\int \left (\frac {3 e^{1-x} x}{(2+x) \log \left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )}-\frac {e^{1-x} x^3}{(2+x) \log \left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )}-\frac {2 e^{1-x} \log (2+x)}{(2+x) \log \left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )}+\frac {e^{1-x} x \log (2+x)}{(2+x) \log \left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )}+\frac {e^{1-x} x^2 \log (2+x)}{(2+x) \log \left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )}\right ) \, dx\\ &=2 \int \left (\frac {e^{1-x} x}{\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right ) \log ^2\left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )}-\frac {e^{1-x} \log (2+x)}{\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right ) \log ^2\left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )}\right ) \, dx-2 \int \left (\frac {32 e^{1-x} x}{\left (16+7 x+x^2\right ) \log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right ) \log ^2\left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )}+\frac {7 e^{1-x} x^2}{\left (16+7 x+x^2\right ) \log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right ) \log ^2\left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )}-\frac {32 e^{1-x} \log (2+x)}{\left (16+7 x+x^2\right ) \log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right ) \log ^2\left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )}-\frac {7 e^{1-x} x \log (2+x)}{\left (16+7 x+x^2\right ) \log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right ) \log ^2\left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )}\right ) \, dx-2 \int \frac {e^{1-x} \log (2+x)}{(2+x) \log \left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )} \, dx+3 \int \frac {e^{1-x} x}{(2+x) \log \left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )} \, dx-\int \frac {e^{1-x} x^3}{(2+x) \log \left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )} \, dx+\int \frac {e^{1-x} x \log (2+x)}{(2+x) \log \left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )} \, dx+\int \frac {e^{1-x} x^2 \log (2+x)}{(2+x) \log \left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.26, size = 35, normalized size = 0.97 \begin {gather*} \frac {e^{1-x} x (x-\log (2+x))}{\log \left (\log \left (\frac {x^2}{\left (16+7 x+x^2\right )^2}\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.90, size = 53, normalized size = 1.47 \begin {gather*} \frac {x^{2} e^{\left (-x + 1\right )} - x e^{\left (-x + 1\right )} \log \left (x + 2\right )}{\log \left (\log \left (\frac {x^{2}}{x^{4} + 14 \, x^{3} + 81 \, x^{2} + 224 \, x + 256}\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*} \mathit {sage}_{0} x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.39, size = 195, normalized size = 5.42
method | result | size |
risch | \(\frac {x \left (x -\ln \left (2+x \right )\right ) {\mathrm e}^{1-x}}{\ln \left (2 \ln \relax (x )-2 \ln \left (x^{2}+7 x +16\right )+\frac {i \pi \,\mathrm {csgn}\left (i \left (x^{2}+7 x +16\right )^{2}\right ) \left (-\mathrm {csgn}\left (i \left (x^{2}+7 x +16\right )^{2}\right )+\mathrm {csgn}\left (i \left (x^{2}+7 x +16\right )\right )\right )^{2}}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i x^{2}}{\left (x^{2}+7 x +16\right )^{2}}\right ) \left (-\mathrm {csgn}\left (\frac {i x^{2}}{\left (x^{2}+7 x +16\right )^{2}}\right )+\mathrm {csgn}\left (i x^{2}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i x^{2}}{\left (x^{2}+7 x +16\right )^{2}}\right )+\mathrm {csgn}\left (\frac {i}{\left (x^{2}+7 x +16\right )^{2}}\right )\right )}{2}\right )}\) | \(195\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 1.06, size = 47, normalized size = 1.31 \begin {gather*} \frac {x^{2} e - x e \log \left (x + 2\right )}{{\left (i \, \pi + \log \relax (2)\right )} e^{x} + e^{x} \log \left (\log \left (x^{2} + 7 \, x + 16\right ) - \log \relax (x)\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.16, size = 82, normalized size = 2.28 \begin {gather*} \frac {x^2\,{\mathrm {e}}^{1-x}}{\ln \left (\ln \left (x^2\right )-\ln \left (x^4+14\,x^3+81\,x^2+224\,x+256\right )\right )}-\frac {x\,\ln \left (x+2\right )\,{\mathrm {e}}^{1-x}}{\ln \left (\ln \left (x^2\right )-\ln \left (x^4+14\,x^3+81\,x^2+224\,x+256\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 3.15, size = 39, normalized size = 1.08 \begin {gather*} \frac {\left (x^{2} - x \log {\left (x + 2 \right )}\right ) e^{1 - x}}{\log {\left (\log {\left (\frac {x^{2}}{x^{4} + 14 x^{3} + 81 x^{2} + 224 x + 256} \right )} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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