Optimal. Leaf size=35 \[ \left (-e^{-x+\frac {9 \left (2+\frac {2 x}{3}\right )}{4 \log (2-x)}}+\frac {x}{2}\right )^2 \]
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Rubi [F] time = 7.35, 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^{-2 x} \left (e^{2 x} \left (-2 x+x^2\right ) \log ^2(2-x)+e^{\frac {9+3 x}{\log (2-x)}} \left (-18-6 x+(-12+6 x) \log (2-x)+(8-4 x) \log ^2(2-x)\right )+e^{\frac {9+3 x}{2 \log (2-x)}} \left (e^x \left (9 x+3 x^2\right )+e^x \left (6 x-3 x^2\right ) \log (2-x)+e^x \left (4-6 x+2 x^2\right ) \log ^2(2-x)\right )\right )}{(-4+2 x) \log ^2(2-x)} \, 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 {x}{2}+\frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} \left (9+3 x+6 \log (2-x)-3 x \log (2-x)-4 \log ^2(2-x)+2 x \log ^2(2-x)\right )}{(2-x) \log ^2(2-x)}+\frac {\exp \left (-2 x+\frac {9+3 x+2 x \log (2-x)}{2 \log (2-x)}\right ) \left (-9 x-3 x^2-6 x \log (2-x)+3 x^2 \log (2-x)-4 \log ^2(2-x)+6 x \log ^2(2-x)-2 x^2 \log ^2(2-x)\right )}{2 (2-x) \log ^2(2-x)}\right ) \, dx\\ &=\frac {x^2}{4}+\frac {1}{2} \int \frac {\exp \left (-2 x+\frac {9+3 x+2 x \log (2-x)}{2 \log (2-x)}\right ) \left (-9 x-3 x^2-6 x \log (2-x)+3 x^2 \log (2-x)-4 \log ^2(2-x)+6 x \log ^2(2-x)-2 x^2 \log ^2(2-x)\right )}{(2-x) \log ^2(2-x)} \, dx+\int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} \left (9+3 x+6 \log (2-x)-3 x \log (2-x)-4 \log ^2(2-x)+2 x \log ^2(2-x)\right )}{(2-x) \log ^2(2-x)} \, dx\\ &=\frac {x^2}{4}+\frac {1}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} \left (-3 x (3+x)+3 (-2+x) x \log (2-x)-2 \left (2-3 x+x^2\right ) \log ^2(2-x)\right )}{(2-x) \log ^2(2-x)} \, dx+\int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} \left (3 (3+x)-3 (-2+x) \log (2-x)+2 (-2+x) \log ^2(2-x)\right )}{(2-x) \log ^2(2-x)} \, dx\\ &=\frac {x^2}{4}+\frac {1}{2} \int \left (2 e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} (-1+x)+\frac {3 e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x (3+x)}{(-2+x) \log ^2(2-x)}-\frac {3 e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x}{\log (2-x)}\right ) \, dx+\int \left (-2 e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}-\frac {3 e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} (3+x)}{(-2+x) \log ^2(2-x)}+\frac {3 e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log (2-x)}\right ) \, dx\\ &=\frac {x^2}{4}+\frac {3}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x (3+x)}{(-2+x) \log ^2(2-x)} \, dx-\frac {3}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x}{\log (2-x)} \, dx-2 \int e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} \, dx-3 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} (3+x)}{(-2+x) \log ^2(2-x)} \, dx+3 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log (2-x)} \, dx+\int e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} (-1+x) \, dx\\ &=\frac {x^2}{4}+\frac {3}{2} \int \left (\frac {5 e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log ^2(2-x)}+\frac {10 e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{(-2+x) \log ^2(2-x)}+\frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x}{\log ^2(2-x)}\right ) \, dx-\frac {3}{2} \int \left (\frac {2 e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log (2-x)}-\frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} (2-x)}{\log (2-x)}\right ) \, dx-2 \int e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} \, dx-3 \int \left (\frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log ^2(2-x)}+\frac {5 e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{(-2+x) \log ^2(2-x)}\right ) \, dx+3 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log (2-x)} \, dx+\int \left (-e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}+e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x\right ) \, dx\\ &=\frac {x^2}{4}+\frac {3}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x}{\log ^2(2-x)} \, dx+\frac {3}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} (2-x)}{\log (2-x)} \, dx-2 \int e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} \, dx-3 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log ^2(2-x)} \, dx+3 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log (2-x)} \, dx-3 \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log (2-x)} \, dx+\frac {15}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log ^2(2-x)} \, dx-15 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{(-2+x) \log ^2(2-x)} \, dx+15 \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{(-2+x) \log ^2(2-x)} \, dx-\int e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} \, dx+\int e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x \, dx\\ &=\frac {x^2}{4}+\frac {3}{2} \int \left (\frac {2 e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log ^2(2-x)}-\frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} (2-x)}{\log ^2(2-x)}\right ) \, dx+\frac {3}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} (2-x)}{\log (2-x)} \, dx-2 \int e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} \, dx-3 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log ^2(2-x)} \, dx+3 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log (2-x)} \, dx-3 \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log (2-x)} \, dx+\frac {15}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log ^2(2-x)} \, dx-15 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{(-2+x) \log ^2(2-x)} \, dx+15 \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{(-2+x) \log ^2(2-x)} \, dx-\int e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} \, dx+\int e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x \, dx\\ &=\frac {x^2}{4}-\frac {3}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} (2-x)}{\log ^2(2-x)} \, dx+\frac {3}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} (2-x)}{\log (2-x)} \, dx-2 \int e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} \, dx-3 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log ^2(2-x)} \, dx+3 \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log ^2(2-x)} \, dx+3 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log (2-x)} \, dx-3 \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log (2-x)} \, dx+\frac {15}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log ^2(2-x)} \, dx-15 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{(-2+x) \log ^2(2-x)} \, dx+15 \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{(-2+x) \log ^2(2-x)} \, dx-\int e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} \, dx+\int e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.20, size = 36, normalized size = 1.03 \begin {gather*} \frac {1}{4} e^{-2 x} \left (-2 e^{\frac {3 (3+x)}{2 \log (2-x)}}+e^x x\right )^2 \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.73, size = 50, normalized size = 1.43 \begin {gather*} \frac {1}{4} \, {\left (x^{2} e^{\left (2 \, x\right )} - 4 \, x e^{\left (x + \frac {3 \, {\left (x + 3\right )}}{2 \, \log \left (-x + 2\right )}\right )} + 4 \, e^{\left (\frac {3 \, {\left (x + 3\right )}}{\log \left (-x + 2\right )}\right )}\right )} e^{\left (-2 \, x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.61, size = 59, normalized size = 1.69 \begin {gather*} \frac {1}{4} \, x^{2} - x e^{\left (-\frac {2 \, x \log \left (-x + 2\right ) - 3 \, x - 9}{2 \, \log \left (-x + 2\right )}\right )} + e^{\left (-\frac {2 \, x \log \left (-x + 2\right ) - 3 \, x - 9}{\log \left (-x + 2\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.56, size = 60, normalized size = 1.71
method | result | size |
risch | \(\frac {x^{2}}{4}+{\mathrm e}^{-\frac {2 x \ln \left (2-x \right )-3 x -9}{\ln \left (2-x \right )}}-x \,{\mathrm e}^{-\frac {2 x \ln \left (2-x \right )-3 x -9}{2 \ln \left (2-x \right )}}\) | \(60\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {{\mathrm {e}}^{-2\,x}\,\left ({\mathrm {e}}^{\frac {2\,\left (\frac {3\,x}{2}+\frac {9}{2}\right )}{\ln \left (2-x\right )}}\,\left (\left (4\,x-8\right )\,{\ln \left (2-x\right )}^2+\left (12-6\,x\right )\,\ln \left (2-x\right )+6\,x+18\right )-{\mathrm {e}}^{\frac {\frac {3\,x}{2}+\frac {9}{2}}{\ln \left (2-x\right )}}\,\left ({\mathrm {e}}^x\,\left (2\,x^2-6\,x+4\right )\,{\ln \left (2-x\right )}^2+{\mathrm {e}}^x\,\left (6\,x-3\,x^2\right )\,\ln \left (2-x\right )+{\mathrm {e}}^x\,\left (3\,x^2+9\,x\right )\right )+{\mathrm {e}}^{2\,x}\,{\ln \left (2-x\right )}^2\,\left (2\,x-x^2\right )\right )}{{\ln \left (2-x\right )}^2\,\left (2\,x-4\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.84, size = 46, normalized size = 1.31 \begin {gather*} \frac {x^{2}}{4} + \left (- x e^{2 x} e^{\frac {\frac {3 x}{2} + \frac {9}{2}}{\log {\left (2 - x \right )}}} + e^{x} e^{\frac {2 \left (\frac {3 x}{2} + \frac {9}{2}\right )}{\log {\left (2 - x \right )}}}\right ) e^{- 3 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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