Optimal. Leaf size=30 \[ \log \left (\frac {2 \left (x+4 \left (3+x^2 \left (e^{-2-x} x+\log (-3+x)\right )\right )\right )}{x}\right ) \]
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Rubi [F] time = 44.69, 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 {36-12 x+4 x^3+e^{-2-x} x \left (-24 x^2+20 x^3-4 x^4\right )+\left (-12 x^2+4 x^3\right ) \log (-3+x)}{-36 x+9 x^2+x^3+e^{-2-x} x \left (-12 x^3+4 x^4\right )+\left (-12 x^3+4 x^4\right ) \log (-3+x)} \, 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^{2+x} \left (-36+12 x-4 x^3-e^{-2-x} x \left (-24 x^2+20 x^3-4 x^4\right )-\left (-12 x^2+4 x^3\right ) \log (-3+x)\right )}{(3-x) x \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )} \, dx\\ &=\int \left (-\frac {-2+x}{x}+\frac {e^{2+x} \left (108-66 x+7 x^2+5 x^3+12 x^2 \log (-3+x)-16 x^3 \log (-3+x)+4 x^4 \log (-3+x)\right )}{(-3+x) x \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )}\right ) \, dx\\ &=-\int \frac {-2+x}{x} \, dx+\int \frac {e^{2+x} \left (108-66 x+7 x^2+5 x^3+12 x^2 \log (-3+x)-16 x^3 \log (-3+x)+4 x^4 \log (-3+x)\right )}{(-3+x) x \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )} \, dx\\ &=-\int \left (1-\frac {2}{x}\right ) \, dx+\int \left (\frac {e^{2+x} \left (108-66 x+7 x^2+5 x^3+12 x^2 \log (-3+x)-16 x^3 \log (-3+x)+4 x^4 \log (-3+x)\right )}{3 (-3+x) \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )}-\frac {e^{2+x} \left (108-66 x+7 x^2+5 x^3+12 x^2 \log (-3+x)-16 x^3 \log (-3+x)+4 x^4 \log (-3+x)\right )}{3 x \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )}\right ) \, dx\\ &=-x+2 \log (x)+\frac {1}{3} \int \frac {e^{2+x} \left (108-66 x+7 x^2+5 x^3+12 x^2 \log (-3+x)-16 x^3 \log (-3+x)+4 x^4 \log (-3+x)\right )}{(-3+x) \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )} \, dx-\frac {1}{3} \int \frac {e^{2+x} \left (108-66 x+7 x^2+5 x^3+12 x^2 \log (-3+x)-16 x^3 \log (-3+x)+4 x^4 \log (-3+x)\right )}{x \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )} \, dx\\ &=-x+2 \log (x)-\frac {1}{3} \int \left (-\frac {66 e^{2+x}}{12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)}+\frac {108 e^{2+x}}{x \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )}+\frac {7 e^{2+x} x}{12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)}+\frac {5 e^{2+x} x^2}{12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)}+\frac {12 e^{2+x} x \log (-3+x)}{12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)}-\frac {16 e^{2+x} x^2 \log (-3+x)}{12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)}+\frac {4 e^{2+x} x^3 \log (-3+x)}{12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)}\right ) \, dx+\frac {1}{3} \int \left (\frac {108 e^{2+x}}{(-3+x) \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )}-\frac {66 e^{2+x} x}{(-3+x) \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )}+\frac {7 e^{2+x} x^2}{(-3+x) \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )}+\frac {5 e^{2+x} x^3}{(-3+x) \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )}+\frac {12 e^{2+x} x^2 \log (-3+x)}{(-3+x) \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )}-\frac {16 e^{2+x} x^3 \log (-3+x)}{(-3+x) \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )}+\frac {4 e^{2+x} x^4 \log (-3+x)}{(-3+x) \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )}\right ) \, dx\\ &=-x+2 \log (x)-\frac {4}{3} \int \frac {e^{2+x} x^3 \log (-3+x)}{12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)} \, dx+\frac {4}{3} \int \frac {e^{2+x} x^4 \log (-3+x)}{(-3+x) \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )} \, dx-\frac {5}{3} \int \frac {e^{2+x} x^2}{12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)} \, dx+\frac {5}{3} \int \frac {e^{2+x} x^3}{(-3+x) \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )} \, dx-\frac {7}{3} \int \frac {e^{2+x} x}{12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)} \, dx+\frac {7}{3} \int \frac {e^{2+x} x^2}{(-3+x) \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )} \, dx-4 \int \frac {e^{2+x} x \log (-3+x)}{12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)} \, dx+4 \int \frac {e^{2+x} x^2 \log (-3+x)}{(-3+x) \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )} \, dx+\frac {16}{3} \int \frac {e^{2+x} x^2 \log (-3+x)}{12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)} \, dx-\frac {16}{3} \int \frac {e^{2+x} x^3 \log (-3+x)}{(-3+x) \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )} \, dx+22 \int \frac {e^{2+x}}{12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)} \, dx-22 \int \frac {e^{2+x} x}{(-3+x) \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )} \, dx+36 \int \frac {e^{2+x}}{(-3+x) \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )} \, dx-36 \int \frac {e^{2+x}}{x \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.11, size = 53, normalized size = 1.77 \begin {gather*} 4 \left (-\frac {x}{4}-\frac {\log (x)}{4}+\frac {1}{4} \log \left (12 e^{2+x}+e^{2+x} x+4 x^3+4 e^{2+x} x^2 \log (-3+x)\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.51, size = 33, normalized size = 1.10 \begin {gather*} \log \relax (x) + \log \left (\frac {4 \, x^{2} e^{\left (-x + \log \relax (x) - 2\right )} + 4 \, x^{2} \log \left (x - 3\right ) + x + 12}{x^{2}}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.17, size = 40, normalized size = 1.33 \begin {gather*} -x + \log \left (4 \, x^{2} e^{\left (x + 2\right )} \log \left (x - 3\right ) + 4 \, x^{3} + x e^{\left (x + 2\right )} + 12 \, e^{\left (x + 2\right )}\right ) - \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 32, normalized size = 1.07
method | result | size |
risch | \(\ln \relax (x )+2+\ln \left (x \,{\mathrm e}^{-x -2}+\frac {4 x^{2} \ln \left (x -3\right )+x +12}{4 x^{2}}\right )\) | \(32\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 46, normalized size = 1.53 \begin {gather*} \log \relax (x) + \log \left (\frac {{\left (4 \, x^{2} e^{\left (x + 2\right )} \log \left (x - 3\right ) + 4 \, x^{3} + {\left (x e^{2} + 12 \, e^{2}\right )} e^{x}\right )} e^{\left (-x - 2\right )}}{4 \, x^{2}}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 8.86, size = 31, normalized size = 1.03 \begin {gather*} \ln \left (\frac {x+4\,x^2\,\ln \left (x-3\right )+4\,x^3\,{\mathrm {e}}^{-x-2}+12}{x^2}\right )+\ln \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.57, size = 31, normalized size = 1.03 \begin {gather*} 2 \log {\relax (x )} + \log {\left (e^{- x - 2} + \frac {4 x^{2} \log {\left (x - 3 \right )} + x + 12}{4 x^{3}} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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