Optimal. Leaf size=29 \[ -4 x+\frac {4 e^{3-x} \log (6)}{\left (e^x-\frac {10}{x}\right ) x} \]
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Rubi [F] time = 1.95, 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 {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, 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^{3-x} \left (-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )\right )}{\left (10-e^x x\right )^2} \, dx\\ &=\int \left (-4-\frac {40 e^{3-x} (1+x) \log (6)}{x \left (-10+e^x x\right )^2}-\frac {4 e^{3-x} (1+2 x) \log (6)}{x \left (-10+e^x x\right )}\right ) \, dx\\ &=-4 x-(4 \log (6)) \int \frac {e^{3-x} (1+2 x)}{x \left (-10+e^x x\right )} \, dx-(40 \log (6)) \int \frac {e^{3-x} (1+x)}{x \left (-10+e^x x\right )^2} \, dx\\ &=-4 x-(4 \log (6)) \int \left (\frac {2 e^{3-x}}{-10+e^x x}+\frac {e^{3-x}}{x \left (-10+e^x x\right )}\right ) \, dx-(40 \log (6)) \int \left (\frac {e^{3-x}}{\left (-10+e^x x\right )^2}+\frac {e^{3-x}}{x \left (-10+e^x x\right )^2}\right ) \, dx\\ &=-4 x-(4 \log (6)) \int \frac {e^{3-x}}{x \left (-10+e^x x\right )} \, dx-(8 \log (6)) \int \frac {e^{3-x}}{-10+e^x x} \, dx-(40 \log (6)) \int \frac {e^{3-x}}{\left (-10+e^x x\right )^2} \, dx-(40 \log (6)) \int \frac {e^{3-x}}{x \left (-10+e^x x\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.99, size = 34, normalized size = 1.17 \begin {gather*} -\frac {4 \left (10 x-e^x x^2+e^{3-x} \log (6)\right )}{10-e^x x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.66, size = 35, normalized size = 1.21 \begin {gather*} -\frac {4 \, {\left (x^{2} e^{\left (2 \, x\right )} - 10 \, x e^{x} - e^{3} \log \relax (6)\right )}}{x e^{\left (2 \, x\right )} - 10 \, e^{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.16, size = 35, normalized size = 1.21 \begin {gather*} -\frac {4 \, {\left (x^{2} e^{\left (2 \, x\right )} - 10 \, x e^{x} - e^{3} \log \relax (6)\right )}}{x e^{\left (2 \, x\right )} - 10 \, e^{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.13, size = 35, normalized size = 1.21
method | result | size |
norman | \(\frac {\left (4 \,{\mathrm e}^{3} \ln \relax (6)+40 \,{\mathrm e}^{x} x -4 \,{\mathrm e}^{2 x} x^{2}\right ) {\mathrm e}^{-x}}{{\mathrm e}^{x} x -10}\) | \(35\) |
risch | \(-4 x -\frac {2 \,{\mathrm e}^{3-x} \ln \relax (2)}{5}-\frac {2 \,{\mathrm e}^{3-x} \ln \relax (3)}{5}+\frac {2 x \left (\ln \relax (2)+\ln \relax (3)\right ) {\mathrm e}^{3}}{5 \left ({\mathrm e}^{x} x -10\right )}\) | \(43\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.49, size = 38, normalized size = 1.31 \begin {gather*} -\frac {4 \, {\left (x^{2} e^{\left (2 \, x\right )} - {\left (\log \relax (3) + \log \relax (2)\right )} e^{3} - 10 \, x e^{x}\right )}}{x e^{\left (2 \, x\right )} - 10 \, e^{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.54, size = 22, normalized size = 0.76 \begin {gather*} \frac {4\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^3\,\ln \relax (6)}{x\,{\mathrm {e}}^x-10}-4\,x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.18, size = 34, normalized size = 1.17 \begin {gather*} - 4 x + \frac {2 x e^{3} \log {\relax (6 )}}{5 x e^{x} - 50} - \frac {2 e^{3} e^{- x} \log {\relax (6 )}}{5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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