Integrand size = 77, antiderivative size = 29 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=-4 x+\frac {4 e^{3-x} \log (6)}{\left (e^x-\frac {10}{x}\right ) x} \]
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\[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=\int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx \]
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Rubi steps \begin{align*} \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{align*}
Time = 2.40 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=-\frac {4 \left (10 x-e^x x^2+e^{3-x} \log (6)\right )}{10-e^x x} \]
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Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21
method | result | size |
norman | \(\frac {\left (4 \,{\mathrm e}^{3} \ln \left (6\right )+40 \,{\mathrm e}^{x} x -4 \,{\mathrm e}^{2 x} x^{2}\right ) {\mathrm e}^{-x}}{{\mathrm e}^{x} x -10}\) | \(35\) |
parallelrisch | \(\frac {\left (-4 \,{\mathrm e}^{-3+x} x^{2} {\mathrm e}^{x}+40 x \,{\mathrm e}^{-3+x}+4 \ln \left (6\right )\right ) {\mathrm e}^{-x +3}}{{\mathrm e}^{x} x -10}\) | \(39\) |
risch | \(-4 x -\frac {2 \,{\mathrm e}^{-x +3} \ln \left (2\right )}{5}-\frac {2 \,{\mathrm e}^{-x +3} \ln \left (3\right )}{5}+\frac {2 x \left (\ln \left (2\right )+\ln \left (3\right )\right ) {\mathrm e}^{3}}{5 \left ({\mathrm e}^{x} x -10\right )}\) | \(43\) |
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Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=-\frac {4 \, {\left (x^{2} e^{\left (2 \, x\right )} - 10 \, x e^{x} - e^{3} \log \left (6\right )\right )}}{x e^{\left (2 \, x\right )} - 10 \, e^{x}} \]
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Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=- 4 x + \frac {2 x e^{3} \log {\left (6 \right )}}{5 x e^{x} - 50} - \frac {2 e^{3} e^{- x} \log {\left (6 \right )}}{5} \]
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Time = 0.34 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=-\frac {4 \, {\left (x^{2} e^{\left (2 \, x\right )} - {\left (\log \left (3\right ) + \log \left (2\right )\right )} e^{3} - 10 \, x e^{x}\right )}}{x e^{\left (2 \, x\right )} - 10 \, e^{x}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (29) = 58\).
Time = 0.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.28 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=-\frac {4 \, {\left ({\left (x - 3\right )}^{2} e^{\left (2 \, x - 3\right )} + 3 \, {\left (x - 3\right )} e^{\left (2 \, x - 3\right )} - 10 \, {\left (x - 3\right )} e^{\left (x - 3\right )} - \log \left (6\right )\right )}}{{\left (x - 3\right )} e^{\left (2 \, x - 3\right )} + 3 \, e^{\left (2 \, x - 3\right )} - 10 \, e^{\left (x - 3\right )}} \]
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Time = 11.89 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=\frac {4\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^3\,\ln \left (6\right )}{x\,{\mathrm {e}}^x-10}-4\,x \]
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