Integrand size = 50, antiderivative size = 22 \[ \int \frac {e^x \left (-10+5 e^{-x^2}\right )+5 x+e^x \left (-10 x+e^{-x^2} \left (5 x-10 x^2\right )\right ) \log (x)}{x} \, dx=5 \left (x-e^x \left (2-e^{-x^2}\right ) \log (x)\right ) \]
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Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.00, number of steps used = 6, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {14, 2326} \[ \int \frac {e^x \left (-10+5 e^{-x^2}\right )+5 x+e^x \left (-10 x+e^{-x^2} \left (5 x-10 x^2\right )\right ) \log (x)}{x} \, dx=\frac {5 e^{x-x^2} \left (x \log (x)-2 x^2 \log (x)\right )}{(1-2 x) x}+5 x-10 e^x \log (x) \]
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Rule 14
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {5 \left (2 e^x-x+2 e^x x \log (x)\right )}{x}-\frac {5 e^{x-x^2} \left (-1-x \log (x)+2 x^2 \log (x)\right )}{x}\right ) \, dx \\ & = -\left (5 \int \frac {2 e^x-x+2 e^x x \log (x)}{x} \, dx\right )-5 \int \frac {e^{x-x^2} \left (-1-x \log (x)+2 x^2 \log (x)\right )}{x} \, dx \\ & = \frac {5 e^{x-x^2} \left (x \log (x)-2 x^2 \log (x)\right )}{(1-2 x) x}-5 \int \left (-1+\frac {2 e^x (1+x \log (x))}{x}\right ) \, dx \\ & = 5 x+\frac {5 e^{x-x^2} \left (x \log (x)-2 x^2 \log (x)\right )}{(1-2 x) x}-10 \int \frac {e^x (1+x \log (x))}{x} \, dx \\ & = 5 x-10 e^x \log (x)+\frac {5 e^{x-x^2} \left (x \log (x)-2 x^2 \log (x)\right )}{(1-2 x) x} \\ \end{align*}
Time = 0.34 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^x \left (-10+5 e^{-x^2}\right )+5 x+e^x \left (-10 x+e^{-x^2} \left (5 x-10 x^2\right )\right ) \log (x)}{x} \, dx=5 \left (x+e^x \left (-2+e^{-x^2}\right ) \log (x)\right ) \]
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Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\left (5 \,{\mathrm e}^{-x \left (-1+x \right )}-10 \,{\mathrm e}^{x}\right ) \ln \left (x \right )+5 x\) | \(22\) |
parallelrisch | \(5 \,{\mathrm e}^{-x^{2}} \ln \left (x \right ) {\mathrm e}^{x}-10 \,{\mathrm e}^{x} \ln \left (x \right )+5 x\) | \(23\) |
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Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {e^x \left (-10+5 e^{-x^2}\right )+5 x+e^x \left (-10 x+e^{-x^2} \left (5 x-10 x^2\right )\right ) \log (x)}{x} \, dx=5 \, {\left (e^{\left (-x^{2}\right )} - 2\right )} e^{x} \log \left (x\right ) + 5 \, x \]
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Time = 4.75 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {e^x \left (-10+5 e^{-x^2}\right )+5 x+e^x \left (-10 x+e^{-x^2} \left (5 x-10 x^2\right )\right ) \log (x)}{x} \, dx=5 x - 10 e^{x} \log {\left (x \right )} + 5 e^{x} e^{- x^{2}} \log {\left (x \right )} \]
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\[ \int \frac {e^x \left (-10+5 e^{-x^2}\right )+5 x+e^x \left (-10 x+e^{-x^2} \left (5 x-10 x^2\right )\right ) \log (x)}{x} \, dx=\int { -\frac {5 \, {\left ({\left ({\left (2 \, x^{2} - x\right )} e^{\left (-x^{2}\right )} + 2 \, x\right )} e^{x} \log \left (x\right ) - {\left (e^{\left (-x^{2}\right )} - 2\right )} e^{x} - x\right )}}{x} \,d x } \]
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Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (-10+5 e^{-x^2}\right )+5 x+e^x \left (-10 x+e^{-x^2} \left (5 x-10 x^2\right )\right ) \log (x)}{x} \, dx=5 \, e^{\left (-x^{2} + x\right )} \log \left (x\right ) - 10 \, e^{x} \log \left (x\right ) + 5 \, x \]
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Timed out. \[ \int \frac {e^x \left (-10+5 e^{-x^2}\right )+5 x+e^x \left (-10 x+e^{-x^2} \left (5 x-10 x^2\right )\right ) \log (x)}{x} \, dx=\int \frac {5\,x+{\mathrm {e}}^x\,\left (5\,{\mathrm {e}}^{-x^2}-10\right )-{\mathrm {e}}^x\,\ln \left (x\right )\,\left (10\,x-{\mathrm {e}}^{-x^2}\,\left (5\,x-10\,x^2\right )\right )}{x} \,d x \]
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