Integrand size = 43, antiderivative size = 24 \[ \int \frac {-11 e x+e^{1+\frac {2+x \log \left (\frac {e^x}{x}\right )}{x}} \left (2+2 x-x^2\right )}{5 x^3} \, dx=\frac {e \left (11-\frac {e^{\frac {2}{x}+x}}{x}\right )}{5 x} \]
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Time = 0.05 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {12, 14, 2326} \[ \int \frac {-11 e x+e^{1+\frac {2+x \log \left (\frac {e^x}{x}\right )}{x}} \left (2+2 x-x^2\right )}{5 x^3} \, dx=\frac {e^{x+\frac {2}{x}+1} \left (2-x^2\right )}{5 \left (1-\frac {2}{x^2}\right ) x^4}+\frac {11 e}{5 x} \]
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Rule 12
Rule 14
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {-11 e x+e^{1+\frac {2+x \log \left (\frac {e^x}{x}\right )}{x}} \left (2+2 x-x^2\right )}{x^3} \, dx \\ & = \frac {1}{5} \int \left (-\frac {11 e}{x^2}-\frac {e^{1+\frac {2}{x}+x} \left (-2-2 x+x^2\right )}{x^4}\right ) \, dx \\ & = \frac {11 e}{5 x}-\frac {1}{5} \int \frac {e^{1+\frac {2}{x}+x} \left (-2-2 x+x^2\right )}{x^4} \, dx \\ & = \frac {11 e}{5 x}+\frac {e^{1+\frac {2}{x}+x} \left (2-x^2\right )}{5 \left (1-\frac {2}{x^2}\right ) x^4} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {-11 e x+e^{1+\frac {2+x \log \left (\frac {e^x}{x}\right )}{x}} \left (2+2 x-x^2\right )}{5 x^3} \, dx=-\frac {e \left (e^{\frac {2}{x}+x}-11 x\right )}{5 x^2} \]
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Time = 0.48 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29
method | result | size |
parallelrisch | \(\frac {-{\mathrm e} \,{\mathrm e}^{\frac {x \ln \left (\frac {{\mathrm e}^{x}}{x}\right )+2}{x}}+11 \,{\mathrm e}}{5 x}\) | \(31\) |
default | \(\frac {11 \,{\mathrm e}}{5 x}-\frac {{\mathrm e} \,{\mathrm e}^{\frac {x \ln \left (\frac {{\mathrm e}^{x}}{x}\right )+2}{x}}}{5 x}\) | \(32\) |
parts | \(\frac {11 \,{\mathrm e}}{5 x}-\frac {{\mathrm e} \,{\mathrm e}^{\frac {x \ln \left (\frac {{\mathrm e}^{x}}{x}\right )+2}{x}}}{5 x}\) | \(32\) |
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Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-11 e x+e^{1+\frac {2+x \log \left (\frac {e^x}{x}\right )}{x}} \left (2+2 x-x^2\right )}{5 x^3} \, dx=\frac {11 \, e - e^{\left (\frac {x \log \left (\frac {e^{x}}{x}\right ) + x + 2}{x}\right )}}{5 \, x} \]
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\[ \int \frac {-11 e x+e^{1+\frac {2+x \log \left (\frac {e^x}{x}\right )}{x}} \left (2+2 x-x^2\right )}{5 x^3} \, dx=- \frac {e \left (\int \frac {11}{x^{2}}\, dx + \int \left (- \frac {2 e^{\frac {2}{x}} e^{x}}{x^{4}}\right )\, dx + \int \left (- \frac {2 e^{\frac {2}{x}} e^{x}}{x^{3}}\right )\, dx + \int \frac {e^{\frac {2}{x}} e^{x}}{x^{2}}\, dx\right )}{5} \]
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Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-11 e x+e^{1+\frac {2+x \log \left (\frac {e^x}{x}\right )}{x}} \left (2+2 x-x^2\right )}{5 x^3} \, dx=\frac {11 \, e}{5 \, x} - \frac {e^{\left (x + \frac {2}{x} + 1\right )}}{5 \, x^{2}} \]
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Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-11 e x+e^{1+\frac {2+x \log \left (\frac {e^x}{x}\right )}{x}} \left (2+2 x-x^2\right )}{5 x^3} \, dx=\frac {11 \, x e - e^{\left (\frac {x^{2} + x + 2}{x}\right )}}{5 \, x^{2}} \]
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Timed out. \[ \int \frac {-11 e x+e^{1+\frac {2+x \log \left (\frac {e^x}{x}\right )}{x}} \left (2+2 x-x^2\right )}{5 x^3} \, dx=\int -\frac {\frac {11\,x\,\mathrm {e}}{5}-\frac {{\mathrm {e}}^{\frac {x\,\ln \left (\frac {{\mathrm {e}}^x}{x}\right )+2}{x}}\,\mathrm {e}\,\left (-x^2+2\,x+2\right )}{5}}{x^3} \,d x \]
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