Integrand size = 20, antiderivative size = 15 \[ \int \frac {e^{-\frac {2 (-4-4 e+e \log (100))}{e}}}{x} \, dx=\frac {e^{8+\frac {8}{e}} \log (x)}{10000} \]
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Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 29} \[ \int \frac {e^{-\frac {2 (-4-4 e+e \log (100))}{e}}}{x} \, dx=\frac {e^{8+\frac {8}{e}} \log (x)}{10000} \]
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Rule 12
Rule 29
Rubi steps \begin{align*} \text {integral}& = \frac {e^{8+\frac {8}{e}} \int \frac {1}{x} \, dx}{10000} \\ & = \frac {e^{8+\frac {8}{e}} \log (x)}{10000} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {2 (-4-4 e+e \log (100))}{e}}}{x} \, dx=\frac {e^{8+\frac {8}{e}} \log (x)}{10000} \]
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Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80
method | result | size |
risch | \(\frac {\ln \left (x \right ) {\mathrm e}^{8+8 \,{\mathrm e}^{-1}}}{10000}\) | \(12\) |
norman | \(\frac {{\mathrm e}^{8} {\mathrm e}^{8 \,{\mathrm e}^{-1}} \ln \left (x \right )}{10000}\) | \(16\) |
default | \({\mathrm e}^{-4 \left ({\mathrm e} \ln \left (10\right )-2 \,{\mathrm e}-2\right ) {\mathrm e}^{-1}} \ln \left (x \right )\) | \(24\) |
parallelrisch | \({\mathrm e}^{-4 \left ({\mathrm e} \ln \left (10\right )-2 \,{\mathrm e}-2\right ) {\mathrm e}^{-1}} \ln \left (x \right )\) | \(24\) |
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none
Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27 \[ \int \frac {e^{-\frac {2 (-4-4 e+e \log (100))}{e}}}{x} \, dx=e^{\left (-4 \, {\left (e \log \left (10\right ) - 2 \, e - 2\right )} e^{\left (-1\right )}\right )} \log \left (x\right ) \]
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Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-\frac {2 (-4-4 e+e \log (100))}{e}}}{x} \, dx=\frac {e^{8} e^{\frac {8}{e}} \log {\left (x \right )}}{10000} \]
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none
Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27 \[ \int \frac {e^{-\frac {2 (-4-4 e+e \log (100))}{e}}}{x} \, dx=e^{\left (-4 \, {\left (e \log \left (10\right ) - 2 \, e - 2\right )} e^{\left (-1\right )}\right )} \log \left (x\right ) \]
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none
Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.33 \[ \int \frac {e^{-\frac {2 (-4-4 e+e \log (100))}{e}}}{x} \, dx=e^{\left (-4 \, {\left (e \log \left (10\right ) - 2 \, e - 2\right )} e^{\left (-1\right )}\right )} \log \left ({\left | x \right |}\right ) \]
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Time = 0.09 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {e^{-\frac {2 (-4-4 e+e \log (100))}{e}}}{x} \, dx=\frac {{\mathrm {e}}^{8\,{\mathrm {e}}^{-1}+8}\,\ln \left (x\right )}{10000} \]
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