Integrand size = 32, antiderivative size = 24 \[ \int \frac {2 x-e^{\frac {1}{x}} x+e^{\frac {1}{x}} (1-x) \log (x)}{4 x} \, dx=1+x+\frac {1}{2} \left (-22-x-\frac {1}{2} e^{\frac {1}{x}} x \log (x)\right ) \]
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Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {12, 14, 2326} \[ \int \frac {2 x-e^{\frac {1}{x}} x+e^{\frac {1}{x}} (1-x) \log (x)}{4 x} \, dx=\frac {x}{2}-\frac {1}{4} e^{\frac {1}{x}} x \log (x) \]
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Rule 12
Rule 14
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \frac {2 x-e^{\frac {1}{x}} x+e^{\frac {1}{x}} (1-x) \log (x)}{x} \, dx \\ & = \frac {1}{4} \int \left (2-\frac {e^{\frac {1}{x}} (x-\log (x)+x \log (x))}{x}\right ) \, dx \\ & = \frac {x}{2}-\frac {1}{4} \int \frac {e^{\frac {1}{x}} (x-\log (x)+x \log (x))}{x} \, dx \\ & = \frac {x}{2}-\frac {1}{4} e^{\frac {1}{x}} x \log (x) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {2 x-e^{\frac {1}{x}} x+e^{\frac {1}{x}} (1-x) \log (x)}{4 x} \, dx=\frac {1}{4} \left (2 x-e^{\frac {1}{x}} x \log (x)\right ) \]
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Time = 0.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.58
method | result | size |
norman | \(\frac {x}{2}-\frac {x \,{\mathrm e}^{\frac {1}{x}} \ln \left (x \right )}{4}\) | \(14\) |
risch | \(\frac {x}{2}-\frac {x \,{\mathrm e}^{\frac {1}{x}} \ln \left (x \right )}{4}\) | \(14\) |
parallelrisch | \(\frac {x}{2}-\frac {x \,{\mathrm e}^{\frac {1}{x}} \ln \left (x \right )}{4}\) | \(14\) |
derivativedivides | \(\frac {x}{2}-\frac {\left (\frac {\left (\ln \left (x \right )+\ln \left (\frac {1}{x}\right )\right ) {\mathrm e}^{\frac {1}{x}}}{x}-\frac {{\mathrm e}^{\frac {1}{x}} \ln \left (\frac {1}{x}\right )}{x}\right ) x^{2}}{4}\) | \(39\) |
default | \(\frac {x}{2}-\frac {\left (\frac {\left (\ln \left (x \right )+\ln \left (\frac {1}{x}\right )\right ) {\mathrm e}^{\frac {1}{x}}}{x}-\frac {{\mathrm e}^{\frac {1}{x}} \ln \left (\frac {1}{x}\right )}{x}\right ) x^{2}}{4}\) | \(39\) |
parts | \(\frac {x}{2}-\frac {\left (\frac {\left (\ln \left (x \right )+\ln \left (\frac {1}{x}\right )\right ) {\mathrm e}^{\frac {1}{x}}}{x}-\frac {{\mathrm e}^{\frac {1}{x}} \ln \left (\frac {1}{x}\right )}{x}\right ) x^{2}}{4}\) | \(39\) |
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Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.54 \[ \int \frac {2 x-e^{\frac {1}{x}} x+e^{\frac {1}{x}} (1-x) \log (x)}{4 x} \, dx=-\frac {1}{4} \, x e^{\frac {1}{x}} \log \left (x\right ) + \frac {1}{2} \, x \]
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Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.58 \[ \int \frac {2 x-e^{\frac {1}{x}} x+e^{\frac {1}{x}} (1-x) \log (x)}{4 x} \, dx=- \frac {x e^{\frac {1}{x}} \log {\left (x \right )}}{4} + \frac {x}{2} \]
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\[ \int \frac {2 x-e^{\frac {1}{x}} x+e^{\frac {1}{x}} (1-x) \log (x)}{4 x} \, dx=\int { -\frac {{\left (x - 1\right )} e^{\frac {1}{x}} \log \left (x\right ) + x e^{\frac {1}{x}} - 2 \, x}{4 \, x} \,d x } \]
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Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.54 \[ \int \frac {2 x-e^{\frac {1}{x}} x+e^{\frac {1}{x}} (1-x) \log (x)}{4 x} \, dx=-\frac {1}{4} \, x e^{\frac {1}{x}} \log \left (x\right ) + \frac {1}{2} \, x \]
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Time = 9.33 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.50 \[ \int \frac {2 x-e^{\frac {1}{x}} x+e^{\frac {1}{x}} (1-x) \log (x)}{4 x} \, dx=-\frac {x\,\left ({\mathrm {e}}^{1/x}\,\ln \left (x\right )-2\right )}{4} \]
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