Integrand size = 17, antiderivative size = 14 \[ \int \frac {-2+x-2 e^{10-2 x} x}{x} \, dx=1+e^{10-2 x}+x-2 \log (x) \]
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Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {14, 2225, 45} \[ \int \frac {-2+x-2 e^{10-2 x} x}{x} \, dx=x+e^{10-2 x}-2 \log (x) \]
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Rule 14
Rule 45
Rule 2225
Rubi steps \begin{align*} \text {integral}& = \int \left (-2 e^{10-2 x}+\frac {-2+x}{x}\right ) \, dx \\ & = -\left (2 \int e^{10-2 x} \, dx\right )+\int \frac {-2+x}{x} \, dx \\ & = e^{10-2 x}+\int \left (1-\frac {2}{x}\right ) \, dx \\ & = e^{10-2 x}+x-2 \log (x) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {-2+x-2 e^{10-2 x} x}{x} \, dx=e^{10-2 x}+x-2 \log (x) \]
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Time = 0.04 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93
method | result | size |
risch | \(x +{\mathrm e}^{-2 x +10}-2 \ln \left (x \right )\) | \(13\) |
norman | \(x +{\mathrm e}^{-2 x +10}-2 \ln \left (x \right )\) | \(15\) |
parallelrisch | \(x +{\mathrm e}^{-2 x +10}-2 \ln \left (x \right )\) | \(15\) |
parts | \(x +{\mathrm e}^{-2 x +10}-2 \ln \left (x \right )\) | \(15\) |
derivativedivides | \(3 \ln \left (x \right )-5+x -5 \ln \left (-x \right )+{\mathrm e}^{-2 x +10}\) | \(22\) |
default | \(3 \ln \left (x \right )-5+x -5 \ln \left (-x \right )+{\mathrm e}^{-2 x +10}\) | \(22\) |
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Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {-2+x-2 e^{10-2 x} x}{x} \, dx=x + e^{\left (-2 \, x + 10\right )} - 2 \, \log \left (x\right ) \]
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Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {-2+x-2 e^{10-2 x} x}{x} \, dx=x + e^{10 - 2 x} - 2 \log {\left (x \right )} \]
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Time = 0.17 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {-2+x-2 e^{10-2 x} x}{x} \, dx=x + e^{\left (-2 \, x + 10\right )} - 2 \, \log \left (x\right ) \]
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Time = 0.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {-2+x-2 e^{10-2 x} x}{x} \, dx=x + e^{\left (-2 \, x + 10\right )} - 2 \, \log \left (x\right ) \]
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Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {-2+x-2 e^{10-2 x} x}{x} \, dx=x+{\mathrm {e}}^{10-2\,x}-2\,\ln \left (x\right ) \]
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