Integrand size = 21, antiderivative size = 16 \[ \int \frac {1}{-2 e-2 x+(e+x) \log (5 e+5 x)} \, dx=\log \left (\frac {93}{7}\right )+\log (2-\log (5 (e+x))) \]
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Time = 0.05 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6873, 2437, 2339, 29} \[ \int \frac {1}{-2 e-2 x+(e+x) \log (5 e+5 x)} \, dx=\log (2-\log (5 (x+e))) \]
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Rule 29
Rule 2339
Rule 2437
Rule 6873
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{(-e-x) (2-\log (5 (e+x)))} \, dx \\ & = -\text {Subst}\left (\int \frac {1}{x (2-\log (5 x))} \, dx,x,e+x\right ) \\ & = \text {Subst}\left (\int \frac {1}{x} \, dx,x,2-\log (5 (e+x))\right ) \\ & = \log (2-\log (5 (e+x))) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.56 \[ \int \frac {1}{-2 e-2 x+(e+x) \log (5 e+5 x)} \, dx=\log (-2+\log (5 (e+x))) \]
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Time = 0.04 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81
method | result | size |
derivativedivides | \(\ln \left (\ln \left (5 \,{\mathrm e}+5 x \right )-2\right )\) | \(13\) |
default | \(\ln \left (\ln \left (5 \,{\mathrm e}+5 x \right )-2\right )\) | \(13\) |
norman | \(\ln \left (\ln \left (5 \,{\mathrm e}+5 x \right )-2\right )\) | \(13\) |
risch | \(\ln \left (\ln \left (5 \,{\mathrm e}+5 x \right )-2\right )\) | \(13\) |
parallelrisch | \(\ln \left (\ln \left (5 \,{\mathrm e}+5 x \right )-2\right )\) | \(13\) |
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none
Time = 0.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {1}{-2 e-2 x+(e+x) \log (5 e+5 x)} \, dx=\log \left (\log \left (5 \, x + 5 \, e\right ) - 2\right ) \]
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Time = 0.08 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {1}{-2 e-2 x+(e+x) \log (5 e+5 x)} \, dx=\log {\left (\log {\left (5 x + 5 e \right )} - 2 \right )} \]
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none
Time = 0.31 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \frac {1}{-2 e-2 x+(e+x) \log (5 e+5 x)} \, dx=\log \left (\log \left (5\right ) + \log \left (x + e\right ) - 2\right ) \]
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none
Time = 0.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {1}{-2 e-2 x+(e+x) \log (5 e+5 x)} \, dx=\log \left (\log \left (5 \, x + 5 \, e\right ) - 2\right ) \]
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Time = 0.55 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {1}{-2 e-2 x+(e+x) \log (5 e+5 x)} \, dx=\ln \left (\ln \left (5\,x+5\,\mathrm {e}\right )-2\right ) \]
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