Integrand size = 38, antiderivative size = 17 \[ \int \frac {-4-e^x+4 x+8 x \log (2 x)}{-e^x-4 x+4 x^2 \log (2 x)} \, dx=\log \left (e^x+4 x (1-x \log (2 x))\right ) \]
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Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6816} \[ \int \frac {-4-e^x+4 x+8 x \log (2 x)}{-e^x-4 x+4 x^2 \log (2 x)} \, dx=\log \left (-4 x^2 \log (2 x)+4 x+e^x\right ) \]
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Rule 6816
Rubi steps \begin{align*} \text {integral}& = \log \left (e^x+4 x-4 x^2 \log (2 x)\right ) \\ \end{align*}
Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-4-e^x+4 x+8 x \log (2 x)}{-e^x-4 x+4 x^2 \log (2 x)} \, dx=\log \left (e^x+4 x-4 x^2 \log (2 x)\right ) \]
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Time = 0.64 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06
method | result | size |
parallelrisch | \(\ln \left (x^{2} \ln \left (2 x \right )-x -\frac {{\mathrm e}^{x}}{4}\right )\) | \(18\) |
derivativedivides | \(\ln \left (4 x^{2} \ln \left (2 x \right )-{\mathrm e}^{x}-4 x \right )\) | \(19\) |
default | \(\ln \left (4 x^{2} \ln \left (2 x \right )-{\mathrm e}^{x}-4 x \right )\) | \(19\) |
norman | \(\ln \left (4 x^{2} \ln \left (2 x \right )-{\mathrm e}^{x}-4 x \right )\) | \(19\) |
risch | \(2 \ln \left (x \right )+\ln \left (\ln \left (2 x \right )-\frac {4 x +{\mathrm e}^{x}}{4 x^{2}}\right )\) | \(23\) |
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none
Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.71 \[ \int \frac {-4-e^x+4 x+8 x \log (2 x)}{-e^x-4 x+4 x^2 \log (2 x)} \, dx=2 \, \log \left (2 \, x\right ) + \log \left (\frac {4 \, x^{2} \log \left (2 \, x\right ) - 4 \, x - e^{x}}{x^{2}}\right ) \]
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Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-4-e^x+4 x+8 x \log (2 x)}{-e^x-4 x+4 x^2 \log (2 x)} \, dx=\log {\left (- 4 x^{2} \log {\left (2 x \right )} + 4 x + e^{x} \right )} \]
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none
Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {-4-e^x+4 x+8 x \log (2 x)}{-e^x-4 x+4 x^2 \log (2 x)} \, dx=\log \left (4 \, x^{2} \log \left (2 \, x\right ) - 4 \, x - e^{x}\right ) \]
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none
Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {-4-e^x+4 x+8 x \log (2 x)}{-e^x-4 x+4 x^2 \log (2 x)} \, dx=\log \left (4 \, x^{2} \log \left (2 \, x\right ) - 4 \, x - e^{x}\right ) \]
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Time = 9.22 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {-4-e^x+4 x+8 x \log (2 x)}{-e^x-4 x+4 x^2 \log (2 x)} \, dx=\ln \left (4\,x^2\,\ln \left (2\,x\right )-{\mathrm {e}}^x-4\,x\right ) \]
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