Integrand size = 22, antiderivative size = 21 \[ \int \frac {4+4 x}{e^9-2 x-x^2+\log (5)} \, dx=\log \left (\frac {1}{2 \left (e^9-2 x-x^2+\log (5)\right )^2}\right ) \]
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Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {642} \[ \int \frac {4+4 x}{e^9-2 x-x^2+\log (5)} \, dx=-2 \log \left (-x^2-2 x+e^9+\log (5)\right ) \]
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Rule 642
Rubi steps \begin{align*} \text {integral}& = -2 \log \left (e^9-2 x-x^2+\log (5)\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {4+4 x}{e^9-2 x-x^2+\log (5)} \, dx=-2 \log \left (e^9-2 x-x^2+\log (5)\right ) \]
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Time = 0.35 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81
method | result | size |
default | \(-2 \ln \left (\ln \left (5\right )+{\mathrm e}^{9}-x^{2}-2 x \right )\) | \(17\) |
norman | \(-2 \ln \left (\ln \left (5\right )+{\mathrm e}^{9}-x^{2}-2 x \right )\) | \(17\) |
risch | \(-2 \ln \left (-\ln \left (5\right )-{\mathrm e}^{9}+x^{2}+2 x \right )\) | \(19\) |
parallelrisch | \(-2 \ln \left (-\ln \left (5\right )-{\mathrm e}^{9}+x^{2}+2 x \right )\) | \(19\) |
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none
Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {4+4 x}{e^9-2 x-x^2+\log (5)} \, dx=-2 \, \log \left (x^{2} + 2 \, x - e^{9} - \log \left (5\right )\right ) \]
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Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {4+4 x}{e^9-2 x-x^2+\log (5)} \, dx=- 2 \log {\left (x^{2} + 2 x - e^{9} - \log {\left (5 \right )} \right )} \]
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none
Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {4+4 x}{e^9-2 x-x^2+\log (5)} \, dx=-2 \, \log \left (x^{2} + 2 \, x - e^{9} - \log \left (5\right )\right ) \]
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none
Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {4+4 x}{e^9-2 x-x^2+\log (5)} \, dx=-2 \, \log \left ({\left | x^{2} + 2 \, x - e^{9} - \log \left (5\right ) \right |}\right ) \]
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Time = 0.09 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {4+4 x}{e^9-2 x-x^2+\log (5)} \, dx=-2\,\ln \left (x^2+2\,x-{\mathrm {e}}^9-\ln \left (5\right )\right ) \]
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