Integrand size = 34, antiderivative size = 21 \[ \int \frac {-10+e^4 (-2-x)-9 x-x^2}{5 x+e^4 x+x^2} \, dx=-23-x-\log \left (2 x^2 \left (5+e^4+x\right )^2\right ) \]
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Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {6, 1607, 1834} \[ \int \frac {-10+e^4 (-2-x)-9 x-x^2}{5 x+e^4 x+x^2} \, dx=-x-2 \log (x)-2 \log \left (x+e^4+5\right ) \]
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Rule 6
Rule 1607
Rule 1834
Rubi steps \begin{align*} \text {integral}& = \int \frac {-10+e^4 (-2-x)-9 x-x^2}{\left (5+e^4\right ) x+x^2} \, dx \\ & = \int \frac {-10+e^4 (-2-x)-9 x-x^2}{x \left (5+e^4+x\right )} \, dx \\ & = \int \left (-1-\frac {2}{x}-\frac {2}{5+e^4+x}\right ) \, dx \\ & = -x-2 \log (x)-2 \log \left (5+e^4+x\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-10+e^4 (-2-x)-9 x-x^2}{5 x+e^4 x+x^2} \, dx=-x-2 \log (x)-2 \log \left (5+e^4+x\right ) \]
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Time = 0.44 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81
method | result | size |
default | \(-x -2 \ln \left (x \right )-2 \ln \left ({\mathrm e}^{4}+x +5\right )\) | \(17\) |
norman | \(-x -2 \ln \left (x \right )-2 \ln \left ({\mathrm e}^{4}+x +5\right )\) | \(17\) |
parallelrisch | \(-x -2 \ln \left (x \right )-2 \ln \left ({\mathrm e}^{4}+x +5\right )\) | \(17\) |
risch | \(-x -2 \ln \left (x^{2}+\left (5+{\mathrm e}^{4}\right ) x \right )\) | \(18\) |
meijerg | \(-\left (5+{\mathrm e}^{4}\right ) \left (\frac {x}{5+{\mathrm e}^{4}}-\ln \left (1+\frac {x}{5+{\mathrm e}^{4}}\right )\right )+\left (-{\mathrm e}^{4}-9\right ) \ln \left (1+\frac {x}{5+{\mathrm e}^{4}}\right )-\frac {2 \,{\mathrm e}^{4} \left (\ln \left (x \right )-\ln \left (5+{\mathrm e}^{4}\right )-\ln \left (1+\frac {x}{5+{\mathrm e}^{4}}\right )\right )}{5+{\mathrm e}^{4}}-\frac {10 \left (\ln \left (x \right )-\ln \left (5+{\mathrm e}^{4}\right )-\ln \left (1+\frac {x}{5+{\mathrm e}^{4}}\right )\right )}{5+{\mathrm e}^{4}}\) | \(112\) |
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Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {-10+e^4 (-2-x)-9 x-x^2}{5 x+e^4 x+x^2} \, dx=-x - 2 \, \log \left (x^{2} + x e^{4} + 5 \, x\right ) \]
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Time = 0.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {-10+e^4 (-2-x)-9 x-x^2}{5 x+e^4 x+x^2} \, dx=- x - 2 \log {\left (x^{2} + x \left (5 + e^{4}\right ) \right )} \]
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Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {-10+e^4 (-2-x)-9 x-x^2}{5 x+e^4 x+x^2} \, dx=-x - 2 \, \log \left (x + e^{4} + 5\right ) - 2 \, \log \left (x\right ) \]
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Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {-10+e^4 (-2-x)-9 x-x^2}{5 x+e^4 x+x^2} \, dx=-x - 2 \, \log \left ({\left | x + e^{4} + 5 \right |}\right ) - 2 \, \log \left ({\left | x \right |}\right ) \]
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Time = 0.14 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-10+e^4 (-2-x)-9 x-x^2}{5 x+e^4 x+x^2} \, dx=-x-2\,\ln \left (x^2+\left ({\mathrm {e}}^4+5\right )\,x\right ) \]
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