Integrand size = 32, antiderivative size = 24 \[ \int \frac {-15+12 e^{-5+x}-24 x}{-16+4 e^{-5+x}-5 x-4 x^2} \, dx=3 \log \left (\frac {2}{3} \left (-4+e^{-5+x}-\frac {5 x}{4}-x^2\right )\right ) \]
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Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {6816} \[ \int \frac {-15+12 e^{-5+x}-24 x}{-16+4 e^{-5+x}-5 x-4 x^2} \, dx=3 \log \left (4 e^5 x^2+5 e^5 x-4 e^x+16 e^5\right ) \]
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Rule 6816
Rubi steps \begin{align*} \text {integral}& = 3 \log \left (16 e^5-4 e^x+5 e^5 x+4 e^5 x^2\right ) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-15+12 e^{-5+x}-24 x}{-16+4 e^{-5+x}-5 x-4 x^2} \, dx=3 \log \left (-16+4 e^{-5+x}-5 x-4 x^2\right ) \]
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Time = 0.12 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75
method | result | size |
parallelrisch | \(3 \ln \left (x^{2}+\frac {5 x}{4}-{\mathrm e}^{-5+x}+4\right )\) | \(18\) |
derivativedivides | \(3 \ln \left (4 \,{\mathrm e}^{-5+x}-4 x^{2}-5 x -16\right )\) | \(20\) |
default | \(3 \ln \left (4 \,{\mathrm e}^{-5+x}-4 x^{2}-5 x -16\right )\) | \(20\) |
norman | \(3 \ln \left (4 x^{2}+5 x -4 \,{\mathrm e}^{-5+x}+16\right )\) | \(20\) |
risch | \(15+3 \ln \left (-x^{2}-\frac {5 x}{4}+{\mathrm e}^{-5+x}-4\right )\) | \(20\) |
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none
Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {-15+12 e^{-5+x}-24 x}{-16+4 e^{-5+x}-5 x-4 x^2} \, dx=3 \, \log \left (-4 \, x^{2} - 5 \, x + 4 \, e^{\left (x - 5\right )} - 16\right ) \]
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Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-15+12 e^{-5+x}-24 x}{-16+4 e^{-5+x}-5 x-4 x^2} \, dx=3 \log {\left (- x^{2} - \frac {5 x}{4} + e^{x - 5} - 4 \right )} \]
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none
Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {-15+12 e^{-5+x}-24 x}{-16+4 e^{-5+x}-5 x-4 x^2} \, dx=3 \, \log \left (4 \, x^{2} + 5 \, x - 4 \, e^{\left (x - 5\right )} + 16\right ) \]
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none
Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-15+12 e^{-5+x}-24 x}{-16+4 e^{-5+x}-5 x-4 x^2} \, dx=3 \, \log \left (-4 \, x^{2} e^{5} - 5 \, x e^{5} - 16 \, e^{5} + 4 \, e^{x}\right ) \]
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Time = 0.14 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-15+12 e^{-5+x}-24 x}{-16+4 e^{-5+x}-5 x-4 x^2} \, dx=3\,\ln \left (\frac {5\,x}{4}-{\mathrm {e}}^{x-5}+x^2+4\right ) \]
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