Integrand size = 51, antiderivative size = 22 \[ \int \frac {-5+e^x (-3+x)+2 x}{-6+e^2 (3-x)+e^x (-3+x)-4 x+2 x^2+(-3+x) \log (3-x)} \, dx=\log \left (-2+e^2-e^x-2 x-\log (3-x)\right ) \]
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Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {6873, 6816} \[ \int \frac {-5+e^x (-3+x)+2 x}{-6+e^2 (3-x)+e^x (-3+x)-4 x+2 x^2+(-3+x) \log (3-x)} \, dx=\log \left (2 x+e^x+\log (3-x)-e^2+2\right ) \]
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Rule 6816
Rule 6873
Rubi steps \begin{align*} \text {integral}& = \int \frac {5-e^x (-3+x)-2 x}{(3-x) \left (e^x+2 \left (1-\frac {e^2}{2}\right )+2 x+\log (3-x)\right )} \, dx \\ & = \log \left (2-e^2+e^x+2 x+\log (3-x)\right ) \\ \end{align*}
Time = 0.64 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-5+e^x (-3+x)+2 x}{-6+e^2 (3-x)+e^x (-3+x)-4 x+2 x^2+(-3+x) \log (3-x)} \, dx=\log \left (2-e^2+e^x+2 x+\log (3-x)\right ) \]
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Time = 1.32 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86
method | result | size |
risch | \(\ln \left (2 x -{\mathrm e}^{2}+{\mathrm e}^{x}+\ln \left (-x +3\right )+2\right )\) | \(19\) |
norman | \(\ln \left ({\mathrm e}^{2}-2 x -\ln \left (-x +3\right )-{\mathrm e}^{x}-2\right )\) | \(21\) |
parallelrisch | \(\ln \left (x -\frac {{\mathrm e}^{2}}{2}+\frac {{\mathrm e}^{x}}{2}+\frac {\ln \left (-x +3\right )}{2}+1\right )\) | \(21\) |
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none
Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {-5+e^x (-3+x)+2 x}{-6+e^2 (3-x)+e^x (-3+x)-4 x+2 x^2+(-3+x) \log (3-x)} \, dx=\log \left (2 \, x - e^{2} + e^{x} + \log \left (-x + 3\right ) + 2\right ) \]
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Time = 0.15 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {-5+e^x (-3+x)+2 x}{-6+e^2 (3-x)+e^x (-3+x)-4 x+2 x^2+(-3+x) \log (3-x)} \, dx=\log {\left (2 x + e^{x} + \log {\left (3 - x \right )} - e^{2} + 2 \right )} \]
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none
Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {-5+e^x (-3+x)+2 x}{-6+e^2 (3-x)+e^x (-3+x)-4 x+2 x^2+(-3+x) \log (3-x)} \, dx=\log \left (2 \, x - e^{2} + e^{x} + \log \left (-x + 3\right ) + 2\right ) \]
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none
Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {-5+e^x (-3+x)+2 x}{-6+e^2 (3-x)+e^x (-3+x)-4 x+2 x^2+(-3+x) \log (3-x)} \, dx=\log \left (2 \, x - e^{2} + e^{x} + \log \left (-x + 3\right ) + 2\right ) \]
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Time = 0.51 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {-5+e^x (-3+x)+2 x}{-6+e^2 (3-x)+e^x (-3+x)-4 x+2 x^2+(-3+x) \log (3-x)} \, dx=\ln \left (2\,x-{\mathrm {e}}^2+\ln \left (3-x\right )+{\mathrm {e}}^x+2\right ) \]
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