Integrand size = 32, antiderivative size = 34 \[ \int \frac {-1-e^2+11 x-5 x^2}{2 e^2-2 x+10 x^2} \, dx=e^{e^3}-x+\frac {1}{2} \left (x-\log \left (\frac {4}{e^2-x+5 x^2}\right )\right ) \]
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Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.68, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1671, 642} \[ \int \frac {-1-e^2+11 x-5 x^2}{2 e^2-2 x+10 x^2} \, dx=\frac {1}{2} \log \left (5 x^2-x+e^2\right )-\frac {x}{2} \]
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Rule 642
Rule 1671
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {1}{2}-\frac {1-10 x}{2 e^2-2 x+10 x^2}\right ) \, dx \\ & = -\frac {x}{2}-\int \frac {1-10 x}{2 e^2-2 x+10 x^2} \, dx \\ & = -\frac {x}{2}+\frac {1}{2} \log \left (e^2-x+5 x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.62 \[ \int \frac {-1-e^2+11 x-5 x^2}{2 e^2-2 x+10 x^2} \, dx=\frac {1}{2} \left (-x+\log \left (e^2-x+5 x^2\right )\right ) \]
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Time = 0.83 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.56
method | result | size |
default | \(-\frac {x}{2}+\frac {\ln \left (5 x^{2}-x +{\mathrm e}^{2}\right )}{2}\) | \(19\) |
risch | \(-\frac {x}{2}+\frac {\ln \left (5 x^{2}-x +{\mathrm e}^{2}\right )}{2}\) | \(19\) |
parallelrisch | \(-\frac {x}{2}+\frac {\ln \left (x^{2}-\frac {x}{5}+\frac {{\mathrm e}^{2}}{5}\right )}{2}\) | \(19\) |
norman | \(-\frac {x}{2}+\frac {\ln \left (2 \,{\mathrm e}^{2}+10 x^{2}-2 x \right )}{2}\) | \(21\) |
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Time = 0.29 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.53 \[ \int \frac {-1-e^2+11 x-5 x^2}{2 e^2-2 x+10 x^2} \, dx=-\frac {1}{2} \, x + \frac {1}{2} \, \log \left (5 \, x^{2} - x + e^{2}\right ) \]
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Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.44 \[ \int \frac {-1-e^2+11 x-5 x^2}{2 e^2-2 x+10 x^2} \, dx=- \frac {x}{2} + \frac {\log {\left (5 x^{2} - x + e^{2} \right )}}{2} \]
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Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.53 \[ \int \frac {-1-e^2+11 x-5 x^2}{2 e^2-2 x+10 x^2} \, dx=-\frac {1}{2} \, x + \frac {1}{2} \, \log \left (5 \, x^{2} - x + e^{2}\right ) \]
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Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.53 \[ \int \frac {-1-e^2+11 x-5 x^2}{2 e^2-2 x+10 x^2} \, dx=-\frac {1}{2} \, x + \frac {1}{2} \, \log \left (5 \, x^{2} - x + e^{2}\right ) \]
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Time = 13.13 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.53 \[ \int \frac {-1-e^2+11 x-5 x^2}{2 e^2-2 x+10 x^2} \, dx=\frac {\ln \left (5\,x^2-x+{\mathrm {e}}^2\right )}{2}-\frac {x}{2} \]
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