Integrand size = 36, antiderivative size = 21 \[ \int \frac {-2 e^3+e^x \left (e^3 x-x^2\right )}{2 e^3 x-2 x^2} \, dx=-1+\frac {e^x}{2}+\log \left (\frac {-e^3+x}{x}\right ) \]
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Time = 0.16 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1607, 6874, 2225, 36, 31, 29} \[ \int \frac {-2 e^3+e^x \left (e^3 x-x^2\right )}{2 e^3 x-2 x^2} \, dx=\frac {e^x}{2}+\log \left (e^3-x\right )-\log (x) \]
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Rule 29
Rule 31
Rule 36
Rule 1607
Rule 2225
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-2 e^3+e^x \left (e^3 x-x^2\right )}{\left (2 e^3-2 x\right ) x} \, dx \\ & = \int \left (\frac {e^x}{2}-\frac {e^3}{\left (e^3-x\right ) x}\right ) \, dx \\ & = \frac {\int e^x \, dx}{2}-e^3 \int \frac {1}{\left (e^3-x\right ) x} \, dx \\ & = \frac {e^x}{2}-\int \frac {1}{e^3-x} \, dx-\int \frac {1}{x} \, dx \\ & = \frac {e^x}{2}+\log \left (e^3-x\right )-\log (x) \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {-2 e^3+e^x \left (e^3 x-x^2\right )}{2 e^3 x-2 x^2} \, dx=\frac {e^x}{2}+\log \left (e^3-x\right )-\log (x) \]
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Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81
method | result | size |
default | \(-\ln \left (x \right )+\ln \left (x -{\mathrm e}^{3}\right )+\frac {{\mathrm e}^{x}}{2}\) | \(17\) |
norman | \(\frac {{\mathrm e}^{x}}{2}-\ln \left (x \right )+\ln \left (-x +{\mathrm e}^{3}\right )\) | \(17\) |
risch | \(-\ln \left (x \right )+\ln \left (x -{\mathrm e}^{3}\right )+\frac {{\mathrm e}^{x}}{2}\) | \(17\) |
parallelrisch | \(-\ln \left (x \right )+\ln \left (x -{\mathrm e}^{3}\right )+\frac {{\mathrm e}^{x}}{2}\) | \(17\) |
parts | \(-\ln \left (x \right )+\ln \left (x -{\mathrm e}^{3}\right )+\frac {{\mathrm e}^{x}}{2}\) | \(17\) |
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Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {-2 e^3+e^x \left (e^3 x-x^2\right )}{2 e^3 x-2 x^2} \, dx=\frac {1}{2} \, e^{x} + \log \left (x - e^{3}\right ) - \log \left (x\right ) \]
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Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {-2 e^3+e^x \left (e^3 x-x^2\right )}{2 e^3 x-2 x^2} \, dx=\left (- \frac {\log {\left (x \right )}}{e^{3}} + \frac {\log {\left (x - e^{3} \right )}}{e^{3}}\right ) e^{3} + \frac {e^{x}}{2} \]
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Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {-2 e^3+e^x \left (e^3 x-x^2\right )}{2 e^3 x-2 x^2} \, dx={\left (e^{\left (-3\right )} \log \left (x - e^{3}\right ) - e^{\left (-3\right )} \log \left (x\right )\right )} e^{3} + \frac {1}{2} \, e^{x} \]
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Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {-2 e^3+e^x \left (e^3 x-x^2\right )}{2 e^3 x-2 x^2} \, dx=\frac {1}{2} \, e^{x} + \log \left (x - e^{3}\right ) - \log \left (x\right ) \]
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Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {-2 e^3+e^x \left (e^3 x-x^2\right )}{2 e^3 x-2 x^2} \, dx=\ln \left (x-{\mathrm {e}}^3\right )+\frac {{\mathrm {e}}^x}{2}-\ln \left (x\right ) \]
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