Integrand size = 26, antiderivative size = 25 \[ \int \frac {12-17 x-e^2 x+e^x \left (x-x^2\right )}{x^3} \, dx=\frac {18+e^2-e^x+2 x-\frac {6+x}{x}}{x} \]
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Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6, 14, 2228, 37} \[ \int \frac {12-17 x-e^2 x+e^x \left (x-x^2\right )}{x^3} \, dx=-\frac {\left (12-\left (17+e^2\right ) x\right )^2}{24 x^2}-\frac {e^x}{x} \]
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Rule 6
Rule 14
Rule 37
Rule 2228
Rubi steps \begin{align*} \text {integral}& = \int \frac {12+\left (-17-e^2\right ) x+e^x \left (x-x^2\right )}{x^3} \, dx \\ & = \int \left (-\frac {e^x (-1+x)}{x^2}+\frac {12-\left (17+e^2\right ) x}{x^3}\right ) \, dx \\ & = -\int \frac {e^x (-1+x)}{x^2} \, dx+\int \frac {12-\left (17+e^2\right ) x}{x^3} \, dx \\ & = -\frac {e^x}{x}-\frac {\left (12-\left (17+e^2\right ) x\right )^2}{24 x^2} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {12-17 x-e^2 x+e^x \left (x-x^2\right )}{x^3} \, dx=-\frac {6+\left (-17-e^2+e^x\right ) x}{x^2} \]
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Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72
method | result | size |
norman | \(\frac {-6+\left ({\mathrm e}^{2}+17\right ) x -{\mathrm e}^{x} x}{x^{2}}\) | \(18\) |
parallelrisch | \(\frac {{\mathrm e}^{2} x -{\mathrm e}^{x} x +17 x -6}{x^{2}}\) | \(19\) |
risch | \(\frac {\left ({\mathrm e}^{2}+17\right ) x -6}{x^{2}}-\frac {{\mathrm e}^{x}}{x}\) | \(21\) |
parts | \(\frac {{\mathrm e}^{2}+17}{x}-\frac {6}{x^{2}}-\frac {{\mathrm e}^{x}}{x}\) | \(22\) |
default | \(-\frac {{\mathrm e}^{x}}{x}-\frac {6}{x^{2}}+\frac {17}{x}+\frac {{\mathrm e}^{2}}{x}\) | \(25\) |
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Time = 0.29 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {12-17 x-e^2 x+e^x \left (x-x^2\right )}{x^3} \, dx=\frac {x e^{2} - x e^{x} + 17 \, x - 6}{x^{2}} \]
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Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {12-17 x-e^2 x+e^x \left (x-x^2\right )}{x^3} \, dx=- \frac {e^{x}}{x} - \frac {x \left (-17 - e^{2}\right ) + 6}{x^{2}} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {12-17 x-e^2 x+e^x \left (x-x^2\right )}{x^3} \, dx=\frac {e^{2}}{x} + \frac {17}{x} - \frac {6}{x^{2}} - {\rm Ei}\left (x\right ) + \Gamma \left (-1, -x\right ) \]
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none
Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {12-17 x-e^2 x+e^x \left (x-x^2\right )}{x^3} \, dx=\frac {x e^{2} - x e^{x} + 17 \, x - 6}{x^{2}} \]
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Time = 12.59 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64 \[ \int \frac {12-17 x-e^2 x+e^x \left (x-x^2\right )}{x^3} \, dx=\frac {x\,\left ({\mathrm {e}}^2-{\mathrm {e}}^x+17\right )-6}{x^2} \]
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