Integrand size = 41, antiderivative size = 23 \[ \int \frac {e^x (2-x)-14 x^3+e^{5-x} \left (14 x^3-14 x^4\right )}{14 x^3} \, dx=-\frac {e^x}{14 x^2}-x+e^{5-x} x \]
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Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {12, 14, 2207, 2225, 2228} \[ \int \frac {e^x (2-x)-14 x^3+e^{5-x} \left (14 x^3-14 x^4\right )}{14 x^3} \, dx=-\frac {e^x}{14 x^2}-e^{5-x} (1-x)+e^{5-x}-x \]
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Rule 12
Rule 14
Rule 2207
Rule 2225
Rule 2228
Rubi steps \begin{align*} \text {integral}& = \frac {1}{14} \int \frac {e^x (2-x)-14 x^3+e^{5-x} \left (14 x^3-14 x^4\right )}{x^3} \, dx \\ & = \frac {1}{14} \int \left (-14-14 e^{5-x} (-1+x)-\frac {e^x (-2+x)}{x^3}\right ) \, dx \\ & = -x-\frac {1}{14} \int \frac {e^x (-2+x)}{x^3} \, dx-\int e^{5-x} (-1+x) \, dx \\ & = -e^{5-x} (1-x)-\frac {e^x}{14 x^2}-x-\int e^{5-x} \, dx \\ & = e^{5-x}-e^{5-x} (1-x)-\frac {e^x}{14 x^2}-x \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^x (2-x)-14 x^3+e^{5-x} \left (14 x^3-14 x^4\right )}{14 x^3} \, dx=-\frac {e^x}{14 x^2}-x+e^{5-x} x \]
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Time = 0.47 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87
method | result | size |
risch | \(x \,{\mathrm e}^{5-x}-x -\frac {{\mathrm e}^{x}}{14 x^{2}}\) | \(20\) |
parallelrisch | \(\frac {14 \,{\mathrm e}^{5-x} x^{3}-14 x^{3}-{\mathrm e}^{x}}{14 x^{2}}\) | \(27\) |
norman | \(\frac {\left (x^{3} {\mathrm e}^{5}-\frac {{\mathrm e}^{2 x}}{14}-{\mathrm e}^{x} x^{3}\right ) {\mathrm e}^{-x}}{x^{2}}\) | \(29\) |
parts | \(-x -{\mathrm e}^{5-x} \left (5-x \right )+5 \,{\mathrm e}^{5-x}-\frac {{\mathrm e}^{x}}{14 x^{2}}\) | \(33\) |
default | \(-x -{\mathrm e}^{5} {\mathrm e}^{-x}-\frac {{\mathrm e}^{x}}{14 x^{2}}-{\mathrm e}^{5} \left (-x \,{\mathrm e}^{-x}-{\mathrm e}^{-x}\right )\) | \(38\) |
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none
Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {e^x (2-x)-14 x^3+e^{5-x} \left (14 x^3-14 x^4\right )}{14 x^3} \, dx=\frac {{\left (14 \, x^{3} e^{5} - 14 \, x^{3} e^{x} - e^{\left (2 \, x\right )}\right )} e^{\left (-x\right )}}{14 \, x^{2}} \]
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Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^x (2-x)-14 x^3+e^{5-x} \left (14 x^3-14 x^4\right )}{14 x^3} \, dx=- x + \frac {14 x^{3} e^{5} e^{- x} - e^{x}}{14 x^{2}} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {e^x (2-x)-14 x^3+e^{5-x} \left (14 x^3-14 x^4\right )}{14 x^3} \, dx={\left (x e^{5} + e^{5}\right )} e^{\left (-x\right )} - x - e^{\left (-x + 5\right )} - \frac {1}{14} \, \Gamma \left (-1, -x\right ) - \frac {1}{7} \, \Gamma \left (-2, -x\right ) \]
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Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {e^x (2-x)-14 x^3+e^{5-x} \left (14 x^3-14 x^4\right )}{14 x^3} \, dx=\frac {14 \, x^{3} e^{\left (-x + 5\right )} - 14 \, x^{3} - e^{x}}{14 \, x^{2}} \]
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Time = 14.06 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {e^x (2-x)-14 x^3+e^{5-x} \left (14 x^3-14 x^4\right )}{14 x^3} \, dx=x\,\left ({\mathrm {e}}^{5-x}-1\right )-\frac {{\mathrm {e}}^x}{14\,x^2} \]
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