Integrand size = 40, antiderivative size = 27 \[ \int \frac {6-6 x-6 x^2+4 x^3+e^x \left (4 x-x^2-x^3\right )}{9 e^3} \, dx=\frac {(2-x) \left (e^x+\frac {3}{x}-x\right ) x^2}{9 e^3} \]
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Leaf count is larger than twice the leaf count of optimal. \(63\) vs. \(2(27)=54\).
Time = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.33, number of steps used = 14, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 1608, 2227, 2207, 2225} \[ \int \frac {6-6 x-6 x^2+4 x^3+e^x \left (4 x-x^2-x^3\right )}{9 e^3} \, dx=\frac {x^4}{9 e^3}-\frac {1}{9} e^{x-3} x^3-\frac {2 x^3}{9 e^3}+\frac {2}{9} e^{x-3} x^2-\frac {x^2}{3 e^3}+\frac {2 x}{3 e^3} \]
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Rule 12
Rule 1608
Rule 2207
Rule 2225
Rule 2227
Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (6-6 x-6 x^2+4 x^3+e^x \left (4 x-x^2-x^3\right )\right ) \, dx}{9 e^3} \\ & = \frac {2 x}{3 e^3}-\frac {x^2}{3 e^3}-\frac {2 x^3}{9 e^3}+\frac {x^4}{9 e^3}+\frac {\int e^x \left (4 x-x^2-x^3\right ) \, dx}{9 e^3} \\ & = \frac {2 x}{3 e^3}-\frac {x^2}{3 e^3}-\frac {2 x^3}{9 e^3}+\frac {x^4}{9 e^3}+\frac {\int e^x x \left (4-x-x^2\right ) \, dx}{9 e^3} \\ & = \frac {2 x}{3 e^3}-\frac {x^2}{3 e^3}-\frac {2 x^3}{9 e^3}+\frac {x^4}{9 e^3}+\frac {\int \left (4 e^x x-e^x x^2-e^x x^3\right ) \, dx}{9 e^3} \\ & = \frac {2 x}{3 e^3}-\frac {x^2}{3 e^3}-\frac {2 x^3}{9 e^3}+\frac {x^4}{9 e^3}-\frac {\int e^x x^2 \, dx}{9 e^3}-\frac {\int e^x x^3 \, dx}{9 e^3}+\frac {4 \int e^x x \, dx}{9 e^3} \\ & = \frac {2 x}{3 e^3}+\frac {4}{9} e^{-3+x} x-\frac {x^2}{3 e^3}-\frac {1}{9} e^{-3+x} x^2-\frac {2 x^3}{9 e^3}-\frac {1}{9} e^{-3+x} x^3+\frac {x^4}{9 e^3}+\frac {2 \int e^x x \, dx}{9 e^3}+\frac {\int e^x x^2 \, dx}{3 e^3}-\frac {4 \int e^x \, dx}{9 e^3} \\ & = -\frac {4}{9} e^{-3+x}+\frac {2 x}{3 e^3}+\frac {2}{3} e^{-3+x} x-\frac {x^2}{3 e^3}+\frac {2}{9} e^{-3+x} x^2-\frac {2 x^3}{9 e^3}-\frac {1}{9} e^{-3+x} x^3+\frac {x^4}{9 e^3}-\frac {2 \int e^x \, dx}{9 e^3}-\frac {2 \int e^x x \, dx}{3 e^3} \\ & = -\frac {2}{3} e^{-3+x}+\frac {2 x}{3 e^3}-\frac {x^2}{3 e^3}+\frac {2}{9} e^{-3+x} x^2-\frac {2 x^3}{9 e^3}-\frac {1}{9} e^{-3+x} x^3+\frac {x^4}{9 e^3}+\frac {2 \int e^x \, dx}{3 e^3} \\ & = \frac {2 x}{3 e^3}-\frac {x^2}{3 e^3}+\frac {2}{9} e^{-3+x} x^2-\frac {2 x^3}{9 e^3}-\frac {1}{9} e^{-3+x} x^3+\frac {x^4}{9 e^3} \\ \end{align*}
Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {6-6 x-6 x^2+4 x^3+e^x \left (4 x-x^2-x^3\right )}{9 e^3} \, dx=\frac {(-2+x) x \left (-3-e^x x+x^2\right )}{9 e^3} \]
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Time = 0.16 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41
method | result | size |
default | \(\frac {{\mathrm e}^{-3} \left (6 x +2 \,{\mathrm e}^{x} x^{2}-{\mathrm e}^{x} x^{3}-3 x^{2}-2 x^{3}+x^{4}\right )}{9}\) | \(38\) |
parallelrisch | \(\frac {{\mathrm e}^{-3} \left (6 x +2 \,{\mathrm e}^{x} x^{2}-{\mathrm e}^{x} x^{3}-3 x^{2}-2 x^{3}+x^{4}\right )}{9}\) | \(38\) |
risch | \(\frac {{\mathrm e}^{-3} x^{4}}{9}-\frac {2 \,{\mathrm e}^{-3} x^{3}}{9}-\frac {{\mathrm e}^{-3} x^{2}}{3}+\frac {2 \,{\mathrm e}^{-3} x}{3}+\frac {\left (-x^{3}+2 x^{2}\right ) {\mathrm e}^{-3+x}}{9}\) | \(45\) |
parts | \(\frac {2 \,{\mathrm e}^{-3} \left (\frac {1}{2} x^{4}-x^{3}-\frac {3}{2} x^{2}+3 x \right )}{9}-\frac {{\mathrm e}^{-3} \left (-2 \,{\mathrm e}^{x} x^{2}+{\mathrm e}^{x} x^{3}\right )}{9}\) | \(47\) |
norman | \(\frac {2 \,{\mathrm e}^{-3} x}{3}-\frac {{\mathrm e}^{-3} x^{2}}{3}-\frac {2 \,{\mathrm e}^{-3} x^{3}}{9}+\frac {{\mathrm e}^{-3} x^{4}}{9}+\frac {2 x^{2} {\mathrm e}^{-3} {\mathrm e}^{x}}{9}-\frac {{\mathrm e}^{-3} x^{3} {\mathrm e}^{x}}{9}\) | \(58\) |
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Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {6-6 x-6 x^2+4 x^3+e^x \left (4 x-x^2-x^3\right )}{9 e^3} \, dx=\frac {1}{9} \, {\left (x^{4} - 2 \, x^{3} - 3 \, x^{2} - {\left (x^{3} - 2 \, x^{2}\right )} e^{x} + 6 \, x\right )} e^{\left (-3\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (22) = 44\).
Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.89 \[ \int \frac {6-6 x-6 x^2+4 x^3+e^x \left (4 x-x^2-x^3\right )}{9 e^3} \, dx=\frac {x^{4}}{9 e^{3}} - \frac {2 x^{3}}{9 e^{3}} - \frac {x^{2}}{3 e^{3}} + \frac {2 x}{3 e^{3}} + \frac {\left (- x^{3} + 2 x^{2}\right ) e^{x}}{9 e^{3}} \]
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Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {6-6 x-6 x^2+4 x^3+e^x \left (4 x-x^2-x^3\right )}{9 e^3} \, dx=\frac {1}{9} \, {\left (x^{4} - 2 \, x^{3} - 3 \, x^{2} - {\left (x^{3} - 2 \, x^{2}\right )} e^{x} + 6 \, x\right )} e^{\left (-3\right )} \]
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Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {6-6 x-6 x^2+4 x^3+e^x \left (4 x-x^2-x^3\right )}{9 e^3} \, dx=\frac {1}{9} \, {\left (x^{4} - 2 \, x^{3} - 3 \, x^{2} - {\left (x^{3} - 2 \, x^{2}\right )} e^{x} + 6 \, x\right )} e^{\left (-3\right )} \]
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Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {6-6 x-6 x^2+4 x^3+e^x \left (4 x-x^2-x^3\right )}{9 e^3} \, dx=-\frac {x\,{\mathrm {e}}^{-3}\,\left (x-2\right )\,\left (x\,{\mathrm {e}}^x-x^2+3\right )}{9} \]
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