Integrand size = 54, antiderivative size = 26 \[ \int \frac {-6 x-16 x^3+e^{10} \left (1-6 x+4 x^2-16 x^3\right )+e^5 \left (12 x+32 x^3\right )}{1-2 e^5+e^{10}} \, dx=\left (x^2+\frac {4 x^4}{3}\right ) \left (-3+\frac {x}{\left (x-\frac {x}{e^5}\right )^2}\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(130\) vs. \(2(26)=52\).
Time = 0.02 (sec) , antiderivative size = 130, normalized size of antiderivative = 5.00, number of steps used = 4, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {12} \[ \int \frac {-6 x-16 x^3+e^{10} \left (1-6 x+4 x^2-16 x^3\right )+e^5 \left (12 x+32 x^3\right )}{1-2 e^5+e^{10}} \, dx=-\frac {4 e^{10} x^4}{\left (1-e^5\right )^2}+\frac {8 e^5 x^4}{\left (1-e^5\right )^2}-\frac {4 x^4}{\left (1-e^5\right )^2}+\frac {4 e^{10} x^3}{3 \left (1-e^5\right )^2}-\frac {3 e^{10} x^2}{\left (1-e^5\right )^2}+\frac {6 e^5 x^2}{\left (1-e^5\right )^2}-\frac {3 x^2}{\left (1-e^5\right )^2}+\frac {e^{10} x}{\left (1-e^5\right )^2} \]
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Rule 12
Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (-6 x-16 x^3+e^{10} \left (1-6 x+4 x^2-16 x^3\right )+e^5 \left (12 x+32 x^3\right )\right ) \, dx}{1-2 e^5+e^{10}} \\ & = -\frac {3 x^2}{\left (1-e^5\right )^2}-\frac {4 x^4}{\left (1-e^5\right )^2}+\frac {e^5 \int \left (12 x+32 x^3\right ) \, dx}{\left (1-e^5\right )^2}+\frac {e^{10} \int \left (1-6 x+4 x^2-16 x^3\right ) \, dx}{\left (1-e^5\right )^2} \\ & = \frac {e^{10} x}{\left (1-e^5\right )^2}-\frac {3 x^2}{\left (1-e^5\right )^2}+\frac {6 e^5 x^2}{\left (1-e^5\right )^2}-\frac {3 e^{10} x^2}{\left (1-e^5\right )^2}+\frac {4 e^{10} x^3}{3 \left (1-e^5\right )^2}-\frac {4 x^4}{\left (1-e^5\right )^2}+\frac {8 e^5 x^4}{\left (1-e^5\right )^2}-\frac {4 e^{10} x^4}{\left (1-e^5\right )^2} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(66\) vs. \(2(26)=52\).
Time = 0.01 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.54 \[ \int \frac {-6 x-16 x^3+e^{10} \left (1-6 x+4 x^2-16 x^3\right )+e^5 \left (12 x+32 x^3\right )}{1-2 e^5+e^{10}} \, dx=\frac {e^{10} x-3 x^2+6 e^5 x^2-3 e^{10} x^2+\frac {4 e^{10} x^3}{3}-4 x^4+8 e^5 x^4-4 e^{10} x^4}{\left (-1+e^5\right )^2} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(55\) vs. \(2(25)=50\).
Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.15
method | result | size |
norman | \(\frac {\left (-4 \,{\mathrm e}^{5}+4\right ) x^{4}+\left (-3 \,{\mathrm e}^{5}+3\right ) x^{2}+\frac {{\mathrm e}^{10} x}{{\mathrm e}^{5}-1}+\frac {4 \,{\mathrm e}^{10} x^{3}}{3 \left ({\mathrm e}^{5}-1\right )}}{{\mathrm e}^{5}-1}\) | \(56\) |
gosper | \(-\frac {x \left (12 x^{3} {\mathrm e}^{10}-24 x^{3} {\mathrm e}^{5}-4 x^{2} {\mathrm e}^{10}+12 x^{3}+9 x \,{\mathrm e}^{10}-18 x \,{\mathrm e}^{5}-3 \,{\mathrm e}^{10}+9 x \right )}{3 \left ({\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1\right )}\) | \(68\) |
default | \(\frac {-4 x^{4} {\mathrm e}^{10}+\frac {4 x^{3} {\mathrm e}^{10}}{3}+8 x^{4} {\mathrm e}^{5}-3 x^{2} {\mathrm e}^{10}-4 x^{4}+x \,{\mathrm e}^{10}+6 x^{2} {\mathrm e}^{5}-3 x^{2}}{{\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1}\) | \(72\) |
parallelrisch | \(\frac {-4 x^{4} {\mathrm e}^{10}+\frac {4 x^{3} {\mathrm e}^{10}}{3}+8 x^{4} {\mathrm e}^{5}-3 x^{2} {\mathrm e}^{10}-4 x^{4}+x \,{\mathrm e}^{10}+6 x^{2} {\mathrm e}^{5}-3 x^{2}}{{\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1}\) | \(72\) |
risch | \(-\frac {4 x^{4} {\mathrm e}^{10}}{{\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1}+\frac {4 x^{3} {\mathrm e}^{10}}{3 \left ({\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1\right )}+\frac {8 x^{4} {\mathrm e}^{5}}{{\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1}-\frac {3 x^{2} {\mathrm e}^{10}}{{\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1}-\frac {4 x^{4}}{{\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1}+\frac {x \,{\mathrm e}^{10}}{{\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1}+\frac {6 x^{2} {\mathrm e}^{5}}{{\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1}-\frac {3 x^{2}}{{\mathrm e}^{10}-2 \,{\mathrm e}^{5}+1}\) | \(131\) |
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Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (27) = 54\).
Time = 0.23 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.31 \[ \int \frac {-6 x-16 x^3+e^{10} \left (1-6 x+4 x^2-16 x^3\right )+e^5 \left (12 x+32 x^3\right )}{1-2 e^5+e^{10}} \, dx=-\frac {12 \, x^{4} + 9 \, x^{2} + {\left (12 \, x^{4} - 4 \, x^{3} + 9 \, x^{2} - 3 \, x\right )} e^{10} - 6 \, {\left (4 \, x^{4} + 3 \, x^{2}\right )} e^{5}}{3 \, {\left (e^{10} - 2 \, e^{5} + 1\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (20) = 40\).
Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {-6 x-16 x^3+e^{10} \left (1-6 x+4 x^2-16 x^3\right )+e^5 \left (12 x+32 x^3\right )}{1-2 e^5+e^{10}} \, dx=- 4 x^{4} + \frac {4 x^{3} e^{10}}{- 6 e^{5} + 3 + 3 e^{10}} - 3 x^{2} + \frac {x e^{10}}{- 2 e^{5} + 1 + e^{10}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (27) = 54\).
Time = 0.17 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.31 \[ \int \frac {-6 x-16 x^3+e^{10} \left (1-6 x+4 x^2-16 x^3\right )+e^5 \left (12 x+32 x^3\right )}{1-2 e^5+e^{10}} \, dx=-\frac {12 \, x^{4} + 9 \, x^{2} + {\left (12 \, x^{4} - 4 \, x^{3} + 9 \, x^{2} - 3 \, x\right )} e^{10} - 6 \, {\left (4 \, x^{4} + 3 \, x^{2}\right )} e^{5}}{3 \, {\left (e^{10} - 2 \, e^{5} + 1\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (27) = 54\).
Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.31 \[ \int \frac {-6 x-16 x^3+e^{10} \left (1-6 x+4 x^2-16 x^3\right )+e^5 \left (12 x+32 x^3\right )}{1-2 e^5+e^{10}} \, dx=-\frac {12 \, x^{4} + 9 \, x^{2} + {\left (12 \, x^{4} - 4 \, x^{3} + 9 \, x^{2} - 3 \, x\right )} e^{10} - 6 \, {\left (4 \, x^{4} + 3 \, x^{2}\right )} e^{5}}{3 \, {\left (e^{10} - 2 \, e^{5} + 1\right )}} \]
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Time = 11.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {-6 x-16 x^3+e^{10} \left (1-6 x+4 x^2-16 x^3\right )+e^5 \left (12 x+32 x^3\right )}{1-2 e^5+e^{10}} \, dx=-4\,x^4+\frac {4\,{\mathrm {e}}^{10}\,x^3}{3\,{\left ({\mathrm {e}}^5-1\right )}^2}-3\,x^2+\frac {{\mathrm {e}}^{10}\,x}{{\left ({\mathrm {e}}^5-1\right )}^2} \]
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