Integrand size = 48, antiderivative size = 19 \[ \int \frac {45-4 x-30 x^2+5 x^4+e^{2+x} \left (18-12 x^2+2 x^4\right )}{18-12 x^2+2 x^4} \, dx=-7+e^{2+x}+\frac {5 x}{2}+\frac {1}{-3+x^2} \]
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Leaf count is larger than twice the leaf count of optimal. \(52\) vs. \(2(19)=38\).
Time = 0.12 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.74, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {28, 6874, 2225, 205, 213, 267, 294, 327} \[ \int \frac {45-4 x-30 x^2+5 x^4+e^{2+x} \left (18-12 x^2+2 x^4\right )}{18-12 x^2+2 x^4} \, dx=-\frac {15 x}{4 \left (3-x^2\right )}-\frac {1}{3-x^2}+\frac {5 x^3}{4 \left (3-x^2\right )}+\frac {15 x}{4}+e^{x+2} \]
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Rule 28
Rule 205
Rule 213
Rule 267
Rule 294
Rule 327
Rule 2225
Rule 6874
Rubi steps \begin{align*} \text {integral}& = 2 \int \frac {45-4 x-30 x^2+5 x^4+e^{2+x} \left (18-12 x^2+2 x^4\right )}{\left (-6+2 x^2\right )^2} \, dx \\ & = 2 \int \left (\frac {e^{2+x}}{2}+\frac {45}{4 \left (-3+x^2\right )^2}-\frac {x}{\left (-3+x^2\right )^2}-\frac {15 x^2}{2 \left (-3+x^2\right )^2}+\frac {5 x^4}{4 \left (-3+x^2\right )^2}\right ) \, dx \\ & = -\left (2 \int \frac {x}{\left (-3+x^2\right )^2} \, dx\right )+\frac {5}{2} \int \frac {x^4}{\left (-3+x^2\right )^2} \, dx-15 \int \frac {x^2}{\left (-3+x^2\right )^2} \, dx+\frac {45}{2} \int \frac {1}{\left (-3+x^2\right )^2} \, dx+\int e^{2+x} \, dx \\ & = e^{2+x}-\frac {1}{3-x^2}-\frac {15 x}{4 \left (3-x^2\right )}+\frac {5 x^3}{4 \left (3-x^2\right )}-\frac {15}{4} \int \frac {1}{-3+x^2} \, dx+\frac {15}{4} \int \frac {x^2}{-3+x^2} \, dx-\frac {15}{2} \int \frac {1}{-3+x^2} \, dx \\ & = e^{2+x}+\frac {15 x}{4}-\frac {1}{3-x^2}-\frac {15 x}{4 \left (3-x^2\right )}+\frac {5 x^3}{4 \left (3-x^2\right )}+\frac {15}{4} \sqrt {3} \tanh ^{-1}\left (\frac {x}{\sqrt {3}}\right )+\frac {45}{4} \int \frac {1}{-3+x^2} \, dx \\ & = e^{2+x}+\frac {15 x}{4}-\frac {1}{3-x^2}-\frac {15 x}{4 \left (3-x^2\right )}+\frac {5 x^3}{4 \left (3-x^2\right )} \\ \end{align*}
Time = 0.45 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {45-4 x-30 x^2+5 x^4+e^{2+x} \left (18-12 x^2+2 x^4\right )}{18-12 x^2+2 x^4} \, dx=\frac {1}{2} \left (2 e^{2+x}+5 x+\frac {2}{-3+x^2}\right ) \]
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Time = 0.73 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84
method | result | size |
risch | \(\frac {5 x}{2}+\frac {1}{x^{2}-3}+{\mathrm e}^{2+x}\) | \(16\) |
parts | \(\frac {5 x}{2}+\frac {1}{x^{2}-3}+{\mathrm e}^{2+x}\) | \(16\) |
norman | \(\frac {x^{2} {\mathrm e}^{2+x}-\frac {15 x}{2}+\frac {5 x^{3}}{2}-3 \,{\mathrm e}^{2+x}+1}{x^{2}-3}\) | \(33\) |
parallelrisch | \(\frac {2 x^{2} {\mathrm e}^{2+x}+5 x^{3}-6 \,{\mathrm e}^{2+x}-15 x +2}{2 x^{2}-6}\) | \(35\) |
derivativedivides | \(-\frac {13 x}{12 \left (\left (2+x \right )^{2}-7-4 x \right )}+\frac {\frac {22 x}{3}+11}{\left (2+x \right )^{2}-7-4 x}+\frac {-90-\frac {105 x}{2}}{\left (2+x \right )^{2}-7-4 x}-\frac {20 \left (-\frac {15}{2}-\frac {13 x}{3}\right )}{\left (2+x \right )^{2}-7-4 x}+5+\frac {5 x}{2}+\frac {-70-\frac {485 x}{12}}{\left (2+x \right )^{2}-7-4 x}-\frac {{\mathrm e}^{2+x} x}{6 \left (\left (2+x \right )^{2}-7-4 x \right )}+\frac {4 \,{\mathrm e}^{2+x} \left (3+2 x \right )}{3 \left (\left (2+x \right )^{2}-7-4 x \right )}-\frac {3 \,{\mathrm e}^{2+x} \left (12+7 x \right )}{\left (2+x \right )^{2}-7-4 x}+\frac {4 \,{\mathrm e}^{2+x} \left (45+26 x \right )}{3 \left (\left (2+x \right )^{2}-7-4 x \right )}+{\mathrm e}^{2+x}-\frac {{\mathrm e}^{2+x} \left (168+97 x \right )}{6 \left (\left (2+x \right )^{2}-7-4 x \right )}\) | \(212\) |
default | \(-\frac {13 x}{12 \left (\left (2+x \right )^{2}-7-4 x \right )}+\frac {\frac {22 x}{3}+11}{\left (2+x \right )^{2}-7-4 x}+\frac {-90-\frac {105 x}{2}}{\left (2+x \right )^{2}-7-4 x}-\frac {20 \left (-\frac {15}{2}-\frac {13 x}{3}\right )}{\left (2+x \right )^{2}-7-4 x}+5+\frac {5 x}{2}+\frac {-70-\frac {485 x}{12}}{\left (2+x \right )^{2}-7-4 x}-\frac {{\mathrm e}^{2+x} x}{6 \left (\left (2+x \right )^{2}-7-4 x \right )}+\frac {4 \,{\mathrm e}^{2+x} \left (3+2 x \right )}{3 \left (\left (2+x \right )^{2}-7-4 x \right )}-\frac {3 \,{\mathrm e}^{2+x} \left (12+7 x \right )}{\left (2+x \right )^{2}-7-4 x}+\frac {4 \,{\mathrm e}^{2+x} \left (45+26 x \right )}{3 \left (\left (2+x \right )^{2}-7-4 x \right )}+{\mathrm e}^{2+x}-\frac {{\mathrm e}^{2+x} \left (168+97 x \right )}{6 \left (\left (2+x \right )^{2}-7-4 x \right )}\) | \(212\) |
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Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \frac {45-4 x-30 x^2+5 x^4+e^{2+x} \left (18-12 x^2+2 x^4\right )}{18-12 x^2+2 x^4} \, dx=\frac {5 \, x^{3} + 2 \, {\left (x^{2} - 3\right )} e^{\left (x + 2\right )} - 15 \, x + 2}{2 \, {\left (x^{2} - 3\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {45-4 x-30 x^2+5 x^4+e^{2+x} \left (18-12 x^2+2 x^4\right )}{18-12 x^2+2 x^4} \, dx=\frac {5 x}{2} + e^{x + 2} + \frac {1}{x^{2} - 3} \]
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Time = 0.31 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {45-4 x-30 x^2+5 x^4+e^{2+x} \left (18-12 x^2+2 x^4\right )}{18-12 x^2+2 x^4} \, dx=\frac {5}{2} \, x + \frac {1}{x^{2} - 3} + e^{\left (x + 2\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (16) = 32\).
Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.79 \[ \int \frac {45-4 x-30 x^2+5 x^4+e^{2+x} \left (18-12 x^2+2 x^4\right )}{18-12 x^2+2 x^4} \, dx=\frac {5 \, x^{3} + 2 \, x^{2} e^{\left (x + 2\right )} - 15 \, x - 6 \, e^{\left (x + 2\right )} + 2}{2 \, {\left (x^{2} - 3\right )}} \]
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Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {45-4 x-30 x^2+5 x^4+e^{2+x} \left (18-12 x^2+2 x^4\right )}{18-12 x^2+2 x^4} \, dx=\frac {5\,x}{2}+{\mathrm {e}}^{x+2}+\frac {2}{2\,x^2-6} \]
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