Integrand size = 149, antiderivative size = 21 \[ \int \frac {64+48 e^3 x+12 e^6 x^2+e^9 x^3+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (144 x^3+18 e^3 x^4\right )}{-320+64 x+e^9 (-5+x) x^3+e^6 x^2 (-60+12 x)+e^3 x (-240+48 x)+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (576+432 e^3 x+108 e^6 x^2+9 e^9 x^3\right )} \, dx=\log \left (-5+9 e^{\frac {x^4}{\left (4+e^3 x\right )^2}}+x\right ) \]
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Time = 0.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {6820, 6816} \[ \int \frac {64+48 e^3 x+12 e^6 x^2+e^9 x^3+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (144 x^3+18 e^3 x^4\right )}{-320+64 x+e^9 (-5+x) x^3+e^6 x^2 (-60+12 x)+e^3 x (-240+48 x)+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (576+432 e^3 x+108 e^6 x^2+9 e^9 x^3\right )} \, dx=\log \left (-9 e^{\frac {x^4}{\left (e^3 x+4\right )^2}}-x+5\right ) \]
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Rule 6816
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {-64-48 e^3 x-12 e^6 x^2-e^9 x^3-144 e^{\frac {x^4}{\left (4+e^3 x\right )^2}} x^3-18 e^{3+\frac {x^4}{\left (4+e^3 x\right )^2}} x^4}{\left (5-9 e^{\frac {x^4}{\left (4+e^3 x\right )^2}}-x\right ) \left (4+e^3 x\right )^3} \, dx \\ & = \log \left (5-9 e^{\frac {x^4}{\left (4+e^3 x\right )^2}}-x\right ) \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(57\) vs. \(2(21)=42\).
Time = 0.13 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.71 \[ \int \frac {64+48 e^3 x+12 e^6 x^2+e^9 x^3+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (144 x^3+18 e^3 x^4\right )}{-320+64 x+e^9 (-5+x) x^3+e^6 x^2 (-60+12 x)+e^3 x (-240+48 x)+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (576+432 e^3 x+108 e^6 x^2+9 e^9 x^3\right )} \, dx=\log \left (5-9 e^{\frac {48}{e^{12}}-\frac {8 x}{e^9}+\frac {x^2}{e^6}+\frac {256}{e^{12} \left (4+e^3 x\right )^2}-\frac {256}{e^{12} \left (4+e^3 x\right )}}-x\right ) \]
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Time = 20.81 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38
method | result | size |
norman | \(\ln \left (x +9 \,{\mathrm e}^{\frac {x^{4}}{x^{2} {\mathrm e}^{6}+8 x \,{\mathrm e}^{3}+16}}-5\right )\) | \(29\) |
parallelrisch | \(\ln \left (9 \,{\mathrm e}^{\frac {x^{4}}{x^{2} {\mathrm e}^{6}+8 \,{\mathrm e}^{3+\ln \left (x \right )}+16}}+x -5\right )\) | \(30\) |
risch | \({\mathrm e}^{3} {\mathrm e}^{-9} x^{2}-8 x \,{\mathrm e}^{-9}+\frac {\left (-256 x -768 \,{\mathrm e}^{-3}\right ) {\mathrm e}^{-9}}{x^{2} {\mathrm e}^{6}+8 x \,{\mathrm e}^{3}+16}-\frac {x^{4}}{x^{2} {\mathrm e}^{6}+8 x \,{\mathrm e}^{3}+16}+\ln \left ({\mathrm e}^{\frac {x^{4}}{x^{2} {\mathrm e}^{6}+8 x \,{\mathrm e}^{3}+16}}-\frac {5}{9}+\frac {x}{9}\right )\) | \(87\) |
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Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {64+48 e^3 x+12 e^6 x^2+e^9 x^3+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (144 x^3+18 e^3 x^4\right )}{-320+64 x+e^9 (-5+x) x^3+e^6 x^2 (-60+12 x)+e^3 x (-240+48 x)+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (576+432 e^3 x+108 e^6 x^2+9 e^9 x^3\right )} \, dx=\log \left (x + 9 \, e^{\left (\frac {x^{4}}{x^{2} e^{6} + 8 \, x e^{3} + 16}\right )} - 5\right ) \]
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Time = 0.37 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {64+48 e^3 x+12 e^6 x^2+e^9 x^3+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (144 x^3+18 e^3 x^4\right )}{-320+64 x+e^9 (-5+x) x^3+e^6 x^2 (-60+12 x)+e^3 x (-240+48 x)+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (576+432 e^3 x+108 e^6 x^2+9 e^9 x^3\right )} \, dx=\log {\left (\frac {x}{9} + e^{\frac {x^{4}}{x^{2} e^{6} + 8 x e^{3} + 16}} - \frac {5}{9} \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (20) = 40\).
Time = 0.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 5.14 \[ \int \frac {64+48 e^3 x+12 e^6 x^2+e^9 x^3+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (144 x^3+18 e^3 x^4\right )}{-320+64 x+e^9 (-5+x) x^3+e^6 x^2 (-60+12 x)+e^3 x (-240+48 x)+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (576+432 e^3 x+108 e^6 x^2+9 e^9 x^3\right )} \, dx=\frac {x^{3} e^{9} - 4 \, x^{2} e^{6} - 32 \, x e^{3} - 256}{x e^{15} + 4 \, e^{12}} + \log \left (\frac {1}{9} \, {\left ({\left (x - 5\right )} e^{\left (8 \, x e^{\left (-9\right )} + \frac {256}{x e^{15} + 4 \, e^{12}}\right )} + 9 \, e^{\left (x^{2} e^{\left (-6\right )} + \frac {256}{x^{2} e^{18} + 8 \, x e^{15} + 16 \, e^{12}} + 48 \, e^{\left (-12\right )}\right )}\right )} e^{\left (-x^{2} e^{\left (-6\right )} - 48 \, e^{\left (-12\right )}\right )}\right ) \]
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Time = 0.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {64+48 e^3 x+12 e^6 x^2+e^9 x^3+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (144 x^3+18 e^3 x^4\right )}{-320+64 x+e^9 (-5+x) x^3+e^6 x^2 (-60+12 x)+e^3 x (-240+48 x)+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (576+432 e^3 x+108 e^6 x^2+9 e^9 x^3\right )} \, dx=\log \left (x + 9 \, e^{\left (\frac {x^{4}}{x^{2} e^{6} + 8 \, x e^{3} + 16}\right )} - 5\right ) \]
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Time = 14.67 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {64+48 e^3 x+12 e^6 x^2+e^9 x^3+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (144 x^3+18 e^3 x^4\right )}{-320+64 x+e^9 (-5+x) x^3+e^6 x^2 (-60+12 x)+e^3 x (-240+48 x)+e^{\frac {x^4}{16+8 e^3 x+e^6 x^2}} \left (576+432 e^3 x+108 e^6 x^2+9 e^9 x^3\right )} \, dx=\ln \left (x+9\,{\mathrm {e}}^{\frac {x^4}{{\left (x\,{\mathrm {e}}^3+4\right )}^2}}-5\right ) \]
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