Integrand size = 108, antiderivative size = 27 \[ \int \frac {e^{\frac {-60+12 e^4+120 x+3 x \log (x)}{-10 x+2 e^4 x+20 x^2}} \left (-300-12 e^8+e^4 (120-237 x)+1185 x-1170 x^2-30 x^2 \log (x)\right )}{50 x^2+2 e^8 x^2-200 x^3+200 x^4+e^4 \left (-20 x^2+40 x^3\right )} \, dx=1+e^{\frac {3}{2} \left (\frac {4}{x}+\frac {\log (x)}{-5+e^4+10 x}\right )} \]
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\[ \int \frac {e^{\frac {-60+12 e^4+120 x+3 x \log (x)}{-10 x+2 e^4 x+20 x^2}} \left (-300-12 e^8+e^4 (120-237 x)+1185 x-1170 x^2-30 x^2 \log (x)\right )}{50 x^2+2 e^8 x^2-200 x^3+200 x^4+e^4 \left (-20 x^2+40 x^3\right )} \, dx=\int \frac {\exp \left (\frac {-60+12 e^4+120 x+3 x \log (x)}{-10 x+2 e^4 x+20 x^2}\right ) \left (-300-12 e^8+e^4 (120-237 x)+1185 x-1170 x^2-30 x^2 \log (x)\right )}{50 x^2+2 e^8 x^2-200 x^3+200 x^4+e^4 \left (-20 x^2+40 x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {-60+12 e^4+120 x+3 x \log (x)}{-10 x+2 e^4 x+20 x^2}\right ) \left (-300-12 e^8+e^4 (120-237 x)+1185 x-1170 x^2-30 x^2 \log (x)\right )}{\left (50+2 e^8\right ) x^2-200 x^3+200 x^4+e^4 \left (-20 x^2+40 x^3\right )} \, dx \\ & = \int \frac {3 e^{6/x} x^{-2+\frac {3}{2 \left (-5+e^4+10 x\right )}} \left (-4 e^8-e^4 (-40+79 x)-5 \left (20-79 x+78 x^2\right )-10 x^2 \log (x)\right )}{2 \left (5-e^4-10 x\right )^2} \, dx \\ & = \frac {3}{2} \int \frac {e^{6/x} x^{-2+\frac {3}{2 \left (-5+e^4+10 x\right )}} \left (-4 e^8-e^4 (-40+79 x)-5 \left (20-79 x+78 x^2\right )-10 x^2 \log (x)\right )}{\left (5-e^4-10 x\right )^2} \, dx \\ & = \frac {3}{2} \int \left (\frac {e^{6/x} \left (20-4 e^4-39 x\right ) x^{-2+\frac {3}{2 \left (-5+e^4+10 x\right )}}}{-5+e^4+10 x}-\frac {10 e^{6/x} x^{\frac {3}{2 \left (-5+e^4+10 x\right )}} \log (x)}{\left (-5+e^4+10 x\right )^2}\right ) \, dx \\ & = \frac {3}{2} \int \frac {e^{6/x} \left (20-4 e^4-39 x\right ) x^{-2+\frac {3}{2 \left (-5+e^4+10 x\right )}}}{-5+e^4+10 x} \, dx-15 \int \frac {e^{6/x} x^{\frac {3}{2 \left (-5+e^4+10 x\right )}} \log (x)}{\left (-5+e^4+10 x\right )^2} \, dx \\ & = \frac {3}{2} \int \left (-\frac {39}{10} e^{6/x} x^{-2+\frac {3}{2 \left (-5+e^4+10 x\right )}}+\frac {e^{6/x} \left (5-e^4\right ) x^{-2+\frac {3}{2 \left (-5+e^4+10 x\right )}}}{10 \left (-5+e^4+10 x\right )}\right ) \, dx+15 \int \frac {\int \frac {e^{6/x} x^{\frac {3}{2 \left (-5+e^4+10 x\right )}}}{\left (-5+e^4+10 x\right )^2} \, dx}{x} \, dx-(15 \log (x)) \int \frac {e^{6/x} x^{\frac {3}{2 \left (-5+e^4+10 x\right )}}}{\left (-5+e^4+10 x\right )^2} \, dx \\ & = -\left (\frac {117}{20} \int e^{6/x} x^{-2+\frac {3}{2 \left (-5+e^4+10 x\right )}} \, dx\right )+15 \int \frac {\int \frac {e^{6/x} x^{\frac {3}{2 \left (-5+e^4+10 x\right )}}}{\left (-5+e^4+10 x\right )^2} \, dx}{x} \, dx+\frac {1}{20} \left (3 \left (5-e^4\right )\right ) \int \frac {e^{6/x} x^{-2+\frac {3}{2 \left (-5+e^4+10 x\right )}}}{-5+e^4+10 x} \, dx-(15 \log (x)) \int \frac {e^{6/x} x^{\frac {3}{2 \left (-5+e^4+10 x\right )}}}{\left (-5+e^4+10 x\right )^2} \, dx \\ \end{align*}
Time = 3.61 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\frac {-60+12 e^4+120 x+3 x \log (x)}{-10 x+2 e^4 x+20 x^2}} \left (-300-12 e^8+e^4 (120-237 x)+1185 x-1170 x^2-30 x^2 \log (x)\right )}{50 x^2+2 e^8 x^2-200 x^3+200 x^4+e^4 \left (-20 x^2+40 x^3\right )} \, dx=e^{6/x} x^{\frac {3}{2 \left (-5+e^4+10 x\right )}} \]
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Time = 1.42 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78
method | result | size |
risch | \(x^{\frac {3}{2 \left (10 x +{\mathrm e}^{4}-5\right )}} {\mathrm e}^{\frac {6}{x}}\) | \(21\) |
parallelrisch | \({\mathrm e}^{\frac {\frac {3 x \ln \left (x \right )}{2}+6 \,{\mathrm e}^{4}+60 x -30}{x \left (10 x +{\mathrm e}^{4}-5\right )}}\) | \(29\) |
norman | \(\frac {\left ({\mathrm e}^{4}-5\right ) x \,{\mathrm e}^{\frac {3 x \ln \left (x \right )+12 \,{\mathrm e}^{4}+120 x -60}{2 x \,{\mathrm e}^{4}+20 x^{2}-10 x}}+10 x^{2} {\mathrm e}^{\frac {3 x \ln \left (x \right )+12 \,{\mathrm e}^{4}+120 x -60}{2 x \,{\mathrm e}^{4}+20 x^{2}-10 x}}}{x \left (10 x +{\mathrm e}^{4}-5\right )}\) | \(90\) |
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Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\frac {-60+12 e^4+120 x+3 x \log (x)}{-10 x+2 e^4 x+20 x^2}} \left (-300-12 e^8+e^4 (120-237 x)+1185 x-1170 x^2-30 x^2 \log (x)\right )}{50 x^2+2 e^8 x^2-200 x^3+200 x^4+e^4 \left (-20 x^2+40 x^3\right )} \, dx=e^{\left (\frac {3 \, {\left (x \log \left (x\right ) + 40 \, x + 4 \, e^{4} - 20\right )}}{2 \, {\left (10 \, x^{2} + x e^{4} - 5 \, x\right )}}\right )} \]
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Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {e^{\frac {-60+12 e^4+120 x+3 x \log (x)}{-10 x+2 e^4 x+20 x^2}} \left (-300-12 e^8+e^4 (120-237 x)+1185 x-1170 x^2-30 x^2 \log (x)\right )}{50 x^2+2 e^8 x^2-200 x^3+200 x^4+e^4 \left (-20 x^2+40 x^3\right )} \, dx=e^{\frac {3 x \log {\left (x \right )} + 120 x - 60 + 12 e^{4}}{20 x^{2} - 10 x + 2 x e^{4}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (22) = 44\).
Time = 0.53 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.33 \[ \int \frac {e^{\frac {-60+12 e^4+120 x+3 x \log (x)}{-10 x+2 e^4 x+20 x^2}} \left (-300-12 e^8+e^4 (120-237 x)+1185 x-1170 x^2-30 x^2 \log (x)\right )}{50 x^2+2 e^8 x^2-200 x^3+200 x^4+e^4 \left (-20 x^2+40 x^3\right )} \, dx=e^{\left (-\frac {60 \, e^{4}}{10 \, x {\left (e^{4} - 5\right )} + e^{8} - 10 \, e^{4} + 25} + \frac {3 \, \log \left (x\right )}{2 \, {\left (10 \, x + e^{4} - 5\right )}} + \frac {300}{10 \, x {\left (e^{4} - 5\right )} + e^{8} - 10 \, e^{4} + 25} + \frac {60}{10 \, x + e^{4} - 5} + \frac {6 \, e^{4}}{x {\left (e^{4} - 5\right )}} - \frac {30}{x {\left (e^{4} - 5\right )}}\right )} \]
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\[ \int \frac {e^{\frac {-60+12 e^4+120 x+3 x \log (x)}{-10 x+2 e^4 x+20 x^2}} \left (-300-12 e^8+e^4 (120-237 x)+1185 x-1170 x^2-30 x^2 \log (x)\right )}{50 x^2+2 e^8 x^2-200 x^3+200 x^4+e^4 \left (-20 x^2+40 x^3\right )} \, dx=\int { -\frac {3 \, {\left (10 \, x^{2} \log \left (x\right ) + 390 \, x^{2} + {\left (79 \, x - 40\right )} e^{4} - 395 \, x + 4 \, e^{8} + 100\right )} e^{\left (\frac {3 \, {\left (x \log \left (x\right ) + 40 \, x + 4 \, e^{4} - 20\right )}}{2 \, {\left (10 \, x^{2} + x e^{4} - 5 \, x\right )}}\right )}}{2 \, {\left (100 \, x^{4} - 100 \, x^{3} + x^{2} e^{8} + 25 \, x^{2} + 10 \, {\left (2 \, x^{3} - x^{2}\right )} e^{4}\right )}} \,d x } \]
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Time = 9.38 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.44 \[ \int \frac {e^{\frac {-60+12 e^4+120 x+3 x \log (x)}{-10 x+2 e^4 x+20 x^2}} \left (-300-12 e^8+e^4 (120-237 x)+1185 x-1170 x^2-30 x^2 \log (x)\right )}{50 x^2+2 e^8 x^2-200 x^3+200 x^4+e^4 \left (-20 x^2+40 x^3\right )} \, dx=x^{\frac {3}{20\,x+2\,{\mathrm {e}}^4-10}}\,{\mathrm {e}}^{\frac {6\,{\mathrm {e}}^4}{x\,{\mathrm {e}}^4-5\,x+10\,x^2}}\,{\mathrm {e}}^{\frac {60}{10\,x+{\mathrm {e}}^4-5}}\,{\mathrm {e}}^{-\frac {30}{x\,{\mathrm {e}}^4-5\,x+10\,x^2}} \]
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