Integrand size = 79, antiderivative size = 23 \[ \int \frac {e^{e^{\frac {-137-34 x-x^2}{5+x}}} \left (e \left (25+10 x+x^2\right )+e^{1+\frac {-137-34 x-x^2}{5+x}} \left (-33 x-10 x^2-x^3\right )\right )}{300+120 x+12 x^2} \, dx=\frac {1}{12} e^{1+e^{-29-x+\frac {8}{5+x}}} x \]
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Leaf count is larger than twice the leaf count of optimal. \(70\) vs. \(2(23)=46\).
Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 3.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {27, 12, 2326} \[ \int \frac {e^{e^{\frac {-137-34 x-x^2}{5+x}}} \left (e \left (25+10 x+x^2\right )+e^{1+\frac {-137-34 x-x^2}{5+x}} \left (-33 x-10 x^2-x^3\right )\right )}{300+120 x+12 x^2} \, dx=\frac {e^{e^{-\frac {x^2+34 x+137}{x+5}}+1} \left (x^3+10 x^2+33 x\right )}{12 (x+5)^2 \left (\frac {2 (x+17)}{x+5}-\frac {x^2+34 x+137}{(x+5)^2}\right )} \]
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Rule 12
Rule 27
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{e^{\frac {-137-34 x-x^2}{5+x}}} \left (e \left (25+10 x+x^2\right )+e^{1+\frac {-137-34 x-x^2}{5+x}} \left (-33 x-10 x^2-x^3\right )\right )}{12 (5+x)^2} \, dx \\ & = \frac {1}{12} \int \frac {e^{e^{\frac {-137-34 x-x^2}{5+x}}} \left (e \left (25+10 x+x^2\right )+e^{1+\frac {-137-34 x-x^2}{5+x}} \left (-33 x-10 x^2-x^3\right )\right )}{(5+x)^2} \, dx \\ & = \frac {e^{1+e^{-\frac {137+34 x+x^2}{5+x}}} \left (33 x+10 x^2+x^3\right )}{12 (5+x)^2 \left (\frac {2 (17+x)}{5+x}-\frac {137+34 x+x^2}{(5+x)^2}\right )} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {e^{e^{\frac {-137-34 x-x^2}{5+x}}} \left (e \left (25+10 x+x^2\right )+e^{1+\frac {-137-34 x-x^2}{5+x}} \left (-33 x-10 x^2-x^3\right )\right )}{300+120 x+12 x^2} \, dx=\frac {1}{12} e^{1+e^{-\frac {137+34 x+x^2}{5+x}}} x \]
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Time = 0.87 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {x \,{\mathrm e}^{1+{\mathrm e}^{-\frac {x^{2}+34 x +137}{5+x}}}}{12}\) | \(23\) |
parallelrisch | \(\frac {x \,{\mathrm e} \,{\mathrm e}^{{\mathrm e}^{-\frac {x^{2}+34 x +137}{5+x}}}}{12}\) | \(23\) |
norman | \(\frac {\frac {5 x \,{\mathrm e} \,{\mathrm e}^{{\mathrm e}^{\frac {-x^{2}-34 x -137}{5+x}}}}{12}+\frac {x^{2} {\mathrm e} \,{\mathrm e}^{{\mathrm e}^{\frac {-x^{2}-34 x -137}{5+x}}}}{12}}{5+x}\) | \(56\) |
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Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {e^{e^{\frac {-137-34 x-x^2}{5+x}}} \left (e \left (25+10 x+x^2\right )+e^{1+\frac {-137-34 x-x^2}{5+x}} \left (-33 x-10 x^2-x^3\right )\right )}{300+120 x+12 x^2} \, dx=\frac {1}{12} \, x e^{\left (e^{\left (-\frac {x^{2} + 33 \, x + 132}{x + 5} - 1\right )} + 1\right )} \]
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Timed out. \[ \int \frac {e^{e^{\frac {-137-34 x-x^2}{5+x}}} \left (e \left (25+10 x+x^2\right )+e^{1+\frac {-137-34 x-x^2}{5+x}} \left (-33 x-10 x^2-x^3\right )\right )}{300+120 x+12 x^2} \, dx=\text {Timed out} \]
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Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{e^{\frac {-137-34 x-x^2}{5+x}}} \left (e \left (25+10 x+x^2\right )+e^{1+\frac {-137-34 x-x^2}{5+x}} \left (-33 x-10 x^2-x^3\right )\right )}{300+120 x+12 x^2} \, dx=\frac {1}{12} \, x e^{\left (e^{\left (-x + \frac {8}{x + 5} - 29\right )} + 1\right )} \]
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\[ \int \frac {e^{e^{\frac {-137-34 x-x^2}{5+x}}} \left (e \left (25+10 x+x^2\right )+e^{1+\frac {-137-34 x-x^2}{5+x}} \left (-33 x-10 x^2-x^3\right )\right )}{300+120 x+12 x^2} \, dx=\int { \frac {{\left ({\left (x^{2} + 10 \, x + 25\right )} e - {\left (x^{3} + 10 \, x^{2} + 33 \, x\right )} e^{\left (-\frac {x^{2} + 34 \, x + 137}{x + 5} + 1\right )}\right )} e^{\left (e^{\left (-\frac {x^{2} + 34 \, x + 137}{x + 5}\right )}\right )}}{12 \, {\left (x^{2} + 10 \, x + 25\right )}} \,d x } \]
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Time = 14.56 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {e^{e^{\frac {-137-34 x-x^2}{5+x}}} \left (e \left (25+10 x+x^2\right )+e^{1+\frac {-137-34 x-x^2}{5+x}} \left (-33 x-10 x^2-x^3\right )\right )}{300+120 x+12 x^2} \, dx=\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^{-\frac {34\,x}{x+5}}\,{\mathrm {e}}^{-\frac {x^2}{x+5}}\,{\mathrm {e}}^{-\frac {137}{x+5}}}\,\mathrm {e}}{12} \]
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