Integrand size = 106, antiderivative size = 28 \[ \int \frac {e^x \left (25+10 x+21 x^2+4 x^3+4 x^4\right )+e^{\frac {e^3+5 x+x^2+2 x^3}{5+x+2 x^2}} \left (25+e^3 (-1-4 x)+10 x+21 x^2+4 x^3+4 x^4\right )}{25+10 x+21 x^2+4 x^3+4 x^4} \, dx=-1-e^5+e^x+e^{x+\frac {e^3}{5+x+2 x^2}} \]
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Time = 0.48 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {6820, 2225, 6838} \[ \int \frac {e^x \left (25+10 x+21 x^2+4 x^3+4 x^4\right )+e^{\frac {e^3+5 x+x^2+2 x^3}{5+x+2 x^2}} \left (25+e^3 (-1-4 x)+10 x+21 x^2+4 x^3+4 x^4\right )}{25+10 x+21 x^2+4 x^3+4 x^4} \, dx=e^{\frac {e^3}{2 x^2+x+5}+x}+e^x \]
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Rule 2225
Rule 6820
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \int \left (e^x+\frac {e^{x+\frac {e^3}{5+x+2 x^2}} \left (25-e^3+2 \left (5-2 e^3\right ) x+21 x^2+4 x^3+4 x^4\right )}{\left (5+x+2 x^2\right )^2}\right ) \, dx \\ & = \int e^x \, dx+\int \frac {e^{x+\frac {e^3}{5+x+2 x^2}} \left (25-e^3+2 \left (5-2 e^3\right ) x+21 x^2+4 x^3+4 x^4\right )}{\left (5+x+2 x^2\right )^2} \, dx \\ & = e^x+e^{x+\frac {e^3}{5+x+2 x^2}} \\ \end{align*}
Time = 0.42 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {e^x \left (25+10 x+21 x^2+4 x^3+4 x^4\right )+e^{\frac {e^3+5 x+x^2+2 x^3}{5+x+2 x^2}} \left (25+e^3 (-1-4 x)+10 x+21 x^2+4 x^3+4 x^4\right )}{25+10 x+21 x^2+4 x^3+4 x^4} \, dx=e^x+e^{x+\frac {e^3}{5+x+2 x^2}} \]
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Time = 1.65 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07
method | result | size |
risch | \({\mathrm e}^{x}+{\mathrm e}^{\frac {{\mathrm e}^{3}+2 x^{3}+x^{2}+5 x}{2 x^{2}+x +5}}\) | \(30\) |
parallelrisch | \({\mathrm e}^{x}+{\mathrm e}^{\frac {{\mathrm e}^{3}+2 x^{3}+x^{2}+5 x}{2 x^{2}+x +5}}\) | \(30\) |
parts | \(\frac {{\mathrm e}^{\frac {{\mathrm e}^{3}+2 x^{3}+x^{2}+5 x}{2 x^{2}+x +5}} x +2 \,{\mathrm e}^{\frac {{\mathrm e}^{3}+2 x^{3}+x^{2}+5 x}{2 x^{2}+x +5}} x^{2}+5 \,{\mathrm e}^{\frac {{\mathrm e}^{3}+2 x^{3}+x^{2}+5 x}{2 x^{2}+x +5}}}{2 x^{2}+x +5}+{\mathrm e}^{x}\) | \(103\) |
norman | \(\frac {{\mathrm e}^{x} x +{\mathrm e}^{\frac {{\mathrm e}^{3}+2 x^{3}+x^{2}+5 x}{2 x^{2}+x +5}} x +2 \,{\mathrm e}^{x} x^{2}+2 \,{\mathrm e}^{\frac {{\mathrm e}^{3}+2 x^{3}+x^{2}+5 x}{2 x^{2}+x +5}} x^{2}+5 \,{\mathrm e}^{x}+5 \,{\mathrm e}^{\frac {{\mathrm e}^{3}+2 x^{3}+x^{2}+5 x}{2 x^{2}+x +5}}}{2 x^{2}+x +5}\) | \(115\) |
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Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^x \left (25+10 x+21 x^2+4 x^3+4 x^4\right )+e^{\frac {e^3+5 x+x^2+2 x^3}{5+x+2 x^2}} \left (25+e^3 (-1-4 x)+10 x+21 x^2+4 x^3+4 x^4\right )}{25+10 x+21 x^2+4 x^3+4 x^4} \, dx=e^{x} + e^{\left (\frac {2 \, x^{3} + x^{2} + 5 \, x + e^{3}}{2 \, x^{2} + x + 5}\right )} \]
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Time = 0.32 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {e^x \left (25+10 x+21 x^2+4 x^3+4 x^4\right )+e^{\frac {e^3+5 x+x^2+2 x^3}{5+x+2 x^2}} \left (25+e^3 (-1-4 x)+10 x+21 x^2+4 x^3+4 x^4\right )}{25+10 x+21 x^2+4 x^3+4 x^4} \, dx=e^{x} + e^{\frac {2 x^{3} + x^{2} + 5 x + e^{3}}{2 x^{2} + x + 5}} \]
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Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {e^x \left (25+10 x+21 x^2+4 x^3+4 x^4\right )+e^{\frac {e^3+5 x+x^2+2 x^3}{5+x+2 x^2}} \left (25+e^3 (-1-4 x)+10 x+21 x^2+4 x^3+4 x^4\right )}{25+10 x+21 x^2+4 x^3+4 x^4} \, dx=e^{\left (x + \frac {e^{3}}{2 \, x^{2} + x + 5}\right )} + e^{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (24) = 48\).
Time = 0.41 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.89 \[ \int \frac {e^x \left (25+10 x+21 x^2+4 x^3+4 x^4\right )+e^{\frac {e^3+5 x+x^2+2 x^3}{5+x+2 x^2}} \left (25+e^3 (-1-4 x)+10 x+21 x^2+4 x^3+4 x^4\right )}{25+10 x+21 x^2+4 x^3+4 x^4} \, dx={\left (e^{\left (x + 3\right )} + e^{\left (\frac {10 \, x^{3} - 2 \, x^{2} e^{3} + 5 \, x^{2} - x e^{3} + 25 \, x}{5 \, {\left (2 \, x^{2} + x + 5\right )}} + \frac {1}{5} \, e^{3} + 3\right )}\right )} e^{\left (-3\right )} \]
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Time = 8.81 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.25 \[ \int \frac {e^x \left (25+10 x+21 x^2+4 x^3+4 x^4\right )+e^{\frac {e^3+5 x+x^2+2 x^3}{5+x+2 x^2}} \left (25+e^3 (-1-4 x)+10 x+21 x^2+4 x^3+4 x^4\right )}{25+10 x+21 x^2+4 x^3+4 x^4} \, dx={\mathrm {e}}^x+{\mathrm {e}}^{\frac {x^2}{2\,x^2+x+5}}\,{\mathrm {e}}^{\frac {2\,x^3}{2\,x^2+x+5}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^3}{2\,x^2+x+5}}\,{\mathrm {e}}^{\frac {5\,x}{2\,x^2+x+5}} \]
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