Integrand size = 58, antiderivative size = 18 \[ \int e^{-17+e^{\frac {6 x+e^{17} \left (-5+4 x+x^2\right )}{e^{17}}}+\frac {6 x+e^{17} \left (-5+4 x+x^2\right )}{e^{17}}} \left (6+e^{17} (4+2 x)\right ) \, dx=5+e^{e^{-5+x \left (4+\frac {6}{e^{17}}+x\right )}} \]
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\[ \int e^{-17+e^{\frac {6 x+e^{17} \left (-5+4 x+x^2\right )}{e^{17}}}+\frac {6 x+e^{17} \left (-5+4 x+x^2\right )}{e^{17}}} \left (6+e^{17} (4+2 x)\right ) \, dx=\int \exp \left (-17+e^{\frac {6 x+e^{17} \left (-5+4 x+x^2\right )}{e^{17}}}+\frac {6 x+e^{17} \left (-5+4 x+x^2\right )}{e^{17}}\right ) \left (6+e^{17} (4+2 x)\right ) \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \exp \left (\frac {-22 e^{17}+e^{12+4 x+\frac {6 x}{e^{17}}+x^2}+6 \left (1+\frac {2 e^{17}}{3}\right ) x+e^{17} x^2}{e^{17}}\right ) \left (2 \left (3+2 e^{17}\right )+2 e^{17} x\right ) \, dx \\ & = \int \exp \left (-22+e^{-5+4 x+\frac {6 x}{e^{17}}+x^2}+4 \left (1+\frac {3}{2 e^{17}}\right ) x+x^2\right ) \left (2 \left (3+2 e^{17}\right )+2 e^{17} x\right ) \, dx \\ & = \int \left (2 \exp \left (-22+e^{-5+4 x+\frac {6 x}{e^{17}}+x^2}+4 \left (1+\frac {3}{2 e^{17}}\right ) x+x^2\right ) \left (3+2 e^{17}\right )+2 \exp \left (-5+e^{-5+4 x+\frac {6 x}{e^{17}}+x^2}+4 \left (1+\frac {3}{2 e^{17}}\right ) x+x^2\right ) x\right ) \, dx \\ & = 2 \int \exp \left (-5+e^{-5+4 x+\frac {6 x}{e^{17}}+x^2}+4 \left (1+\frac {3}{2 e^{17}}\right ) x+x^2\right ) x \, dx+\left (2 \left (3+2 e^{17}\right )\right ) \int \exp \left (-22+e^{-5+4 x+\frac {6 x}{e^{17}}+x^2}+4 \left (1+\frac {3}{2 e^{17}}\right ) x+x^2\right ) \, dx \\ \end{align*}
Time = 0.68 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int e^{-17+e^{\frac {6 x+e^{17} \left (-5+4 x+x^2\right )}{e^{17}}}+\frac {6 x+e^{17} \left (-5+4 x+x^2\right )}{e^{17}}} \left (6+e^{17} (4+2 x)\right ) \, dx=e^{e^{-5+4 x+\frac {6 x}{e^{17}}+x^2}} \]
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Time = 0.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28
method | result | size |
derivativedivides | \({\mathrm e}^{{\mathrm e}^{\left (\left (x^{2}+4 x -5\right ) {\mathrm e}^{17}+6 x \right ) {\mathrm e}^{-17}}}\) | \(23\) |
default | \({\mathrm e}^{{\mathrm e}^{\left (\left (x^{2}+4 x -5\right ) {\mathrm e}^{17}+6 x \right ) {\mathrm e}^{-17}}}\) | \(23\) |
norman | \({\mathrm e}^{{\mathrm e}^{\left (\left (x^{2}+4 x -5\right ) {\mathrm e}^{17}+6 x \right ) {\mathrm e}^{-17}}}\) | \(23\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{\left (\left (x^{2}+4 x -5\right ) {\mathrm e}^{17}+6 x \right ) {\mathrm e}^{-17}}}\) | \(23\) |
risch | \({\mathrm e}^{{\mathrm e}^{\left ({\mathrm e}^{17} x^{2}+4 \,{\mathrm e}^{17} x -5 \,{\mathrm e}^{17}+6 x \right ) {\mathrm e}^{-17}}}\) | \(25\) |
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Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (15) = 30\).
Time = 0.26 (sec) , antiderivative size = 61, normalized size of antiderivative = 3.39 \[ \int e^{-17+e^{\frac {6 x+e^{17} \left (-5+4 x+x^2\right )}{e^{17}}}+\frac {6 x+e^{17} \left (-5+4 x+x^2\right )}{e^{17}}} \left (6+e^{17} (4+2 x)\right ) \, dx=e^{\left (-{\left ({\left (x^{2} + 4 \, x - 5\right )} e^{17} + 6 \, x\right )} e^{\left (-17\right )} + {\left ({\left (x^{2} + 4 \, x - 22\right )} e^{17} + 6 \, x + e^{\left ({\left ({\left (x^{2} + 4 \, x - 5\right )} e^{17} + 6 \, x\right )} e^{\left (-17\right )} + 17\right )}\right )} e^{\left (-17\right )} + 17\right )} \]
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Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int e^{-17+e^{\frac {6 x+e^{17} \left (-5+4 x+x^2\right )}{e^{17}}}+\frac {6 x+e^{17} \left (-5+4 x+x^2\right )}{e^{17}}} \left (6+e^{17} (4+2 x)\right ) \, dx=e^{e^{\frac {6 x + \left (x^{2} + 4 x - 5\right ) e^{17}}{e^{17}}}} \]
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Time = 0.49 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int e^{-17+e^{\frac {6 x+e^{17} \left (-5+4 x+x^2\right )}{e^{17}}}+\frac {6 x+e^{17} \left (-5+4 x+x^2\right )}{e^{17}}} \left (6+e^{17} (4+2 x)\right ) \, dx=e^{\left (e^{\left (x^{2} + 6 \, x e^{\left (-17\right )} + 4 \, x - 5\right )}\right )} \]
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\[ \int e^{-17+e^{\frac {6 x+e^{17} \left (-5+4 x+x^2\right )}{e^{17}}}+\frac {6 x+e^{17} \left (-5+4 x+x^2\right )}{e^{17}}} \left (6+e^{17} (4+2 x)\right ) \, dx=\int { 2 \, {\left ({\left (x + 2\right )} e^{17} + 3\right )} e^{\left ({\left ({\left (x^{2} + 4 \, x - 5\right )} e^{17} + 6 \, x\right )} e^{\left (-17\right )} + e^{\left ({\left ({\left (x^{2} + 4 \, x - 5\right )} e^{17} + 6 \, x\right )} e^{\left (-17\right )}\right )} - 17\right )} \,d x } \]
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Time = 8.83 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int e^{-17+e^{\frac {6 x+e^{17} \left (-5+4 x+x^2\right )}{e^{17}}}+\frac {6 x+e^{17} \left (-5+4 x+x^2\right )}{e^{17}}} \left (6+e^{17} (4+2 x)\right ) \, dx={\mathrm {e}}^{{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{6\,x\,{\mathrm {e}}^{-17}}} \]
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