Integrand size = 140, antiderivative size = 33 \[ \int \frac {e^{\frac {2 \left (-8 e^{\frac {9-6 e^5 x+e^{10} x^2}{x^2}}+13 x\right )}{x}+\frac {9-6 e^5 x+e^{10} x^2}{x^2}} \left (288-96 e^5 x+16 x^2\right )+e^{\frac {-8 e^{\frac {9-6 e^5 x+e^{10} x^2}{x^2}}+13 x}{x}+\frac {9-6 e^5 x+e^{10} x^2}{x^2}} \left (4608-1536 e^5 x+256 x^2\right )}{x^4} \, dx=\left (16+e^{5+\frac {8 \left (-e^{\frac {\left (3-e^5 x\right )^2}{x^2}}+x\right )}{x}}\right )^2 \]
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\[ \int \frac {e^{\frac {2 \left (-8 e^{\frac {9-6 e^5 x+e^{10} x^2}{x^2}}+13 x\right )}{x}+\frac {9-6 e^5 x+e^{10} x^2}{x^2}} \left (288-96 e^5 x+16 x^2\right )+e^{\frac {-8 e^{\frac {9-6 e^5 x+e^{10} x^2}{x^2}}+13 x}{x}+\frac {9-6 e^5 x+e^{10} x^2}{x^2}} \left (4608-1536 e^5 x+256 x^2\right )}{x^4} \, dx=\int \frac {\exp \left (\frac {2 \left (-8 e^{\frac {9-6 e^5 x+e^{10} x^2}{x^2}}+13 x\right )}{x}+\frac {9-6 e^5 x+e^{10} x^2}{x^2}\right ) \left (288-96 e^5 x+16 x^2\right )+\exp \left (\frac {-8 e^{\frac {9-6 e^5 x+e^{10} x^2}{x^2}}+13 x}{x}+\frac {9-6 e^5 x+e^{10} x^2}{x^2}\right ) \left (4608-1536 e^5 x+256 x^2\right )}{x^4} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {288 \exp \left (26 \left (1+\frac {e^{10}}{26}\right )+\frac {9}{x^2}-\frac {6 e^5}{x}-\frac {16 e^{\frac {\left (-3+e^5 x\right )^2}{x^2}}}{x}\right )}{x^4}+\frac {4608 \exp \left (13 \left (1+\frac {e^{10}}{13}\right )+\frac {9}{x^2}-\frac {6 e^5}{x}-\frac {8 e^{\frac {\left (-3+e^5 x\right )^2}{x^2}}}{x}\right )}{x^4}-\frac {96 \exp \left (31 \left (1+\frac {e^{10}}{31}\right )+\frac {9}{x^2}-\frac {6 e^5}{x}-\frac {16 e^{\frac {\left (-3+e^5 x\right )^2}{x^2}}}{x}\right )}{x^3}-\frac {1536 \exp \left (18 \left (1+\frac {e^{10}}{18}\right )+\frac {9}{x^2}-\frac {6 e^5}{x}-\frac {8 e^{\frac {\left (-3+e^5 x\right )^2}{x^2}}}{x}\right )}{x^3}+\frac {16 \exp \left (26 \left (1+\frac {e^{10}}{26}\right )+\frac {9}{x^2}-\frac {6 e^5}{x}-\frac {16 e^{\frac {\left (-3+e^5 x\right )^2}{x^2}}}{x}\right )}{x^2}+\frac {256 \exp \left (13 \left (1+\frac {e^{10}}{13}\right )+\frac {9}{x^2}-\frac {6 e^5}{x}-\frac {8 e^{\frac {\left (-3+e^5 x\right )^2}{x^2}}}{x}\right )}{x^2}\right ) \, dx \\ & = 16 \int \frac {\exp \left (26 \left (1+\frac {e^{10}}{26}\right )+\frac {9}{x^2}-\frac {6 e^5}{x}-\frac {16 e^{\frac {\left (-3+e^5 x\right )^2}{x^2}}}{x}\right )}{x^2} \, dx-96 \int \frac {\exp \left (31 \left (1+\frac {e^{10}}{31}\right )+\frac {9}{x^2}-\frac {6 e^5}{x}-\frac {16 e^{\frac {\left (-3+e^5 x\right )^2}{x^2}}}{x}\right )}{x^3} \, dx+256 \int \frac {\exp \left (13 \left (1+\frac {e^{10}}{13}\right )+\frac {9}{x^2}-\frac {6 e^5}{x}-\frac {8 e^{\frac {\left (-3+e^5 x\right )^2}{x^2}}}{x}\right )}{x^2} \, dx+288 \int \frac {\exp \left (26 \left (1+\frac {e^{10}}{26}\right )+\frac {9}{x^2}-\frac {6 e^5}{x}-\frac {16 e^{\frac {\left (-3+e^5 x\right )^2}{x^2}}}{x}\right )}{x^4} \, dx-1536 \int \frac {\exp \left (18 \left (1+\frac {e^{10}}{18}\right )+\frac {9}{x^2}-\frac {6 e^5}{x}-\frac {8 e^{\frac {\left (-3+e^5 x\right )^2}{x^2}}}{x}\right )}{x^3} \, dx+4608 \int \frac {\exp \left (13 \left (1+\frac {e^{10}}{13}\right )+\frac {9}{x^2}-\frac {6 e^5}{x}-\frac {8 e^{\frac {\left (-3+e^5 x\right )^2}{x^2}}}{x}\right )}{x^4} \, dx \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.61 \[ \int \frac {e^{\frac {2 \left (-8 e^{\frac {9-6 e^5 x+e^{10} x^2}{x^2}}+13 x\right )}{x}+\frac {9-6 e^5 x+e^{10} x^2}{x^2}} \left (288-96 e^5 x+16 x^2\right )+e^{\frac {-8 e^{\frac {9-6 e^5 x+e^{10} x^2}{x^2}}+13 x}{x}+\frac {9-6 e^5 x+e^{10} x^2}{x^2}} \left (4608-1536 e^5 x+256 x^2\right )}{x^4} \, dx=e^{13-\frac {16 e^{\frac {\left (-3+e^5 x\right )^2}{x^2}}}{x}} \left (e^{13}+32 e^{\frac {8 e^{\frac {\left (-3+e^5 x\right )^2}{x^2}}}{x}}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. \(62\) vs. \(2(30)=60\).
Time = 7.97 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.91
method | result | size |
risch | \({\mathrm e}^{\frac {-16 \,{\mathrm e}^{\frac {x^{2} {\mathrm e}^{10}-6 x \,{\mathrm e}^{5}+9}{x^{2}}}+26 x}{x}}+32 \,{\mathrm e}^{\frac {-8 \,{\mathrm e}^{\frac {x^{2} {\mathrm e}^{10}-6 x \,{\mathrm e}^{5}+9}{x^{2}}}+13 x}{x}}\) | \(63\) |
parts | \({\mathrm e}^{\frac {-16 \,{\mathrm e}^{\frac {x^{2} {\mathrm e}^{10}-6 x \,{\mathrm e}^{5}+9}{x^{2}}}+26 x}{x}}+32 \,{\mathrm e}^{\frac {-8 \,{\mathrm e}^{\frac {x^{2} {\mathrm e}^{10}-6 x \,{\mathrm e}^{5}+9}{x^{2}}}+13 x}{x}}\) | \(68\) |
parallelrisch | \({\mathrm e}^{-\frac {2 \left (8 \,{\mathrm e}^{\frac {x^{2} {\mathrm e}^{10}-6 x \,{\mathrm e}^{5}+9}{x^{2}}}-13 x \right )}{x}}+32 \,{\mathrm e}^{-\frac {8 \,{\mathrm e}^{\frac {x^{2} {\mathrm e}^{10}-6 x \,{\mathrm e}^{5}+9}{x^{2}}}-13 x}{x}}\) | \(70\) |
norman | \(\frac {x^{3} {\mathrm e}^{\frac {-16 \,{\mathrm e}^{\frac {x^{2} {\mathrm e}^{10}-6 x \,{\mathrm e}^{5}+9}{x^{2}}}+26 x}{x}}+32 x^{3} {\mathrm e}^{\frac {-8 \,{\mathrm e}^{\frac {x^{2} {\mathrm e}^{10}-6 x \,{\mathrm e}^{5}+9}{x^{2}}}+13 x}{x}}}{x^{3}}\) | \(79\) |
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Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (29) = 58\).
Time = 0.30 (sec) , antiderivative size = 130, normalized size of antiderivative = 3.94 \[ \int \frac {e^{\frac {2 \left (-8 e^{\frac {9-6 e^5 x+e^{10} x^2}{x^2}}+13 x\right )}{x}+\frac {9-6 e^5 x+e^{10} x^2}{x^2}} \left (288-96 e^5 x+16 x^2\right )+e^{\frac {-8 e^{\frac {9-6 e^5 x+e^{10} x^2}{x^2}}+13 x}{x}+\frac {9-6 e^5 x+e^{10} x^2}{x^2}} \left (4608-1536 e^5 x+256 x^2\right )}{x^4} \, dx={\left (e^{\left (\frac {2 \, {\left (x^{2} e^{10} + 13 \, x^{2} - 6 \, x e^{5} - 8 \, x e^{\left (\frac {x^{2} e^{10} - 6 \, x e^{5} + 9}{x^{2}}\right )} + 9\right )}}{x^{2}}\right )} + 32 \, e^{\left (\frac {x^{2} e^{10} + 13 \, x^{2} - 6 \, x e^{5} - 8 \, x e^{\left (\frac {x^{2} e^{10} - 6 \, x e^{5} + 9}{x^{2}}\right )} + 9}{x^{2}} + \frac {x^{2} e^{10} - 6 \, x e^{5} + 9}{x^{2}}\right )}\right )} e^{\left (-\frac {2 \, {\left (x^{2} e^{10} - 6 \, x e^{5} + 9\right )}}{x^{2}}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (26) = 52\).
Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.76 \[ \int \frac {e^{\frac {2 \left (-8 e^{\frac {9-6 e^5 x+e^{10} x^2}{x^2}}+13 x\right )}{x}+\frac {9-6 e^5 x+e^{10} x^2}{x^2}} \left (288-96 e^5 x+16 x^2\right )+e^{\frac {-8 e^{\frac {9-6 e^5 x+e^{10} x^2}{x^2}}+13 x}{x}+\frac {9-6 e^5 x+e^{10} x^2}{x^2}} \left (4608-1536 e^5 x+256 x^2\right )}{x^4} \, dx=e^{\frac {2 \cdot \left (13 x - 8 e^{\frac {x^{2} e^{10} - 6 x e^{5} + 9}{x^{2}}}\right )}{x}} + 32 e^{\frac {13 x - 8 e^{\frac {x^{2} e^{10} - 6 x e^{5} + 9}{x^{2}}}}{x}} \]
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Time = 0.48 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.58 \[ \int \frac {e^{\frac {2 \left (-8 e^{\frac {9-6 e^5 x+e^{10} x^2}{x^2}}+13 x\right )}{x}+\frac {9-6 e^5 x+e^{10} x^2}{x^2}} \left (288-96 e^5 x+16 x^2\right )+e^{\frac {-8 e^{\frac {9-6 e^5 x+e^{10} x^2}{x^2}}+13 x}{x}+\frac {9-6 e^5 x+e^{10} x^2}{x^2}} \left (4608-1536 e^5 x+256 x^2\right )}{x^4} \, dx={\left (e^{26} + 32 \, e^{\left (\frac {8 \, e^{\left (-\frac {6 \, e^{5}}{x} + \frac {9}{x^{2}} + e^{10}\right )}}{x} + 13\right )}\right )} e^{\left (-\frac {16 \, e^{\left (-\frac {6 \, e^{5}}{x} + \frac {9}{x^{2}} + e^{10}\right )}}{x}\right )} \]
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\[ \int \frac {e^{\frac {2 \left (-8 e^{\frac {9-6 e^5 x+e^{10} x^2}{x^2}}+13 x\right )}{x}+\frac {9-6 e^5 x+e^{10} x^2}{x^2}} \left (288-96 e^5 x+16 x^2\right )+e^{\frac {-8 e^{\frac {9-6 e^5 x+e^{10} x^2}{x^2}}+13 x}{x}+\frac {9-6 e^5 x+e^{10} x^2}{x^2}} \left (4608-1536 e^5 x+256 x^2\right )}{x^4} \, dx=\int { \frac {16 \, {\left ({\left (x^{2} - 6 \, x e^{5} + 18\right )} e^{\left (\frac {2 \, {\left (13 \, x - 8 \, e^{\left (\frac {x^{2} e^{10} - 6 \, x e^{5} + 9}{x^{2}}\right )}\right )}}{x} + \frac {x^{2} e^{10} - 6 \, x e^{5} + 9}{x^{2}}\right )} + 16 \, {\left (x^{2} - 6 \, x e^{5} + 18\right )} e^{\left (\frac {13 \, x - 8 \, e^{\left (\frac {x^{2} e^{10} - 6 \, x e^{5} + 9}{x^{2}}\right )}}{x} + \frac {x^{2} e^{10} - 6 \, x e^{5} + 9}{x^{2}}\right )}\right )}}{x^{4}} \,d x } \]
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Time = 14.48 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.64 \[ \int \frac {e^{\frac {2 \left (-8 e^{\frac {9-6 e^5 x+e^{10} x^2}{x^2}}+13 x\right )}{x}+\frac {9-6 e^5 x+e^{10} x^2}{x^2}} \left (288-96 e^5 x+16 x^2\right )+e^{\frac {-8 e^{\frac {9-6 e^5 x+e^{10} x^2}{x^2}}+13 x}{x}+\frac {9-6 e^5 x+e^{10} x^2}{x^2}} \left (4608-1536 e^5 x+256 x^2\right )}{x^4} \, dx=32\,{\mathrm {e}}^{-\frac {8\,{\mathrm {e}}^{-\frac {6\,{\mathrm {e}}^5}{x}}\,{\mathrm {e}}^{\frac {9}{x^2}}\,{\mathrm {e}}^{{\mathrm {e}}^{10}}}{x}}\,{\mathrm {e}}^{13}+{\mathrm {e}}^{-\frac {16\,{\mathrm {e}}^{-\frac {6\,{\mathrm {e}}^5}{x}}\,{\mathrm {e}}^{\frac {9}{x^2}}\,{\mathrm {e}}^{{\mathrm {e}}^{10}}}{x}}\,{\mathrm {e}}^{26} \]
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