Integrand size = 151, antiderivative size = 29 \[ \int \frac {8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5+e^{e^x} \left (8 x^3-8 e^x x^4+e^{\frac {1+16 x^3}{16 x^2}} \left (1-8 x^3+8 e^x x^3\right )\right )}{8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5} \, dx=-\frac {7}{4}+x-\frac {e^{e^x}}{-e^{\frac {1}{16 x^2}+x}+x} \]
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\[ \int \frac {8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5+e^{e^x} \left (8 x^3-8 e^x x^4+e^{\frac {1+16 x^3}{16 x^2}} \left (1-8 x^3+8 e^x x^3\right )\right )}{8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5} \, dx=\int \frac {8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5+e^{e^x} \left (8 x^3-8 e^x x^4+e^{\frac {1+16 x^3}{16 x^2}} \left (1-8 x^3+8 e^x x^3\right )\right )}{8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5+e^{e^x} \left (8 x^3-8 e^x x^4+e^{\frac {1+16 x^3}{16 x^2}} \left (1-8 x^3+8 e^x x^3\right )\right )}{8 \left (e^{\frac {1}{16 x^2}+x}-x\right )^2 x^3} \, dx \\ & = \frac {1}{8} \int \frac {8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5+e^{e^x} \left (8 x^3-8 e^x x^4+e^{\frac {1+16 x^3}{16 x^2}} \left (1-8 x^3+8 e^x x^3\right )\right )}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2 x^3} \, dx \\ & = \frac {1}{8} \int \left (8 e^{-\frac {1}{16 x^2}} \left (e^{e^x}+e^{\frac {1}{16 x^2}}\right )-\frac {e^{e^x} \left (-1-8 x^2+8 x^3\right )}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2 x^2}-\frac {e^{e^x-\frac {1}{16 x^2}} \left (-e^{\frac {1}{16 x^2}}+8 e^{\frac {1}{16 x^2}} x^3-8 x^4\right )}{\left (e^{\frac {1}{16 x^2}+x}-x\right ) x^3}\right ) \, dx \\ & = -\left (\frac {1}{8} \int \frac {e^{e^x} \left (-1-8 x^2+8 x^3\right )}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2 x^2} \, dx\right )-\frac {1}{8} \int \frac {e^{e^x-\frac {1}{16 x^2}} \left (-e^{\frac {1}{16 x^2}}+8 e^{\frac {1}{16 x^2}} x^3-8 x^4\right )}{\left (e^{\frac {1}{16 x^2}+x}-x\right ) x^3} \, dx+\int e^{-\frac {1}{16 x^2}} \left (e^{e^x}+e^{\frac {1}{16 x^2}}\right ) \, dx \\ & = -\left (\frac {1}{8} \int \left (-\frac {8 e^{e^x}}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2}-\frac {e^{e^x}}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2 x^2}+\frac {8 e^{e^x} x}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2}\right ) \, dx\right )-\frac {1}{8} \int \frac {e^{e^x} \left (-1+8 x^3-8 e^{-\frac {1}{16 x^2}} x^4\right )}{\left (e^{\frac {1}{16 x^2}+x}-x\right ) x^3} \, dx+\int \left (1+e^{e^x-\frac {1}{16 x^2}}\right ) \, dx \\ & = x+\frac {1}{8} \int \frac {e^{e^x}}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2 x^2} \, dx-\frac {1}{8} \int \left (\frac {8 e^{e^x}}{e^{\frac {1}{16 x^2}+x}-x}-\frac {e^{e^x}}{\left (e^{\frac {1}{16 x^2}+x}-x\right ) x^3}-\frac {8 e^{e^x-\frac {1}{16 x^2}} x}{e^{\frac {1}{16 x^2}+x}-x}\right ) \, dx+\int e^{e^x-\frac {1}{16 x^2}} \, dx+\int \frac {e^{e^x}}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2} \, dx-\int \frac {e^{e^x} x}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2} \, dx \\ & = x+\frac {1}{8} \int \frac {e^{e^x}}{\left (e^{\frac {1}{16 x^2}+x}-x\right ) x^3} \, dx+\frac {1}{8} \int \frac {e^{e^x}}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2 x^2} \, dx+\int e^{e^x-\frac {1}{16 x^2}} \, dx+\int \frac {e^{e^x}}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2} \, dx-\int \frac {e^{e^x}}{e^{\frac {1}{16 x^2}+x}-x} \, dx-\int \frac {e^{e^x} x}{\left (e^{\frac {1}{16 x^2}+x}-x\right )^2} \, dx+\int \frac {e^{e^x-\frac {1}{16 x^2}} x}{e^{\frac {1}{16 x^2}+x}-x} \, dx \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5+e^{e^x} \left (8 x^3-8 e^x x^4+e^{\frac {1+16 x^3}{16 x^2}} \left (1-8 x^3+8 e^x x^3\right )\right )}{8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5} \, dx=\frac {1}{8} \left (\frac {8 e^{e^x}}{e^{\frac {1}{16 x^2}+x}-x}+8 x\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. \(48\) vs. \(2(22)=44\).
Time = 6.15 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.69
method | result | size |
parallelrisch | \(\frac {8 x^{2}-8 x \,{\mathrm e}^{\frac {16 x^{3}+1}{16 x^{2}}}-8 \,{\mathrm e}^{{\mathrm e}^{x}}}{8 x -8 \,{\mathrm e}^{\frac {16 x^{3}+1}{16 x^{2}}}}\) | \(49\) |
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Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (22) = 44\).
Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5+e^{e^x} \left (8 x^3-8 e^x x^4+e^{\frac {1+16 x^3}{16 x^2}} \left (1-8 x^3+8 e^x x^3\right )\right )}{8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5} \, dx=\frac {x^{2} - x e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}}\right )} - e^{\left (e^{x}\right )}}{x - e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}}\right )}} \]
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Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5+e^{e^x} \left (8 x^3-8 e^x x^4+e^{\frac {1+16 x^3}{16 x^2}} \left (1-8 x^3+8 e^x x^3\right )\right )}{8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5} \, dx=x - \frac {e^{e^{x}}}{x - e^{\frac {x^{3} + \frac {1}{16}}{x^{2}}}} \]
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\[ \int \frac {8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5+e^{e^x} \left (8 x^3-8 e^x x^4+e^{\frac {1+16 x^3}{16 x^2}} \left (1-8 x^3+8 e^x x^3\right )\right )}{8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5} \, dx=\int { \frac {8 \, x^{5} - 16 \, x^{4} e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}}\right )} + 8 \, x^{3} e^{\left (\frac {16 \, x^{3} + 1}{8 \, x^{2}}\right )} - {\left (8 \, x^{4} e^{x} - 8 \, x^{3} - {\left (8 \, x^{3} e^{x} - 8 \, x^{3} + 1\right )} e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}}\right )}\right )} e^{\left (e^{x}\right )}}{8 \, {\left (x^{5} - 2 \, x^{4} e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}}\right )} + x^{3} e^{\left (\frac {16 \, x^{3} + 1}{8 \, x^{2}}\right )}\right )}} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (22) = 44\).
Time = 0.33 (sec) , antiderivative size = 326, normalized size of antiderivative = 11.24 \[ \int \frac {8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5+e^{e^x} \left (8 x^3-8 e^x x^4+e^{\frac {1+16 x^3}{16 x^2}} \left (1-8 x^3+8 e^x x^3\right )\right )}{8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5} \, dx=\frac {8 \, x^{5} e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}}\right )} - 8 \, x^{4} e^{\left (\frac {16 \, x^{3} + 1}{8 \, x^{2}}\right )} - 8 \, x^{4} e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}}\right )} + 8 \, x^{3} e^{\left (x + \frac {16 \, x^{3} + 16 \, x^{2} e^{x} + 1}{16 \, x^{2}}\right )} - 8 \, x^{3} e^{\left (x + \frac {16 \, x^{3} + 1}{16 \, x^{2}} + e^{x}\right )} - 8 \, x^{3} e^{\left (\frac {16 \, x^{3} + 16 \, x^{2} e^{x} + 1}{16 \, x^{2}}\right )} + 8 \, x^{3} e^{\left (\frac {16 \, x^{3} + 1}{8 \, x^{2}}\right )} + 8 \, x^{2} e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}} + e^{x}\right )} - x^{2} e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}}\right )} + x e^{\left (\frac {16 \, x^{3} + 1}{8 \, x^{2}}\right )} + e^{\left (\frac {16 \, x^{3} + 16 \, x^{2} e^{x} + 1}{16 \, x^{2}}\right )}}{8 \, x^{4} e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}}\right )} - 8 \, x^{3} e^{\left (\frac {16 \, x^{3} + 1}{8 \, x^{2}}\right )} - 8 \, x^{3} e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}}\right )} + 8 \, x^{2} e^{\left (\frac {16 \, x^{3} + 1}{8 \, x^{2}}\right )} - x e^{\left (\frac {16 \, x^{3} + 1}{16 \, x^{2}}\right )} + e^{\left (\frac {16 \, x^{3} + 1}{8 \, x^{2}}\right )}} \]
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Time = 9.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5+e^{e^x} \left (8 x^3-8 e^x x^4+e^{\frac {1+16 x^3}{16 x^2}} \left (1-8 x^3+8 e^x x^3\right )\right )}{8 e^{\frac {1+16 x^3}{8 x^2}} x^3-16 e^{\frac {1+16 x^3}{16 x^2}} x^4+8 x^5} \, dx=x-\frac {{\mathrm {e}}^{{\mathrm {e}}^x}}{x-{\mathrm {e}}^{x+\frac {1}{16\,x^2}}} \]
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