Integrand size = 160, antiderivative size = 33 \[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\frac {1}{16} e^{e^{25 x^2 \log ^2\left (\frac {x}{x+\frac {x}{e^3-x}}\right )}} \]
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\[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\int \frac {\exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^3 \left (1+e^3\right )-8 \left (1+2 e^3\right ) x+8 x^2} \, dx \\ & = \int \left (-\frac {25 \exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )}{8 \left (e^3-x\right ) \left (1+e^3-x\right )}+\frac {25}{8} \exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) x \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) \, dx \\ & = -\left (\frac {25}{8} \int \frac {\exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )}{\left (e^3-x\right ) \left (1+e^3-x\right )} \, dx\right )+\frac {25}{8} \int \exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) x \log ^2\left (\frac {e^3-x}{1+e^3-x}\right ) \, dx \\ & = \frac {25}{8} \int \exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) x \log ^2\left (\frac {e^3-x}{1+e^3-x}\right ) \, dx-\frac {25}{8} \int \left (\exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) \log \left (\frac {e^3-x}{1+e^3-x}\right )+\frac {\exp \left (6+e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) \log \left (\frac {e^3-x}{1+e^3-x}\right )}{e^3-x}-\frac {\exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) \left (1+e^3\right )^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )}{1+e^3-x}\right ) \, dx \\ & = -\left (\frac {25}{8} \int \exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) \log \left (\frac {e^3-x}{1+e^3-x}\right ) \, dx\right )-\frac {25}{8} \int \frac {\exp \left (6+e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) \log \left (\frac {e^3-x}{1+e^3-x}\right )}{e^3-x} \, dx+\frac {25}{8} \int \exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) x \log ^2\left (\frac {e^3-x}{1+e^3-x}\right ) \, dx+\frac {1}{8} \left (25 \left (1+e^3\right )^2\right ) \int \frac {\exp \left (e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right ) \log \left (\frac {e^3-x}{1+e^3-x}\right )}{1+e^3-x} \, dx \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\frac {1}{16} e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}} \]
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Time = 30.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88
method | result | size |
risch | \(\frac {{\mathrm e}^{{\mathrm e}^{25 x^{2} \ln \left (\frac {-x +{\mathrm e}^{3}}{{\mathrm e}^{3}-x +1}\right )^{2}}}}{16}\) | \(29\) |
parallelrisch | \(\frac {{\mathrm e}^{{\mathrm e}^{25 x^{2} \ln \left (\frac {-x +{\mathrm e}^{3}}{{\mathrm e}^{3}-x +1}\right )^{2}}}}{16}\) | \(29\) |
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Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\frac {1}{16} \, e^{\left (e^{\left (25 \, x^{2} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right )^{2}\right )}\right )} \]
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Time = 3.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\frac {e^{e^{25 x^{2} \log {\left (\frac {- x + e^{3}}{- x + 1 + e^{3}} \right )}^{2}}}}{16} \]
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Time = 0.56 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.64 \[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\frac {1}{16} \, e^{\left (e^{\left (25 \, x^{2} \log \left (x - e^{3}\right )^{2} - 50 \, x^{2} \log \left (x - e^{3}\right ) \log \left (x - e^{3} - 1\right ) + 25 \, x^{2} \log \left (x - e^{3} - 1\right )^{2}\right )}\right )} \]
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\[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\int { -\frac {25 \, {\left (x^{2} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right ) - {\left (x^{3} - x^{2} + x e^{6} - {\left (2 \, x^{2} - x\right )} e^{3}\right )} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right )^{2}\right )} e^{\left (25 \, x^{2} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right )^{2} + e^{\left (25 \, x^{2} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right )^{2}\right )}\right )}}{8 \, {\left (x^{2} - {\left (2 \, x - 1\right )} e^{3} - x + e^{6}\right )}} \,d x } \]
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Time = 14.48 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^{25\,x^2\,{\ln \left (-\frac {x-{\mathrm {e}}^3}{{\mathrm {e}}^3-x+1}\right )}^2}}}{16} \]
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