Integrand size = 146, antiderivative size = 25 \[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=(-225+x)^{\frac {e^4}{e^{2/x}-e^{5+x}}} \]
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\[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=\int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^4 (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} \left (\left (e^{2/x}-e^{5+x}\right ) x^2+(-225+x) \left (2 e^{2/x}+e^{5+x} x^2\right ) \log (-225+x)\right )}{\left (e^{2/x}-e^{5+x}\right )^2 x^2} \, dx \\ & = e^4 \int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} \left (\left (e^{2/x}-e^{5+x}\right ) x^2+(-225+x) \left (2 e^{2/x}+e^{5+x} x^2\right ) \log (-225+x)\right )}{\left (e^{2/x}-e^{5+x}\right )^2 x^2} \, dx \\ & = e^4 \int \left (\frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} \left (-450+2 x-225 x^2+x^3\right ) \log (-225+x)}{\left (e^{2/x}-e^{5+x}\right )^2 x^2}-\frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} (-1-225 \log (-225+x)+x \log (-225+x))}{e^{2/x}-e^{5+x}}\right ) \, dx \\ & = e^4 \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} \left (-450+2 x-225 x^2+x^3\right ) \log (-225+x)}{\left (e^{2/x}-e^{5+x}\right )^2 x^2} \, dx-e^4 \int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} (-1-225 \log (-225+x)+x \log (-225+x))}{e^{2/x}-e^{5+x}} \, dx \\ & = -\left (e^4 \int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} (-1+(-225+x) \log (-225+x))}{e^{2/x}-e^{5+x}} \, dx\right )-e^4 \int \frac {-225 \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx-450 \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x^2} \, dx+2 \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x} \, dx+\int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx}{-225+x} \, dx+\left (e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx+\left (2 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x} \, dx-\left (225 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx-\left (450 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x^2} \, dx \\ & = -\left (e^4 \int \left (-\frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{e^{2/x}-e^{5+x}}-\frac {225 (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} \log (-225+x)}{e^{2/x}-e^{5+x}}+\frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x \log (-225+x)}{e^{2/x}-e^{5+x}}\right ) \, dx\right )-e^4 \int \left (\frac {-225 \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx-450 \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x^2} \, dx+2 \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x} \, dx}{-225+x}+\frac {\int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx}{-225+x}\right ) \, dx+\left (e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx+\left (2 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x} \, dx-\left (225 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx-\left (450 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x^2} \, dx \\ & = e^4 \int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{e^{2/x}-e^{5+x}} \, dx-e^4 \int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x \log (-225+x)}{e^{2/x}-e^{5+x}} \, dx-e^4 \int \frac {-225 \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx-450 \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x^2} \, dx+2 \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x} \, dx}{-225+x} \, dx-e^4 \int \frac {\int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx}{-225+x} \, dx+\left (225 e^4\right ) \int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} \log (-225+x)}{e^{2/x}-e^{5+x}} \, dx+\left (e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx+\left (2 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x} \, dx-\left (225 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx-\left (450 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x^2} \, dx \\ & = e^4 \int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{e^{2/x}-e^{5+x}} \, dx-e^4 \int \left (-\frac {225 \left (\int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx+2 \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x^2} \, dx\right )}{-225+x}+\frac {2 \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x} \, dx}{-225+x}\right ) \, dx-e^4 \int \frac {\int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx}{-225+x} \, dx+e^4 \int \frac {\int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{e^{2/x}-e^{5+x}} \, dx}{-225+x} \, dx-\left (225 e^4\right ) \int \frac {\int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{e^{2/x}-e^{5+x}} \, dx}{-225+x} \, dx+\left (e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx-\left (e^4 \log (-225+x)\right ) \int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{e^{2/x}-e^{5+x}} \, dx+\left (2 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x} \, dx-\left (225 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx+\left (225 e^4 \log (-225+x)\right ) \int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{e^{2/x}-e^{5+x}} \, dx-\left (450 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x^2} \, dx \\ & = e^4 \int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{e^{2/x}-e^{5+x}} \, dx-e^4 \int \frac {\int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx}{-225+x} \, dx+e^4 \int \frac {\int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{e^{2/x}-e^{5+x}} \, dx}{-225+x} \, dx-\left (2 e^4\right ) \int \frac {\int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x} \, dx}{-225+x} \, dx-\left (225 e^4\right ) \int \frac {\int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{e^{2/x}-e^{5+x}} \, dx}{-225+x} \, dx+\left (225 e^4\right ) \int \frac {\int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx+2 \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x^2} \, dx}{-225+x} \, dx+\left (e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx-\left (e^4 \log (-225+x)\right ) \int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{e^{2/x}-e^{5+x}} \, dx+\left (2 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x} \, dx-\left (225 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx+\left (225 e^4 \log (-225+x)\right ) \int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{e^{2/x}-e^{5+x}} \, dx-\left (450 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x^2} \, dx \\ & = e^4 \int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{e^{2/x}-e^{5+x}} \, dx-e^4 \int \frac {\int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx}{-225+x} \, dx+e^4 \int \frac {\int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{e^{2/x}-e^{5+x}} \, dx}{-225+x} \, dx-\left (2 e^4\right ) \int \frac {\int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x} \, dx}{-225+x} \, dx-\left (225 e^4\right ) \int \frac {\int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{e^{2/x}-e^{5+x}} \, dx}{-225+x} \, dx+\left (225 e^4\right ) \int \left (\frac {\int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx}{-225+x}+\frac {2 \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x^2} \, dx}{-225+x}\right ) \, dx+\left (e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx-\left (e^4 \log (-225+x)\right ) \int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{e^{2/x}-e^{5+x}} \, dx+\left (2 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x} \, dx-\left (225 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx+\left (225 e^4 \log (-225+x)\right ) \int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{e^{2/x}-e^{5+x}} \, dx-\left (450 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x^2} \, dx \\ & = e^4 \int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{e^{2/x}-e^{5+x}} \, dx-e^4 \int \frac {\int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx}{-225+x} \, dx+e^4 \int \frac {\int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{e^{2/x}-e^{5+x}} \, dx}{-225+x} \, dx-\left (2 e^4\right ) \int \frac {\int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x} \, dx}{-225+x} \, dx+\left (225 e^4\right ) \int \frac {\int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx}{-225+x} \, dx-\left (225 e^4\right ) \int \frac {\int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{e^{2/x}-e^{5+x}} \, dx}{-225+x} \, dx+\left (450 e^4\right ) \int \frac {\int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x^2} \, dx}{-225+x} \, dx+\left (e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx-\left (e^4 \log (-225+x)\right ) \int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}} x}{e^{2/x}-e^{5+x}} \, dx+\left (2 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x} \, dx-\left (225 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2} \, dx+\left (225 e^4 \log (-225+x)\right ) \int \frac {(-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{e^{2/x}-e^{5+x}} \, dx-\left (450 e^4 \log (-225+x)\right ) \int \frac {e^{2/x} (-225+x)^{-1+\frac {e^4}{e^{2/x}-e^{5+x}}}}{\left (e^{2/x}-e^{5+x}\right )^2 x^2} \, dx \\ \end{align*}
Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \]
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Time = 0.46 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96
\[\left (x -225\right )^{-\frac {{\mathrm e}^{4}}{{\mathrm e}^{5+x}-{\mathrm e}^{\frac {2}{x}}}}\]
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Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=\frac {1}{{\left (x - 225\right )}^{\frac {e^{\left (x + 17\right )}}{e^{\left (2 \, x + 18\right )} - e^{\left (\frac {x^{2} + 5 \, x + 2}{x} + 8\right )}}}} \]
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Time = 5.60 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=e^{- \frac {e^{4} \log {\left (x - 225 \right )}}{- e^{\frac {2}{x}} + e^{x + 5}}} \]
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Time = 0.65 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=\frac {1}{{\left (x - 225\right )}^{\frac {e^{4}}{e^{\left (x + 5\right )} - e^{\frac {2}{x}}}}} \]
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\[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=\int { -\frac {x^{2} e^{\left (x + 9\right )} - x^{2} e^{\left (\frac {2}{x} + 4\right )} - {\left ({\left (x^{3} - 225 \, x^{2}\right )} e^{\left (x + 9\right )} + 2 \, {\left (x - 225\right )} e^{\left (\frac {2}{x} + 4\right )}\right )} \log \left (x - 225\right )}{{\left ({\left (x^{3} - 225 \, x^{2}\right )} e^{\left (2 \, x + 10\right )} - 2 \, {\left (x^{3} - 225 \, x^{2}\right )} e^{\left (x + \frac {2}{x} + 5\right )} + {\left (x^{3} - 225 \, x^{2}\right )} e^{\frac {4}{x}}\right )} {\left (x - 225\right )}^{\frac {e^{4}}{e^{\left (x + 5\right )} - e^{\frac {2}{x}}}}} \,d x } \]
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Time = 10.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {(-225+x)^{-\frac {e^4}{-e^{2/x}+e^{5+x}}} \left (e^{4+\frac {2}{x}} x^2-e^{9+x} x^2+\left (e^{4+\frac {2}{x}} (-450+2 x)+e^{9+x} \left (-225 x^2+x^3\right )\right ) \log (-225+x)\right )}{e^{5+\frac {2}{x}+x} \left (450 x^2-2 x^3\right )+e^{4/x} \left (-225 x^2+x^3\right )+e^{10+2 x} \left (-225 x^2+x^3\right )} \, dx=\frac {1}{{\left (x-225\right )}^{\frac {{\mathrm {e}}^4}{{\mathrm {e}}^{x+5}-{\mathrm {e}}^{2/x}}}} \]
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