Integrand size = 184, antiderivative size = 27 \[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx=(4-x) \left (-e^x+\log \left (\frac {\log (\log (x (3+x) (5+x)))}{x}\right )\right ) \]
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\[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx=\int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{x \left (15+8 x+x^2\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx \\ & = \int \left (e^x (-3+x)-\frac {17}{15+8 x+x^2}-\frac {60}{x \left (15+8 x+x^2\right )}+\frac {4 x}{15+8 x+x^2}+\frac {x^2}{15+8 x+x^2}+\frac {49}{(3+x) (5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}+\frac {60}{x (3+x) (5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}-\frac {4 x}{(3+x) (5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}-\frac {3 x^2}{(3+x) (5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}-\log \left (\frac {\log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}{x}\right )\right ) \, dx \\ & = -\left (3 \int \frac {x^2}{(3+x) (5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx\right )+4 \int \frac {x}{15+8 x+x^2} \, dx-4 \int \frac {x}{(3+x) (5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx-17 \int \frac {1}{15+8 x+x^2} \, dx+49 \int \frac {1}{(3+x) (5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx-60 \int \frac {1}{x \left (15+8 x+x^2\right )} \, dx+60 \int \frac {1}{x (3+x) (5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx+\int e^x (-3+x) \, dx+\int \frac {x^2}{15+8 x+x^2} \, dx-\int \log \left (\frac {\log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}{x}\right ) \, dx \\ & = -e^x (3-x)+x-x \log \left (\frac {\log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}{x}\right )-3 \int \left (\frac {1}{\log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}+\frac {9}{2 (3+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}-\frac {25}{2 (5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}\right ) \, dx-4 \int \frac {1}{x} \, dx-4 \int \frac {-8-x}{15+8 x+x^2} \, dx-4 \int \left (-\frac {3}{2 (3+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}+\frac {5}{2 (5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}\right ) \, dx-6 \int \frac {1}{3+x} \, dx-\frac {17}{2} \int \frac {1}{3+x} \, dx+\frac {17}{2} \int \frac {1}{5+x} \, dx+10 \int \frac {1}{5+x} \, dx+49 \int \left (\frac {1}{2 (3+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}-\frac {1}{2 (5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}\right ) \, dx+60 \int \left (\frac {1}{15 x \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}-\frac {1}{6 (3+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}+\frac {1}{10 (5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}\right ) \, dx-\int e^x \, dx+\int \frac {-15-8 x}{15+8 x+x^2} \, dx+\int \left (-1+\frac {15+16 x+3 x^2}{\left (15+8 x+x^2\right ) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}\right ) \, dx \\ & = -e^x-e^x (3-x)-4 \log (x)-\frac {29}{2} \log (3+x)+\frac {37}{2} \log (5+x)-x \log \left (\frac {\log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}{x}\right )-3 \int \frac {1}{\log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx+4 \int \frac {1}{x \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx+\frac {9}{2} \int \frac {1}{3+x} \, dx-6 \int \frac {1}{5+x} \, dx+6 \int \frac {1}{(3+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx+6 \int \frac {1}{(5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx+10 \int \frac {1}{3+x} \, dx-10 \int \frac {1}{(3+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx-10 \int \frac {1}{(5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx-\frac {25}{2} \int \frac {1}{5+x} \, dx-\frac {27}{2} \int \frac {1}{(3+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx+\frac {49}{2} \int \frac {1}{(3+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx-\frac {49}{2} \int \frac {1}{(5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx+\frac {75}{2} \int \frac {1}{(5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx+\int \frac {15+16 x+3 x^2}{\left (15+8 x+x^2\right ) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx \\ & = -e^x-e^x (3-x)-4 \log (x)-x \log \left (\frac {\log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}{x}\right )-3 \int \frac {1}{\log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx+4 \int \frac {1}{x \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx+6 \int \frac {1}{(3+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx+6 \int \frac {1}{(5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx-10 \int \frac {1}{(3+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx-10 \int \frac {1}{(5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx-\frac {27}{2} \int \frac {1}{(3+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx+\frac {49}{2} \int \frac {1}{(3+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx-\frac {49}{2} \int \frac {1}{(5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx+\frac {75}{2} \int \frac {1}{(5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx+\int \left (\frac {3}{\log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}-\frac {3}{(3+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}-\frac {5}{(5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}\right ) \, dx \\ & = -e^x-e^x (3-x)-4 \log (x)-x \log \left (\frac {\log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}{x}\right )-3 \int \frac {1}{(3+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx+4 \int \frac {1}{x \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx-5 \int \frac {1}{(5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx+6 \int \frac {1}{(3+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx+6 \int \frac {1}{(5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx-10 \int \frac {1}{(3+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx-10 \int \frac {1}{(5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx-\frac {27}{2} \int \frac {1}{(3+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx+\frac {49}{2} \int \frac {1}{(3+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx-\frac {49}{2} \int \frac {1}{(5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx+\frac {75}{2} \int \frac {1}{(5+x) \log \left (x \left (15+8 x+x^2\right )\right ) \log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )} \, dx \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx=e^x (-4+x)-4 \log (x)+4 \log \left (\log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )\right )-x \log \left (\frac {\log \left (\log \left (x \left (15+8 x+x^2\right )\right )\right )}{x}\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.35 (sec) , antiderivative size = 733, normalized size of antiderivative = 27.15
\[\text {Expression too large to display}\]
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Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx={\left (x - 4\right )} e^{x} - {\left (x - 4\right )} \log \left (\frac {\log \left (\log \left (x^{3} + 8 \, x^{2} + 15 \, x\right )\right )}{x}\right ) \]
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Timed out. \[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx=\text {Timed out} \]
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Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx={\left (x - 4\right )} e^{x} + {\left (x - 4\right )} \log \left (x\right ) - {\left (x - 4\right )} \log \left (\log \left (\log \left (x + 5\right ) + \log \left (x + 3\right ) + \log \left (x\right )\right )\right ) \]
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\[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx=\int { -\frac {{\left (x^{3} + 8 \, x^{2} + 15 \, x\right )} \log \left (x^{3} + 8 \, x^{2} + 15 \, x\right ) \log \left (\frac {\log \left (\log \left (x^{3} + 8 \, x^{2} + 15 \, x\right )\right )}{x}\right ) \log \left (\log \left (x^{3} + 8 \, x^{2} + 15 \, x\right )\right ) + 3 \, x^{3} - {\left (x^{3} + 4 \, x^{2} + {\left (x^{4} + 5 \, x^{3} - 9 \, x^{2} - 45 \, x\right )} e^{x} - 17 \, x - 60\right )} \log \left (x^{3} + 8 \, x^{2} + 15 \, x\right ) \log \left (\log \left (x^{3} + 8 \, x^{2} + 15 \, x\right )\right ) + 4 \, x^{2} - 49 \, x - 60}{{\left (x^{3} + 8 \, x^{2} + 15 \, x\right )} \log \left (x^{3} + 8 \, x^{2} + 15 \, x\right ) \log \left (\log \left (x^{3} + 8 \, x^{2} + 15 \, x\right )\right )} \,d x } \]
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Time = 9.37 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.81 \[ \int \frac {60+49 x-4 x^2-3 x^3+\left (-60-17 x+4 x^2+x^3+e^x \left (-45 x-9 x^2+5 x^3+x^4\right )\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )+\left (-15 x-8 x^2-x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right ) \log \left (\frac {\log \left (\log \left (15 x+8 x^2+x^3\right )\right )}{x}\right )}{\left (15 x+8 x^2+x^3\right ) \log \left (15 x+8 x^2+x^3\right ) \log \left (\log \left (15 x+8 x^2+x^3\right )\right )} \, dx=4\,\ln \left (\ln \left (\ln \left (x^3+8\,x^2+15\,x\right )\right )\right )-4\,\ln \left (x\right )+{\mathrm {e}}^x\,\left (x-4\right )-\frac {\ln \left (\frac {\ln \left (\ln \left (x^3+8\,x^2+15\,x\right )\right )}{x}\right )\,\left (x^4+8\,x^3+15\,x^2\right )}{x\,\left (x^2+8\,x+15\right )} \]
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