$$ \displaystyle\int \cos\left(\dfrac{{\pi}{\cdot}{x}^{2}}{240}\right)\, \mathrm d x = -{{\sqrt{15}\,\left(i\,\mathrm{erf}\left({{\sqrt{\pi}\,\left(i+1 \right)\,x}\over{4\sqrt{2}\,\sqrt{15}}}\right)-\mathrm{erf} \left({{\sqrt{\pi}\,\left(i+1\right)\,x}\over{4\sqrt{2}\, \sqrt{15}}}\right)+i\,\mathrm{erf}\left({{\sqrt{\pi}\,\left(i-1 \right)\,x}\over{4\sqrt{2}\,\sqrt{15}}}\right)+\mathrm{erf} \left({{\sqrt{\pi}\,\left(i-1\right)\,x}\over{4\sqrt{2}\, \sqrt{15}}}\right)-i\,\mathrm{erf}\left({{\sqrt{\pi}\,\sqrt{-i}\,x }\over{4\,\sqrt{15}}}\right)-\mathrm{erf}\left({{\sqrt{\pi}\,\sqrt{- i}\,x}\over{4\,\sqrt{15}}}\right)+i\,\mathrm{erf}\left({{\left(-1 \right)^{{{1}\over{4}}}\,\sqrt{\pi}\,x}\over{4\,\sqrt{15}}}\right)- \mathrm{erf}\left({{\left(-1\right)^{{{1}\over{4}}}\,\sqrt{\pi}\,x }\over{4\,\sqrt{15}}}\right)\right)}\over{2\sqrt{2}}} $$

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