超楕円積分の縮約の例

n=6の場合
\begin{gather*} \int_{0}^{x}\frac{\mathrm{d}t}{\sqrt{1+t^{6}}}=\frac{1}{2\sqrt[4]{3}}\int_{0}^{\frac{2\cdot3^{1/4}x\sqrt{1+x^{2}}}{1+(1+\sqrt{3})x^{2}}} \frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\left[1-({\raise{2px}\scriptsize{\frac{1+\sqrt{3}}{2\sqrt{2}}}})^{2}t^{2}\right]}}\\ -\sqrt{\frac{1+\sqrt{3}}{2}}\le x\le\sqrt{\frac{1+\sqrt{3}}{2}} \end{gather*}
n=8の場合
\begin{gather*} \int_{0}^{x}\frac{1}{\sqrt{1+t^{8}}}dt\\ =\frac{1}{2\sqrt{2+\sqrt{2}}}\int_{0}^{\frac{\sqrt{2+\sqrt{2}}x}{1+x^{2}}}\frac{{\mathrm{d}}t}{\sqrt{(1-t^{2})(1-(\sqrt{2}-1)^{2}t^{2})}}\\ +\frac{1}{\sqrt{4+2\sqrt{2}}}\int_{0}^{\frac{1+\sqrt{2}x^{2}+x^{4}-\sqrt{1+x^{8}}}{x\sqrt{2(2-\sqrt{2})(1+\sqrt{2}x^{2}+x^{4})}}}\frac{{\mathrm{d}}t}{\sqrt{(1-t^{2})(1-(\sqrt{2}-1)^{2}t^{2})}}\\ \\ (0< x<1) \end{gather*}

n=12の場合
\begin{gather*} \int_{0}^{x}\frac{{\mathrm{d}}t}{\sqrt{1+t^{12}}}\\ =\frac{1}{\sqrt{6\sqrt{3}}}\int_{0}^{X_{1}}\frac{{\mathrm{d}}t}{\sqrt{\left(1-t^{2}\right)\left(1-\frac{1}{2}t^{2}\right)}}+\frac{1}{\sqrt{2\sqrt{3}}}\int_{0}^{X_{2}}\frac{{\mathrm{d}}t}{\sqrt{\left(1-t^{2}\right)\left(1-\frac{1}{2}t^{2}\right)}}\\ \\ \left\{ \begin{split} X_{1}=\sqrt{\tfrac{2\sqrt{3}x^{2}+\sqrt{(1+x^{4})(1+\sqrt{3}x^{2}+x^{4})}-\sqrt{(1+x^{4})(1-\sqrt{3}x^{2}+x^{4})}}{\sqrt{3}x^{2}+\sqrt{1-x^{4}+x^{8}}+\sqrt{(1+x^{4})(1+\sqrt{3}x^{2}+x^{4})}}}\\ X_{2}=\sqrt{\tfrac{2\sqrt{3}x^{2}-\sqrt{(1+x^{4})(1+\sqrt{3}x^{2}+x^{4})}+\sqrt{(1+x^{4})(1-\sqrt{3}x^{2}+x^{4})}}{\sqrt{3}x^{2}+\sqrt{1-x^{4}+x^{8}}+\sqrt{(1+x^{4})(1-\sqrt{3}x^{2}+x^{4})}}} \end{split} \right.\\ \\ (0< x<1) \end{gather*}
n=20の場合
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