超楕円積分の縮約の例

n=6の場合
$$\begin{array}{c}{\int }_0^x\frac{\mathrm{d}t}{\sqrt{1+t^6}}=\frac{1}{2\sqrt[4]{3}}{\int }_0^{\frac{2\cdot 3^{1/4}x\sqrt{1+x^2}}{1+(1+\sqrt{3})x^2}}\frac{\mathrm{d}t}{\sqrt{(1-t^2)[1-(\frac{1+\sqrt{3}}{2\sqrt{2}}{)}^2{t}^2]}}\\ -\sqrt{\frac{1+\sqrt{3}}{2}}\le x\le \sqrt{\frac{1+\sqrt{3}}{2}}\end{array}$$
n=8の場合
$$\begin{array}{c}{\int }_0^x\frac{1}{\sqrt{1+t^8}}dt\\ =\frac{1}{2\sqrt{2+\sqrt{2}}}{\int }_0^{\frac{\sqrt{2+\sqrt{2}}x}{1+x^2}}\frac{\mathrm{d}t}{\sqrt{(1-t^2)(1-(\sqrt{2}-1{)}^2{t}^2)}}\\ +\frac{1}{\sqrt{4+2\sqrt{2}}}{\int }_0^{\frac{1+\sqrt{2}{x}^2+x^4-\sqrt{1+x^8}}{x\sqrt{2(2-\sqrt{2})(1+\sqrt{2}{x}^2+x^4)}}}\frac{\mathrm{d}t}{\sqrt{(1-t^2)(1-(\sqrt{2}-1{)}^2{t}^2)}}\\ \\ (0<x<1)\end{array}$$

n=12の場合
$$\begin{array}{c}{\int }_0^x\frac{\mathrm{d}t}{\sqrt{1+t^{12}}}\\ =\frac{1}{\sqrt{6\sqrt{3}}}{\int }_0^{X_1}\frac{\mathrm{d}t}{\sqrt{(1-t^2)(1-\frac{1}{2}{t}^2)}}+\frac{1}{\sqrt{2\sqrt{3}}}{\int }_0^{X_2}\frac{\mathrm{d}t}{\sqrt{(1-t^2)(1-\frac{1}{2}{t}^2)}}\\ \\ \{\begin{array}{r}{X}_1=\sqrt{\frac{2\sqrt{3}{x}^2+\sqrt{(1+x^4)(1+\sqrt{3}{x}^2+x^4)}-\sqrt{(1+x^4)(1-\sqrt{3}{x}^2+x^4)}}{\sqrt{3}{x}^2+\sqrt{1-x^4+x^8}+\sqrt{(1+x^4)(1+\sqrt{3}{x}^2+x^4)}}}\\ X_2=\sqrt{\frac{2\sqrt{3}{x}^2-\sqrt{(1+x^4)(1+\sqrt{3}{x}^2+x^4)}+\sqrt{(1+x^4)(1-\sqrt{3}{x}^2+x^4)}}{\sqrt{3}{x}^2+\sqrt{1-x^4+x^8}+\sqrt{(1+x^4)(1-\sqrt{3}{x}^2+x^4)}}}\end{array}\\ \\ (0<x<1)\end{array}$$
n=20の場合
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