Entry 121

This is the case \(a=\frac13\) of $${_2F_1\left(a ,a ;a +\tfrac12;-u\right)}=2^{a}\frac{\Gamma\big(a+\tfrac12\big)}{\sqrt\pi\,\Gamma(a)}\int_0^\infty\frac{dx}{(1+2u+\cosh x)^a}$$ we have $$ \frac{1}{48^{1/4}\,K(k_3)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[3]{x^2+\color{blue}{4}x^3}}=\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-\color{blue}{4}\big)= \frac3{5^{5/6}}$$

$$ \frac{1}{48^{1/4}\,K(k_3)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[3]{x^2+\color{blue}{27}x^3}}=\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-\color{blue}{27}\big)=\frac{4}{7}$$ (To be continued.) 

Entry 120

This is the case \(a=\frac14\) of $${_2F_1\left(a ,a ;a +\tfrac12;-u\right)}=2^{a}\frac{\Gamma\big(a+\tfrac12\big)}{\sqrt\pi\,\Gamma(a)}\int_0^\infty\frac{dx}{(1+2u+\cosh x)^a}$$ we have $$\frac{1}{2\sqrt2\,K(k_1)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+\color{blue}{3}x^4}}=\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-\color{blue}3\big) = \frac{2}{3^{3/4}}$$

$$\frac{1}{2\sqrt2\,K(k_1)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+\color{blue}{80}x^4}}=\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-\color{blue}{80}\big) = \frac35$$ (To be continued.) 

Entry 119

This is the case \(a=\frac16\) of $${_2F_1\left(a ,a ;a +\tfrac12;-u\right)}=2^{a}\frac{\Gamma\big(a+\tfrac12\big)}{\sqrt\pi\,\Gamma(a)}\int_0^\infty\frac{dx}{(1+2u+\cosh x)^a}$$ we have $$\frac{1}{\color{red}{432}^{1/4}\,K(k_3)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[6]{x^5+\color{blue}{\tfrac{125}3}x^6}}=\,_2F_1\big(\tfrac16,\tfrac16;\tfrac23;-\color{blue}{\tfrac{125}{3}})=\frac{2}{3^{5/6}}$$

$$\frac{1}{\color{red}{432}^{1/4}\,K(k_3)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[6]{x^5+\color{blue}{2^7\phi^9}\, x^6}}=\,_2F_1\big(\tfrac16,\tfrac16;\tfrac23;-\color{blue}{2^7\phi^9})=\frac{3}{5^{5/6}}\phi^{-1}$$ with golden ratio \(phi\). (To be continued.) 

Entry 118

If one has a palindromic quartic of form $$z^4-abz^3+(a^2+b^2-2)z^2-abz+1=0$$ then its roots can be factored as roots \((x,y)\) of quadratics $$x^2+ax+1=0\\ y^2+by+1= 0$$ $$z = xy = \left(\tfrac{-a+\sqrt{a^2-4}}2\right) \left(\tfrac{-b+\sqrt{b^2-4}}2\right)$$ hence are products of quadratic units. (To be continued.)

Entry 117

For fundamental discriminants \(d=4m\) with class number \(h(-d)=16\), there are exactly 60 \(m\) that are even. The largest is \(m = 3502 = 2\times17\times103\) hence has \(2^3 = 8\) divisors. But this set has no \(m\) with 32 divisors so it seems one can't express their modular lambda function \(\lambda(\sqrt{-m})\) with 16 quadratic units. (Unlike for \(h(-d)=8\) where we can express a few \(\lambda(\sqrt{-m})\) with 8 quadratic units.) However, using another function, we can have four quartic units. Given the nome \(q = e^{\pi i\tau}\), \(\tau=\sqrt{-n}\), and the Ramanujan G and g functions $$\begin{align}2^{1/4}G_n &= q^{-\frac{1}{24}}\prod_{k>0}(1+q^{2k-1}) = \frac{\eta^2(\tau)}{\eta\big(\tfrac{\tau}{2}\big)\eta(2\tau)}\\ 2^{1/4}g_n &= q^{-\frac{1}{24}}\prod_{k>0}(1-q^{2k-1}) = \frac{\eta\big(\tfrac{\tau}{2}\big)}{\eta(\tau)}\end{align}$$ discussed in Entry 116. Then $$\color{red}u  = (g_{3502})^4 = \small\left(\frac{\;\eta\big(\tfrac{\tau}{2}\big)}{2^{1/4}\eta(\tau)}\right)^4 = \big(a+\sqrt{a^2-1}\big)^2 \big(b+\sqrt{b^2-1}\big)^2 \big(c+\sqrt{c^2-1}\big) \big(d+\sqrt{d^2-1}\big) \approx 1.43\times10^{13}$$ where \(\tau = \sqrt{-3502}\) and \((a,b,c,d)\) are $$\begin{align}a &= \tfrac{1}{2}(23+4\sqrt{34})\\ b &= \tfrac{1}{2}(19\sqrt{2}+7\sqrt{17})\\ c &= (429+304\sqrt{2})\\ d &= \tfrac{1}{2}(627+442\sqrt{2})\end{align}$$A version of this was first found by Daniel Shanks in 1980 (Quartic Approximations for Pi) but this one is slightly different with smaller integers since I simplified the first two expressions as squares. These radicals imply a very close approximation to pi,$$\pi \approx \frac{1}{\sqrt{3502}}\ln\big((2\color{red}u)^6+24\big)$$ which differs by just \(10^{-161}.\)