On $\big(\tfrac{1+\sqrt{5}}{2}\big)^{12}=\small 161+72\sqrt{5}$ and $\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt{161+72\sqrt{5}\,x}}$

(This summarizes scattered results from here, here, here and elsewhere. See also this older post.)

I. Cubic

Define $\beta= \tfrac{\Gamma\big(\tfrac56\big)}{\Gamma\big(\tfrac13\big)\sqrt{\pi}}= \frac{1}{48^{1/4}\,K(k_3)}$. Then we have the nice evaluations,

\begin{aligned}\frac{3}{5^{5/6}} &=\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-4\big)\\ &=\beta\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt{x^2+4x^3}}\\[1.7mm] &=\beta\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small2/3} \sqrt{\color{blue}{9+4\sqrt{5}}\,x}}\\[1.7mm] &=2^{1/3}\,\beta\,\int_0^\infty\frac{dx}{\sqrt{9+\cosh x}} \end{aligned}\tag1 and, \begin{aligned}\frac{4}{7} &=\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-27\big)\\ &=\beta\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt{x^2+27x^3}}\\[1.7mm] &=\beta\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small2/3} \sqrt{\color{blue}{55+12\sqrt{21}}\,x}}\\[1.7mm] &=2^{1/3}\,\beta\,\int_0^\infty\frac{dx}{\sqrt{55+\cosh x}} \end{aligned}\tag2 Note the powers of fundamental units, $$U_{5}^6 = \big(\tfrac{1+\sqrt{5}}{2}\big)^6=\color{blue}{9+4\sqrt{5}}$$ $$U_{21}^3 = \big(\tfrac{5+\sqrt{21}}{2}\big)^3=\color{blue}{55+12\sqrt{21}}$$ Those two instances can't be coincidence.

II. Quartic

Define $\gamma= \tfrac{\sqrt{2\pi}}{\Gamma^2\big(\tfrac14\big)}= \frac{1}{2\sqrt2\,K(k_1)}=\frac1{2L}$ with lemniscate constant $L$. Then we have the nice,

\begin{aligned}\frac{2}{3^{3/4}} &=\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-3\big)\\ &=\gamma\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt{x^3+3x^4}}\\[1.7mm] &\overset{\color{red}?}=\gamma\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt{\color{blue}{7+4\sqrt{3}}\,x}}\\[1.7mm] &=2^{1/4}\,\gamma\,\int_0^\infty\frac{dx}{\sqrt{7+\cosh x}} \end{aligned}\tag3 and, \begin{aligned}\frac{3}{5}&=\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-80\big)\\ &=\gamma\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt{x^3+80x^4}}\\[1.7mm] &\overset{\color{red}?}=\gamma\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt{\color{blue}{161+72\sqrt{5}}\,x}}\\[1.7mm] &=2^{1/4}\,\gamma\,\int_0^\infty\frac{dx}{\sqrt{161+\cosh x}} \end{aligned}\tag4

with $a=161$ given by Noam Elkies in this comment. (For $4$th roots, I just assumed the equality using the blue radicals based on the ones for cube roots.) Note again the powers of fundamental units, $$U_{3}^2 = \big(2+\sqrt3\big)^2=\color{blue}{7+4\sqrt{3}}$$ $$U_{5}^{12} = \big(\tfrac{1+\sqrt{5}}{2}\big)^{12}=\color{blue}{161+72\sqrt{5}}$$ Just like for the cube roots version, these can't be coincidence.

Questions:

Is it true these observations can be explained by, let $b=2a+1$, then,

$$\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt{x^2+ax^3}}=\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small2/3} \sqrt{b+\sqrt{b^2-1}\,x}}=2^{1/3}\int_0^\infty\frac{dx}{\sqrt{b+\cosh x}}$$

$$\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt{x^3+ax^4}}=\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt{b+\sqrt{b^2-1}\,x}}=2^{1/4}\int_0^\infty\frac{dx}{\sqrt{b+\cosh x}}$$

Starting from $$\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-a\big)=\gamma\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt{x^3+ax^4}},$$ $$(b+\sqrt{b^2-1})^{-1/4}\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{1}{2};\tfrac{2\sqrt{b^2-1}}{b+\sqrt{b^2-1}}\big)={\gamma}\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt{{b+\sqrt{b^2-1}}\,x}},$$ (with $\gamma$ defined above) and applying transformations 2.11(4), 2.10(6), 2.11(2) from Erdelyi, Higher transcendental functions, vol. I, to the second hypergeometric function one gets \begin{align} (b+\sqrt{b^2-1})^{-1/4}\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{1}{2};\tfrac{2\sqrt{b^2-1}}{b+\sqrt{b^2-1}}\big)&=b^{-1/4}\,_2F_1\big(\tfrac{1}{8},\tfrac{5}{8};\tfrac{3}{4};\tfrac{{b^2-1}}{b^2}\big)\\ &=\,_2F_1\big(\tfrac{1}{8},\tfrac{1}{8};\tfrac{3}{4};1-b^2\big)\\ &=\,_2F_1\big(\tfrac{1}{8},\tfrac{1}{8};\tfrac{3}{4};-4a(1+a)\big)\\ &=\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-a\big), \end{align} where $b=2a+1$, thus proving that $$\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt{x^3+ax^4}}=\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt{b+\sqrt{b^2-1}\,x}}.$$

More generally application of the same series of transformations gives $${(b+\sqrt{b^2-1})^{-\alpha } \, _2F_1\left(\alpha ,\alpha ;2 \alpha ;\tfrac{2 \sqrt{b^2-1}}{b+\sqrt{b^2-1}}\right)}={\, _2F_1\left(\alpha ,\alpha ;\alpha +\tfrac{1}{2};-a\right)},$$ i.e. $$\int_0^1 \frac{dx}{\sqrt{1-x}\,x^{1-\alpha}(1+ax)^{\alpha}}=\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{1-\alpha} (b+\sqrt{b^2-1}\,x)^{\alpha}}.$$ When $\alpha=1/3$ this answers the related question.

Formula 2.12(10) from Erdelyi, Higher transcendental functions, vol. I answers the second equality, namely $${\, _2F_1\left(\alpha ,\alpha ;\alpha +\tfrac{1}{2};-a\right)}=2^{\alpha}\frac{\Gamma(\alpha+1/2)}{\sqrt{\pi}\Gamma(\alpha)}\int_0^\infty\frac{dx}{(b+\cosh x)^\alpha}.$$

• @TitoPiezasIII , isn't it better to combine the cubic and quartic cases nicely in one question putting them side by side, even at the cost of making the question too long? Do you know what is the significance of this pattern? – Nemo Dec 5 '16 at 11:39
• Very well, they have been combined. I have a thing for finding patterns anyway. – Tito Piezas III Dec 5 '16 at 12:07
• Do you know a transformation between $_2F_1\big(\tfrac13,\tfrac13, \tfrac56,\alpha\big)$ and $_2F_1\big(\tfrac13,\tfrac23, \tfrac56,\beta\big)$? – Tito Piezas III Dec 18 '16 at 19:33
• No, I don't know. – Nemo Dec 18 '16 at 21:59

Too long for a comment : In general, for strictly positive values of n we have

\begin{align} \sqrt[n]2\int_0^\infty\frac{dx}{\sqrt[n]{\cosh2t~+~\cosh x}} ~&=~ \int_0^1\frac{dx}{\sqrt{1-x}\cdot\sqrt[n]{x^{n-1}~+~x^n\cdot\sinh^2t}} \\\\ ~&=~ \int_{-1}^1\frac{dx}{\sqrt[n]{(1-x^2)^{n-1}}\cdot\sqrt[n]{\cosh2t~+~x\cdot\sinh2t}} \end{align}

• +1 Thanks. This will help categorize related integrals together. – Tito Piezas III Dec 9 '16 at 5:38