On $\big(\tfrac{1+\sqrt{5}}{2}\big)^{12}=\small 161+72\sqrt{5}$ and $\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt[4]{161+72\sqrt{5}\,x}}$
Starting from $$ \,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-a\big)=\gamma\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+ax^4}}, $$ $$ (b+\sqrt{b^2-1})^{-1/4}\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{1}{2};\tfrac{2\sqrt{b^2-1}}{b+\sqrt{b^2-1}}\big)={\gamma}\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt[4]{{b+\sqrt{b^2-1}}\,x}}, $$ (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 $$ \int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+ax^4}}=\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt[4]{b+\sqrt{b^2-1}\,x}}. $$
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 $$ {\, _2F_1\left(\alpha ,\alpha ;\alpha +\tfrac{1}{2};-a\right)}=2^{\alpha}\frac{\Gamma(\alpha+1/2)}{\sqrt{\pi}\Gamma(\alpha)}\int_0^\infty\frac{dx}{(b+\cosh x)^\alpha}. $$
Too long for a comment : In general, for strictly positive values of n we have
$$\begin{align} \sqrt[n]2\int_0^\infty\frac{dx}{\sqrt[n]{\cosh2t~+~\cosh x}} ~&=~ \int_0^1\frac{dx}{\sqrt{1-x}\cdot\sqrt[n]{x^{n-1}~+~x^n\cdot\sinh^2t}} \\\\ ~&=~ \int_{-1}^1\frac{dx}{\sqrt[n]{(1-x^2)^{n-1}}\cdot\sqrt[n]{\cosh2t~+~x\cdot\sinh2t}} \end{align}$$