We will evaluate two algebraic integrals by rewriting them as incomplete beta functions. The first yields to a pair of substitutions; the second leads to the addition and duplication formulas for the Weierstrass elliptic function.
\[
\begin{align}
\int_0^1\frac{dx}{\sqrt{1-x}\sqrt[6]{9-x}\sqrt[3]x}&=\frac\pi{\sqrt3} \label{hello-world-eq-1}\\
\int_0^1\frac{dx}{\sqrt{1-x}\sqrt[4]x\sqrt[4]{2-x\sqrt3}} &= \frac{2\sqrt{2}\pi}{3\sqrt[8]{3}} \label{hello-world-eq-2}
\end{align}
\]
These problems are from a math.stackexchange.com question posted by Vladimir Reshetnikov.
Alternate forms
The equations \(\eqref{hello-world-eq-1}\) and \(\eqref{hello-world-eq-2}\) can be expressed in terms of the incomplete beta function using some elementary hypergeometric identities. This makes our problems equivalent to proving that
\[
\begin{align}
B\left(\frac{1}{9};\frac{1}{6},\frac{1}{3}\right) &= \frac{\sqrt{3}}{2^{\frac{4}{3}}\pi}\Gamma\left(\frac{1}{3}\right)^3 \label{hello-world-eq-3}\\
B\left(\frac{\sqrt{3}}{2};\frac{1}{4},\frac{1}{4}\right) &= \frac{2}{3\sqrt{\pi}}\Gamma\left(\frac{1}{4}\right)^2 \label{hello-world-eq-4}
\end{align}
\]
Proof of \(\eqref{hello-world-eq-3}\)
\[
B\left(\frac{1}{9};\frac{1}{6},\frac{1}{3}\right) = \int_0^{\frac{1}{9}}x^{-\frac{5}{6}}\left( 1-x\right)^{-\frac{2}{3}} dx
\]
Substitute \(x=\frac{y^3}{1+y^3}\) in the above integral to get
\[
B\left(\frac{1}{9};\frac{1}{6},\frac{1}{3}\right) =3\int_0^{\frac{1}{2}}\frac{1}{\sqrt{y+y^4}}dy
\]
We can now transform this integral into a beta integral using the substitution \(y=\frac{1-z}{2+z}\).
\[
B\left(\frac{1}{9};\frac{1}{6},\frac{1}{3}\right) =3\int_0^1\frac{1}{\sqrt{1-z^3}}dz = B\left( \frac{1}{3},\frac{1}{2}\right)=\frac{\sqrt{3}}{2^{\frac{4}{3}}\pi}\Gamma\left(\frac{1}{3}\right)^3
\]
This proves equations \(\eqref{hello-world-eq-1}\) and \(\eqref{hello-world-eq-3}\).
Proof of \(\eqref{hello-world-eq-4}\)
The proof of integrals \(\eqref{hello-world-eq-2}\) and \(\eqref{hello-world-eq-4}\) is more involved. Using the substitution \(x=\frac{4t^2}{1+4t^2}\), the integral transforms into the Weierstrass form:
\[
B\left(\frac{\sqrt{3}}{2};\frac{1}{4},\frac{1}{4}\right) =\int_0^{\frac{\sqrt{3}}{2}}x^{-\frac{3}{4}}(1-x)^{-\frac{3}{4}}dx=2\sqrt{2} \int_0^\alpha \frac{dt}{\sqrt{4t^3+t}}
\]
where \(\alpha=\frac{\sqrt{3+2\sqrt{3}}}{2}\). Let \(\wp(z)\) denote the Weierstrass Elliptic function with invariants \(g_2=-1\) and \(g_3=0\). \(\wp(z)\) satisfies the differential equation:
\[
\wp'(z)^2=4\wp(z)^3+\wp(z)
\]
The half periods are \(\omega_1=\frac{1+i}{4}L\) and \(\omega_2=\frac{-1+i}{4}L\) where \(L=\frac{1}{\sqrt{2\pi}}\Gamma\left(\frac{1}{4}\right)^2\) is the Lemniscate constant. Now, we can express our integral as follows:
\[
B\left(\frac{\sqrt{3}}{2};\frac{1}{4},\frac{1}{4}\right) =2\sqrt{2}\left(\wp^{-1}(0)-\wp^{-1}(\alpha) \right)
\]
Note that the following identity is easy to prove using the beta function:
\[
\int_0^\infty \frac{dt}{\sqrt{4t^3+t}}=\frac{1}{2\sqrt{2\pi}}\Gamma\left(\frac{1}{4}\right)^2=\frac{L}{2}
\]
Therefore, \(\wp^{-1}(0)=\frac{L}{2}\). To prove the claim we only need to prove that \(\wp\left(\frac{L}{6}\right)=\alpha\). Using the addition theorem of \(\wp(z)\) and the fact that \(\wp\left(\frac{L}{2}\right)=0\) we get:
\[
\begin{equation}\label{hello-world-eq-5}
\begin{aligned}
\wp\left(\frac{L}{6}\right) &= \wp\left(\frac{L}{2}-\frac{L}{3}\right)\\
&= \frac{1}{4}\left(\frac{\wp'\left(\frac{L}{3}\right)}{\wp\left(\frac{L}{3}\right)} \right)^2-\wp\left(\frac{L}{3}\right) \\
&= \frac{4\wp\left( \frac{L}{3}\right)^3+\wp\left(\frac{L}{3}\right)}{4\wp\left(\frac{L}{3}\right)^2}-\wp\left(\frac{L}{3}\right)\\
&= \frac{1}{4\wp\left(\frac{L}{3}\right)}
\end{aligned}
\end{equation}
\]
On the other hand, by the duplication theorem:
\[
\begin{equation}\label{hello-world-eq-6}
\begin{aligned} \wp\left(\frac{L}{3}\right)&= \frac{1}{4}\frac{\left(6\wp\left(\frac{L}{6}\right)^2+\frac{1}{2}\right)^2}{4\wp\left(\frac{L}{6}\right)^3+\wp\left(\frac{L}{6}\right)}-2\wp\left(\frac{L}{6}\right) \end{aligned}
\end{equation}
\]
Combine equations \(\eqref{hello-world-eq-5}\) and \(\eqref{hello-world-eq-6}\) to get:
\[
\frac{1+4\wp\left(\frac{L}{6}\right)^2}{\left(6\wp\left(\frac{L}{6}\right)^2+\frac{1}{2} \right)^2-32\wp\left(\frac{L}{6}\right)^4-8\wp\left(\frac{L}{6}\right)^2}=1 \\
\implies 16\wp\left(\frac{L}{6}\right)^4-24\wp\left(\frac{L}{6}\right)^2-3=0
\]
Since \(\wp\left( \frac{L}{6}\right)\) is positive, we conclude that
\[
\wp\left(\frac{L}{6}\right)=\frac{\sqrt{3+2\sqrt{3}}}{2}
\]
This completes our proof of equation \(\eqref{hello-world-eq-4}\) and \(\eqref{hello-world-eq-2}\).