I have three problems of the form
$\int_0^{\frac{1}{3}} f(x) \, dx$,
where
$f(x)=-2 \sqrt{9 x+1} \sqrt{21 x-4 \sqrt{3} \sqrt{x (9 x+1)}+1}$, and $f(x)=-42 x \sqrt{9 x+1} \sqrt{21 x-4 \sqrt{3} \sqrt{x (9 x+1)}+1}$ and $f(x)=8 \sqrt{3} \sqrt{9 x+1} \sqrt{x (9 x+1)} \sqrt{21 x-4 \sqrt{3} \sqrt{x (9 x+1)}+1}$.
Mathematica does not solve these integration problems, but readily does the problems of the form $\int_0^{\frac{1}{3}} f(x)^2 \, dx$, yielding $\frac{2 \sinh ^{-1}\left(\sqrt{3}\right)}{9 \sqrt{3}}$, $\frac{49}{648} \left(6+\sqrt{3} \sinh ^{-1}\left(\sqrt{3}\right)\right)$ and $\frac{4}{3}-\frac{\log \left(7+4 \sqrt{3}\right)}{9 \sqrt{3}}$, respectively.
Can these observations be employed in any manner in solving the original three problems, or can these three problems be solved directly (transformation of variables,...), despite Mathematica's failure to do so?
Of course, the indefinite integrations $\int f(x)^2 \, dx$ can also be solved, yielding $4 \left(63 x^3+15 x^2-\frac{\sqrt{x (9 x+1)} \left(216 x^2+42 x+1\right)}{6 \sqrt{3}}+x+\frac{\log \left(18 x+6 \sqrt{x (9 x+1)}+1\right)}{36 \sqrt{3}}\right)$ and $1764 \left(\frac{189 x^5}{5}+\frac{15 x^4}{2}+\frac{x^3}{3}-\frac{\sqrt{x (9 x+1)} \left(279936 x^4+42768 x^3+216 x^2-30 x+5\right)}{12960 \sqrt{3}}+\frac{\log \left(18 x+6 \sqrt{x (9 x+1)}+1\right)}{15552 \sqrt{3}}\right)$ and $192 \left(\frac{1701 x^5}{5}+\frac{459 x^4}{4}+13 x^3+\frac{x^2}{2}-\frac{\sqrt{x (9 x+1)} \left(279936 x^4+81648 x^3+6696 x^2+30 x-5\right)}{1440 \sqrt{3}}-\frac{\log \left(18 x+6 \sqrt{x (9 x+1)}+1\right)}{1728 \sqrt{3}}\right)$, respectively.
Per the comment of "uniquesolution", the NIntegrate command with WorkingPrecision->32 gives -0.19559399958899403861456602832005, -0.41995514862765510309782111510810, and 0.58218377818878373916321174945773, respectively. From the quantum-information-theoretic context, from which these problems stem https://arxiv.org/abs/1905.09228, I strongly suspect/hope that these numerical results correspond to rather simple exact expressions (possibly involving inverse hyperbolic functions, such as $\sinh ^{-1}\left(\sqrt{3}\right))$. But the WolframAlpha site https://www.wolframalpha.com (my usual recourse in such matters) does not seem to discern them.
The third numerical value given in the previous paragraph is clearly equal to the first value plus $\frac{7}{9}$. So, we have (taking the difference of the two integrands) that $\int_0^{\frac{1}{3}} 2 \sqrt{9 x+1} \sqrt{21 x-4 \sqrt{3} \sqrt{x (9 x+1)}+1} \left(4 \sqrt{3} \sqrt{x (9 x+1)}+1\right) \, dx= \frac{7}{9}$. Can this be formally demonstrated? (See https://mathoverflow.net/questions/338679/prove-that-a-certain-integration-yields-the-value-frac79 for an affirmative answer to this question, and an apparent route to solving the three initial problems of this question.)
$$\DeclareMathOperator{\arcoth}{arcoth}$$ Let $$\begin{align} 3x &= u^2 &9x+1&=v^2\text{,} \end{align}$$ and stereographically project the curve $v^2-3u^2=1$ about $(u,v)=(0,1)$ so that $$\frac{v-1}{u}=\frac{3u}{v+1}=m\text{.}$$ Then the integrands reduce to rational expressions in $m$, and the integrals are given by $$-\frac{1}{27}\left(3 +2\sqrt{3}\arcoth{\sqrt{3}}\right)$$ $$-\frac{7}{81}\left(6 -\sqrt{3}\arcoth{\sqrt{3}}\right)$$ $$\frac{2}{27}\left(9 -\sqrt{3}\arcoth{\sqrt{3}}\right)$$