Let us denote by $J$ the integral to be computed. We have
$$
\begin{aligned}
J&=\int_{\sqrt{5/3}}^{\sqrt2}
\frac{\arctan x}{( 2x^2-3 )\sqrt{3x^2-5}}
\; dx
\\
&=
\int_{\sqrt{1/2}}^{\sqrt{3/5}}
x\cdot\frac{\arctan (1/x)}{( 2-3x^2 )\sqrt{3 - 5x^2}}
\; dx
=
\int_{\sqrt{1/2}}^{\sqrt{3/5}}
x\cdot \frac{\frac\pi 2-\arctan x}{( 2-3x^2 )\sqrt{3 - 5x^2}}
\; dx
\\
&=
\frac\pi2\cdot\frac 12\cdot
\int_{1/2}^{3/5}
\frac{du}{( 2-3u )\sqrt{3 - 5u}}
\; du
-
\frac 12
\int_{1/2}^{3/5}
\frac{\arctan \sqrt u}{( 2-3u )\sqrt{3 - 5u}}
\; du
\\
&=\frac \pi4\left(
\frac{\sqrt2}3-\frac{2\sqrt3}9\arctan\sqrt{\frac32}
\right)
-
\frac 1{6\sqrt 5}
\underbrace{
\color{blue}{
\int_{1/2}^{3/5}
\frac{\arctan \sqrt u}{\left( \frac 23-u \right)\sqrt{\frac 35 - u}}
\; du}}
_{=:\color{blue}{K}}
\ .
\end{aligned}
$$
So let us compute $K$. Recall from a related post the
excellent presentation of pisco the way to compute such integrals.
He also cites the book of Lewin,
Polylogarihm and Associated Functions, Leonard Lewin, page 115-117.
We have with the conventions and definitions in loc. cit. the relation
$$
\color{blue}{\int_0^x\frac{\arctan\sqrt t}{(a-t)\sqrt{b-t}}\; dt}
=
\frac 1{\sqrt{a-b}}
S\left(\
\arctan \sqrt{\frac{b-x}{a-b}}\ ,\
\arctan \sqrt{\frac{b+1}{a-b}}\ ,\
\arctan \sqrt{\frac1a}\
\right)
\ ,
$$
where $S(\alpha, \beta,\gamma)$ is defined and computed as follows.
First of all associate the following "modulus" $k=k(\alpha,\beta,\gamma)$.
$$
\begin{aligned}
d_1 &= \sqrt{\cos^2\alpha \cos^2\gamma-\cos^2\beta}\ ,\\
d_2 &= \sin\alpha \sin\gamma\ ,\\
k &=\frac{d_1-d_2}{d_1+d_2}\ .
\end{aligned}
$$
Then set for this value of $k=k(\alpha,\beta,\gamma)$:
$$
\begin{aligned}
S(\alpha,\beta,\gamma)
&=f(\alpha,k)-f(\beta,k)+f(\gamma,k)-f(0,k)-\alpha^2+\beta^2-\gamma^2
\ ,\\
f(\alpha,k)
&=\sum_{n\ge 1}\frac1{n^2}k^n\cos(2n\alpha)
=\Re\sum_{n\ge 1}\frac1{n^2}(ke^{2\alpha i})^n
=\Re\operatorname{Li}_2\Big(ke^{2\alpha i}\Big)
\ .
\end{aligned}
$$
Then $S$ has the following properties, see again loc. cit.:
$$
\begin{aligned}
S(0,\beta,\gamma) &= \pi(\beta-\gamma)\ ,\\
S(\alpha,\pi-2\alpha,\alpha) &= 6\cdot S\left(\alpha,\frac\pi 3,\frac \pi6\right)
\ .
\end{aligned}
$$
In our case we have
- $a=\frac 23$, $b=\frac35$, $x_1=\frac 12$, $x_2=\frac 35=b$,
- so $a-b=\frac1{15}$, $b-x_1=\frac 1{10}$, $b-x_2=0$,
- and associate $\alpha=\alpha_1=\arctan\sqrt{\frac{b-x_1}{a-b}}=\arctan\sqrt{\frac{1/10}{1/15}}=\arctan\sqrt{\frac 32}$, later we also need $\cos^\alpha=\frac 1{1+\tan^2\alpha}=\frac 1{5/2}=\frac 25$, and $\sin^\alpha=\frac 35$,
- $\alpha_2=\arctan\sqrt{\frac{b-x_2}{a-b}}=\arctan 0=0$,
- $\beta=\arctan\sqrt{\frac{b+1}{a-b}}=\arctan\sqrt{\frac{8/5}{1/15}}=\arctan\sqrt{24}=\pi-2\alpha$,
- $\gamma=\arctan\sqrt{\frac1a}=\arctan\sqrt{\frac32}$.
and have to compute
$$
\begin{aligned}
\color{blue}{K}&=
\frac 1{\sqrt{a-b}}
\left[\
S\left(\
\arctan \sqrt{\frac{b-x}{a-b}}\ ,\
\arctan \sqrt{\frac{b+1}{a-b}}\ ,\
\arctan \sqrt{\frac1a}\
\right)\
\right]_{x=x_1}^{x=x_2}
\\
&=
\sqrt{15}S\left(\ 0\ ,\ \arctan \sqrt {24}\ ,\ \sqrt{\frac32}\ \right)
\\
&\qquad
-\sqrt{15}S\left(\ \arctan\sqrt{\frac32}\ ,\arctan \sqrt {24}\ ,\ \arctan\sqrt{\frac32}\ \right)
\\
&=\sqrt{15}(S_1-S_2)\ ,\\[3mm]
S_1 &:=S(\alpha_1,\beta,\gamma)=S(0,\beta,\gamma)=\pi(\beta-\gamma)=\pi\left(\arctan \sqrt{24} - \arctan\sqrt{\frac32}\right)
\\
&=\pi\arctan\frac{3\sqrt 6}{14}\ ,
\\
S_2 &=S(\alpha_2,\beta,\gamma)=S(\alpha,2\pi-\alpha,\alpha)=
6\cdot S\left(\alpha,\frac\pi 3,\frac \pi6\right)\ ,
\\
&\qquad \text{ and only $S_2$ has to be computed to complete.}
\\
&\qquad \text{ Above, there are two way to do this. Same modulus $k$.}
\\
k\left(\alpha,\beta,\gamma\right)
&%=\frac{d_1-d_2}{d_1+d_2}
=\frac
{\sqrt{\frac25\cdot\frac 25-\frac 1{25}}-\frac 35}
{\sqrt{\frac25\cdot\frac 25-\frac 1{25}}+\frac 35}
=
\frac{\sqrt 3-3}{\sqrt 3+3}
=-\frac 12(\sqrt 3-1)^2=\sqrt 3-2\ ,
\\
k\left(\alpha,\frac\pi 3,\frac \pi6\right)
&=\frac
{\sqrt{\frac25\cdot\frac 34-\frac 14}-\frac 12\cdot\sqrt{\frac 35}}
{\sqrt{\frac25\cdot\frac 34-\frac 14}+\frac 12\cdot\sqrt{\frac 35}}
=
\frac{\sqrt{\frac 15}-\sqrt{\frac 35}}{\sqrt{\frac 15}+\sqrt{\frac 35}}
=-\frac 12(\sqrt 3-1)^2=\sqrt 3-2\ ,
\end{aligned}
$$
We have now an explicit expression in terms of dilogarithmic values and "easier" data for the integral $K=\sqrt{15}(S_1-S_2)$,
thus also for $J$. Here, $S_1$ is also simple, but $S_2$ involves the (real part of the) dilogarithm $\operatorname{Li}_2$ computed in points like
$$
(\sqrt 3-2)\cdot\left(\sqrt{\frac 25}-i\sqrt{\frac 35}\right)^2\ ,\\
(\sqrt 3-2)\cdot\frac 12(-1+\sqrt 3)\ ,\qquad
(\sqrt 3-2)\cdot\frac 12(1+i\sqrt 3)\ ,\qquad
(\sqrt 3-2)\cdot 1\ .
$$
And there is no K-theoretic trick to get something like a multiple of $\pi$.
I have to stop here, computations were done without double check, there may be some, but the computational line applies at any rate.
