Integration problem: int ln (sin sqrt(x)+ cos sqrt(x))dx

Integration problem: $\int \mathrm{ln}\left(\mathrm{sin}\left(\sqrt{x}\right)+\mathrm{cos}\left(\sqrt{x}\right)\right)dx$
I really don't know how to start solving this problem; any tips or solutions will be greatly appreciated.
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Cagliusov8
By setting $x={u}^{2}$ we have:
$I=\int 2u\mathrm{log}\left(\mathrm{sin}u+\mathrm{cos}u\right)\phantom{\rule{thinmathspace}{0ex}}du={u}^{2}\mathrm{log}\left(\mathrm{sin}u+\mathrm{cos}u\right)-\int {u}^{2}\frac{1-\mathrm{tan}u}{1+\mathrm{tan}u}\phantom{\rule{thinmathspace}{0ex}}du$
and, by putting $v=\frac{\pi }{4}-v$
$-\int {u}^{2}\frac{1-\mathrm{tan}u}{1+\mathrm{tan}u}\phantom{\rule{thinmathspace}{0ex}}du=\int {\left(\frac{\pi }{4}-v\right)}^{2}\mathrm{tan}v\phantom{\rule{thinmathspace}{0ex}}dv$
Now we may exploit $\int \mathrm{tan}v\phantom{\rule{thinmathspace}{0ex}}dv=-\mathrm{log}\mathrm{cos}v$, so the last integral just depends on:
$\int v\mathrm{log}\left(\mathrm{cos}v\right)\phantom{\rule{thinmathspace}{0ex}}dv$
that, however, is not an elementary function, but a combination of a logarithm, a dilogarithm and a trilogarithm multiplied by powers of $v$: just write $\mathrm{cos}v$ as $\frac{{e}^{iv}+{e}^{-iv}}{2}$, exploit the Taylor series of $\mathrm{log}\left(1+z\right)$ and integrate termwise.