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Is there a closed form for

$$\int_{\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{2}}} \frac{\tan ^{-1}\left(\frac{\sqrt{-2 m^2+1}}{m^2}\right)}{a^2 m^2+1} \, dm $$ where $a$ is parameter.

Integrating by parts: $$\int_{1/3\,\sqrt {3}}^{1/2\,\sqrt {2}}\!{\frac {m\arctan \left( am \right) }{\sqrt {-2\,{m}^{2}+1} \left( -{m}^{2}+1 \right) }}\,{\rm d} m $$

and by replacing $m=sin(t)$

$$\int_{\arcsin \left( 1/3\,\sqrt {3} \right) }^{\pi/4}\!{\frac {\arctan \left( a\sin \left( t \right) \right) \sin \left( t \right) }{\cos \left( t \right) \sqrt {2\, \left( \cos \left( t \right) \right) ^{2 }-1}}}\,{\rm d}t $$

and......

I tried everything, but I still can not solve it. Any Ideas?

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    What is the source of such problem? Can you give us a taste of the *everything* you have tried? Have you tried integration by parts and/or differentiation under the integral sign? – 2017-02-07
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    Yes I tired integration by parts and/or differentiation under the integral sign,but can not solve :( – 2017-02-07
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    By integration by parts it is enough to compute $$ \int_{1/\sqrt{3}}^{1/\sqrt{2}}\frac{m\arctan(am)}{(1-m^2)\sqrt{1-2m^2}}\,dm $$ and by replacing $m$ with $\sin\theta$ we get something resembling Coxeter/Ahmed's integrals. – 2017-02-07

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