<s>[INST] How do i evaluate this integral $ \int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx $? Is there some one show me how do i evaluate this integral :$$ \int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx $$
Note :By mathematica,the result is :
$\frac{Gamma\left(\frac1 4\right)Gamma\left(\frac5 4\right)}{\sqrt{\pi}}-\sqrt{2} Hypergeometric2F1\left(\frac1 4,\frac3 4,\frac5 4,\frac1 4\right).$
and i think it elliptic integral .
Thank you for any kind of help 
 [/INST] This is not an answer but it is too long for a comment.
As I wrote in comment, there is something wrong somewhere since $$\int\frac{\sqrt{\tan x}}{\sin x}dx=-2 \sqrt{\cos (x)} \, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};\cos
   ^2(x)\right)$$ and $$I=\int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx=\frac{\Gamma \left(\frac{1}{4}\right) \Gamma \left(\frac{5}{4}\right)}{\sqrt{\pi
   }}-\sqrt{2} \, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};\frac{1}{4}\right)\approx 0.379133$$ which does not match with Mercy's result $$\ln\left(\frac{1+\sqrt2}{\sqrt3}\right) \approx 0.332067$$
Reusing all steps done for Mercy, I thing that in fact, we should arrive to $$I=\int_1^{\sqrt{3}}\frac{1}{\sqrt t \,\sqrt{1+t^2}}\,dt=\frac{4 \Gamma \left(\frac{5}{4}\right)^2}{\sqrt{\pi }}-\frac{2 \,
   _2F_1\left(\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{1}{3}\right)}{\sqrt[4]{3}}\approx 0.379133$$ which is still a nightmare ( the Inverse Symbolic Calculator did not find any solution for this number).
Totally unable to find any closed form derivative, I came back to the original problem and expanded the integrand as a Taylor series built at $x=\frac \pi 4$. This gave $$\frac{\sqrt{\tan x}}{\sin x}=\sqrt{2}+\sqrt{2} \left(x-\frac{\pi }{4}\right)^2+\frac{7 \left(x-\frac{\pi
   }{4}\right)^4}{3 \sqrt{2}}+\frac{139 \left(x-\frac{\pi }{4}\right)^6}{45
   \sqrt{2}}+\frac{5473 \left(x-\frac{\pi }{4}\right)^8}{1260
   \sqrt{2}}+\cdots$$ Integrating between the given bounds leads to $$I\approx \frac{\pi }{6 \sqrt{2}}+\frac{\pi ^3}{2592 \sqrt{2}}+\frac{7 \pi ^5}{3732480
   \sqrt{2}}+\frac{139 \pi ^7}{11287019520 \sqrt{2}}+\frac{5473 \pi
   ^9}{58511909191680 \sqrt{2}}+\cdots$$ which gives the correct value for six significant figures.
Edit
For sure, the exact solution exits but it involves elliptic integrals. Using the tangent half-angle substitution $t=\tan(\frac x2)$ and the fiven bounds, the result is $$I=-2 i \sqrt{2} \left(F\left(\left.\sin
   ^{-1}\left(\sqrt[4]{3}\right)\right|-1\right)-F\left(\left.\sin
   ^{-1}\left(\sqrt{1+\sqrt{2}}\right)\right|-1\right)\right)$$ where appear  elliptic integrals of the first kind.
Edit
If we change variable $x=y+\frac \pi 4$, expanding the trigonometric functions and simplifying, we can arrive at $$I=\int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx=\sqrt{2}\int_{0}^{\pi /12}\frac{dy}{\sqrt{\cos (2 y)}}=\sqrt{2}\, F\left(\left.\frac{\pi }{12}\right|2\right)$$
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