Is there a closed form for $\int_0^{\pi/2} \frac{e^{-x}\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx $? Is there any closed formula for $\int_{0}^{\pi/2} \dfrac{e^{-x}\sqrt{\cos x}\ dx}{\sqrt{\cos x} + \sqrt{\sin x}}?$
I know $\int_{0}^{\pi/2} \dfrac{\sqrt{\cos x}\ dx}{\sqrt{\cos x} + \sqrt{\sin x}} = \dfrac{\pi}{4},$ replacing $x$ by $\frac{\pi}{2} - y$.
 A: Here is an experimental result using a special case of Lauricella D from the DLMF’s hyperelliptic $\text R$ function which has a multiple series expansion, using the Pochhammer symbol $(u)_v$, from

Is this Lauricella $\text F_\text D$ to hypergeometric R, from DLMF, conversion formula correct?:

$$\text B(-a,(-a)’)\text R_a(b_1,\dots,b_n;z_1,\dots z_n)=\int_0^\infty t^{(-a)’-1}\prod_{j=1}^n(t+z_j)^{-b_j}dt=\sum_{m_1\ge0}\cdots\sum_{m_n\ge0}\frac{(-a)_{\sum_{j=1}^n m_j}}{\left(\sum\limits_{j=1}^nb_j\right)_{\sum_{j=1}^n m_j}}\prod_{j=1}^n\frac{(b_j)_{m_j}(1-x_j)^{m_j}}{m_j!},(-a)’=a+\sum_{j=1}^n b_j$$
Using
$$I=1-(-1)^\frac i2\int_0^\infty t^0(t-\sqrt i)^\frac i2(t+\sqrt i)^\frac i2(t-\sqrt{-i})^{-\frac i2}(t+\sqrt{-i})^{-\frac i2}(t+1)^{-2}dt$$
and $(-a)’=1=a-\frac i2-\frac i2+\frac i2+\frac i2+2\implies a=-1$
Therefore?:
$$I=1-i^i\text R_{-1}\left(-\frac i2,-\frac i2,\frac i2,\frac i2,2;-\sqrt i,\sqrt i,-\sqrt{-i},\sqrt {-i},1\right)$$
One problem is $x_5=1$ gives $(1-x_5)^{m_5}=0^{m_5}$ in the series expansion. Fortunately, the function is homogenous:
$$1-(i^{-i})^{-1}\text R_{-1}\left(-\frac i2,-\frac i2,\frac i2,\frac i2,2;-\sqrt i,\sqrt i,-\sqrt{-i},\sqrt {-i},1\right) = 1-\text R_{-1}\left(-\frac i2,-\frac i2,\frac i2,\frac i2,2;i^{\frac 52-i},i^{\frac12-i},i^{\frac32-i},i^{-\frac12-i},i^{-i}\right) $$
Now for the possible series expansion
