$ \int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x $ I am trying to evaluate this antiderivative $$
\int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x
$$
What i have done:
$$
\begin{split}
I
 &= \int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x \\
 &= \int \frac{2+\sqrt{x+1}-\sqrt{3-x}}{4+x+1+4 \sqrt{x+1}-3+x} \cdot d x\\
 &= \int \frac{2+\sqrt{x+1}-\sqrt{3-x}}{2+2 x+4 \sqrt{x+1}} \cdot d x\\&=\int \frac{1}{1+x+2 \sqrt{x+1}} \cdot d x\\
 &\quad +\int \frac{\sqrt{x+1}}{2(1+x)+4 \sqrt{x+1}}\cdot d x \\
 &\quad -\int\frac{\sqrt{3-x}}{2(1+x)+4 \sqrt{x+1}} \cdot d x
\end{split}
$$
From this last line i can evaluate the first two antiderivatives but the last one $$\int\frac{\sqrt{3-x}}{2(1+x)+4 \sqrt{x+1}} \cdot dx$$ seems a bit hard for me two evaluate, i don't find the good substitution. I am opened to any suggestion. Thanks in advance!

Too large for a comment in the other answer.
Let us first try to identify the correct substitution
For real calculus, we need $x+1\ge0\text{ and }3-x\ge0\iff-1\le x\le3$
$$\iff-\dfrac12-\dfrac12\le\dfrac{x-1}2\le\dfrac32-\dfrac12$$
WLOG $\dfrac{x-1}2=\cos2t, 0\le2t\le\pi\implies x=2\cos2t+1$
$\sqrt{x+1}=+(2\cos t),\sqrt{3-x}=+(2\sin t)$ as $0\le t\le\dfrac\pi2$
Finally use $(\sin t+\cos t+1)(\sin t+\cos t-1)=(\sin t+\cos t)^2-1=?$
