What is the closed form of :$\int\frac{x+2}{\left(x^2+x+2\right)\left(\sqrt{x^2+2x+3}\right)}\,dx$ over $\mathbb{R}$? I have took many times to get antiderivative of the below integral A in $\mathbb{R}$ , but i didn't succeed however it is a constant from $-\infty \to +\infty$ and it's equal :$1.4389$ , but i'm sure that it has a closed form where inverse symbole calculator didn't give me anything , then Is there any way help me to get it's antiderivative over $\mathbb{R}$ ?
$$A=\int_{\mathbb{R}}\frac{x+2}{\left(x^2+x+2\right)\left(\sqrt{x^2+2x+3}\right)}\,dx$$

Using Mathematica with Rubi 4.15.1 we can find very simple form  antiderivative:
$$\int \frac{x+2}{\left(x^2+x+2\right) \sqrt{x^2+2 x+3}} \, dx=\\-\sqrt{\frac{1}{14} \left(11+8 \sqrt{2}\right)} \tan ^{-1}\left(\frac{1+2 \sqrt{2}-\left(5+3
   \sqrt{2}\right) x}{\sqrt{7 \left(11+8 \sqrt{2}\right)} \sqrt{3+2 x+x^2}}\right)+\sqrt{\frac{1}{14} \left(-11+8 \sqrt{2}\right)} \tanh ^{-1}\left(\frac{1-2
   \sqrt{2}-\left(5-3 \sqrt{2}\right) x}{\sqrt{7 \left(-11+8 \sqrt{2}\right)} \sqrt{3+2 x+x^2}}\right)$$
MMA code:
HoldForm[Integrate[(x + 2)/((x^2 + x + 2)*Sqrt[x^2 + 2 x + 3]), 
x] == -Sqrt[1/14 (11 + 8 Sqrt[2])] ArcTan[(
  1 + 2 Sqrt[2] - (5 + 3 Sqrt[2]) x)/(
  Sqrt[7 (11 + 8 Sqrt[2])] Sqrt[3 + 2 x + x^2])] + 
Sqrt[1/14 (-11 + 8 Sqrt[2])]
  ArcTanh[(1 - 2 Sqrt[2] - (5 - 3 Sqrt[2]) x)/(
  Sqrt[7 (-11 + 8 Sqrt[2])] Sqrt[3 + 2 x + x^2])]] // TeXForm


and define integral:
$$\color{blue}{\int_{-\infty }^{\infty } \frac{x+2}{\left(x^2+x+2\right) \sqrt{x^2+2 x+3}} \, dx=\sqrt{\frac{2}{7} \left(11+8 \sqrt{2}\right)} \tan ^{-1}\left(\sqrt{\frac{1}{7}
   \left(1+2 \sqrt{2}\right)}\right)-\sqrt{\frac{1}{14} \left(-11+8 \sqrt{2}\right)} \ln \left(1+\frac{1}{\sqrt{2}}+\sqrt{\frac{1}{2}+\sqrt{2}}\right)}$$
MMA code:
HoldForm[Integrate[(x + 2)/((x^2 + x + 2)*Sqrt[x^2 + 2 x + 3]), {x, -Infinity, Infinity}] == 
Sqrt[2/7 (11 + 8 Sqrt[2])] ArcTan[Sqrt[1/7 (1 + 2 Sqrt[2])]] - 
Sqrt[1/14 (-11 + 8 Sqrt[2])]Log[1 + 1/Sqrt[2] + Sqrt[1/2 + Sqrt[2]]]] // TeXForm

