How to integrate $\int_{0}^{1} \frac{1-x}{1+x} \frac{dx}{\sqrt{x^4 + ax^2 + 1}}$?

The question is how to show the identity

$$ \int_{0}^{1} \frac{1-x}{1+x} \cdot \frac{dx}{\sqrt{x^4 + ax^2 + 1}} = \frac{1}{\sqrt{a+2}} \log\left( 1 + \frac{\sqrt{a+2}}{2} \right), \tag{$a>-2$} $$

I checked this numerically for several cases, but even Mathematica 11 could not manage this symbolically for general $a$, except for some special cases like $a = 0, 1, 2$.

Addendum. Here are some backgrounds and my ideas:

• This integral came from my personal attempt to find the pattern for the integral

  $$ J(a, b) := \int_{0}^{1} \frac{1-x}{1+x} \cdot \frac{dx}{\sqrt{1 + ax^2 + bx^4}}. $$

  This drew my attention as we have the following identity

  $$ \int_{0}^{\infty} \frac{x}{x+1} \cdot \frac{dx}{\sqrt{4x^4 + 8x^3 + 12x^2 + 8x + 1}} = J(6,-3), $$

  where the LHS is the integral from this question. So establishing the claim in this question amounts to showing that $J(6,-3) = \frac{1}{2}\log 3 - \frac{1}{3}\log 2$, though I am skeptical that $J(a, b)$ has a nice closed form for every pair of parameters $(a, b)$.

• A possible idea is to write

  \begin{align*} &\int_{0}^{1} \frac{1-x}{1+x} \cdot \frac{dx}{\sqrt{x^4 + ax^2 + 1}} \\ &\hspace{5em}= \int_{0}^{1} \frac{(x^{-2} + 1) - 2x^{-1}}{x^{-1} - x} \cdot \frac{dx}{\sqrt{(x^{-1} - x)^2 + a + 2}} \end{align*}

  This follows from a simple algebraic manipulation. This suggests that we might be able to apply Glasser's master theorem, though in a less trivial way.

I do not believe that this is particularly hard, but I literally have not enough time to think about this now. So I guess it is a good time to seek help.

Enforcing the substitution $x^{-1}-x=u$ in your last integral we get:

$$ \int_{0}^{+\infty}\left(-1+\frac{2}{\sqrt{4+u^2}}\right)\frac{du}{u\sqrt{u^2+a+2}} $$ and by setting $u=\sqrt{a+2}\sinh\theta$ we get: $$ \frac{1}{\sqrt{a+2}}\int_{0}^{+\infty}\left(-1+\frac{2}{\sqrt{4+(a+2)\sinh^2\theta}}\right)\frac{d\theta}{\sinh\theta}$$ We may get rid of the last term through the "hyperbolic Weierstrass substitution" $$ \theta = 2\,\text{arctanh}(e^{-v}) = \log\left(\frac{e^v+1}{e^v-1}\right)$$ that wizardly gives $$ \frac{1}{\sqrt{a+2}}\int_{0}^{+\infty}\left(-1+\frac{2}{\sqrt{4+\frac{a+2}{\sinh^2 v}}}\right)\,dv$$ i.e., finally, a manageable integral through differentiation under the integral sign.
This proves OP's initial identity. Beers on me.

$$\text{let } \ \frac{1-x}{1+x}=t \Rightarrow x=\frac{1-t}{1+t}\Rightarrow dx=-\frac{2}{(1+t)^2}dt\ \text{ then:}$$ $$\int_0^1 \frac{1-x}{1+x}\frac{dx}{\sqrt{x^4+ax^2+1}}=\int_0^1 \frac{2t}{\sqrt{(a+2)t^4-2(a-6)t^2+(a+2)}}dt$$ $$\overset{t^2=x}=\frac{1}{\sqrt{a+2}}\int_0^1 \frac{dx}{\sqrt{x^2-2\left(\frac{a-6}{a+2}\right)x+1}}=\boxed{\frac{1}{\sqrt{a+2}}\ln \left(1+\frac{\sqrt{a+2}}{2}\right),\quad a>-2}$$