Definite integration of an algebraic expression 
Evaluate $$\int_{0}^{1}\frac{1-x}{1+x}\frac{dx}{\sqrt{x+x^2+x^3}}$$

I think none of the properties of definite integral will be useful here so I think I will have to integrate. But I am unable to do so. Some hints on how to integrate. Thanks.
 A: Let $u = \sqrt{x+1+x^{-1}}$, we have $2 u du = (1 - x^{-2})dx$ and
$$
\int_0^1 \frac{1-x}{1+x}\frac{dx}{\sqrt{x+x^2+x^3}}
= \int_\infty^\sqrt{3} \frac{1-x}{1+x}\frac{2u du}{(x-x^{-1})u}
= 2 \int_\sqrt{3}^\infty \frac{du}{x+2+x^{-1}}\\
= 2 \int_\sqrt{3}^\infty \frac{du}{u^2+1}
= 2 \left[\tan^{-1}u\right]_{\sqrt{3}}^\infty
= 2 \left(\frac{\pi}{2} - \frac{\pi}{3}\right) = \frac{\pi}{3}
$$
A: Mathematica gives the indefinite integral:
$\int {1-x \over 1+x}{1 \over \sqrt{x+x^2+x^3}} dx =$
$-\frac{2 \sqrt[6]{-1} x^{3/2} \left(i \sqrt{1-\frac{(-1)^{2/3}}{x}}
   \sqrt{\frac{x+\sqrt[3]{-1}}{x}} F\left(i \sinh
   ^{-1}\left(\frac{(-1)^{5/6}}{\sqrt{x}}\right)|(-1)^{2/3}\right)+\frac{2
   \left(1+\sqrt[3]{-1}\right)
   \sqrt{\frac{\sqrt{x}-\sqrt[3]{-1}+1}{\left(1+\sqrt[3]{-1}\right) \left(\sqrt[3]{-1}
   \sqrt{x}-1\right)}} \sqrt{\frac{(-1)^{2/3} \sqrt{x}-1}{\left(1+\sqrt[3]{-1}\right)
   \left(\sqrt[3]{-1} \sqrt{x}-1\right)}} \sqrt{-\frac{(-1)^{2/3}
   \sqrt{x}+1}{\sqrt{x}+\sqrt[3]{-1}-1}} \left(\sqrt[3]{-1} \sqrt{x}-1\right)^2
   \left(\left(\sqrt[3]{-1}+i\right) F\left(\left.\sin
   ^{-1}\left(\sqrt{\frac{\sqrt{x}-\sqrt[3]{-1}+1}{\left(1+\sqrt[3]{-1}\right)
   \left(\sqrt[3]{-1} \sqrt{x}-1\right)}}\right)\right|-3\right)-2 \sqrt[3]{-1} \Pi
   \left(\frac{-3 i+(1+2 i) \sqrt{3}}{(2+i)+\sqrt{3}};\left.\sin
   ^{-1}\left(\sqrt{\frac{\sqrt{x}-\sqrt[3]{-1}+1}{\left(1+\sqrt[3]{-1}\right)
   \left(\sqrt[3]{-1} \sqrt{x}-1\right)}}\right)\right|-3\right)\right)}{x}-\frac{2
   \left(1+\sqrt[3]{-1}\right)
   \sqrt{\frac{\sqrt{x}-\sqrt[3]{-1}+1}{\left(1+\sqrt[3]{-1}\right) \left(\sqrt[3]{-1}
   \sqrt{x}-1\right)}} \sqrt{\frac{(-1)^{2/3} \sqrt{x}-1}{\left(1+\sqrt[3]{-1}\right)
   \left(\sqrt[3]{-1} \sqrt{x}-1\right)}} \sqrt{-\frac{(-1)^{2/3}
   \sqrt{x}+1}{\sqrt{x}+\sqrt[3]{-1}-1}} \left(\sqrt[3]{-1} \sqrt{x}-1\right)^2
   \left(\left(\sqrt[3]{-1}-i\right) F\left(\left.\sin
   ^{-1}\left(\sqrt{\frac{\sqrt{x}-\sqrt[3]{-1}+1}{\left(1+\sqrt[3]{-1}\right)
   \left(\sqrt[3]{-1} \sqrt{x}-1\right)}}\right)\right|-3\right)-2 \sqrt[3]{-1} \Pi
   \left(-\frac{i \left(3+(2+i) \sqrt{3}\right)}{(-2+i)+\sqrt{3}};\left.\sin
   ^{-1}\left(\sqrt{\frac{\sqrt{x}-\sqrt[3]{-1}+1}{\left(1+\sqrt[3]{-1}\right)
   \left(\sqrt[3]{-1}
   \sqrt{x}-1\right)}}\right)\right|-3\right)\right)}{x}\right)}{\sqrt{x
   \left(x^2+x+1\right)}}$
which very strongly suggests that there's no simple method "by hand."
You can also do a Taylor series:
$\frac{1-x}{(x+1) \sqrt{x^3+x^2+x}} \approx \frac{1}{\sqrt{x}}-\frac{5 \sqrt{x}}{2}+\frac{23 x^{3/2}}{8}-\frac{37
   x^{5/2}}{16}+\frac{203 x^{7/2}}{128}-\frac{355 x^{9/2}}{256}+O\left(x^{11/2}\right)$
and then integrate each term, but that will likely not give an adequate solution.
