Set R = 4 x3 + 5 x2 + 10 for brevity. Maybe based on Liouville's theorem, we can guess
$$\int \sqrt{R} \left( x^5 + x^4 + x^2 \right) = PR^{3/2} + c$$
where P is a polynomial of x and c is the constant of integration. Differentiate on both sides, we get
$$\sqrt{R} \left( x^5 + x^4 + x^2 \right) = P'R^{3/2} + PR' \sqrt{R}.$$
Rationalize the formula
$$\left( x^5 + x^4 + x^2 \right) = \left( 4x^3 + 5x^2 + 10 \right) P' + \left(
12x^2 + 10x \right) P.$$
P is cubic, so set P = a3x3 + a2x2 + a1x + a0 and solve the linear system. The system is overdeterined, but luckily there is still a solution
$$P = \frac{x^3}{30}.$$
As a result,
$$\int \sqrt{4x^3 + 5x^2 + 10} \left( x^5 + x^4 + x^2 \right) = \frac{x^3 \left( 4x^3 + 5x^2 + 10 \right)^{3/2}}{30} + c.$$
Therefore
$$\int_0^1 \sqrt{4x^3 + 5x^2 + 10} \left( x^5 + x^4 + x^2 \right) = \frac{19^{3/2}}{30}.$$