Double integral $x^2(x^2+y^2+1)^{3/2}$

Question

How to evaluate $\int_{-1}^1 \int_{-1}^1 x^2(x^2+y^2+1)^{3/2} \ dx \ dy$ ? I tried substituting $x^2+y^2=tan^2(u)$ but it becomes more complicated. Using Wolfram Alpha Widget the answer is $3.64092$. Any idea? Thanks.

Answer 1

The symmetry of the integrand implies $$I = \int_{y=-1}^1 \int_{x=-1}^1 x^2 (x^2 + y^2 + 1)^{3/2} \, dx \, dy = \int_{x=-1}^1 \int_{y=-1}^1 y^2 (x^2 + y^2 + 1)^{3/2} \, dy \, dx.$$ Thus $$2I = \int_{y=-1}^1 \int_{x=-1}^1 (x^2 + y^2)(x^2 + y^2 + 1)^{3/2} \, dx \, dy.$$ Now perform the coordinate transformation $$(x,y) = (r \cos \theta, r \sin \theta)$$ to obtain $$2I = 8 \int_{\theta = 0}^{\pi/4} \int_{r = 0}^{\sec \theta} r^2 (r^2 + 1)^{3/2} r \, dr \, d\theta.$$ Since $$\begin{align*} \int r^3 (r^2 + 1)^{3/2} \, dr &= \frac{1}{2} \int r^2 (r^2 + 1)^{3/2} 2r \, dr \\ &= \frac{1}{2} \int (u-1)u^{3/2} \, du \\ &= \frac{1}{2} \left( \frac{2}{7} u^{7/2} - \frac{2}{5} u^{5/2} \right) + C \\ &= \frac{(r^2+1)^{7/2}}{7} - \frac{(r^2+1)^{5/2}}{5} + C, \end{align*}$$ we obtain $$ I = 4 \int_{\theta=0}^{\pi/4} \frac{(\sec^2 \theta + 1)^{7/2} - 1}{7} - \frac{(\sec^2 \theta + 1)^{5/2} - 1}{5} \, d\theta.$$ At this point, I don't have a simple way to evaluate this integral; however, the value provided is $$I = \frac{2}{105}\left(96 \sqrt{3} + \pi + 33 \coth^{-1} \sqrt{3} \right) \approx 3.640919532942123602084014417112851989279\ldots.$$