$\int x^{2}\sqrt{a^{2}+x^{2}}\,dx$. Is there another way to solve it faster?

$$x^2 \sqrt{a^2+x^2};\;\, x\to a \sinh u$$ $$dx=\cosh u\,du$$ $$\int x^2 \sqrt{a^2+x^2}\,dx=\int (a^2 \sinh ^2 u)(a \cosh u )\sqrt{a^2 \sinh ^2 u+a^2}\,du=$$ $$=a^4\int \sinh^2 u\cosh^2 u\,du=\frac{a^4}{4}\int\sinh^2 2u\,du=\frac{a^4}{8} \int (\cosh 4 u-1) \, du=$$ $$=\frac{1}{8} a^4 \left(\frac{1}{4} \sinh 4 u-u\right)+C=\frac{1}{8} \left(x \sqrt{a^2+x^2} \left(a^2+2 x^2\right)-a^4 \text{arcsinh}\left(\frac{x}{a}\right)\right)+C$$

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Useful formulas

$\cosh^2 u -\sinh^2 u=1$

$\sinh 2u=2\sinh u\cosh u$

$\cosh 4u =\sinh ^2 2u +\cosh ^2 2u=2\sinh^2 2u+1\to \sinh^2 2u = \frac{1}{2}(\cosh4u - 1)$

$x=a\sinh u\to u=\text{arcsinh}\left(\frac{x}{a}\right)$

$\sinh 4u = 2\sinh 2u \cosh 2u = 4\sinh u\cosh u(\cosh^2 u+ \sinh^2 u)$