$\int x^{2}\sqrt{a^{2}+x^{2}}\,dx$. Is there another way to solve it faster? I have to calculate this integral:

\begin{align} \int x^{2}\sqrt{a^{2}+x^{2}}\,dx \qquad\text{with}
\quad a \in \mathbb{R} \end{align}

My attempt:
Using, trigonometric substitution
\begin{align}
\tan \theta &= \frac{x}{a}\\ \Longrightarrow  \ x&=a \tan \theta\\ \Longrightarrow  \ dx&=a \sec^{2}\theta\\ \Longrightarrow \ x^{2}&=a^{2}\tan^{2}\theta
\end{align}
Thus,
\begin{align}
\int x^{2}\sqrt{a^{2}+x^{2}}\,dx&=\int a^2 \tan^{2}\theta \sqrt{a^2+a^2\tan^{2}\theta}\ a\sec^{2}\theta\, d \theta\\&=a^{3}\int \tan^{2}\theta \sqrt{a^{2}(1+\tan^{2}\theta)}\sec^{2}\theta\, d\theta\\&=a^{3}\int \tan^{2}\theta \sqrt{a^{2}(\sec^{2}\theta)}\sec^{2}\theta \, d\theta\\&=a^{4}\int (1-\sec^{2}\theta)\sec^{3}\theta \, d\theta\\&=a^{4}\underbrace{\int \sec^{3}\theta \, d\theta}_{\text{solve by parts}}-a^{4}\underbrace{\int \sec^{5}\theta \, d\theta}_{\text{solve by parts}}
\end{align}
My doubt is: Is there any other way to solve it faster? Because by parts is a large process to solve each one. I really appreciate your help

Integrate by parts
\begin{align}
I & =\int x^{2}\sqrt{a^{2}+x^{2}}\,dx
=\int \frac x3d(a^2+x^2)^{3/2} \\
&=\frac13x( a^2+x^2)^{3/2}-\frac13I-\frac{a^2}3\underset{=J}{\int \sqrt{a^2+x^2}dx} \\
 &=\frac14x( a^2+x^2)^{3/2}-\frac{a^2}4J\tag 1
\end{align}
Integrate $J$ by parts again
\begin{align}
J & =\int \sqrt{a^{2}+x^{2}}\,dx=x\sqrt{ a^2+x^2}-J-\int \frac{a^2}{\sqrt{a^2+x^2}}dx\\
 &=\frac12  x\sqrt{ a^2+x^2}-\frac{a^2}2\sinh^{-1}\frac xa \tag2
\end{align}
Plug (2) into (1) to obtain
$$I= \frac14x( a^2+x^2)^{3/2}- \frac{a^2}8 x\sqrt{a^2+x^2} - \frac{a^4}8\sinh^{-1}\frac xa +C
$$
