$\int x^2\sqrt{4\+9 x^2}\mathit{\ⅆ}x$

$\= \int {\left(\frac{2}{3}\tan \left(\mathrm{\θ}\right)\right)}^2\left(2 \sec \left(\mathrm{\θ}\right)\right)\left(\frac{2}{3}{\sec }^2\left(\mathrm{\θ}\right) d\mathrm{\θ}\right)$

$\=\frac{16}{27}\int {\sec }^3\left(\mathrm{\θ}\right){\tan }^2\left(\mathrm{\θ}\right) d\mathrm{\θ}$

$\=\frac{16}{27}\int {\sec }^3\left(\mathrm{\θ}\right)\left({\sec }^2\left(\mathrm{\θ}\right)-1\right) d\mathrm{\θ}$

$\=\frac{16}{27}\left(\int {\sec }^5\left(\mathrm{\θ}\right) d\mathrm{\θ}-\int {\sec }^3\left(\mathrm{\θ}\right) d\mathrm{\θ}\right)$

$\=\frac{16}{27}\left(\frac{1}{4}{\sec }^3\left(\mathrm{\θ}\right)\tan \left(\mathrm{\θ}\right)\+\frac{3}{4}\int {\sec }^3\left(\mathrm{\θ}\right) d\mathrm{\θ}-\int {\sec }^3\left(\mathrm{\θ}\right) d\mathrm{\θ}\right)$

$\=\frac{16}{27}\left(\frac{1}{4}{\sec }^3\left(\mathrm{\θ}\right)\tan \left(\mathrm{\θ}\right)-\frac{1}{4}\int {\sec }^3\left(\mathrm{\θ}\right) d\mathrm{\θ}\right)$

$\=\frac{4}{27}{\sec }^3\left(\mathrm{\θ}\right)\tan \left(\mathrm{\θ}\right)-\frac{4}{27}\(\sec \left(\mathrm{\θ}\right)\tan \left(\mathrm{\θ}\right)\+\ln \(\left|\sec \left(\mathrm{\θ}\right)\+\tan \left(\mathrm{\θ}\right)\right|\)\)\/2$

$\=\frac{4}{27}{\sec }^3\left(\mathrm{\θ}\right)\tan \left(\mathrm{\θ}\right)-\frac{2}{27}\sec \left(\mathrm{\θ}\right)\tan \left(\mathrm{\θ}\right)-\frac{2}{27}\ln \left(\sec \left(\mathrm{\θ}\right)\+\tan \left(\mathrm{\θ}\right)\right)$

$\=\frac{2}{27}\sec \left(\mathrm{\θ}\right)\tan \left(\mathrm{\θ}\right)\left(2 {\sec }^2\left(\mathrm{\θ}\right)-1\right)-\frac{2}{27}\ln \left(\sec \left(\mathrm{\theta }\right)\+\tan \left(\mathrm{\theta }\right)\right)$

$\=\frac{2}{27}\left(\frac{\sqrt{4\+9 x^2}}{2}\right)\left(\frac{3}{2}x\right)\left(2 \frac{\sqrt{4\+9 x^2}}{4}-1\right)-\frac{2}{27}\ln \left(\frac{\sqrt{4\+9 x^2}}{2}\+\frac{3}{2}x\right)$

$\=\frac{x}{36}\sqrt{4\+9 x^2}\left(2\+9 x^2\right)-\frac{2}{27}\ln \left(\frac{\sqrt{4\+9 x^2}}{2}\+\frac{3}{2}x\right)$

Evaluate the given integral

$\int x^2\sqrt{4\+9 x^2}\mathit{\ⅆ}x$ = $\frac{1}{36}x{\left(9x^2\+4\right)}^{3\/2}-\frac{1}{18}x\sqrt{9x^2\+4}-\frac{2}{27}\mathrm{arcsinh}\left(\frac{3}{2}x\right)$$$$$

Initialization

$\mathrm{with}\left(\mathrm{IntegrationTools}\right)\:$

$Q≔\int x^2\sqrt{4\+9 x^2}\mathit{\ⅆ}x\:$

Change variables as per Table 6.3.1

$q_1≔\mathrm{Change}\left(Q\,x\=\frac{2}{3}\tan \left(\mathrm{\θ}\right)\right)$

$\∫\frac{8}{27}{\tan \left(\mathrm{\θ}\right)}^2\sqrt{4{\tan \left(\mathrm{\θ}\right)}^2\+4}\left(1\+{\tan \left(\mathrm{\θ}\right)}^2\right)\ⅆ\mathrm{\θ}$

$q_2≔\mathrm{simplify}\left(q_1\right) assuming \mathrm{\θ}\colon\colon \mathrm{RealRange}\left(-\frac{\mathrm{\π}}{2}\,\frac{\mathrm{\π}}{2}\right)$

$\frac{16}{27}\∫\frac{{\sin \left(\mathrm{\θ}\right)}^2}{{\cos \left(\mathrm{\θ}\right)}^5}\ⅆ\mathrm{\θ}$

$q_3≔\mathrm{value}\left(q_2\right)$

$\frac{4}{27}\frac{{\sin \left(\mathrm{\θ}\right)}^3}{{\cos \left(\mathrm{\θ}\right)}^4}\+\frac{2}{27}\frac{{\sin \left(\mathrm{\θ}\right)}^3}{{\cos \left(\mathrm{\θ}\right)}^2}\+\frac{2}{27}\sin \left(\mathrm{\θ}\right)-\frac{2}{27}\ln \left(\sec \left(\mathrm{\θ}\right)\+\tan \left(\mathrm{\θ}\right)\right)$

$\mathrm{eval}\left(q_3\,\mathrm{\θ}\=\arctan \left(\frac{3}{2}x\right)\right)$

$\frac{1}{4}\sqrt{9x^2\+4}{x}^3\+\frac{1}{2}\frac{x^3}{\sqrt{9x^2\+4}}\+\frac{2}{9}\frac{x}{\sqrt{9x^2\+4}}-\frac{2}{27}\ln \left(\frac{1}{2}\sqrt{9x^2\+4}\+\frac{3}{2}x\right)$

$\begin{array}{cccc}\multicolumn{4}{c}{\int x^2\sqrt{9x^2+4}\ⅆx}\\ & \text{=}& -\frac{\int u^3\ⅆu}{432}+\frac{\int \frac{32u^4-256}{u^5}\ⅆu}{432}\hfill & \left[\mathrm{change}\,u=\sqrt{9x^2+4}-3x\,u\right]\\ & \text{=}& -\frac{u^4}{1728}+\frac{\int \frac{32u^4-256}{u^5}\ⅆu}{432}\hfill & \left[\mathrm{power}\right]\\ & \text{=}& -\frac{u^4}{1728}+\frac{2\int \frac{1}{u}\ⅆu}{27}-\frac{16\int \frac{1}{u^5}\ⅆu}{27}\hfill & \left[\mathrm{rewrite}\,\frac{32u^4-256}{u^5}=\frac{32}{u}-\frac{256}{u^5}\right]\\ & \text{=}& -\frac{u^4}{1728}+\frac{2\ln \left(u\right)}{27}-\frac{16\int \frac{1}{u^5}\ⅆu}{27}\hfill & \left[\mathrm{power}\right]\\ & \text{=}& -\frac{u^4}{1728}+\frac{2\ln \left(u\right)}{27}+\frac{4}{27u^4}\hfill & \left[\mathrm{power}\right]\end{array}$

Table 6.3.14(a)   Maple's stepwise approach to the given integral

$\begin{array}{c}\int x^2\sqrt{9x^2+4}\ⅆx\hfill \\ & \text{=}& \frac{16\int {\tan \left(\mathrm{\theta }\right)}^2{\sec \left(\mathrm{\theta }\right)}^3\ⅆ\mathrm{\theta }}{27}\hfill & \left[\mathrm{change}\,x=\frac{2\tan \left(\mathrm{\theta }\right)}{3}\right]\\ & \text{=}& \frac{16\int {\sec \left(\mathrm{\theta }\right)}^3\left({\sec \left(\mathrm{\theta }\right)}^2-1\right)\ⅆ\mathrm{\theta }}{27}\hfill & \left[\mathrm{rewrite}\,{\tan \left(\mathrm{\theta }\right)}^2={\sec \left(\mathrm{\theta }\right)}^2-1\right]\\ & \text{=}& \frac{16\int {\sec \left(\mathrm{\theta }\right)}^5\ⅆ\mathrm{\theta }}{27}-\frac{16\int {\sec \left(\mathrm{\theta }\right)}^3\ⅆ\mathrm{\theta }}{27}\hfill & \left[\mathrm{rewrite}\,{\sec \left(\mathrm{\theta }\right)}^3\left({\sec \left(\mathrm{\theta }\right)}^2-1\right)={\sec \left(\mathrm{\theta }\right)}^5-{\sec \left(\mathrm{\theta }\right)}^3\right]\end{array}$

Table 6.3.14(b)   Initial steps in an annotated stepwise solution via Integration Methods tutor