<s>[INST] Solving $\int \frac{x^3}{(4x^2 + 9)^\frac{3}{2}} dx$ The integral I must solve is this:
$$\begin{equation} \int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx \tag{1}\end{equation}$$
Now I know that I'm supposed to convert the denominator into some form of $\sqrt{a^2 + x^2}$ in order to apply the substitution $x = a \ tan \ \theta$. Now I can change the form of the denominator to become easy enough to substitute as follows:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +x^2\bigg)\bigg) ^\frac{3}{2} $$
Which makes it clear that $x$ needs to be substituted as  $x = \frac{3}{2} \tan \theta $, and $dx = \frac{3}{2} \sec^2 \theta $
for later use. At this point I can represent $(1)$ in terms of my substituted trignometric function. The only problem comes with the denominator where I get stuck on the power. Here is how I went about solving it:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\left(\frac{3}{2} \tan \theta\right)^2 \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\frac{9}{4} \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \frac{9}{4}\bigg (\ 1 + \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$ 9^\frac{3}{2}\bigg( \ 1 + \tan^2 \theta \bigg) ^\frac{3}{2} $$
$$ 27\ ( \sec^2 \theta ) ^\frac{3}{2} $$
Now I have no idea how to evaluate this power of $sec$. The author says that it changes into $sec^3 \theta$ but I just can't fathom how that would go about. If what I understand is correct, the power it is raised to would be added to it's own making it $ \sec^\frac{7}{2} \theta$. My question is that how exactly is my reasoning wrong here?
 [/INST] $$\int \frac{x^3}{\left(4x^2+9\right)^{\frac{3}{2}}}dx$$
if $4x^2+9=t^2$ as in the answer of the user @Anton Vrdoljak you have:
$$=\int \frac{t-9}{32t^{\frac{3}{2}}}\ dt=\frac{1}{32}\cdot \int \left(\frac{1}{t^{\frac{1}{2}}}-\frac{9}{t^{\frac{3}{2}}}\right)dt=\frac{1}{32}\left(2t^{\frac{1}{2}}-\left(-\frac{18}{t^{\frac{1}{2}}}\right)\right)+k, \quad k\in\Bbb R$$
And at the end:
$$=\frac{1}{32}\left(2\left(4x^2+9\right)^{\frac{1}{2}}-\left(-\frac{18}{\left(4x^2+9\right)^{\frac{1}{2}}}\right)\right)+k, \quad k\in\Bbb R$$
$$=\frac{2x^2+9}{8\left(4x^2+9\right)^{\frac{1}{2}}}+k, \quad k\in\Bbb R$$
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