At this point we have an answer in terms of the variable $\theta$, but we require the variable $x$. We will use the triangle below to convert back to $x$.

From the triangle, we see that $\sin (\theta ) = \sqrt {1 - x^2}$, and we substitute to get the final answer: $$\int x^3 \sqrt {1-x^2} \; dx = \frac 15 \sin ^5(\theta ) - \frac 13 \sin ^3(\theta ) + C = \frac 15 (1-x^2)^{5/2} - \frac 13 (1-x^2)^{3/2} + C.$$

From the triangle above, we can see that $\tan (\theta ) = \frac {\sqrt {4-x^2}}{x}$ so that: $$\int \frac {\sqrt {4-x^2}}{x^2} \; dx = \theta -\tan (\theta ) + C = \cos ^{-1}\left (\frac {x}{2}\right ) -\frac {\sqrt {4-x^2}}{x} + + C.$$

From the triangle above, we can see that $\sec (\theta ) = \frac {3}{2x}$ and $\tan (\theta ) = \frac {\sqrt {9-4x^2}}{2x}$ so that: $$\int \frac {1}{x\sqrt {9-4x^2}} \; dx = -\frac 13 \ln |\sec (\theta ) + \tan (\theta )| + C = - \frac 13\ln \bigg |\frac {3 + \sqrt {9-4x^2}}{2x}\bigg | + C.$$

The integral becomes \begin{align*} \int \frac{1}{\sqrt{x^2-1}}\; dx &= \int \frac{\sec (\theta )\tan (\theta )}{\sqrt{\sec ^2(\theta )-1}}\; d\theta \\[6pt] &= \int \frac{\sec (\theta )\tan (\theta )}{\sqrt{\tan ^2(\theta )}}\; d\theta \\[6pt] &= \int \frac{\sec (\theta )\tan (\theta )}{\tan (\theta )}\; d\theta \\[6pt] &= \int \sec (\theta )\; d\theta \\[6pt] &= \ln |\sec (\theta )+ \tan (\theta )| + C. \end{align*}

From the triangle above, we can see that $\sec (\theta ) = x$ and $\tan (\theta ) = \sqrt {x^2-1}$ so that: $$\int \frac {1}{\sqrt {x^2-1}} \; dx = \ln |\sec (\theta )+ \tan (\theta )| + C = \ln |x+\sqrt {x^2-1}| + C.$$

The integral becomes \begin{align*} \int x^3\sqrt{x^2-4}\; dx &= \int 8\sec ^3(\theta )\sqrt{4\sec ^2(\theta )-4}\cdot 2\sec (\theta )\tan (\theta )\, d\theta \\[6pt] &= 16\int \sec ^4(\theta )\sqrt{4\tan ^2(\theta )}\cdot \tan (\theta )\, d\theta \\[6pt] &= 16\int \sec ^4(\theta )\cdot 2\tan (\theta )\cdot \tan (\theta )\, d\theta \\[6pt] &= 32\int \sec ^4(\theta )\tan ^2(\theta )\, d\theta \;\; \text{(even power of secant)} \\[6pt] &= 32\int \sec ^2(\theta ) \cdot \tan ^2(\theta )\cdot \sec ^2(\theta )\, d\theta \\[6pt] &= 32\int [1 + \tan ^2(\theta )] \cdot \tan ^2(\theta )\cdot \sec ^2(\theta )\, d\theta \\[6pt] \text{Let } \; & u = \tan (\theta ), \; du = \sec ^2(\theta ) \, d\theta .\\[6pt] &= 32\int \left (1 + u^2\right )u^2\, du\\[6pt] &= 32\int \left (u^2 + u^4\right ) \, du\\[6pt] &= \frac{32}{3} u^3 + \frac{32}{5} u^5 + C\\[6pt] &= \frac{32}{3} \tan ^3(\theta ) + \frac{32}{5} \tan ^5(\theta ) + C \end{align*}

From the triangle above, we can see that $\tan (\theta ) = \frac {\sqrt {x^2-4}}{2}$ so that: $$\int x^3\sqrt {x^2-4} \; dx = \frac {32}{3} \tan ^3(\theta ) + \frac {32}{5} \tan ^5(\theta ) + C = \frac 43 (x^2 - 4)^{3/2} + \frac {1}{5} (x^2 - 4)^{5/2} + C.$$

The integral becomes \begin{align*} \int \frac{1}{x\sqrt{x^2-16}}\; dx &= \int \frac{4\sec (\theta )\tan (\theta )}{4\sec (\theta )\sqrt{16\sec ^2(\theta )-16}}\; d\theta \\[6pt] &= \int \frac{\tan (\theta )}{\sqrt{16\tan ^2(\theta )}}\; d\theta \\[6pt] &= \int \frac{\tan (\theta )}{4\tan (\theta )}\; d\theta \\[6pt] &= \int \frac 14 \; d\theta \\[6pt] &= \frac 14 \theta + C\\[6pt] &= \frac 14 \sec ^{-1}\left (\frac{x}{4}\right )+C. \end{align*}

The integral becomes \begin{align*} \int \frac{1}{x^2\sqrt{9x^2-4}}\; dx &= \int \frac{(2/3)\sec (\theta )\tan (\theta )}{(2/3)^2\sec ^2(\theta )\sqrt{4\sec ^2(\theta )-4}}\; d\theta \\[6pt] &= \frac 32\int \frac{\tan (\theta )}{\sec (\theta )\sqrt{4\tan ^2(\theta )}}\; d\theta \\[6pt] &= \frac 32\int \frac{\tan (\theta )}{2\sec (\theta )\tan (\theta )}\; d\theta \\[6pt] &= \frac 34 \int \frac{1}{\sec (\theta )} \; d\theta \\[6pt] &= \frac 34 \int \cos (\theta ) \; d\theta \\[6pt] &= \frac 34 \sin (\theta ) + C. \end{align*}

From the triangle above, we can see that $\sin (\theta ) = \frac {\sqrt {9x^2-4}}{3x}$ so that: $$\int \frac {1}{x^2\sqrt {9x^2-4}} \; dx = \frac 34 \sin (\theta ) + C = \frac {\sqrt {9x^2 - 4}}{4x} + C.$$

The integral becomes \begin{align*} \int \frac{1}{x^3\sqrt{x^2-25}}\; dx &= \int \frac{5\sec (\theta )\tan (\theta )}{5^3\sec ^3(\theta )\sqrt{25\sec ^2(\theta )-25}}\; d\theta \\[6pt] &= \frac{1}{25}\int \frac{\tan (\theta )}{\sec ^2(\theta )\sqrt{25\tan ^2(\theta )}}\; d\theta \\[6pt] &= \frac{1}{25}\int \frac{\tan (\theta )}{5\sec ^2(\theta )\tan (\theta )}\; d\theta \\[6pt] &= \frac{1}{125} \int \frac{1}{\sec ^2(\theta )} \; d\theta \\[6pt] &= \frac{1}{125} \int \cos ^2(\theta ) \; d\theta \\[6pt] &= \frac{1}{125} \int \frac{1+\cos (2\theta )}{2} \; d\theta \\[6pt] &= \frac{1}{250}\left [\theta + \frac 12 \sin (2\theta )\right ] + C\\[6pt] &= \frac{1}{250} \theta + \frac{1}{500} \cdot 2\sin (\theta )\cos (\theta ) + C. \end{align*}

From the triangle above, we can see that $\sin (\theta ) = \frac {\sqrt {x^2-25}}{x}$ so that: \begin{align*} \int \frac{1}{x^3\sqrt{x^2-25}} \; dx &= \frac{\theta }{250} + \frac{\sin (\theta )\cos (\theta )}{250} + C \\ &= \frac{1}{250} \sec ^{-1}\left (\frac{x}{5}\right ) + \frac{1}{250} \cdot \frac{\sqrt{x^2 - 25}}{x}\cdot \frac{5}{x} + C\\ &= \frac{1}{250} \sec ^{-1}\left (\frac{x}{5}\right ) + \frac{\sqrt{x^2 - 25}}{50x^2}+ C. \end{align*}

From the triangle above, we see that $\sec (\theta )= \sqrt {1+x^2}$. Back-substituting gives $$\int x^3\sqrt {1+x^2} \; dx = \frac 15 \sec ^5(\theta ) - \frac 13 \sec ^3(\theta ) + C = \frac 15 (1+x^2)^{5/2} - \frac 13 (1+x^2)^{3/2} + C$$

From the triangle above, we can see that $\sec (\theta ) = \frac {\sqrt {9+x^2}}{3}$ and so

$$\int \frac {x^3}{\sqrt {9 + x^2}} \; dx = 9\sec ^3(\theta ) - 27 \sec (\theta ) + C = \frac 13 (9+x^2)^{3/2} - 9(9+x^2)^{1/2} + C.$$

From the triangle above, we can see that $\sec (\theta ) = \frac {\sqrt {16+9x^2}}{4 }$. Back substitution gives \begin{align*} \int \sqrt{16 + 9x^2} \; dx &= \frac{8}{3} \sec (x) \tan (x) + \frac{8}{3} \ln |\sec (x) + \tan (x)| + C\\ &= \frac{8}{3} \cdot \frac{\sqrt{16+9x^2}}{4 } \cdot \frac{3x}{4} + \frac{8}{3} \ln \bigg |\frac{\sqrt{16+9x^2}}{4 } + \frac{3x}{4}\bigg | + C\\ &= \frac 12 x\sqrt{16+9x^2} + \frac 83 \ln \left (3x + \sqrt{16+9x^2}\right ) + C\\ \end{align*}

Note that the absolute value bars were removed because $3x + \sqrt {16+9x^2} > 0$ for any $x$. Also note that the $1/4$ was removed from inside the log because $\ln (u/v) = \ln (u) - \ln (v)$ and $-\frac 83\ln (4)$ can be combined with the constant of integration.