How find $\int(x^7/8+x^5/4+x^3/2+x)\big((1-x^2/2)^2-x^2\big)^{-\frac{3}{2}}dx$ How can I compute the following integral:
$$\int \dfrac{\frac{x^7}{8}+\frac{x^5}{4}+\frac{x^3}{2}+x}{\left(\left(1-\frac{x^2}{2}\right)^2-x^2\right)^{\frac{3}{2}}}dx$$
According to Wolfram Alpha, the answer is
$$\frac{x^4 - 32x^2 + 20}{2\sqrt{x^4 -8x^2 + 4}} + 7\log\bigl(-x^2 - \sqrt{x^4 - 8x^2 + 4} + 4\bigr) + C$$
but how do I get it by hand?
 A: Perform the substitutions $y=\frac{x^2}{2}, y=2+\sqrt{3}\sec{\theta}$ to get:
\begin{align}
&\int{\frac{\frac{x^7}{8}+\frac{x^5}{4}+\frac{x^3}{2}+x}{((1-\frac{x^2}{2})^2-x^2)^{\frac{3}{2}}} dx} \\
& =\int{\frac{y^3+y^2+y+1}{(y^2-4y+1)^{\frac{3}{2}}} dy} \\
&=\int{\left(\frac{y+5}{\sqrt{(y-2)^2-3}}+\frac{20y-4}{\sqrt{((y-2)^2-3)^3}}\right) dy} \\
&=\int{\left[\frac{7+\sqrt{3}\sec{\theta}}{\sqrt{3}\tan{\theta}}+\frac{36+20\sqrt{3}\sec{\theta}}{3\sqrt{3}\tan^3{\theta}}\right](\sqrt{3}\sec{\theta}\tan{\theta}) d\theta} \\
&=\int{\left(7\sec{\theta}+\sqrt{3}\sec^2{\theta}+12\csc{\theta}\cot{\theta}+\frac{20}{\sqrt{3}}\csc^2{\theta}\right) d\theta} \\
&=7\ln{|\tan{\theta}+\sec{\theta}|}+\sqrt{3}\tan{\theta}-12\csc{\theta}-\frac{20}{\sqrt{3}}\cot{\theta}+c \\
&=7\ln{|\frac{\sqrt{y^2-4y+1}+y-2}{\sqrt{3}}|}+\sqrt{y^2-4y+1}-\frac{12(y-2)}{\sqrt{y^2-4y+1}}-\frac{20}{\sqrt{y^2-4y+1}}+c \\
&=7\ln{|\frac{\sqrt{y^2-4y+1}+y-2}{\sqrt{3}}|}+\frac{y^2-16y+5}{\sqrt{y^2-4y+1}}+c \\
&=7\ln{|\sqrt{x^4-8x^2+4}+x^2-4|}+\frac{x^4-32x^2+20}{2\sqrt{x^4-8x^2+4}}+c'
\end{align}
where $c,c'$ are arbitrary constants of integration.
