Evaluating indefinite integrals of the form $\int \frac{x^2 \,dx}{a x^5 + b}$ 
Evaluate the indefinite integral
$$\int \frac{x^2 \,dx}{a x^5 + b},$$
for real parameters $a, b \neq 0$.

No apparent substitutions simplify the expression (if the exponent of $x$ in the denominator were an integral multiple of $3$, the form of the integrand would suggest the substitution $u = x^3$, $du = 3 x^2 \,dx$, but the exponent is not). Applying integration by parts with $u = \frac{1}{a x^5 + b}$, $dv = x^2 dx$ is straightforward, but it produces an integrand with a much larger degree in the denominator and so appears only to make the situation worse. Applying integration by parts instead with $dw = \frac{x^k dx}{a x^5 + b}$ results in integrating $\frac{x^k dx}{a x^5 + b}$, which, except when $k \equiv 4 \pmod 5$ (which doesn't appear immediately fruitful), doesn't appear much easier than the given integral.

Old post, but the original author bumped it so...  
We want to scale the problem so we can concentrate on
$$\int\frac{x^2dx}{x^5+1}=\sum_{n=0}^4\int\frac{A_ndx}{x-x_n}=\sum_{n=0}^4A_n\ln(x-x_n)+C_1$$
Here $x_n^5=-1=e^{i(2n+1)\pi}$ so $x_n=e^{\frac{i(2n+1)\pi}5}$,
$$A_n=\lim_{x\rightarrow x_n}\frac{(x-x_n)x^2}{x^5+1}=\frac{x_n^2}{5x_n^4}=-\frac15x_n^3$$
$x_0^3=x_1=\cos\frac{3\pi}5+i\sin\frac{3\pi}5$; $x_1^3=x_4=x_0^*$; $x_2^3=x_2=-1$; $x_3^3=x_0=\cos\frac{\pi}5+i\sin\frac{\pi}5$; $x_4^3=x_3=x_1^*$, $x-x_n=r_ne^{i\theta_n}$ where
$$r_n=|x-x_n|=\sqrt{|x-x_n|^2}=\sqrt{x^2-2x\cos\frac{(2n+1)\pi}5+1}$$
$$\theta_n=\operatorname{atan2}\left(-\sin\frac{(2n+1)\pi}5,x-\cos\frac{(2n+1)\pi}5\right)=\tan^{-1}\left(\frac{x-\cos\frac{(2n+1)\pi}5}{\sin\frac{(2n+1)\pi}5}\right)-\frac{\pi}2$$
So
$$\begin{align}\int\frac{x^2dx}{x^5+1}&=-\frac15\left\{\cos\frac{3\pi}5\ln\left(x^2-2x\cos\frac{\pi}5+1\right)-2\sin\frac{3\pi}5\tan^{-1}\left(\frac{x-\cos\frac{\pi}5}{\sin\frac{\pi}5}\right)\right.\\
&\quad\left.+\cos\frac{\pi}5\ln\left(x^2-2x\cos\frac{3\pi}5+1\right)+2\sin\frac{\pi}5\tan^{-1}\left(\frac{x-\cos\frac{3\pi}5}{\sin\frac{3\pi}5}\right)\right.\\
&\quad\left.-\ln|x+1|\right\}+C\\
&=-\frac15\left\{\frac{1-\sqrt5}4\ln\left(x^2-\frac{1+\sqrt5}2x+1\right)-\frac{\sqrt{10+2\sqrt5}}2\tan^{-1}\left(\frac{4x-\sqrt5-1}{\sqrt{10-2\sqrt5}}\right)\right.\\
&\quad\left.+\frac{1+\sqrt5}4\ln\left(x^2+\frac{\sqrt5-1}2x+1\right)+\frac{\sqrt{10-2\sqrt5}}2\tan^{-1}\left(\frac{4x+\sqrt5-1}{\sqrt{10+2\sqrt5}}\right)\right.\\
&\quad\left.-\ln|x+1|\right\}+C\\
&=-\frac15\left\{-\frac1{2\phi}\ln\left(x^2-\phi x+1\right)-5^{1/4}\phi^{1/2}\tan^{-1}\left(\frac{2x-\phi}{5^{1/4}\phi^{-1/2}}\right)\right.\\
&\quad\left.+\frac{\phi}2\ln\left(x^2+\frac1{\phi}x+1\right)+5^{1/4}\phi^{-1/2}\tan^{-1}\left(\frac{2x+\frac1{\phi}}{5^{1/4}\phi^{1/2}}\right)-\ln|x+1|\right\}+C\end{align}$$
So if we organize our arithmetic reasonably, the operation isn't all that bad.
