Integral $\int( 3x^2 +5x + 1 )\sqrt{2x^2 + 2x + 1}dx$ I tried to integrate
$$\int( 3x^2 +5x + 1 )\sqrt{2x^2 + 2x + 1}dx$$
by multipling by $\sqrt{2x^2 + 2x + 1}$ in the numerator and the denominator to break it into $5$ fractions
The answer is possible but it's too long
Another better solution
 A: Notice that
$$
\left(3x^2 + 5x +1\right)\sqrt{2x^2 + 2x+1} = \frac{1}{4\sqrt{2}}\left(3(2x+1)^2  + 4(2x+1) -3\right)\sqrt{(2x+1)^2+1}
$$
under substitution $2x +1 = \tan(\alpha)$ simplifies your integral to
\begin{align*}
&\int\left(3x^2 + 5x +1\right)\sqrt{2x^2 + 2x+1} \, \mathrm{d}x \\
 &=\frac{1}{8\sqrt{2}} \int \left(3 \tan^2(\alpha) + 4\tan(\alpha) -3\right)\sec^3(\alpha) \, \mathrm{d}\alpha \\
& = \frac{3}{8\sqrt{2}} \int \left(\sec^2(\alpha)  -2\right)\sec^3(\alpha) \, \mathrm{d}\alpha  + \frac{1}{2\sqrt{2}} \int \left[\tan(\alpha)\sec(\alpha)\right]\sec^2(\alpha) \, \mathrm{d}\alpha\\
\end{align*}
Since the last integral is immediate under susbtitution $u = \sec(\alpha)$, you just need to solve $\int \sec^n(\alpha) \mathrm{d} \alpha$ for $n =3,5$. For this use the reduction formula for secant until you get to $
\int\sec(x) \mathrm{d}x = \ln|\sec(x) + \tan(x)| $, and afer returning the resulting expression to be written in terms of $x$ this concludes the problem.
A: A systematic approach after $t=2x+1$
\begin{align}
I=&\int (3x^2 +5x + 1 )\sqrt{2x^2 + 2x + 1}\>dx\\
= &\frac1{8\sqrt2}\int (4t+3)(t^2+1)^{3/2}-6 (t^2+1)^{1/2}\>dt\tag1
\end{align}
Integrate below by parts
\begin{align}
&\int (t^2+1)^{3/2}dt=\frac14t(t^2+1)^{3/2}+\frac34\int(t^2+1)^{1/2}dt\\
 &\int (t^2+1)^{1/2}dt=\frac12 t(t^2+1)^{1/2}+\frac12\sinh^{-1}t
\end{align}
Plug into (1) to arrive at
$$I= \frac1{8\sqrt2}\left[\bigg(\frac34t + \frac43 \bigg)(t^2+1)^{3/2}-\frac{15}8 t(t^2+1)^{1/2}-\frac{15}8\sinh^{-1}t\right]
$$
