Periodic or not Periodic signal
If a signal is periodic then its derivative will also be periodic 

So $$x(t)=sin((\sqrt{2t}+5)+sin(\pi t))-1$$ its derivative is 
$$x'(t)=\frac{cos((\sqrt{2t}+sin(\pi t)+5))}{\sqrt{2t}}+\pi cos(\pi t)cos((\sqrt2t+sin(\pi t)+5))$$

$$x'(t)=x_1(t)+x_2(t)$$
where \$x_1(t)=\frac{cos((\sqrt{2t}+sin(\pi t)+5))}{\sqrt{2t}}\$ and \$x_2(t)=\pi cos(\pi t)cos((\sqrt2t+sin(\pi t)+5))\$
Now \$x_1(t)\$ is aperiodic because wolframalpha result
$$\lim_{t \to \infty}\frac{cos((\sqrt{2t}+sin(\pi t)+5))}{\sqrt{2t}}=0$$

Now $$x_2(t)=\pi cos(\pi t)cos((\sqrt2t+sin(\pi t)+5))$$
$$x_2(t)=\frac{\pi}{2}{[cos((\pi t-\sqrt2t-sin(\pi t)-5))+cos((\pi t+\sqrt2t+sin(\pi t)+5))]}$$

Now \$x_2(t)\$ can also be checked with above property of derivative individually that will come as aperiodic signal

So \$x'(t)\$ is sum of two aperiodic signal that will be a aperiodic signal.

Hence \$x(t)\$ is aperiodic signal