The answer is 'no'. Making the substitution $$ x = \frac{(t-1)(t-5)(t^2+2t+5)}{16t^2}, $$ one finds $$ {\textstyle\sqrt{x+\sqrt{x+\sqrt{x+1}}}\,\mathrm{d}x} = \frac{(t^2-2t+5)(t^2-5)\sqrt{t^4{-}2t^2{-}40t+25}\ \mathrm{d}t}{32t^4}. $$ Denote the right hand side of the above equation by $\beta$. Now, setting $$ Q(t) = \frac{(25-25t-14t^2-3t^3+t^4)}{96t^3}, $$ one finds that $$ \beta = \mathrm{d}\left(Q(t)\sqrt{t^4{-}2t^2{-}40t+25}\right) + \frac{(10+15t-2t^2+t^3)\,\mathrm{d}t}{8t \sqrt{t^4-2t^2-40t+25}}\, $$ The second term on the right hand side is in normal form for differentials on the elliptic curve $s^2 = t^4-2t^2-40t+25$ that are odd with respect to the involution $(t,s)\mapsto(t,-s)$, and an application of Liouville's Theorem (on integration in elementary terms) shows that this term is not integrable in elementary terms. Hence $\beta$ is not, and hence the original integrand is not.

The answer is 'no'. Making the substitution $$ x = \frac{(t-1)(t-5)(t^2+2t+5)}{16t^2}, $$ one finds $$ {\textstyle\sqrt{x+\sqrt{x+\sqrt{x+1}}}\,\mathrm{d}x} = \frac{(t^2-2t+5)(t^2-5)\sqrt{t^4{-}2t^2{-}40t+25}\ \mathrm{d}t}{32t^4}. $$ Denote the right hand side of the above equation by $\beta$. Now, setting $$ Q(t) = \frac{(25-25t-14t^2-3t^3+t^4)}{96t^3}, $$ one finds that $$ \beta = \mathrm{d}\left(Q(t)\sqrt{t^4{-}2t^2{-}40t+25}\right) + \frac{(10+15t-2t^2+t^3)\,\mathrm{d}t}{8t \sqrt{t^4-2t^2-40t+25}}\, $$ The second term on the right hand side is in normal form for differentials on the elliptic curve $s^2 = t^4-2t^2-40t+25$ that are odd with respect to the involution $(t,s)\mapsto(t,-s)$. Since this term represents a differential that has two poles of order $2$ (over the two points where $t=\infty$) and two poles of order $1$ (over the two points where $t=0$), an application of Liouville's Theorem (on integration in elementary terms, with the differential field taken to be the field of meromorphic functions on the elliptic curve) shows that this term is not integrable in elementary terms. Hence $\beta$ is not, and hence the original integrand is not.

The answer is 'no'. If you make the substitution $$ x = \frac{(t-1)(t-5)(t^2+2t+5)}{16t^2}, $$ one gets $$ {\textstyle\sqrt{x+\sqrt{x+\sqrt{x+1}}}\,\mathrm{d}x} = \frac{(t^2-2t+5)(t^2-5)\sqrt{t^4{-}2t^2{-}40t+25}\ \mathrm{d}t}{32t^4}. $$ Call the right hand side of the above equation $\beta$. If we now set $$ Q(t) = \frac{(25-25t-14t^2-3t^3+t^4)}{96t^3} $$ then $$ \beta = \mathrm{d}\left(Q(t)\sqrt{t^4{-}2t^2{-}40t+25}\right) + \frac{(10+15t-2t^2+t^3)\,\mathrm{d}t}{8t \sqrt{t^4-2t^2-40t+25}}\, $$ The second term on the right hand side is now in normal form for differentials on the elliptic curve $s^2 = t^4-2t^2-40t+25$ that are odd with respect to the involution $(t,s)\mapsto(t,-s)$, and an application of Liouville's Theorem (on integration in elementary terms) shows that this term is not integrable in elementary terms. Hence $\beta$ is not, and hence the original integrand is not.

The answer is 'no'. Making the substitution $$ x = \frac{(t-1)(t-5)(t^2+2t+5)}{16t^2}, $$ one finds $$ {\textstyle\sqrt{x+\sqrt{x+\sqrt{x+1}}}\,\mathrm{d}x} = \frac{(t^2-2t+5)(t^2-5)\sqrt{t^4{-}2t^2{-}40t+25}\ \mathrm{d}t}{32t^4}. $$ Denote the right hand side of the above equation by $\beta$. Now, setting $$ Q(t) = \frac{(25-25t-14t^2-3t^3+t^4)}{96t^3}, $$ one finds that $$ \beta = \mathrm{d}\left(Q(t)\sqrt{t^4{-}2t^2{-}40t+25}\right) + \frac{(10+15t-2t^2+t^3)\,\mathrm{d}t}{8t \sqrt{t^4-2t^2-40t+25}}\, $$ The second term on the right hand side is in normal form for differentials on the elliptic curve $s^2 = t^4-2t^2-40t+25$ that are odd with respect to the involution $(t,s)\mapsto(t,-s)$, and an application of Liouville's Theorem (on integration in elementary terms) shows that this term is not integrable in elementary terms. Hence $\beta$ is not, and hence the original integrand is not.