Start with the transformed integral

$$\begin{align*} \int_0^z\frac{\mathrm du}{\sqrt{a(u+1)^3+b}}&=\int_1^{z+1}\frac{\mathrm du}{\sqrt{au^3+b}}\\ &=\frac{\sqrt[3]{b/a}}{\sqrt{b}}\int_{\frac1{\sqrt[3]{b/a}}}^{\frac{z+1}{\sqrt[3]{b/a}}}\frac{\mathrm du}{\sqrt{u^3+1}} \end{align*}$$

Now, I had already dealt with that elliptic integral in this math.SE answer, so I'm not repeating the work here; suffice it to say that we now have an explicit expression for your integral in terms of EllipticF[]:

With[{a = 4/3, b = 5/2, z = 7},
     {(* integral *)
      NIntegrate[1/Sqrt[a (t + 1)^3 + b], {t, 0, z}, WorkingPrecision -> 20],
      (* closed form solution *)
      N[((b/a)^(1/3)/(3^(1/4) Sqrt[b]))
        (EllipticF[ArcCos[2 Sqrt[3]/(1 + Sqrt[3] + (z + 1)/(b/a)^(1/3)) - 1], (2 + Sqrt[3])/4] -
         EllipticF[ArcCos[2 Sqrt[3]/(1 + Sqrt[3] + 1/(b/a)^(1/3)) - 1], (2 + Sqrt[3])/4]), 20]}]

   {0.97715948427474779292, 0.97715948427474779300}

Alternatively, since EllipticF[ArcCos[u], m] == InverseJacobiCN[u, m], you can also use this function:

With[{a = 4/3, b = 5/2, z = 7},
 N[((b/a)^(1/3)/(3^(1/4) Sqrt[b]))
   (InverseJacobiCN[2 Sqrt[3]/(1 + Sqrt[3] + (z + 1)/(b/a)^(1/3)) - 1, (2 + Sqrt[3])/4] - 
    InverseJacobiCN[2 Sqrt[3]/(1 + Sqrt[3] + 1/(b/a)^(1/3)) - 1, (2 + Sqrt[3])/4]), 20]]

Start with the transformed integral

$$\begin{align*} \int_0^z\frac{\mathrm du}{\sqrt{a(u+1)^3+b}}&=\int_1^{z+1}\frac{\mathrm du}{\sqrt{au^3+b}}\\ &=\frac{\sqrt[3]{b/a}}{\sqrt{b}}\int_{\frac1{\sqrt[3]{b/a}}}^{\frac{z+1}{\sqrt[3]{b/a}}}\frac{\mathrm du}{\sqrt{u^3+1}} \end{align*}$$

Now, I had already dealt with that elliptic integral in this math.SE answer, so I'm not repeating the work here; suffice it to say that we now have an explicit expression for your integral in terms of EllipticF[]:

With[{a = 4/3, b = 5/2, z = 7},
     {(* integral *)
      NIntegrate[1/Sqrt[a (t + 1)^3 + b], {t, 0, z}, WorkingPrecision -> 20],
      (* closed form solution *)
      N[((b/a)^(1/3)/(3^(1/4) Sqrt[b]))
        (EllipticF[ArcCos[2 Sqrt[3]/(1 + Sqrt[3] + (z + 1)/(b/a)^(1/3)) - 1], (2 + Sqrt[3])/4] -
         EllipticF[ArcCos[2 Sqrt[3]/(1 + Sqrt[3] + 1/(b/a)^(1/3)) - 1], (2 + Sqrt[3])/4]), 20]}]

   {0.97715948427474779292, 0.97715948427474779300}

Alternatively, since EllipticF[ArcCos[u], m] == InverseJacobiCN[u, m], you can also use this function:

With[{a = 4/3, b = 5/2, z = 7},
 N[((b/a)^(1/3)/(3^(1/4) Sqrt[b]))
   (InverseJacobiCN[2 Sqrt[3]/(1 + Sqrt[3] + (z + 1)/(b/a)^(1/3)) - 1, (2 + Sqrt[3])/4] - 
    InverseJacobiCN[2 Sqrt[3]/(1 + Sqrt[3] + 1/(b/a)^(1/3)) - 1, (2 + Sqrt[3])/4]), 20]]