I have a problem to solve, and i have a solution, but not sure if it is right one. E.g. i have never used that $a > 0.$ Can you please look at it and correct it if something went wrong. Thanks.
So, i have $\kappa=\frac{a}{a^2+b^2}$ and $\tau=\frac{b}{a^2+b^2}, a > 0%.$ I think i need to solve this system $$ \begin{cases} \mathbf{t}'=\frac{a}{a^2+b^2}\mathbf{n}\\ \mathbf{n}'=-\frac{a}{a^2+b^2}\mathbf{t}+\frac{b}{a^2+b^2}\mathbf{b}\\ \mathbf{b}'=-\frac{b}{a^2+b^2}\mathbf{n} \end{cases} $$ Substituting first and third eqs into second i got: $$\mathbf{n}'=-\frac{1}{a^2+b^2}\int\mathbf{n}ds$$ $$\mathbf{n}''=-\frac{1}{a^2+b^2}\mathbf{n}$$
So, $$\mathbf{n_1} = \left(C_1\sin\frac{s}{\sqrt{a^2+b^2}} + C_2s + C_3, C_4\cos\frac{s}{\sqrt{a^2+b^2}}+C_5s+C_6\right)$$ $$\mathbf{n_2} = \left(K_1\cos\frac{s}{\sqrt{a^2+b^2}} + K_2s + K_3, K_4\sin\frac{s}{\sqrt{a^2+b^2}}+K_5s+K_6\right)$$ $$\mathbf{n_{general}}=\mathbf{n_1}+\mathbf{n_2}$$
And as i have $\mathbf{n},$ i can integrate it over $s$ twice and by this get frightening formula for $\mathbf{r}.$
Update
When the constant curvature $\kappa$ vanishes the curve $\gamma$ in question is a line, which has no torsion.
Therefore let's assume $\kappa>0$, and put $\sqrt{\kappa^2+\tau^2}=:\lambda>0$. For the "reconstruction" of $\gamma$ from $\kappa$ and $\tau$ we have to keep track of the involved vectors ${\bf t}$, ${\bf n}$, and ${\bf b}:={\bf t}\times{\bf n}$. From these we define the auxiliary vectors $${\bf p}:={1\over\lambda}(-\kappa{\bf t}+\tau{\bf b}),\qquad {\bf q}:={\bf n}\times{\bf p}\ .$$ Then ${\bf p}$ is a unit vector orthogonal to ${\bf n}$; therefore $({\bf n},{\bf p},{\bf q})$ is an orthonormal frame. It follows that $${\bf t}=\langle{\bf t},{\bf n}\rangle{\bf n}+\langle{\bf t},{\bf p}\rangle{\bf p}+\langle{\bf t},{\bf q}\rangle{\bf q}={1\over\lambda}(-\kappa{\bf p}+\tau{\bf q})\ ,\tag{1}$$ and Frenet's formulas give $$\eqalign{\dot{\bf n}&=\lambda{\bf p}\cr \dot{\bf p}&={1\over\lambda}(-\kappa^2-\tau^2){\bf n}=-\lambda{\bf n}\cr \dot{\bf q}&={\bf 0}\ .\cr}\tag{2}$$ In particular, the vector ${\bf q}$ is constant. Choosing the coordinate system in ${\Bbb R}^3$ such that $${\bf n}(0)=(1,0,0),\quad {\bf p}(0)=(0,1,0),\quad{\bf q}=(0,0,1)$$ one immediately derives from $(2)$ that necessarily $${\bf n}(s)=\bigl(\cos(\lambda s),\sin(\lambda s),0\bigr),\qquad{\bf p}(s)=\bigl(-\sin(\lambda s),\cos(\lambda s),0\bigr)\ .$$ Plugging this into $(1)$ we obtain $${\bf t}(s)=\left({\kappa\over\lambda}\sin(\lambda s),\ -{\kappa\over\lambda}\cos(\lambda s),\ {\tau\over\lambda} \right)\ ,$$ and one more integration gives $${\bf x}(s)=\left(-{\kappa\over\lambda^2}\cos(\lambda s),\ -{\kappa\over\lambda^2}\sin(\lambda s),\ {\tau\over\lambda}s \right)\qquad(-\infty<s<\infty)\ ,$$ up to an additive constant. This is a helix winding around the $x_3$-axis when $\tau\ne0$ , resp., the universal covering of a circle in the $(x_1,x_2)$-plane when $\tau=0$.