Let $\zeta_1=-i\left(10-2\sqrt{5}\right)^{1/2}$, $\zeta_2=i\left(10-2\sqrt{5}\right)^{1/2}$, $\zeta_3=-i\left(10+2\sqrt{5}\right)^{1/2}$, $\zeta_4=i\left(10+2\sqrt{5}\right)^{1/2}$ be the solution set of the polynomial equation $\zeta^4+20\zeta^2+80=0$ and $[\varepsilon]\equiv \exp(i\pi\varepsilon)$.
A $1$-parametric, non-affine complex Hadamard matrix found by W. Bruzda in January 2018 [165] reads: $$\begin{aligned} T_8^{(1)}(x)&={\rm EXP} \left(\frac{i \pi}{10} \left[\begin{array}{rrrrrrrr} \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet\\ \bullet & \bullet & \bullet & \bullet & \bullet & 3 & 15 & 18\\ \bullet & 8 & 8 & 8 & 8 & 13 & 15 & 8\\ \bullet & 2 & 2 & 2 & 2 & 17 & 5 & 2\\ \bullet & \bullet & \bullet & \bullet & \bullet & 7 & 5 & 12\\ \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & 10 & 10\\ \bullet & \bullet & \bullet & \bullet & \bullet & 10 & \bullet & 10\\ \bullet & \bullet & \bullet & \bullet & \bullet & 10 & 10 & \bullet\\ \end{array}\right]\right)\circ \left[\begin{array}{rrrrrrrr} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ 1 & x & y & z & -\frac{y z}{x} & 1 & 1 & 1 \\ 1 & -\frac{u}{y} & \frac{u}{x} & -\frac{u z}{x y} & -\frac{u z}{x^2} & 1 & 1 & 1\\ 1 & y & x & -\frac{y z}{x} & z & 1 & 1 & 1\\ 1 & \frac{u}{x} & -\frac{u}{y} & -\frac{u z}{x^2} & -\frac{u z}{ x y} & 1 & 1 & 1\\ 1 & -i\frac{x}{z} & i\frac{x}{z} & -i\frac{z}{x} & i\frac{z}{x} & 1 & 1 & 1\\ 1 & u & -u & -\frac{u z^2}{x^2} & \frac{u z^2}{x^2} & 1 & 1 & 1\\ 1 & i\frac{u z}{x} & i\frac{u z}{x} & -i\frac{u z}{x} & -i\frac{u z}{x} & 1 & 1 & 1 \\ \end{array}\right]\end{aligned},$$
where $\circ$ is the entrywise matrix product and
$$\begin{aligned} \chi_1&=5-3\sqrt{5}+4\sqrt{5-2\sqrt{5}}+(1+i)\frac{1}{x}+\zeta_2,\\ \chi_2&=45-9i-(17-7i)\sqrt{5}-2\sqrt{620-500i-(285-234i)\sqrt{5}}-(i+1)\frac{\chi_1}{x},\\ \chi_3&=10i-10+(6-2i)\sqrt{5}+4\sqrt{10-10i-(5-4i)\sqrt{5}}+(1+i)\frac{\chi_2}{x},\\ \chi_4&=\sqrt{5}-1+i\sqrt{2}\sqrt{5+\sqrt{5}}+(1+i)\frac{\chi_3}{x},\\ \chi_5&=(1+i)\sqrt{1-\sqrt{5}+\zeta_4}-(1-i)\left(2x^2+\overline{\sqrt{\chi_4}}\right)-x\left(1-4i+\sqrt{5}+\zeta_1\right),\\ \chi_6&=\sqrt{5}-1-i-\sqrt{5-2\sqrt{5}}+(1-i)x, \end{aligned}$$(bar over the square root of $\chi_4$ denotes its complex conjugate)
$$ y=\frac{1}{4}\frac{\chi_5}{\chi_6}, $$ $$ z=\frac{x}{x-y}\left((1+i)\left[\frac{4}{5}\right]-x-y-1+i\right), $$and eventually
$$ u=\frac{i}{z}\frac{x^2-zx+xy+yz}{1+i-\left[\frac{3}{10}\right]-\left[\frac{4}{5}\right]}. $$Unimodular number $x=[\gamma]$ is a free non-affine parameter for $\gamma\in\left(2/5, 4/5\right)\cup\left(1, 7/5\right)$.
Generic defect is $d\Big(T_8^{(1)}\Big) =3$.
For $\gamma\in\{1/2,1,13/10\}$ one recovers a Butson type matrix $B_{8A}$ $\in BH(8,20)$ with defect $7$, while for $\gamma=4/5$ one recovers another Butson type matrix $B_{8B}$ $\in BH(8,20)$ with defect $11$.