Owen's $T$ function is given by $$ T(h,a) = \frac{1}{2 \pi} \int_0^{a} \frac{\mathrm{e}^{-\frac{h^2}{2} (1+x^2)}}{1 + x^2} \mathrm{d}x $$ There are some properties of the function, which one can find in the Wikipedia.
one of the interesting properties is as $$ T(h,a) + T(ha,\frac{1}{a}) = \frac{1}{4}\Bigg(\mathrm{sign}(a) - \mathrm{erf}\Big(\frac{h}{\sqrt{2}} \Big) \ \mathrm{erf}\Big(\frac{ha}{\sqrt{2}} \Big) \Bigg) $$
where the $\mathrm{sign}(a)$ is the Sign function, and the $\mathrm{erf}$ is the Error function.
Now my question is that there is such a similar relation for the following Owen's $T$ functions. $$ T(h, c \frac{a}{h}+b)+T(a, c \frac{h}{a} + b) = ? $$
I would highly appreciate it if one could help me find an expression of the above Owen's $T$ functions.
Edit
Since this question did not get attention, here I open the first known relation link. Hopefully, this helps
$$ T(h,a) = \frac{\arctan(a)}{2 \pi} - \int_0^{h} \Phi^{\prime}(x) \Phi(ax) \mathrm{d}x + \frac{1}{2}\Phi(h)-\frac{1}{4} $$ and $$ T(ha,\frac{1}{a}) = \frac{\arctan(\frac{1}{a})}{2 \pi} - \int_0^{ha} \Phi^{\prime}(x) \Phi(\frac{x}{a}) \mathrm{d}x + \frac{1}{2}\Phi(ha)-\frac{1}{4} $$ therefore $$ T(h,a)+T(ha,\frac{1}{a}) = \frac{\arctan(a)+\arctan(\frac{1}{a})}{2 \pi} - \int_0^{h} \Phi^{\prime}(x) \Phi(ax) \mathrm{d}x - \int_0^{ha} \Phi^{\prime}(x) \Phi(\frac{x}{a}) \mathrm{d}x + \frac{1}{2}\Phi(a)+ \frac{1}{2}\Phi(ha)-\frac{1}{2} $$ by changing variable, and using integral by part $$ \int_0^{ha} \Phi^{\prime}(x) \Phi(\frac{x}{a}) \mathrm{d}x \longrightarrow_{x\rightarrow x/a} a \int_0^{h} \Phi^{\prime}(ax) \Phi(x) \mathrm{d}x \longrightarrow_{\mbox{int by parts} } a\Big( \frac{1}{a} \Phi(ha) \Phi(h) - \frac{1}{a} \int_0^{h} \Phi^{\prime}(x) \Phi(a x) \mathrm{d}x \Big) $$ thus $$ T(h,a)+T(ha,\frac{1}{a}) = \frac{\arctan(a)+\arctan(\frac{1}{a})}{2 \pi} - \Phi(ha) \Phi(h)+ \frac{1}{2}\Phi(a)+ \frac{1}{2}\Phi(ha)-\frac{1}{2} $$ where $\Phi(x)$ is the cumulative distribution function (CDF) $$ \Phi(x) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{x} \mathrm{e}^{-t^2/2} dt = \frac{1}{2} \Big(1 + \mathrm{erf}\Big(\frac{x}{\sqrt{2}} \Big) \Big) $$ and $$ \arctan(a)+\arctan(\frac{1}{a}) =\frac{\pi}{2} \mathrm{sign}(a) $$
Edit2
Now for the new expression $$ T(h, c \frac{a}{h}+b) = \frac{\arctan(c \frac{a}{h}+b)}{2 \pi} - \int_0^{h} \Phi^{\prime}(x) \Phi\Big((c \frac{a}{h}+b)x\Big) \mathrm{d}x + \frac{1}{2}\Phi(h)-\frac{1}{4} $$ and $$ T(a,c \frac{h}{a} + b) = \frac{\arctan(c \frac{h}{a} + b)}{2 \pi} - \int_0^{a} \Phi^{\prime}(x) \Phi\Big((c \frac{h}{a}+b)x\Big) \mathrm{d}x + \frac{1}{2}\Phi(a)-\frac{1}{4} $$ thus, finding a relation between these integrations, as it is in the simple one $$ \int_0^{a} \Phi^{\prime}(x) \Phi\Big((c \frac{h}{a}+b)x\Big) \mathrm{d}x + \int_0^{h} \Phi^{\prime}(x) \Phi\Big((c \frac{a}{h}+b)x\Big) \mathrm{d}x = ? $$