I do not wish to $\LaTeX$ out all the intermediate steps for these integrations, but if you like I can edit my answer to tell you each step along the way. Having said that, let's get down to business.
$I_1(p)$ integration
I will split $I_1$ into 2 parts.
$$I_1^a(p) = p^2\int_{0}^1 dx \int_0^{1-x} dy \frac{x y (-x-y+1)}{(x y+(-x-y+1) (x+y))^3}$$
and
$$I_1^b(p) = m^2\int_{0}^1 dx \int_0^{1-x} dy \frac{1}{(x y+(-x-y+1) (x+y))^2}$$
As you can see, I have performed the initial integral over $z$, but have strictly followed the definitions provided.
This integral can be done. It can be shown (after much algebra and substitutions) that
$$I_a^1(p) =p^2 \int_0^1 dx\ \frac{3 x}{(x-1) (3 x+1)^2}-\frac{1}{(x-1) (3 x+1)^2} -\frac{8 x \tanh ^{-1}\left(\frac{1-x}{\sqrt{-3 x^2+2 x+1}}\right)}{(x-1) (3 x+1)^2 \sqrt{-3 x^2+2 x+1}}$$
which, I'll admit is a lot less messy than I thought it'd be. The integration at this point is actually somewhat standard. At any rate, it is found after integrating over $x$ that
\begin{align*}
&I_a^1(p)= p^2 \int_0^1 dx\ \frac{3 x}{(x-1) (3 x+1)^2}-\frac{1}{(x-1) (3 x+1)^2} -\frac{8 x \tanh ^{-1}\left(\frac{1-x}{\sqrt{-3 x^2+2 x+1}}\right)}{(x-1) (3 x+1)^2 \sqrt{-3 x^2+2 x+1}}\\
&=p^2\bigg{(} \frac{\sqrt{-3 x^2+2 x+1} \tanh ^{-1}\left(\frac{1-x}{\sqrt{-3 x^2+2 x+1}}\right)}{4 (1-x)}+\frac{\sqrt{-3 x^2+2 x+1} \tanh ^{-1}\left(\frac{1-x}{\sqrt{-3 x^2+2 x+1}}\right)}{3 (3 x+1)^2} + \cdots\\
&-\frac{7 \sqrt{-3 x^2+2 x+1} \tanh ^{-1}\left(\frac{1-x}{\sqrt{-3 x^2+2 x+1}}\right)}{12 (3 x+1)}-\frac{1}{3 (3 x+1)}\bigg{)} \bigg\rvert_0^1\\
& \implies \boxed{I_1^a(p)= \frac{p^2}{2}}
\end{align*}
This integral had no right to simplify so nicely, but sometimes the math gods smile upon us.
Now, we work ok $I_1^b(p)$. After performing the $z$ integration we have that
$$I_1^b(p) = m^2\int_0^1 dx\ \frac{8 \tanh ^{-1}\left(\frac{1-x}{\sqrt{-3 x^2+2 x+1}}\right)}{\left(-3 x^2+2 x+1\right)^{3/2}}+\frac{2}{-3 x^3+2 x^2+x}$$
Although this looks simpler than the other one $I_1^a$ there are divergences lurking. Namely on the endpoints. Therefore, I will use a small parameter $\epsilon$ in an attempt to tame it. We arrive at
\begin{align*}
&I_1^b(p, \epsilon) = m^2\int_\epsilon^{1-\epsilon}dx \ \frac{8 \tanh ^{-1}\left(\frac{1-x}{\sqrt{-3 x^2+2 x+1}}\right)}{\left(-3 x^2+2 x+1\right)^{3/2}}+\frac{2}{-3 x^3+2 x^2+x}\\
&= 2 m^2\left(\frac{(1-3 \epsilon ) \tanh ^{-1}\left(\frac{1-\epsilon }{\sqrt{-3 \epsilon ^2+2 \epsilon +1}}\right)}{\sqrt{-3 \epsilon ^2+2 \epsilon +1}}+ \underbrace{\log (1-\epsilon )}_{\to 0 \ as\ \epsilon \to 0}-\log (\epsilon )- \underbrace{\frac{(3 \epsilon -2) \tanh ^{-1}\left(\frac{\epsilon }{\sqrt{\epsilon (4- 3 \epsilon )}}\right)}{\sqrt{\epsilon (4-3 \epsilon)}}}_{\to \frac{1}{2}, \ as\ \epsilon \to 0}\right)
\end{align*}
The divergences live in the other two terms. So we may substitute in the limits when $\epsilon \to 0$ since that's what we will eventually take. We arrive at
\begin{align*}
&I_1^b(p, \epsilon) = m^2\int_\epsilon^{1-\epsilon}dx \ \frac{8 \tanh ^{-1}\left(\frac{1-x}{\sqrt{-3 x^2+2 x+1}}\right)}{\left(-3 x^2+2 x+1\right)^{3/2}}+\frac{2}{-3 x^3+2 x^2+x}\\
& \boxed{I_1^b(p, \epsilon) = 2m^2\frac{(1-3 \epsilon ) \tanh ^{-1}\left(\frac{1-\epsilon }{\sqrt{-3 \epsilon ^2+2 \epsilon +1}}\right)}{\sqrt{-3 \epsilon ^2+2 \epsilon +1}}- 2m^2\log (\epsilon )- m^2}
\end{align*}
I request that OP tries to expand this in small $\epsilon$ before I do, as I still have another integral to do
$I_2(p)$ integration
I must admit, this one pushed back quite a bit. Nevertheless, onward we must go.
The integral to overcome is
$$I_2(p) = \int_0^1 dx dy dz \delta(x+y+z-1) \frac{\gamma p^2 +m^2}{(\alpha \beta)^2}\ln\left(\frac{\sqrt{\alpha \beta}}{\gamma p^2 +m^2}\right). $$
We split it into two integrals
$$I_2^a = \frac{1}{2} \int_0^1 dx dy dz \delta(x+y+z-1) \frac{\gamma p^2 +m^2}{(\alpha \beta)^2}\ln\left(\alpha \beta\right) $$
and
$$ I_2^b = -\int_0^1 dx dy dz \delta(x+y+z-1) \frac{\gamma p^2 +m^2}{(\alpha \beta)^2}\ln\left(\gamma p^2 +m^2\right)$$
so that our desired integral is the sum.
We know that we can find the indefinite integral of the function multiplying the log, so we close our eyes and hope for the best.
$I_2^a$ integration
Integrating over $z$ we get that
$$ I_2^a = \frac{1}{2}\int_0^1dx \int_0^{1-x} dy \frac{m^2}{(x y+(-x-y+1) (x+y))^2}+\frac{p^2 (x y (-x-y+1))}{(x y+(-x-y+1) (x+y))^3}\times\\
\log\left(x y+(-x-y+1) (x+y) \right)$$
We integrate by parts in $y$; taking $u$ as the log and $dv$ as the coefficient on the log. Then, we have that
$$ v= m^2 \left(-\frac{-x-2 y+1}{\left(-3 x^2+2 x+1\right) \left((1-x) y+(1-x) x-y^2\right)}-\frac{4 \tanh ^{-1}\left(\frac{-x-2 y+1}{\sqrt{-3 x^2+2 x+1}}\right)}{\left(-3 x^2+2 x+1\right)^{3/2}}\right)+p^2 \left(\frac{x^2 (-x-2 y+1)}{2 (3 x+1) \left((1-x) y+(1-x) x-y^2\right)^2}-\frac{4 x \tanh ^{-1}\left(\frac{-x-2 y+1}{\sqrt{-3 x^2+2 x+1}}\right)}{(1-x) (3 x+1)^2 \sqrt{-3 x^2+2 x+1}}-\frac{x (-x-2 y+1)}{(1-x) (3 x+1)^2 \left((1-x) y+(1-x) x-y^2\right)}\right) $$
and
$$ du = dy \frac{x+2 y-1}{x^2+x (y-1)+(y-1) y}$$
Our surface terms reduce to
$$uv\bigg\rvert^{1-x}_0 = \frac{2 \log ((1-x) x) \left((3 x-1) \sqrt{-3 x^2+2 x+1}-8 x \tanh ^{-1}\left(\frac{1-x}{\sqrt{-3 x^2+2 x+1}}\right)\right)}{(x-1) (3 x+1)^2 \sqrt{-3 x^2+2 x+1}} $$
And our integral left to be done is
$$\int_0^{1-x} v \ du = \int_0^{1-x} dy\ \Bigg{\{} m^2\left(-\frac{-x-2 y+1}{\left(-3 x^2+2 x+1\right) \left((1-x) y+(1-x) x-y^2\right)}-\frac{4 \tanh ^{-1}\left(\frac{-x-2 y+1}{\sqrt{-3 x^2+2 x+1}}\right)}{\left(-3 x^2+2 x+1\right)^{3/2}}\right)+p^2 \left(\frac{x^2 (-x-2 y+1)}{2 (3 x+1) \left((1-x) y+(1-x) x-y^2\right)^2}-\frac{4 x \tanh ^{-1}\left(\frac{-x-2 y+1}{\sqrt{-3 x^2+2 x+1}}\right)}{(1-x) (3 x+1)^2 \sqrt{-3 x^2+2 x+1}}-\frac{x (-x-2 y+1)}{(1-x) (3 x+1)^2 \left((1-x) y+(1-x) x-y^2\right)}\right)\Bigg{\}} \frac{x+2 y-1}{x^2+x (y-1)+(y-1) y}$$
Believe it or not this can be done, but not entirely with elementary functions. Putting the surface terms back and simplifying we have that
\begin{align*}
&2I_2^a(p) = \int_0^1 dx\\
&\frac{1}{2 (1-x)^{3/2} (3 x+1)^{5/2}}\bigg{\{} 8 \left(m^2 (3 x+1)+p^2 x\right)+\frac{\sqrt{x (2-3 x)+1} \left(4 m^2 (3 x+1)-p^2 (x-3) x\right)}{x}\\
&+16 \tanh ^{-1}\left(\sqrt{\frac{1-x}{3 x+1}}\right) \bigg{(}\log (4)-\log \left[\left(\sqrt{\frac{1-x}{3 x+1}}+1\right) \left(\frac{x-1}{\sqrt{x (2-3 x)+1}}+1\right)\right]\times\\
& \left(m^2 (3 x+1)+p^2 x\right)-2 m^2 (3 x+1)+p^2 (x-2) x\bigg{)}\\
&-4 \log (x(1-x)) \left((3 x-1) \sqrt{x (2-3 x)+1}-8 x \tanh ^{-1}\left(\sqrt{\frac{1-x}{3 x+1}}\right)\right)\\
&+ \text{Li}_2\left(-\frac{x+\sqrt{1-x} \sqrt{3 x+1}+1}{2 x}\right)-\text{Li}_2\left(-\frac{x-\sqrt{1-x} \sqrt{3 x+1}+1}{2 x}\right)
\bigg{\}}
\end{align*}
I imagine there are divergences lurking in here, but you can use numerical integration to check. Do your best to do $I_2^b$ and get back to me.
I think that's enough for one day.