Integrating over Feynman parameters

Question

Inspired by the answers in a past question I asked on techniques for computing integrals using the Feynman parametrization, I became interested in computing integrals over Feynman parameters like so:

$$\int_0^1\int_{0}^{1-u_1} \frac{u_1u_2\sqrt{u_1u_2(1-u_1-u_2)}}{(u_1(1-u_1)\|\textbf{a}\|^2-2u_1u_2\textbf{a}\cdot\textbf{b}+u_2(1-u_2)\|\textbf{b}\|^2)^3} \,du_2du_1$$
$$\int_0^1\int_{0}^{1-u_1} \frac{u_1(u_1-1)\sqrt{u_1u_2(1-u_1-u_2)}}{(u_1(1-u_1)\|\textbf{a}\|^2-2u_1u_2\textbf{a}\cdot\textbf{b}+u_2(1-u_2)\|\textbf{b}\|^2)^3} \,du_2du_1$$
$$\int_0^1\int_{0}^{1-u_1} \frac{u_2(u_2-1)\sqrt{u_1u_2(1-u_1-u_2)}}{(u_1(1-u_1)\|\textbf{a}\|^2-2u_1u_2\textbf{a}\cdot\textbf{b}+u_2(1-u_2)\|\textbf{b}\|^2)^3} \,du_2du_1$$

...where $\textbf{a},\textbf{b}\in\mathbb{R}^3$ are generic vectors.

My question: Does anyone know how to compute the integrals 1. 2. or 3. above or have any useful techniques to share possibly using some of the symmetries present?

Solution Attempt: As these are integrals over a simplex, I have seen related integrals (like the one in the linked question above) performed by making a change of variables like so: $u_1=s(1-t),u_2=(1-s)(1-t), du_2du_1=(1-t)dtds$, so that the integrals are transformed to integrals over a unit square:

1'. $$\int_0^1\int_{0}^{1} \frac{s(1-s)(1-t)\sqrt{ts(1-s)}}{\big(\left(s(1-s)\|\textbf{a}-\textbf{b}\|^2+t\|s\textbf{a}+(1-s)\textbf{b}\|^2 \right)\big)^3} \,dtds$$

2'. $$\int_0^1\int_{0}^{1} \frac{s\big(1-s(1-t)\big)\sqrt{ts(1-s)}}{\big(\left(s(1-s)\|\textbf{a}-\textbf{b}\|^2+t\|s\textbf{a}+(1-s)\textbf{b}\|^2 \right)\big)^3} \,dtds$$

3'. $$\int_0^1\int_{0}^{1} \frac{(1-s)(s(1-t)+t)\sqrt{ts(1-s)}}{\big(\left(s(1-s)\|\textbf{a}-\textbf{b}\|^2+t\|s\textbf{a}+(1-s)\textbf{b}\|^2 \right)\big)^3} \,dtds$$

Mathematica can compute the $t$ integrals in a fairly straightforward way, however it struggles with the $s$ integrals. I tried as well to make a change of variables $s=\frac{u^2}{1+u^2}$ and attempted applying some residue calculations like I've seen applied elsewhere, however I wasn't able to get too far.

Any clues or leads on how Feynman parameters like these are integrated over would be very much appreciated!

(P.S. I can also summarize how I came to these integrals if anyone is interested. It ultimately comes from the Feynman parametrization with multiple denominators where $\alpha_1=\alpha_2=\alpha_3=3/2$)

Carlo Beenakker · Accepted Answer · 2025-12-01 09:53:23Z

The Feynman parameterization refers to an integral expression of a scattering cross section in terms of three vectors that form a triangle, $\textbf{a},\textbf{b},$ and $\textbf{c}=\textbf{a}-\textbf{b}$. I am not aware of a closed form expression for a general triangle, but an answer in terms of elementary functions exists for an isosceles triangle, $\|\textbf{a}\|=\|\textbf{b}\|\equiv M$, $\textbf{a}\cdot\textbf{b}=M\cos\phi$.

For the isosceles case the first integral in the OP has the form $$I_1(M,\phi)=M^{-6}\int_0^1\!\int_{0}^{1-u_1} \frac{u_1u_2\sqrt{u_1u_2(1-u_1-u_2)}} {\big(u_1(1-u_1)-2u_1u_2\cos\phi+u_2(1-u_2)\big)^3}\,du_2\,du_1.$$ A simple answer results for an equilateral triangle ($\phi=\pi/3$), when $I_1=\tfrac{1}{9}\pi M^{-6}.$ The calculation for general $\phi\in(0,\pi)$ is a bit cumbersome (see below), the result is $$I_1(M,\phi) = \frac{\pi}{M^6} \frac{1+2 \sin \phi/2}{64 \sin ^3(\phi/2) \bigl(1+\sin \phi/2\bigr)^2}.$$

The second integral in the OP follows similarly: $$I_2(M,\phi)=M^{-6}\int_0^1\!\int_{0}^{1-u_1} \frac{u_1(u_1-1)\sqrt{u_1u_2(1-u_1-u_2)}} {\big(u_1(1-u_1)-2u_1u_2\cos\phi+u_2(1-u_2)\big)^3}\,du_2\,du_1. $$ It equals $-\tfrac{2}{9}\pi M^{-6}$ for the equilateral case ($\phi=\pi/3$), while for general $\phi$ it is given by $$I_2(M,\phi)=-\frac{\pi}{M^6}\,\frac{1+4\sin^{3}\phi/2 + 6\sin^{2}\phi/2 + 2\sin\phi/2} {64 \sin ^3(\phi/2) \bigl(1+\sin \phi/2\bigr)^2}.$$

The third integral is equal to $I_2$ in the isosceles case.

_{Calculation of the integrals (Mathematica was of little help, I asked AI for advise):

Transform variables to $u_1+u_2=s$, $u_1-u_2=st$,
\begin{eqnarray}
&I_1(M,\phi)=M^{-6}\frac18 \int_0^1 s\sqrt{1-s}\,J(s)\,ds,\;\;J(s)=\int_0^1\frac{(1-t^2)^{3/2}}{(A-Bt^2)^3}\,dt,\\
&A=1-\tfrac12(1+\cos\phi)s,\;\;
B=\tfrac12(1-\cos\phi)s.
\end{eqnarray}
The integral over $t$ is readily evaluated
$$
J(s)=\frac{3\pi}{16}\,A^{-5/2}\big(A-B\big)^{-1/2}.
$$
Since $A-B=1-s$,
the square-root singularity exactly cancels the factor $\sqrt{1-s}$ in the integral over $s$, leaving
$$
I_1(M,\phi)=M^{-6}\frac{3\pi}{128}\int_0^1 s\,A^{-5/2}\,ds,
$$
which evaluates to the formula for $I_1$ given above.
The integral $I_2$ is evaluated similarly: The coordinate transformation gives
\begin{eqnarray}
&I_2(M,\phi)=M^{-6}\frac18 \int_0^1 \sqrt{1-s}\bigl[(s-2)J_1(s)+sJ_2(s)\bigr]\,ds,\\
&J_1(s)=\int_0^1\frac{(1-t^2)^{1/2}}{(A-Bt^2)^3}\,dt,\;\;
J_2(s)=\int_0^1\frac{t^2(1-t^2)^{1/2}}{(A-Bt^2)^3}\,dt.
\end{eqnarray}
Evalation of the integral over $t$ results in
$$
I_2(M,\phi)=-M^{-6}\frac{\pi}{128}\int_0^1 s\,A^{-5/2}\frac{(4+\cos\phi)s^2-(12+\cos\phi)s+8}{1-s}\,ds,
$$
which then produces the formula given above.
If we abandon the isosceles assumption, the same parameter change still allows to carry out the integral over $t$, but then the integral over $s$ contains elliptic functions which prevent further simplification.}

Thank you Carlo - thinking through your technique a bit more now, the isosceles case elucidates a lot I think — guest
– guest, Commented Nov 30 at 19:16
I saw your edit ("If we abandon the isosceles assumption, the same parameter change still allows to carry out the integral over t, but then the integral over s contains elliptic functions which prevent further simplification.") and tried it myself too. Do you mean the integrand over s involves elliptic functions? When I did integral 1) on my side I found the Jacobian of your coordinate change was s/2, and the bounds of integration were $s\in(0,1)$ and $t\in (-1,1)$. The integral over $s\in(0,1)$ integral had a somewhat simplified form, I'll paste it in the next comment: — guest
– guest, Commented Dec 2 at 20:52
when I repeat the same parameterization for $\|a\|=x\|b\|$, so non-isosceles triangle for $x\neq 1$, I find a function $J(s)$ that is an elliptic integral, not reducible to an elementary function of $s$ unless $x=1$; then I am unable to carry out the remaining integration over $s$ in closed form; it is actually quite remarkable that such a relatively simple answer follows in the isosceles case. — Carlo Beenakker
– Carlo Beenakker, Commented Dec 2 at 21:58
your recommendation above worked very well, and I was able to get the 1) integral in closed form for $x\neq 1$ and and generic $\phi$. I am rewarding the bounty to you because your method (+Mathematica) resulted in a closed form (below). I'll write up all the steps below yours soon :) $\mathrm{1 integral} = \frac{\pi \sec ^4\left(\frac{\phi}{2}\right) \left(x^2 \left(\sqrt{-2x\cos(\phi)+x^2+1}-x\right)+x \cos(\phi) \left(-2 \sqrt{-2 x \cos (\phi)+x^2+1}+3 x+3\right)+\sqrt{-2 x \cos (\phi)+x^2+1}-1\right)}{32 (\textbf{b}.\textbf{b})^3 x^3 \left(-2 x \cos (\phi)+x^2+1\right)^{3/2}}$ — guest
– guest, Commented Dec 4 at 20:03
impressive, I had not thought the general case would work out; thank you for the bounty, I enjoyed working on this problem. — Carlo Beenakker
– Carlo Beenakker, Commented Dec 4 at 21:19

Fred Hucht · Accepted Answer · 2025-12-04 21:49:05Z

$\let\vec=\mathbf$I only discuss integral 1'. Note that there are typos in 1'-3', the denominator should start with $s(1-s) \|\vec a-\vec b\|$ instead of $s(s-1) \|\vec a-\vec b\|$.

Define $\hat{\vec s}=\vec a+\vec b$ and $\hat{\vec d}=\vec a-\vec b$. Rotate the coordinate system such that $\hat{\vec d}$ points in the $z$-direction and $\vec a,\vec b$ lie in the $xz$-plane, and rescale by $\ell=\|\hat{\vec d}\|$ such that $\hat{\vec d} \mapsto \vec d=\{0,0,1\}$ and $\hat{\vec s} \mapsto \vec s = \{ s_x,0,s_z\}$. Then, we evaluate the integral over $t$ and substitiute $s=\sin^2(\sigma/2)$ to get \begin{align}\tag{1a}\label{eq:1a} \frac{I_1}{\pi}&=\frac{1}{2\pi\,\ell^6}\int_{-\pi}^{\pi}\frac{\mathrm d\sigma}{\sin^2\sigma} \left( \frac{A-3}{A^{5/2}}\arctan\sqrt A + \frac{A+3}{A^2(A+1)} \right),\\ \text{with }\quad\tag{1b}\label{eq:1b} A&=\frac{s_x^2+s_z^2-2s_z\cos\sigma+\cos^2\sigma}{\sin^2\sigma}. \end{align} While I found no direct evaluation if this integral (yet), it turned out that $I_1/\pi$ for $s_x=1,\ldots,10$ and $s_z=1,\ldots,10$ could be identified as roots of fourth order polynomials with integer coefficients, using high precision integration and Mathematica's ToRadicals@RootApproximant[I1/Pi,4]. The radicals were simple enough that they could be reverse engineered using FindSequenceFunction[] and some NI (natural intelligence).

The result is surprisingly simple if written with the two roots \begin{align} r_1 &= \sqrt{(s_x^2+s_z^2+1)^2-4s_z^2}, \tag{2a}\label{eq:2a}\\ r_2 &= \frac12(s_x^2+s_z^2-1+r_1), \tag{2b}\label{eq:2b} \end{align} and reads \begin{align}\tag{3}\label{eq:3} \frac{I_1}{\pi}&=\frac{1}{\ell^6} \frac{2+(r_2-2)\sqrt{r_2+1}}{r_1 r_2^2}\,. \end{align} Maybe one can guess an integration scheme from this result.

Finally some Mathematica code to check:

Block[{a = {1,2,3}, b = {4,5,6},M,l,s,d,sx,sz,A,Az,r1,r2}, s=a+b; d=a-b;
 M=RotationMatrix[{d, {0, 0, 1}}]; l = Norm[d];
 a=Simplify[M.a/l];b=Simplify[M.b/l];s=Simplify[a+b];d=Simplify[a-b];
 sx = Simplify@Sqrt[s[[1]]^2 + s[[2]]^2]; sz = s[[3]];
 l^-6 {NIntegrate[(u1 u2 Sqrt[u1 u2 (1-u1-u2)])/
   (u1(1-u1) a.a-2u1 u2 a.b + u2(1-u2) b.b)^3, {u1,0,1}, {u2,0,1-u1}],
   Quiet@NIntegrate[(s(1-s)(1-t) Sqrt[(1-s) s t])/
   (s(1-s)(a-b).(a-b) + t(s a+(1-s)b).(s a+(1-s)b))^3, {s,0,1}, {t,0,1}],
   4 Quiet@NIntegrate[(
      Cos[\[Tau]]^3 Sin[\[Sigma]]^4/Sin[\[Tau]]^4)/(Sin[\[Sigma]]^2/
      Sin[\[Tau]]^2 + s.s - 2s[[3]]Cos[\[Sigma]]+Cos[\[Sigma]]^2)^3,
      {\[Sigma],0, 2\[Pi]}, {\[Tau], 0, \[Pi]/2}],
   A = (sx^2 + sz^2 - 2 sz Cos[\[Sigma]] + Cos[\[Sigma]]^2)/Sin[\[Sigma]]^2; 
   NIntegrate[1/(2 Sin[\[Sigma]]^2) ((A-3)/A^(5/2) ArcTan[Sqrt[A]]
   + (A+3)/(A^2(A+1))), {\[Sigma],-\[Pi],\[Pi]}],
   Az = -((1+2(1+2(sx^2+sz^2))z^2+z^4-4 sz(z+z^3))/(1-z^2)^2); 
   NContourIntegrate[-((2 z)/(I (1-z^2)^2)) ((ArcTan[Sqrt[Az]] (Az-3))/Az^(5/2)
    +(Az+3)/(Az^2 (Az+1))), z \[Element] Circle[]],
   N@\[Pi] (2+(r2-2) Sqrt[r2+1])/(r1 r2^2)//.
   {r1 -> Sqrt[(sx^2+sz^2+1)^2-4 sz^2], r2 -> 1/2(sx^2+sz^2-1+r1)}}]

deeply impressive; (I made a mistake in an earlier check, it all works out) — Carlo Beenakker
– Carlo Beenakker, Commented Dec 4 at 21:42

guest · Accepted Answer · 2025-12-09 08:12:02Z

Just to write up how I got the 1' integral to work using Mathematica and inspired by Carlo's idea:

(*Step 1: t-antideriviative: *)
1/(B.B)^3*
 Integrate[((s*(1 - s)*(1 - t)*
      Sqrt[t*s*(1 - s)])/(s*(1 - s)*(x^2 - 2*x*Cos[T] + 1) + 
       t*(s^2*x^2 + (1 - s)^2 + 2*s*(1 - s)*x*Cos[T]))^3), t]

(*Step 2 apply limit t->1 then t->0*)
Limit[1/(B.B)^3 (-1 + 
     s) s (-((Sqrt[-(-1 + s) s t]
           ArcTan[(
           Sqrt[t] Sqrt[-1 + 2 s - s^2 - s^2 x^2 - 2 s x Cos[T] + 
             2 s^2 x Cos[T]])/(
           Sqrt[1 - s] Sqrt[s] Sqrt[-1 - x^2 + 2 x Cos[T]])] (-1 + 
            5 s - 4 s^2 + 3 s x^2 - 4 s^2 x^2 - 8 s x Cos[T] + 
            8 s^2 x Cos[T]))/(4 Sqrt[1 - s] (-1 + s) s^(3/2) Sqrt[
          t] (-1 - x^2 + 2 x Cos[T])^(
          3/2) (-1 + 2 s - s^2 - s^2 x^2 - 2 s x Cos[T] + 
            2 s^2 x Cos[T])^(5/2))) + (Sqrt[-(-1 + s) s t] (1 - s + 
          s x^2))/(2 (-1 + 2 s - s^2 - s^2 x^2 - 2 s x Cos[T] + 
          2 s^2 x Cos[T])^2 (-s + s^2 - t + 2 s t - s^2 t - s x^2 + 
          s^2 x^2 - s^2 t x^2 + 2 s x Cos[T] - 2 s^2 x Cos[T] - 
          2 s t x Cos[T] + 
          2 s^2 t x Cos[T])^2) + (Sqrt[-(-1 + s) s t] (1 + 3 s - 
          4 s^2 + 5 s x^2 - 4 s^2 x^2 - 8 s x Cos[T] + 
          8 s^2 x Cos[T]))/(4 (-1 + s) s (-1 - x^2 + 
          2 x Cos[T]) (-1 + 2 s - s^2 - s^2 x^2 - 2 s x Cos[T] + 
          2 s^2 x Cos[T])^2 (-s + s^2 - t + 2 s t - s^2 t - s x^2 + 
          s^2 x^2 - s^2 t x^2 + 2 s x Cos[T] - 2 s^2 x Cos[T] - 
          2 s t x Cos[T] + 2 s^2 t x Cos[T]))), t -> 1] - 
 Limit[1/(B.B)^3 (-1 + 
     s) s (-((Sqrt[-(-1 + s) s t]
           ArcTan[(
           Sqrt[t] Sqrt[-1 + 2 s - s^2 - s^2 x^2 - 2 s x Cos[T] + 
             2 s^2 x Cos[T]])/(
           Sqrt[1 - s] Sqrt[s] Sqrt[-1 - x^2 + 2 x Cos[T]])] (-1 + 
            5 s - 4 s^2 + 3 s x^2 - 4 s^2 x^2 - 8 s x Cos[T] + 
            8 s^2 x Cos[T]))/(4 Sqrt[1 - s] (-1 + s) s^(3/2) Sqrt[
          t] (-1 - x^2 + 2 x Cos[T])^(
          3/2) (-1 + 2 s - s^2 - s^2 x^2 - 2 s x Cos[T] + 
            2 s^2 x Cos[T])^(5/2))) + (Sqrt[-(-1 + s) s t] (1 - s + 
          s x^2))/(2 (-1 + 2 s - s^2 - s^2 x^2 - 2 s x Cos[T] + 
          2 s^2 x Cos[T])^2 (-s + s^2 - t + 2 s t - s^2 t - s x^2 + 
          s^2 x^2 - s^2 t x^2 + 2 s x Cos[T] - 2 s^2 x Cos[T] - 
          2 s t x Cos[T] + 
          2 s^2 t x Cos[T])^2) + (Sqrt[-(-1 + s) s t] (1 + 3 s - 
          4 s^2 + 5 s x^2 - 4 s^2 x^2 - 8 s x Cos[T] + 
          8 s^2 x Cos[T]))/(4 (-1 + s) s (-1 - x^2 + 
          2 x Cos[T]) (-1 + 2 s - s^2 - s^2 x^2 - 2 s x Cos[T] + 
          2 s^2 x Cos[T])^2 (-s + s^2 - t + 2 s t - s^2 t - s x^2 + 
          s^2 x^2 - s^2 t x^2 + 2 s x Cos[T] - 2 s^2 x Cos[T] - 
          2 s t x Cos[T] + 2 s^2 t x Cos[T]))), t -> 0]

(*Step 3: take the result from above and integrate over s *)

Integrate[
 1/(B.B)^3 (-1 + s) s ((Sqrt[(1 - s) s] (1 - s + s x^2))/(
    2 (-1 + s - s x^2)^2 (-1 + 2 s - s^2 - s^2 x^2 - 2 s x Cos[T] + 
       2 s^2 x Cos[T])^2) - (Sqrt[(1 - s) s]
        ArcTan[Sqrt[-1 + 2 s - s^2 - s^2 x^2 - 2 s x Cos[T] + 
         2 s^2 x Cos[T]]/(
        Sqrt[1 - s] Sqrt[s] Sqrt[-1 - x^2 + 2 x Cos[T]])] (-1 + 5 s - 
         4 s^2 + 3 s x^2 - 4 s^2 x^2 - 8 s x Cos[T] + 
         8 s^2 x Cos[T]))/(4 Sqrt[1 - s] (-1 + s) s^(
       3/2) (-1 - x^2 + 2 x Cos[T])^(
       3/2) (-1 + 2 s - s^2 - s^2 x^2 - 2 s x Cos[T] + 
         2 s^2 x Cos[T])^(
       5/2)) + (Sqrt[(1 - s) s] (1 + 3 s - 4 s^2 + 5 s x^2 - 
         4 s^2 x^2 - 8 s x Cos[T] + 8 s^2 x Cos[T]))/(4 (-1 + 
         s) s (-1 + s - s x^2) (-1 - x^2 + 2 x Cos[T]) (-1 + 2 s - 
         s^2 - s^2 x^2 - 2 s x Cos[T] + 2 s^2 x Cos[T])^2)), s, 
 Assumptions -> (Element[x, Reals] && (x > 0) && 
    Element[T, Reals] && (T >= 0) && (T <= 2*Pi))]

(*Step 4 - FTC...but weird -(F(1) + F(0)) where F(s) is the antiderivative from above...must arise from a choice of branch cut in Integrate*)

-(Limit[(ArcTan[
         Sqrt[-1 + 2 s - s^2 (1 + x^2) + 2 (-1 + s) s x Cos[T]]/(
         Sqrt[-(-1 + s) s]
           Sqrt[-1 - x^2 + 
           2 x Cos[
             T]])] (-(-8 - 21 x^2 - 24 s^2 (1 + 8 x^2 + 4 x^4) + 
              3 s (8 + 39 x^2 + 7 x^4) + 
              8 s^3 (1 + 12 x^2 + 12 x^4 + x^6)) Cos[T] + 
          x (2 (-1 + 2 s) (6 + x^2 - 4 s (3 + 5 x^2) + 
                s^2 (6 + 20 x^2 + 6 x^4)) Cos[2 T] - 
             3 (1 - 8 s + 8 s^2) x (-1 + s + s x^2) Cos[3 T] + 
             2 (-1 + 2 s) (6 + 3 x^2 - 6 s (2 + 3 x^2) + 
                6 s^2 (1 + 3 x^2 + x^4) + 
                2 (-1 + s) s x^2 Cos[4 T]))) Csc[
         T]^4)/(16 x^3 (-1 - x^2 + 2 x Cos[T])^(
        3/2) (-1 + 2 s - s^2 (1 + x^2) + 2 (-1 + s) s x Cos[T])^(
        3/2) (B.B)^3) + (Sqrt[-(-1 + s) s]
         Csc[T]^4 ((
          2 ArcTanh[x/Sqrt[(-1 + s)/
            s]] (1 + x^2 - 2 x Cos[T]) (3 + Cos[2 T]))/
          Sqrt[-1 + s] + (
          8 Cos[T] (-1 - x^2 + 2 x Cos[T]) Log[
            Sqrt[-1 + s] + Sqrt[s]])/Sqrt[-1 + s] - (
          4 Sqrt[s]
            x (-1 + s + s x^2 + (x - 2 s x) Cos[T]) Sin[T]^2)/(
          1 - 2 s + s^2 (1 + x^2) - 2 (-1 + s) s x Cos[T])))/(16 Sqrt[
        s] x^3 (-1 - x^2 + 2 x Cos[T]) (B.B)^3), s -> 1, 
    Assumptions -> (Element[x, Reals] && (x > 0) && 
       Element[T, Reals] && (T >= 0) && (T <= 2*Pi))] + 
   Limit[(ArcTan[
         Sqrt[-1 + 2 s - s^2 (1 + x^2) + 2 (-1 + s) s x Cos[T]]/(
         Sqrt[-(-1 + s) s]
           Sqrt[-1 - x^2 + 
           2 x Cos[
             T]])] (-(-8 - 21 x^2 - 24 s^2 (1 + 8 x^2 + 4 x^4) + 
              3 s (8 + 39 x^2 + 7 x^4) + 
              8 s^3 (1 + 12 x^2 + 12 x^4 + x^6)) Cos[T] + 
          x (2 (-1 + 2 s) (6 + x^2 - 4 s (3 + 5 x^2) + 
                s^2 (6 + 20 x^2 + 6 x^4)) Cos[2 T] - 
             3 (1 - 8 s + 8 s^2) x (-1 + s + s x^2) Cos[3 T] + 
             2 (-1 + 2 s) (6 + 3 x^2 - 6 s (2 + 3 x^2) + 
                6 s^2 (1 + 3 x^2 + x^4) + 
                2 (-1 + s) s x^2 Cos[4 T]))) Csc[
         T]^4)/(16 x^3 (-1 - x^2 + 2 x Cos[T])^(
        3/2) (-1 + 2 s - s^2 (1 + x^2) + 2 (-1 + s) s x Cos[T])^(
        3/2) (B.B)^3) + (Sqrt[-(-1 + s) s]
         Csc[T]^4 ((
          2 ArcTanh[x/Sqrt[(-1 + s)/
            s]] (1 + x^2 - 2 x Cos[T]) (3 + Cos[2 T]))/
          Sqrt[-1 + s] + (
          8 Cos[T] (-1 - x^2 + 2 x Cos[T]) Log[
            Sqrt[-1 + s] + Sqrt[s]])/Sqrt[-1 + s] - (
          4 Sqrt[s]
            x (-1 + s + s x^2 + (x - 2 s x) Cos[T]) Sin[T]^2)/(
          1 - 2 s + s^2 (1 + x^2) - 2 (-1 + s) s x Cos[T])))/(16 Sqrt[
        s] x^3 (-1 - x^2 + 2 x Cos[T]) (B.B)^3), s -> 0, 
    Assumptions -> (Element[x, Reals] && (x > 0) && 
       Element[T, Reals] && (T >= 0) && (T <= 2*Pi))])

Of note in the Mathematica code above: the final result is obtained by summing the limits of the s-antiderivatve obtained in step 3 as s->1 and s->0. I am not sure why this works, I'm assuming it comes from some choice of branch cut in Mathematica's symbolic integration scheme.

The final result after some simplification is then: $\frac{\pi \sec ^4\left(\frac{T}{2}\right) \left(3 x^2 \cos (T)+\left(-2 x \cos (T)+x^2+1\right)^{3/2}+3 x \cos (T)-x^3-1\right)}{32 (B.B)^3 x^3 \left(-2 x \cos (T)+x^2+1\right)^{3/2}}$

(P.S. Please forgive the sloppy code above, it is the only way I can depict how I got the integral working...weirdly, suppressing output with ";", or by assigning symbolic results to variables to make the code more readable, gives superficially different results, and wrong final answers when compared to:

NIntegrate[
 NIntegrate[
  u2*u1*Sqrt[
     u1*u2*(1 - u1 - u2)]/(u1*(1 - u1)*A.A - 2*u1*u2*A.B + 
       u2*(1 - u2)*B.B)^3, {u2, 0, 1 - u1}], {u1, 0, 1}]

)

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