Integrals of the form ${\large\int}_0^\infty\operatorname{arccot}(x)\cdot\operatorname{arccot}(a\,x)\cdot\operatorname{arccot}(b\,x)\ dx$ I'm interested in integrals of the form
$$I(a,b)=\int_0^\infty\operatorname{arccot}(x)\cdot\operatorname{arccot}(a\,x)\cdot\operatorname{arccot}(b\,x)\ dx,\color{#808080}{\text{ for }a>0,\,b>0}\tag1$$
It's known$\require{action}\require{enclose}\texttip{{}^\dagger}{Gradshteyn & Ryzhik, Table of Integrals, Series, and Products, 7th edition, page 599, (4.511)}$ that
$$I(a,0)=\frac{\pi^2}4\left[\ln\left(1+\frac1a\right)+\frac{\ln(1+a)}a\right].\tag2$$
Maple and Mathematica are also able to evaluate
$$I(1,1)=\frac{3\pi^2}4\ln2-\frac{21}8\zeta(3).\tag3$$

Is it possible to find a general closed form for $I(a,1)$? Or, at least, for $I(2,1)$ or $I(3,1)$?
 A: $$\begin{align}I(2,1)&=\frac{\pi^2}3\ln2-\frac{\pi^2}6\ln3+2\ln^22\cdot\ln3-3\ln2\cdot\ln^23+\frac{29}{24}\ln^33\\&+\frac{73}{16}\zeta(3)-2\ln2\cdot\operatorname{Li}_2\left(\tfrac13\right)-\frac{13}4\operatorname{Li}_3\left(\tfrac13\right)-4\operatorname{Li}_3\left(\tfrac23\right)\end{align}$$

$$\begin{align}I(3,1)&=\frac{13\,\pi^2}{12}\ln2-\frac{4\,\pi^2}9\ln3-\frac13\ln2\cdot\ln^23+\frac7{18}\ln^33\\&-\frac{13}8\zeta(3)+\ln3\cdot\operatorname{Li}_2\left(\tfrac13\right)+\frac43\operatorname{Li}_3\left(\tfrac13\right)-\frac23\operatorname{Li}_3\left(\tfrac23\right)\end{align}$$

Update (in response to a comment):
$$\begin{align}&I(\phi,1)=\frac32\ln^32+\frac{\pi^2}{12}\Big[\left(6-3\sqrt5\right)\ln2+\left(3\sqrt5-4\right)\ln\left(1+\sqrt5\right)\Big]+\frac{51-21\sqrt5}{48}\zeta(3)\\&-\frac{\ln\left(1+\sqrt5\right)}2\Bigg[15\ln^22-15\ln\left(1+\sqrt5\right)\ln2+4\ln^2\left(1+\sqrt5\right)+2\operatorname{Li}_2\left(\frac{1-\sqrt5}4\right)\Bigg]\\&-\ln\left(3+\sqrt5\right)\operatorname{Li}_2\left(\sqrt5-2\right)+\frac{11+3\sqrt5}{48}\operatorname{Li}_3\left(9-4\sqrt5\right)-\frac{13+3\sqrt5}6\operatorname{Li}_3\left(\sqrt5-2\right)\end{align}$$
