TITLE: Integrals of the form ${\large\int}_0^\infty\operatorname{arccot}(x)\cdot\operatorname{arccot}(a\,x)\cdot\operatorname{arccot}(b\,x)\ dx$
QUESTION [67 upvotes]: 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 $I(3,1)$?

REPLY [34 votes]: I finally managed to find a general solution to this problem. Although I put a lot of effort into simplification of the result, it is still not as pretty and symmetric as I would like it to be. I hope to improve it later. Sorry for poor typesetting.
Assuming $0<a<b<c,$
$$\int_0^\infty\operatorname{arccot}(ax)\,\operatorname{arccot}(bx)\,\operatorname{arccot}(cx)\,dx=\\
\frac1{24 a b c}\left(2 a (b+2 c) \ln ^3(a)-3 \left(2 b (a+c) \ln (b)+(4 a c-2 b c) \ln (c)+\\
b \left(-2 c \operatorname{artanh}\left(\frac{b}{c}\right)+(c-a) \ln
   (c-a)-(a+c) \ln (a+c)+a \ln \left(c^2-b^2\right)\right)\right) \ln ^2(a)\\
+3 \left(b c \left(\ln ^2(b)+2 \left(2 \ln (a+c)+\ln
   \left(-\frac{c (b+c)}{(a-b) (a-c) (b-c)}\right)\right) \ln (b)+\ln ^2(b-a)\\
-3 \ln ^2(c)+\ln ^2(c-a)-\ln ^2(c-b)+\ln ^2(b+c)-4 \operatorname{artanh}\left(\frac{a}{c}\right) \ln (a+b)+\\
2 \ln (b-a) \ln \left(\frac{c}{c-a}\right)-6 \ln (c) \ln (a+c)+2 \ln (a+c) \ln (c-b)\\
+4 \ln (c)
   (\ln (c-a)+\ln (c-b))-4 \ln (c) \ln (b+c)-2 \ln (c-a) \ln (b+c)\right)\\
+a \left(b \left(2 \ln ^2(b)-2 \ln
   \left(\frac{a^2-c^2}{b^2-c^2}\right) \ln (b)+\ln ^2(a+b)-\ln ^2(c-a)-\ln ^2(a+c)\\
-3 \ln ^2(b+c)-2 \ln (b-a) \ln (c-b)+2 \ln (c-a) (\ln
   (b-a)+\ln (b+c))\\
+2 \ln (a+b) \ln \left(\frac{b+c}{c^2-a^2}\right)+2 \ln (a+c) \ln \left(c^2-b^2\right)\right)-c \left(\ln ^2(b)-2 (\ln
   ((b-a) c)-\ln (a+c)) \ln (b)+\ln ^2(b-a)+\ln ^2(a+b)\\
-3 \ln ^2(c)+\ln ^2(a+c)+\ln ^2(c-b)+2 \ln ^2(b+c)+2 \ln (b-a) \ln
   \left(\frac{c}{c-b}\right)\\
-2 \ln (a+b) \ln (b+c)-2 \ln (a+c) \ln (c (b+c))\right)\right)\right) \ln (a)\\
-2 (a (b-c)+b c) \ln ^3(b)+a c \ln
   ^3(b-a)-b c \ln ^3(b-a)-3 a c \ln ^3(c)+5 b c \ln ^3(c)\\
+2 a b \ln ^3(c-a)-2 b c \ln ^3(c-a)+2 a c \ln ^3(a+c)-2 b c \ln ^3(a+c)-a b \ln
   ^3(c-b)\\
-b c \ln ^3(c-b)-a b \ln ^3(b+c)-3 b c \ln ^3(b+c)-3 b c \ln (b-a) \ln ^2(c)\\
-3 a b \ln ((a-b) (b-c)) \ln ^2(c-a)+3 b c \ln ((a-b)
   (b-c)) \ln ^2(c-a)\\
+3 b c \ln \left(\left(b^2-a^2\right) c\right) \ln ^2(a+c)-3 a c \ln (a+b) \ln ^2(c-b)+3 a c \ln ((b-a) c) \ln
   ^2(c-b)\\
-3 a b (\ln (b-a)-\ln ((a+b) (a+c))) \ln ^2(c-b)+3 b c \ln (c (a+c)) \ln ^2(c-b)\\
-3 a c \ln \left(\frac{b-a}{a+b}\right) \ln
   ^2(b+c)+6 a c \ln (c) \ln ^2(b+c)-12 b c \ln (c) \ln ^2(b+c)\\
+3 a b \ln (a+c) \ln ^2(b+c)+3 b c \left(\ln \left(b^2-a^2\right)+\ln
   (a+c)\right) \ln ^2(b+c)\\
-3 a c \ln ^2(b-a) \ln \left(1-\frac{b}{c}\right)-b c \pi ^2 \left(\ln (b-a)+\ln
   \left(\frac{b-c}{a-c}\right)\right)-5 a c \pi ^2 \ln (c)-5 b c \pi ^2 \ln (c)\\
-3 b c \ln ^2(b-a) \ln \left(-\frac{c}{b-c}\right)-12 b c \ln
   ^2(c) \ln \left(\frac{c-a}{a+c}\right)-12 b c \ln (a+b) \ln (c) \ln \left(\frac{c-a}{a+c}\right)\\
+\pi ^2 a c (\ln (b-a)-\ln (a+c))+3 a b
   \pi ^2 \ln (a+c)+4 b c \pi ^2 \ln (a+c)-6 b c \ln ^2(c) \ln (c-b)\\
+6 a b \ln (b-a) \ln (c-a) \ln (c-b)-6 b c \ln (b-a) \ln (c-a) \ln
   (c-b)\\
-6 a b \ln (a+b) \ln (a+c) \ln (c-b)-6 b c \ln (a+b) \ln (a+c) \ln (c-b)\\
-12 b c \ln (c) \ln (a+c) \ln (c-b)-a b \pi ^2 (\ln
   (c-a)+\ln (c-b))\\
+3 a c \ln ^2(a+b) \ln \left(\frac{c (c-b)}{b+c}\right)+3 a c \ln ^2(c) (\ln (b-a)-\ln (b+c))+9 b c \ln ^2(c) \ln
   (b+c)\\
-6 a c \ln (a+b) \ln (c) \ln (b+c)+6 b c \ln (a+b) \ln (c) \ln (b+c)+12 b c \ln (c) \ln (c-a) \ln (b+c)\\
+5 \pi ^2 a b \ln (b+c)+6
   a c \pi ^2 \ln (b+c)+3 b c \pi ^2 \ln (b+c)-3 a b \ln ^2(a+c) \ln \left(\frac{b+c}{a+b}\right)\\
+6 a c \ln (b-a) \ln (a+c) \ln (c (b+c))-6
   b c \ln (b-a) \ln (a+c) \ln (c (b+c))\\
-3 a c \ln ^2(a+c) (\ln (b-a)+\ln (c (b+c)))-3 b c \ln ^2(a+b) \ln \left(-\frac{c
   (b+c)}{b-c}\right)\\
-3 \ln ^2(b) \left(a \left(c (\ln (b-a)+\ln (c)-2 \ln (a+c)+\ln (b+c))\\
+b \ln
   \left(\frac{b^2-c^2}{a^2-c^2}\right)\right)-b c \ln \left(\frac{(a-b) c (c-b)}{a^2-c^2}\right)\right)-6 a c \ln (b-a) \ln (c) \ln
   \left(c^2-b^2\right)\\
+6 b c \ln (b-a) \ln (c) \ln \left(c^2-b^2\right)\\
-\ln (b) \left(b c \left(3 \left(-\ln ^2(a+b)+2 (\ln ((a+c)
   (b+c))-\ln (c-a)) \ln (a+b)\\
+2 \ln ^2(c)+\ln ^2(c-a)+\ln ^2(a+c)-3 \ln ^2(b+c)-2 \ln (a+c) \ln (c-b)\\
+2 \ln (c-a) \ln (b+c)+2 \ln (c)
   (\ln ((a+c) (b+c))-2 \ln (c-a))\\
-2 \ln (b-a) \ln \left(c^2-a^2\right)+2 \ln (b-a) \ln \left(c^2-b^2\right)\right)+\pi ^2\right)\\
+a \left(c
   \left(6 (\ln (b-a)+\ln (c)-\ln (a+c)) \ln \left(\frac{a+c}{b+c}\right)+\pi ^2\right)\\
+3 b \left(\ln ^2(a+b)+2 \ln
   \left(\frac{b+c}{c^2-a^2}\right) \ln (a+b)-\ln ^2(c-a)-\ln ^2(a+c)-3 \ln ^2(b+c)\\
+2 \ln (b-a) \ln \left(\frac{a-c}{b-c}\right)+2 \ln (c-a)
   \ln (b+c)+2 \ln (a+c) \ln \left(c^2-b^2\right)+2 \pi ^2\right)\right)\right)\\
+6 \left(2 \left(a \ln \left(\frac{a}{c}\right)+b \ln
   \left(\frac{c}{b}\right)\right) \operatorname{Li}_2\left(\frac{a}{c}\right) c+2 \left(a \ln \left(\frac{a}{c}\right)+b \ln
   \left(\frac{c}{b}\right)\right) \operatorname{Li}_2\left(-\frac{c}{a}\right) c-2 b \ln \left(\frac{b}{c}\right) \operatorname{Li}_2\left(\frac{b (a+c)}{a
   (b-c)}\right) c+2 b \ln \left(\frac{b}{c}\right) \operatorname{Li}_2\left(\frac{b (a-c)}{a (b+c)}\right) c-2 a \operatorname{Li}_3\left(\frac{a}{c}\right) c+2
   b \operatorname{Li}_3\left(-\frac{b}{c}\right) c-2 b \operatorname{Li}_3\left(\frac{b}{c}\right) c+2 a \operatorname{Li}_3\left(-\frac{c}{a}\right) c-(a-b)
   \operatorname{Li}_3\left(\frac{(b-a) c}{b (a+c)}\right) c-(a-b) \operatorname{Li}_3\left(\frac{a (c-b)}{(a-b) c}\right) c+(a+b) \operatorname{Li}_3\left(\frac{a
   (c-b)}{(a+b) c}\right) c-(a-b) \operatorname{Li}_3\left(\frac{b-a}{b+c}\right) c+(a+b) \operatorname{Li}_3\left(\frac{a+b}{b+c}\right) c-(a+b)
   \operatorname{Li}_3\left(\frac{a (b+c)}{(a+b) c}\right) c+(a-3 b) \zeta (3) c+2 a b \ln \left(\frac{a}{b}\right) \operatorname{Li}_2\left(\frac{a}{b}\right)+2
   a b \ln \left(\frac{a}{b}\right) \operatorname{Li}_2\left(-\frac{b}{a}\right)+\left(a b \ln \left(\frac{a}{b}\right)+b c \ln
   \left(\frac{b}{c}\right)+a c \ln \left(\frac{c}{a}\right)\right) \operatorname{Li}_2\left(\frac{b-a}{b-c}\right)+\left(a b \ln
   \left(\frac{a}{b}\right)+b c \ln \left(\frac{b}{c}\right)+a c \ln \left(\frac{c}{a}\right)\right)
   \operatorname{Li}_2\left(\frac{b-c}{a+b}\right)-\left(a b \ln \left(\frac{a}{b}\right)+a c \ln \left(\frac{a}{c}\right)-b c \ln
   \left(\frac{b}{c}\right)\right) \operatorname{Li}_2\left(\frac{a+b}{b+c}\right)-\left(a b \ln \left(\frac{a}{b}\right)+a c \ln
   \left(\frac{a}{c}\right)-b c \ln \left(\frac{b}{c}\right)\right) \operatorname{Li}_2\left(\frac{a+c}{b+c}\right)-2 a b
   \operatorname{Li}_3\left(\frac{a}{b}\right)+2 a b \operatorname{Li}_3\left(-\frac{b}{a}\right)-b (a-c) \operatorname{Li}_3\left(\frac{a-b}{a-c}\right)-a (b-c)
   \operatorname{Li}_3\left(\frac{b-a}{b-c}\right)+a (b-c) \operatorname{Li}_3\left(\frac{a+b}{b-c}\right)-b (a-c) \operatorname{Li}_3\left(\frac{a (b-c)}{b
   (a-c)}\right)+b (a+c) \operatorname{Li}_3\left(\frac{a+b}{a+c}\right)+b (a+c) \operatorname{Li}_3\left(\frac{b (a+c)}{a (b-c)}\right)+b (a-c)
   \operatorname{Li}_3\left(\frac{b (a-c)}{a (b+c)}\right)-b (a-c) \operatorname{Li}_3\left(\frac{c-a}{b+c}\right)+b (a+c)
   \operatorname{Li}_3\left(\frac{a+c}{b+c}\right)-a (b+c) \operatorname{Li}_3\left(\frac{a (b+c)}{b (a+c)}\right)\right)\right)$$
Here is the equivalent Mathematica expression. The formula can be proved using differentiation by parameters $a,b,c.$ 
In fact, the integrand even has a closed-form antiderivative in terms of elementary functions, dilogarithms and trilogarithms, but it is too large to put it here.