A list of Hopeful quartets for Int(x^a*(1-x)^b/(1+c*x)^(d+1),x=0..1)/Int(x^a*(1-x)^b/(1+c*x)^(d),x=0..1) By Shalosh B. Ekhad Definition: For rational a,b,c,d, let C(a,b,c,d)=Int(x^a*(1-x)^b/(1+c*x)^(d+1),x=0..1)/Int(x^a*(1-x)^b/(1+c*x)^d,x\ =0..1) There are , 1, inequivalent classes with denominator , 1 Here there are C(0, 0, 1, 0) = arcsinh(3/4) There are , 4, inequivalent classes with denominator , 2 Here there are 1/2 2 C(0, 0, 1, 1/2) = ---- 2 65 C(0, 0, 1/2, 0) = arcsinh(--) 72 1/2 6 C(0, 0, 1/2, 1/2) = ---- 3 C(1/2, 1/2, 1/2, 1/2) There are , 11, inequivalent classes with denominator , 3 Here there are 1/3 2/3 2 2 2 C(0, 0, 1, 1/3) = ------ + ---- - 2/3 3 3 3367 C(0, 0, 1/3, 0) = arcsinh(----) 3456 2/3 1/3 2 6 3 6 C(0, 0, 1/3, 1/3) = ------ + ------ - 6/7 7 7 49 C(0, 0, 2/3, 0) = arctanh(--) 76 1/3 2/3 45 45 C(0, 0, 2/3, 1/3) = ----- + ----- - 3/4 4 20 C(0, 1/3, 1, 2/3) C(0, 1/3, 1/3, 2/3) C(0, 2/3, 1, 1/3) C(0, 2/3, 1/3, 1/3) C(1/3, 1/3, 1/3, 1/3) C(2/3, 2/3, 1/3, 2/3) There are , 21, inequivalent classes with denominator , 4 Here there are 4 3 2 C(0, 0, 1, 1/4) = 3 RootOf(14 _Z + 24 _Z + 12 _Z - 1, index = 1) 4 3 2 C(0, 0, 1/2, 1/4) = RootOf(19 _Z + 120 _Z + 216 _Z - 216, index = 1) 325089 C(0, 0, 1/4, 0) = arcsinh(------) 320000 1/2 2 5 C(0, 0, 1/4, 1/2) = ------ 5 4 3 2 C(0, 0, 1/4, 1/4) = 6 RootOf(305 _Z + 360 _Z + 120 _Z - 4, index = 1) 1/2 2 7 C(0, 0, 3/4, 1/2) = ------ 7 C(0, 1/2, 1, 1/4) C(0, 1/2, 1, 3/4) C(0, 1/2, 1/2, 1/4) C(0, 1/2, 1/2, 3/4) C(0, 1/2, 1/4, 0) C(0, 1/2, 1/4, 1/4) C(0, 1/2, 1/4, 3/4) C(0, 1/4, 1/4, 1/2) C(0, 3/4, 1/4, 1/2) C(1/2, 1/2, 1/4, 1/2) C(1/2, 1/2, 1/4, 1/4) 1/2 2 C(1/4, 1/4, 1, 3/4) = ---- 2 1/2 6 C(1/4, 1/4, 1/2, 3/4) = ---- 3 C(1/4, 1/4, 1/4, 1/4) C(3/4, 3/4, 1/4, 3/4) There are , 33, inequivalent classes with denominator , 5 Here there are 5 4 3 2 C(0, 0, 1, 1/5) = 2 RootOf(15 _Z + 70 _Z + 120 _Z + 80 _Z - 16, index = 1) 5 4 3 C(0, 0, 1, 2/5) = 3/2 RootOf(28 _Z + 60 _Z + 40 _Z - 10 _Z - 3, index = 1) 50700551 C(0, 0, 1/5, 0) = arcsinh(--------) 48600000 C(0, 0, 1/5, 1/5) = 5 4 3 2 2 RootOf(2013 _Z + 13650 _Z + 33000 _Z + 30000 _Z - 10000, index = 1) C(0, 0, 1/5, 2/5) = 5 4 3 1/2 RootOf(364 _Z + 3300 _Z + 9000 _Z - 33750 _Z - 37125, index = 1) 6841 C(0, 0, 2/5, 0) = arctanh(----) 9966 C(0, 0, 2/5, 1/5) = 5 4 3 2 RootOf(1554 _Z + 19075 _Z + 84000 _Z + 140000 _Z - 160000, index = 1) C(0, 0, 2/5, 2/5) = 5 4 3 3 RootOf(85456 _Z + 117600 _Z + 49000 _Z - 4375 _Z - 750, index = 1) 1/3 2/3 16 5 5 5 C(0, 0, 3/5, 0) = arcsinh(------- - ------) 25 64 C(0, 1/5, 1/5, 0) C(0, 1/5, 1/5, 3/5) C(0, 2/5, 1/5, 0) C(0, 2/5, 1/5, 1/5) C(0, 3/5, 1/5, 0) C(0, 3/5, 1/5, 4/5) C(0, 4/5, 1/5, 0) C(0, 4/5, 1/5, 2/5) C(1/5, 1/5, 1, 3/5) C(1/5, 1/5, 1/5, 1/5) C(1/5, 1/5, 1/5, 3/5) C(1/5, 1/5, 2/5, 3/5) C(1/5, 2/5, 1, 4/5) C(1/5, 2/5, 1/5, 4/5) C(1/5, 2/5, 2/5, 4/5) C(1/5, 3/5, 1, 2/5) C(1/5, 3/5, 1/5, 2/5) C(1/5, 3/5, 2/5, 2/5) C(2/5, 2/5, 1/5, 2/5) C(2/5, 4/5, 1, 3/5) C(2/5, 4/5, 1/5, 3/5) C(2/5, 4/5, 2/5, 3/5) C(3/5, 3/5, 1/5, 3/5) C(4/5, 4/5, 1/5, 4/5) To sum up, here are the identified 4-tuples [[C(0,0,1,0), arcsinh(3/4)], [C(0,0,1,1/2), 1/2*2^(1/2)], [C(0,0,1/2,0), arcsinh(65/72)], [C(0,0,1/2,1/2), 1/3*6^(1/2)], [C(0,0,1,1/3), 2/3*2^(1/3)+1/3* 2^(2/3)-2/3], [C(0,0,1/3,0), arcsinh(3367/3456)], [C(0,0,1/3,1/3), 2/7*6^(2/3)+ 3/7*6^(1/3)-6/7], [C(0,0,2/3,0), arctanh(49/76)], [C(0,0,2/3,1/3), 1/4*45^(1/3) +1/20*45^(2/3)-3/4], [C(0,0,1,1/4), 3*RootOf(14*_Z^4+24*_Z^3+12*_Z^2-1,index = 1)], [C(0,0,1/2,1/4), RootOf(19*_Z^4+120*_Z^3+216*_Z^2-216,index = 1)], [C(0,0, 1/4,0), arcsinh(325089/320000)], [C(0,0,1/4,1/2), 2/5*5^(1/2)], [C(0,0,1/4,1/4) , 6*RootOf(305*_Z^4+360*_Z^3+120*_Z^2-4,index = 1)], [C(0,0,3/4,1/2), 2/7*7^(1/ 2)], [C(1/4,1/4,1,3/4), 1/2*2^(1/2)], [C(1/4,1/4,1/2,3/4), 1/3*6^(1/2)], [C(0,0 ,1,1/5), 2*RootOf(15*_Z^5+70*_Z^4+120*_Z^3+80*_Z^2-16,index = 1)], [C(0,0,1,2/5 ), 3/2*RootOf(28*_Z^5+60*_Z^4+40*_Z^3-10*_Z-3,index = 1)], [C(0,0,1/5,0), arcsinh(50700551/48600000)], [C(0,0,1/5,1/5), 2*RootOf(2013*_Z^5+13650*_Z^4+ 33000*_Z^3+30000*_Z^2-10000,index = 1)], [C(0,0,1/5,2/5), 1/2*RootOf(364*_Z^5+ 3300*_Z^4+9000*_Z^3-33750*_Z-37125,index = 1)], [C(0,0,2/5,0), arctanh(6841/ 9966)], [C(0,0,2/5,1/5), RootOf(1554*_Z^5+19075*_Z^4+84000*_Z^3+140000*_Z^2-\ 160000,index = 1)], [C(0,0,2/5,2/5), 3*RootOf(85456*_Z^5+117600*_Z^4+49000*_Z^3 -4375*_Z-750,index = 1)], [C(0,0,3/5,0), arcsinh(16/25*5^(1/3)-5/64*5^(2/3))]] here are the unidentified 4-tuples, that are candidates to first-ever irrat\ ionality proof [C(1/2,1/2,1/2,1/2), C(0,1/3,1,2/3), C(0,1/3,1/3,2/3), C(0,2/3,1,1/3), C(0,2/3, 1/3,1/3), C(1/3,1/3,1/3,1/3), C(2/3,2/3,1/3,2/3), C(0,1/2,1,1/4), C(0,1/2,1,3/4 ), C(0,1/2,1/2,1/4), C(0,1/2,1/2,3/4), C(0,1/2,1/4,0), C(0,1/2,1/4,1/4), C(0,1/ 2,1/4,3/4), C(0,1/4,1/4,1/2), C(0,3/4,1/4,1/2), C(1/2,1/2,1/4,1/2), C(1/2,1/2,1 /4,1/4), C(1/4,1/4,1/4,1/4), C(3/4,3/4,1/4,3/4), C(0,1/5,1/5,0), C(0,1/5,1/5,3/ 5), C(0,2/5,1/5,0), C(0,2/5,1/5,1/5), C(0,3/5,1/5,0), C(0,3/5,1/5,4/5), C(0,4/5 ,1/5,0), C(0,4/5,1/5,2/5), C(1/5,1/5,1,3/5), C(1/5,1/5,1/5,1/5), C(1/5,1/5,1/5, 3/5), C(1/5,1/5,2/5,3/5), C(1/5,2/5,1,4/5), C(1/5,2/5,1/5,4/5), C(1/5,2/5,2/5,4 /5), C(1/5,3/5,1,2/5), C(1/5,3/5,1/5,2/5), C(1/5,3/5,2/5,2/5), C(2/5,2/5,1/5,2/ 5), C(2/5,4/5,1,3/5), C(2/5,4/5,1/5,3/5), C(2/5,4/5,2/5,3/5), C(3/5,3/5,1/5,3/5 ), C(4/5,4/5,1/5,4/5)] --------------------------------- This took, 76.936, seconds .