# New integral with some parameters

I wrote in Mathematica but it is just running without the result?!

FullSimplify[Integrate[1/(Sqrt[(a^2 + b)*x^2 - c] + a *x),
{x, 0, r}, Assumptions -> {a > 0, b > 0, c > 0}]]

• Try to provoke it into using a different method Integrate[1/(Sqrt[(a^2+b)*x^2-c]+a*x), x, Assumptions->{a>0,b>0,c>0}] and not take the time to do FullSimplify Does that help?
– Bill
Nov 4, 2021 at 15:47
• First, you need to consider it is a*x not ax, Second the solution is very long I would like to make it more simple as much as possible I prefer everything there is Arctanh[] Could you see this simplification? Nov 4, 2021 at 15:52
• Integrate[1/(Sqrt[(a^2 + b)*x^2 - c] + a*x), x, GenerateConditions -> False] results in -(1/(2 b))(2 a ArcTanh[(a x)/Sqrt[-c + (a^2 + b) x^2]] + a Log[c - b x^2] - 2 Sqrt[a^2 + b] Log[a^2 x + b x + Sqrt[a^2 + b] Sqrt[-c + (a^2 + b) x^2]]). Next, think of the fundamental theorem of calculus. Nov 4, 2021 at 16:02
• My mistake, my "a x" pasted as "ax" in the comment and I fixed the comment. Next, you were getting no solution at all. I only offered the possibility that you get any answer. Next, I cannot say, but with c>0 it looks possible that the square root could go negative. That might be part of the reason for the complexity of the solution.. Next, are there any singularities between 0 and r that might make the definite integral much harder? Maybe. Don't know. I understand that everyone wants the answer in exactly the form they want and want it NOW. I sometimes want any answer that I can depend on.
– Bill
Nov 4, 2021 at 16:03
• Integrate[1/(Sqrt[(a^2 + b)*x^2 - c] + a*x), x, GenerateConditions -> True] also produces -(1/(2 b))(2 a ArcTanh[(a x)/Sqrt[-c + (a^2 + b) x^2]] + a Log[c - b x^2] - 2 Sqrt[a^2 + b] Log[a^2 x + b x + Sqrt[a^2 + b] Sqrt[-c + (a^2 + b) x^2]]). Nov 4, 2021 at 16:23

Integrate[1/(Sqrt[(a^2 + b)*x^2 - c] + a*x), {x, 0, r},  GenerateConditions -> True]

ConditionalExpression[ 1/(2 b) (-2 a ArcTanh[(a r)/Sqrt[-c + (a^2 + b) r^2]] - 2 Sqrt[a^2 + b] Log[Sqrt[a^2 + b] Sqrt[-c]] + a Log[c] - a Log[c - b r^2] + 2 Sqrt[a^2 + b] Log[a^2 r + b r + Sqrt[a^2 + b] Sqrt[-c + (a^2 + b) r^2]]), (Sqrt[c]/( Sqrt[b] r) != 0 || Sqrt[c]/(Sqrt[b] r) \[NotElement] Reals || Re[Sqrt[c]/(Sqrt[b] r)] < -1) && (Sqrt[c]/( Sqrt[b] r) \[NotElement] Reals || Re[Sqrt[c]/(Sqrt[b] r)] >= 0 || Re[Sqrt[c]/(Sqrt[b] r)] <= -1) && (Sqrt[c]/( Sqrt[b] r) \[NotElement] Reals || Re[Sqrt[c]/(Sqrt[b] r)] < -1 || Re[Sqrt[c]/(Sqrt[b] r)] >= 0) && ((Sqrt[ Im[c]])/(\[Sqrt](-2 Im[a]^2 Im[r] Re[r] + 2 Im[r] (Re[a]^2 + Re[b]) Re[r] + Im[b] (-Im[r]^2 + Re[r]^2) + 2 Im[a] Re[a] (-Im[r]^2 + Re[r]^2))) \[NotElement] Reals || (Im[ c] (Im[r]^2 (Re[a]^2 + Re[b]) + 2 Im[b] Im[r] Re[r] + 4 Im[a] Im[r] Re[a] Re[r] - (Re[a]^2 + Re[b]) Re[r]^2 + Im[a]^2 (-Im[r]^2 + Re[r]^2)))/(2 Im[a]^2 Im[r] Re[r] - 2 Im[r] (Re[a]^2 + Re[b]) Re[r] + Im[b] (Im[r]^2 - Re[r]^2) + 2 Im[a] Re[a] (Im[r]^2 - Re[r]^2)) >= Re[c] || Re[(Sqrt[ Im[c]])/(\[Sqrt](-2 Im[a]^2 Im[r] Re[r] + 2 Im[r] (Re[a]^2 + Re[b]) Re[r] + Im[b] (-Im[r]^2 + Re[r]^2) + 2 Im[a] Re[a] (-Im[r]^2 + Re[r]^2)))] >= 0 || Re[(Sqrt[ Im[c]])/(\[Sqrt](-2 Im[a]^2 Im[r] Re[r] + 2 Im[r] (Re[a]^2 + Re[b]) Re[r] + Im[b] (-Im[r]^2 + Re[r]^2) + 2 Im[a] Re[a] (-Im[r]^2 + Re[r]^2)))] <= -1)]
• Integrate[1/(Sqrt[(a^2 + b)*x^2 - c] + a*x), {x, 0, r}, Assumptions->{a,b,c}>0, GenerateConditions -> True] returns the input. Nov 4, 2021 at 17:00
• While it would serve as a convenient shorthand, I don't think you can use {a,b,c} > 0 in place of {a > 0, b > 0, c > 0}. I'm not sure how Mathematica interprets {a, b, c} > 0 in a conditions statement, except that it does treat it differently from the standard syntax. Compare, for instance, Simplify[Sqrt[a^2]*Sqrt[b^2]*Sqrt[c^2], Assumptions -> {a > 0, b > 0, c > 0}] (⇒ a b c) with Simplify[Sqrt[a^2]*Sqrt[b^2]*Sqrt[c^2], Assumptions -> {a, b, c} > 0] (⇒ Abs[a] Abs[b] Abs[c]). Nov 5, 2021 at 8:02