To expand on my comment: in Gradshteyn and Ryzhik (the seventh edition, at least), they list formula 2.174, which I think is more practical for computational purposes than the direct output of Mathematica. Translated into Mathematica syntax for the OP's specific case, if we have
int[n_] := Integrate[t^n/(t^2 + b t + 1), t]
then there is the useful (inhomogeneous) difference equation
int[n] == t^(n - 1)/(n - 1) - b int[n - 1] - int[n - 2]
This can in fact be fed into either of RSolve[]
or DifferenceRoot[]
to yield a useful solution if desired; RSolve[{int[n] == t^(n - 1)/(n - 1) - b int[n - 1] - int[n - 2], int[0] == (2 ArcTan[(b + 2 t)/Sqrt[4 - b^2]])/Sqrt[4 - b^2], int[1] == -((b ArcTan[(b + 2 t)/Sqrt[4 - b^2]])/Sqrt[4 - b^2]) + 1/2 Log[1 + b t + t^2]}, int[n], n]
will in fact yield a (complicated!) expression involving ${}_2 F_1$ that works for nonnegative integer n
, and DifferenceRoot[Function[{int, n}, {int[n] == t^(n - 1)/(n - 1) - b int[n - 1] - int[n - 2], int[0] == (2 ArcTan[(b + 2 t)/Sqrt[4 - b^2]])/Sqrt[4 - b^2], int[1] == -((b ArcTan[(b + 2 t)/Sqrt[4 - b^2]])/Sqrt[4 - b^2]) + 1/2 Log[1 + b t + t^2]}]]
shows that int[n]
satisfies a four-term homogeneous difference equation, and is already directly useful besides.
int/. n->-1
. This suggests that it is a maths problem to me? $\endgroup$Plot3D[int /. {b -> 1}, {t, 0, 3}, {n, -4, 4}]
reveals weird things going on whenn>-1
$\endgroup$Integrate[x^a, x]
givesx^(a+1)/(a+1)
which clearly isn't valid fora == -1
. $\endgroup$2F1
. Any other suggestions? $\endgroup$SetDelayed
:int[n_, b_, t_] := Integrate[t^n/(t^2 + b*t + 1), t] //FullSimplify
thenint[#, b, t] & /@ Range[-2, 2]
(changeRange
to fit your needs). $\endgroup$