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I am trying to evaluate the integral

$$ \int_0^{2\pi}\frac{D_{11}^2+2D_{12}^2+D_{22}^2+D_{33}^2}{1+e\cos x(t)}dx $$

Where $D_{ij}$ are third derivatives wrt $t$ as defined below.

a[psi_] := k1/(1 + e*Cos[psi]);
d[psi_] := p1/(1 + e*Cos[psi]);
D11 = D[a[x[t]]*(Cos[x[t]]^2 - 1/3), {t, 3}];
D12 = D[a[x[t]]*Cos[x[t]]*Sin[x[t]], {t, 3}];
D22 = D[a[x[t]]*(Sin[x[t]]^2 - 1/3), {t, 3}];
D33 = D[-a[x[t]]/3];

To do this I have written the input

Assuming[Element[k1, Reals] && Element[e, Reals] && 
x'[t] == p2/(d[x[t]]^2) && Element[p2, Reals] && 
Element[p2, Reals], 
Integrate[(D11^2 + 2*D12^2 + D22^2 + D33^2)/(1 + e*Cos[x[t]])^2, {x[
t], 0, 2*Pi}]]

This is taking a very long time to evaluate, is it OK to specify to Mathematica that I have the condition $\dot{x}=\frac{p_2}{d(x(t))^2}$ in this manner? How could this be sped up?

EDIT: I have re-written this without subscripts.

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    $\begingroup$ To warn you at the outset: using subscripts for anything other than pretty formatting will give you nothing but inconvenience; avoid them altogether. $\endgroup$ Commented Oct 11, 2018 at 16:00
  • $\begingroup$ @J.M.issomewhatokay. in the sense that it is a pain to type, certainly. It won't do anything to my calculation though, right? $\endgroup$
    – tbfr416
    Commented Oct 11, 2018 at 16:12
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    $\begingroup$ Subscripts will harm your calculation because they appear to the human as a variable or symbol but to the kernel as a formatting construct. Just use D11 and D23 instead. $\endgroup$
    – bill s
    Commented Oct 11, 2018 at 16:20

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