How to do algebra on unevaluated integrals?

Question

I am working with functions calculated from a set of general basis functions.

f = a[x] + c1*b[x] + c2*c[x];
g = Expand[f*f];
h = Integrate[g, {x, -Infinity, Infinity}];

This much works fine, and I get a nice expression with products of combinations of a[x], b[x], and c[x] in the integrand.  But now I need to calculate the values of c1 and c2 that optimize h and the optimal value of h.  This fails:

cs = {c1, c2};
s = Solve[{D[h, #] & /@ cs == 0}, cs]
(* Solve::nsmet: This system cannot be solved with the methods available to Solve. >> *)

Is there any hack to make this work?

EVEN better would be if I could get the answer in tidy bracket notation, where for example  denotes the definite integral of a[x]*b[x].

Michael E2 · Accepted Answer

Similar idea to belisarius, except in V10 we can inactivate Integrate to keep it from evaluating or even trying to evaluate:
h = Inactive[Integrate][g, {x, -Infinity, Infinity}]

It is not necessary in this example, as belisarius' answer shows, but one of its intended uses is to do algebra/calculus on integrals and derivatives.  Inactive can be removed easily with
Activate[h]

The function linearExpand expands its argument according to linearity properties.  Factors/terms that do not depend on x are treated as constants (see update below for a more general approach).
Clear[linearExpand];
linearExpand[e_] := e //. {int : Inactive[Integrate][_Plus, _] :> Distribute[int], 
    Inactive[Integrate][integrand_Times, dom : {x_, _, _} | x_] :> 
     With[{dependencies = Internal`DependsOnQ[#, x] & /@ List @@ integrand},
      Pick[integrand, dependencies, False] *
       Inactive[Integrate][Pick[integrand, dependencies, True], dom]
      ]};

OP's sample problem:
Solve[D[h, #] == 0 & /@ cs // linearExpand, cs]

D[h, #] == 0 & /@ cs // linearExpand

For what it's worth...
...here's a general linearity expander.  Considers factors that do not depend on x, which may be a list of symbols, as constants.
linearExpand[e_, x_, head_] := 
  e //. {op : head[arg_Plus, __] :> Distribute[op], 
    head[arg1_Times, rest__] :> 
     With[{dependencies = Internal`DependsOnQ[#, x] & /@ List @@ arg1},
      Pick[arg1, dependencies, False] head[
        Pick[arg1, dependencies, True], rest]
      ]};

Examples:
linearExpand[D[h, #] == 0 & /@ cs, x, Inactive[Integrate]]
(* same as above *)

linearExpand[foo[(a[x] + c b[y]) (2 a[x] - c b[y]) // Expand, randomarg], x, foo]
(*  -c^2 b[y]^2 foo[1, randomarg] +
     c b[y] foo[a[x], randomarg] + 
     2 foo[a[x]^2, randomarg]        *)

linearExpand[foo[(a[x] + c b[y]) (2 a[x] - c b[y]) // Expand, randomarg], {x, y}, foo]
(*  2 foo[a[x]^2, randomarg] +
    c foo[a[x] b[y], randomarg] - 
    c^2 foo[b[y]^2, randomarg]     *)

Dr. belisarius · Answer

f = a[x] + c1*b[x] + c2*c[x];
g = Expand[f*f];

(* we need to get the constants out of the integrals first*)

h = Distribute@Integrate[g, {x, -∞, ∞}] //. Integrate[q1___  r__  q2___, {v_, s__}] /; 
                                        FreeQ[{r}, v] :> r Integrate[q1 q2, {v, s}];

s= Solve[And @@ Thread[D[h, #] & /@ {c1, c2} == 0], {c1, c2}]

(* now we go to bra-ket notation *)

bkRulez = {Integrate[a_ [x] b_[x], {x, -∞, ∞}]     ->   AngleBracket[a, b], 
           Integrate[Power[a_ [x], 2], {x, -∞, ∞}] ->   AngleBracket[a, a]}
Column @@ (s /. bkRulez) // TeXForm

$$begin{array}{l}
 text{c1}to -frac{langle a,crangle 
   langle b,crangle -langle
   c,crangle  langle a,brangle
   }{langle b,crangle ^2-langle
   b,brangle  langle c,crangle } 
 text{c2}to -frac{langle b,brangle 
   langle a,crangle -langle
   a,brangle  langle b,crangle
   }{langle b,brangle  langle
   c,crangle -langle b,crangle ^2} 
end{array}$$

Jerry Guern · Answer

I played with the commands suggested above and found that Distributed[] solved the first half of the problem but only for indefinite integrals:

f = a[x] + c1*b[x] + c2*c[x];
g = Expand[f*f];
h = Distribute[Integrate[g, x]];
cs = {c1, c2};
Solve[{D[h, #] & /@ cs == 0}, cs]

I worry when my solution looks too simple...  Are the more complicated answers above handling some problem I'm not aware of?

How to do algebra on unevaluated integrals?

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