Four variable boolean algebra simplification

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I'm having trouble simplifying this expression further strictly with Boolean identities. However, using algorithmic simplifiers I've found that it's possible to simplify it further. Would anyone be able to help? I'm looking for the minimal form of the boolean function. Thanks in advance. (+ is OR, multiplication is AND and ! is NOT)

F = (!A!B!C) + (!A(B)!D) + (A(!B)D) + (!A(C)D) + (AB!C)

Actual simplest form: (!A!B!D) + (AB!C) + (!B(D)) + (!A(BC))