Can C++ enable_if have a default implementation?

I want to write a function print which behaves differently according to the type of its argument.

Here is my implementation:

template <typename T, typename std::enable_if<std::is_array<T>::value, int>::type = 0>
void print(const T &v) {
  std::cout << "array: ";
  for (const auto &e : v) {
    std::cout << e << ", ";
  }
  std::cout << std::endl;
}

template <typename T, typename std::enable_if<std::is_integral<T>::value, int>::type = 0>
void print(const T &v) {
  std::cout << "integral: " << v << std::endl;
}

template <typename T, typename std::enable_if<!(std::is_array<T>::value || std::is_integral<T>::value), int>::type = 0>
void print(const T &v) {
  std::cout << "default: " << v << std::endl;
}

This code works as expected, but the conditions in the last specification are too complicated.

Is there any solution to simplify the last one?