Time complexity calculation
Suppose that $n = 2^k$, and that the logarithm is base 2. Then
$$
T(2^k) = \log (2^k) + 2 \log(2^{k-1}) + 4 \log(2^{k-2}) + \cdots + 2^{k-1} \log 2 + 2^k T(1) \\ = k + 2(k-1) + 4(k-2) + \cdots + 2^{k-1}(k-(k-1)) + 2^k T(1) \\ =
(1+2+4+\cdots + 2^{k-1})k - (2^1 \cdot 1 + 2^2 \cdot 2 + \cdots + 2^{k-1}(k-1)) + 2^kT(1).
$$
The first summand is equal to $(2^k-1)k$. The second is equal to
$$
(2^1 + \cdots + 2^{k-1}) + (2^2 + \cdots + 2^{k-1}) + \cdots + (2^{k-1}) = \\
(2^k-2^1) + (2^k-2^2) + \cdots + (2^k-2^{k-1}) = \\
(k-1)2^k - (2^1 + 2^2 + \cdots + 2^{k-1}) = \\
(k-1)2^k - (2^k-2) = (k-2)2^k + 2.
$$
In total, we get
$$
T(2^k) = (k2^k - k) - (k2^k - 2^{k+1} + 2) + 2^k T(1) = 2^{k+1} - k - 2 + 2^k T(1).
$$
In terms of $n$, this is
$$
T(n) = (2 + T(1))n - \log n - 2.
$$