Complexity of multiplying two integers of size $n$ and $m$
The multiplication of two integers of size $n$ and size $m$ in base-2 with $n\le m$ can be done in time 
$$O(m\cdot\log n\cdot\log\log n)$$ 
where the product will be given in base-2 as well.

Here is a sketch of the simple idea to prove the above claim. 



Let positive integer $a$ has $n$ digits and positive integer $b$ have $m$ digits in base 2. For simplicity, let $m=kn$ for some integer $k$. 
$$b=\sum_{i=0}^{k-1}\sum_{j=0}^{n-1}b_{i,j}2^{in+j}=\sum_{i=0}^{k-1}2^{in}b_i$$ 
where $0\le b_{i,j}\le1$, $b_i=\sum_{j=0}^{n-1}b_{i,j}2^j$. In other words, $b_{k-1}\cdots b_1b_0$ is $b$ in base-$2^n$.
So $$ab= \sum_{i=0}^{k-1}2^{in}(ab_i).$$

Since both $a_i$ and $b_i$ have $n$ digits, the product $ab_i$ can be computed in $O(n\cdot\log n\cdot\log\log n)$ (using FFT method). Multiplying by factor $k=\dfrac mn$, we see that all summands in the above equality and hence $ab$ can be computed in $O(m\cdot\log n\cdot\log\log n)$. Note that the computation cost of various bookkeeping and additions is not substantial enough to invalidate that bound. 



This idea of splitting the bits of the bigger number by the length of the bits of the smaller number is articulated here first in Yuval's comment. That idea is, apparently, folklore among users experienced in algorithms. 

If we have a smaller bound for the multiplication of the integers of the same size, we will get a smaller bound for the multiplication of integers of different sizes accordingly. For example, an algorithm by Harvey and van der Hoeven gives bound $O(n\log n\cdot 2^{2\log ^{*}n})$. So we will have bound $O(m\log n\cdot 2^{2\log ^{*}n})$ accordingly.