Inverse DTFT of left-sided sequence
There seems to be a mistake in your derivations, the following code shows the correct one for $|\alpha| &lt; 1$ ,  as indicated by @Matt.L $: 

N = 128;         % simulation length (choose large enough)
n = -N : 2;      % simulation time span

C = 1.5;          % set some values
D = 2.78;
a = 0.33;         % a is for alpha

% x = -(n-2).*C.*(a.^(-n+1)).*stepfun(-n+2,0) + ...
%    -(n-1).*D.*(a.^(-n)).*stepfun(-n+1,0);  

x = zeros(1,130);  % use the explicit code as stepfun() has problem with n.
for n = -N:1
    x(n+N+1) = -C*(n-2)*(a^(-n+1)) - D*(n-1)*(a^(-n));
end    

M=2000;          % Select a FFT length for demonstration purposes
F=fft(x,M);      % compute the DFT of x[n] so obtained

w = linspace(0,2*pi-2*pi/M,M);              % choose DFT frequency samples
H = (C*exp(-1j*w)+D)./((1-a*exp(1j*w)).^2); % Evaluate the given DTFT

figure,plot(w,abs(H));title('H'); % compare the magnitudes of DTFTs
figure,plot(w,abs(F));title('F'); % Phases will be different as this
                                  % is an acausal sequence.


Derivation of the result for the left-sided sequence  $|\alpha|&lt; 1$ case) is as follows:

Intuition plays a central for getting the answer as clean as possible which involves inverse Fourier transforms. In this example we shall use those three fundamental Fourier transform properties.

1- $x[n] \leftrightarrow X(\omega) \longrightarrow y[n]=x[-n] \leftrightarrow Y(\omega) = X(-\omega) $

2- $x[n] \leftrightarrow X(\omega) \longrightarrow y[n]=nx[n] \leftrightarrow Y(\omega) = j \frac{d X(\omega)}{d\omega} $

3- $x[n] \leftrightarrow X(\omega) \longrightarrow y[n]=x[n-d] \leftrightarrow Y(\omega) = e^{-j \omega d} X(\omega)$

Begin by $$x[n] = a^n u[n]  \longleftrightarrow X(\omega) = \frac{1}{1 - a e^{-j \omega} }$$

Let $y[n]=x[-n]=a^{-n} u[-n]$ then $Y(\omega)= \frac{1}{1-a e^{j \omega}}$

Apply second rule:
$z[n] = n y[n] = n a^{-n} u[-n]$ then $Z(\omega) = \frac{-a e^{j \omega}}{(1-a e^{j \omega})^2}$

Apply linearity:
$  w[n] =  -a^{-1} z[n] = - n a^{-1} a^{-n} u[-n]$ then $ W(\omega) = \frac{e^{j\omega}}{(1-a e^{j \omega})^2}$

Apply shift property on $w[n]$ to get rid of the numerator exponential.
$v[n] = w[n-1] = - (n-1) a^{-1} a^{-(n-1)} u[-(n-1)]$ then $ V(\omega) =\frac{1}{(1-a e^{j \omega})^2}$

Now apply shift property and linearity to $v[n]$ such that the resulting DTFT will be $$H(\omega) = \frac{ C \cdot e^{-j \omega} + D} {(1-a e^{j \omega})^2} = \frac{ C \cdot e^{-j \omega}} {(1-a e^{j \omega})^2} + \frac{ D} {(1-a e^{j \omega})^2}$$

which implies that 
$$h[n] =  C v[n-1] + D v[n] $$ hence we get $h[n]$ as:

$$h[n] = C \left( - ((n-1)-1) a^{-1} a^{-((n-1)-1)} u[-((n-1)-1)] \right) - D \left( (n-1) a^{-1} a^{-(n-1)} u[-(n-1)] \right) $$

simplify to get the result:
$$ h[n] = C (-n+2) a^{-n+1} u[-n+2] + D (-n+1) a^{-n} u[-n+1] $$