Unclear result from Leibniz Integral Rule

Question

I was trying to answer this question: $\frac{\mathrm d}{\mathrm dt}\int_{t-\mathrm d1}^t \int_h^t f(s) \,\mathrm ds\,\mathrm dh$. I got the accepted answer by hand (using the Leibniz integral rule) but why does Mathematica (and Maple) give the seemingly wrong $\mathrm d_{{1}}f \left( t \right) \color{orange}{-}\int_{t-\mathrm d_{{1}}}^{t}\!f \left( s \right) \,{\rm d}s=\mathrm d_{{1}}f \left( t \right) \color{orange}{-}F \left( t \right) \color{orange}{+} F \left( t-\mathrm d _{{1}} \right)$ instead of $\mathrm d_{{1}}f \left( t \right) \color{lime}{+}\int_{t-\mathrm d_{{1}}}^{t}\!f \left( s \right) \,{\rm d}s=\mathrm d_{{1}}f \left( t \right) \color{lime}{+}F \left( t \right) \color{lime}{-} F \left( t-\mathrm d _{{1}} \right)$

The Mathematica code I used is D[Integrate[Integrate[f[s], {s, h, t}], {h, t - d1, t}], t].

Answer 1

Let the inner integral result be $F(t)-F(h)$ where $F$ is the antiderivative of $f$. Then break $\int_{t-d1}^{t}F(t)-F(h)\mathrm{dh}$ into $\int_{t-d1}^{t}F(t)\mathrm{dh}$ and $-\int_{t-d1}^{t} F(h)\mathrm{dh}$.

The first integral is just $hF(t)\Big|_{t-d1}^{t}=t F(t)-(t-d1)F(t)=d1 F(t)$.

The second integral is $-\int_{t-d1}^{t} F(h)\mathrm{dh}=G(t-d1)-G(t)$ where $G$ is the antiderivative of $F$.

Differentiate with respect to $t$ and we get: $$ \frac{d}{dt}\left[d1 F(t) +G(t-d1)-G(t)\right]=d1 F'(t)+G'(t-d1)-G'(t) \\ =d1f(t)+\left[F(t-d1)-F(t)\right]=d1f(t)-\left[F(t)-F(t-d1)\right] $$ I added the brackets for emphasis. Notice how $\left[F(t)-F(t-d1)\right]$ is just $\int_{t-d1}^{t}f(h)\mathrm{dh}$ we set up at the start. Therefore without needing the Leibniz rule at all Mathematica is correct in saying: D[Integrate[Integrate[f[s], {s, h, t}], {h, t - d1, t}], t] == d1*f[t] - Integrate[f[s], {s, -d1 + t, t}]