We shall prove a more general result. Let $\varphi\in C^{(1)}[0,\infty) \land \big(|\varphi(t)|<1,\,\forall t\in[0,\infty)\big)$. Let $f(t,x) = \varphi\big(\frac{t^3}3+xt\big)$. We will show the differentiation under the integral works for all $x\ge 0$. $I(x)=\int_0^\infty f(t,x)dt$. We need to show
$$\lim_{\delta\to 0}\int_0^\infty \frac{f(t,x+\delta)-f(t,x)}{\delta}dt = \int_0^\infty \lim_{\delta\to 0}\frac{f(t,x+\delta)-f(t,x)}{\delta}dt. \tag1$$
Define
\begin{align}
I_1(x) &:= \int_0^1\varphi\big(\frac{t^3}3+xt\big)\,dt \\
I_2(x) &:= \int_1^\infty\varphi\big(\frac{t^3}3+xt\big)\,dt
\end{align}
Consider $I_2$ first.
Let $u:=\frac{t^3}3+xt$. Surely $u>0$.
$$u<\frac{(t+x)^3}3\Longleftrightarrow (3u)^\frac13-x<t,\,\forall (t>1\land x\ge0). \tag2$$
Now $(u,x)$ is the independent variable, while $t$ becomes a function of $(u,x)$ and is denoted as $t(u,x)$.
$$1 = (t^2+x)\frac{\partial t}{\partial u}$$
$$0=t^2\frac{\partial t}{\partial x}+t+x\frac{\partial t}{\partial x}\Longleftrightarrow \frac{\partial t}{\partial u}=-\frac{t}{t^2+x}$$
$$\frac{\partial}{\partial x}\frac1{t(u,x)^2+x}=-\frac{2t\frac{\partial t}{\partial x}+1}{(t^2+x)^2}=\frac1{(t^2+x)^2}\frac{t^2-x}{t^2+x},$$
$$I_2(x)=\int_{\frac13+x}^\infty \frac{\varphi(u)}{t^2+x}du,$$
and
$$\frac{I_2(x+\delta)-I_2(x)}\delta=J_{2,1}-J_{2,2}$$
where
\begin{align}
J_{2,1}&:=\int_{\frac13+x}^\infty\varphi(u)g(u,x,\delta)du \\
J_{2,2}&:= \frac1\delta\int_{\frac13+x}^{\frac13+x+\delta}\frac{\varphi(u)}{t(u,x+\delta)^2+x+\delta}\,du,
\end{align}
where
\begin{align}
g(u,x,\delta) &:= \frac1\delta\Big(\frac1{t(u,x+\delta)^2+x+\delta}-\frac1{t(u,x)+x}\Big) \\
&=\frac1{(t(u,x+\theta\delta)^2+x+\theta\delta)^2}\frac{t(u,x+\theta\delta)^2-(x+\theta\delta)}{t(u,x+\theta\delta)^2+(x+\theta\delta)}
\end{align}
for some function $\theta:=\theta(u,x,\delta)\in[0,1]$ with the second equation coming from Taylor expansion to the first order.
By inequality (2), for some positive function $C$, $\forall u>C(x)$,
$$t(u,x)^2+x>((3u)^\frac13-x)^2+x>u^\frac23,$$
$$0<\frac{t(u,x)^2-x}{t(u,x)^2+x}\le 1$$
$\forall u>C(u+\Delta),\ \forall \delta<\Delta$, for some $\Delta>0$,
$$0<g(u,x,\delta)<\frac1{u^\frac43}$$
So
$$\int_{C(x+\Delta)}^\infty g(u,x,\delta)\varphi(u)du<\int_{C(x+\Delta)}^\infty\frac1{u^\frac43}du<\infty$$
For $g(u,x,\delta)$ is continuous for $(u,\delta)$ on a compact set $\big[\frac13+x,C(x+\Delta)\big]\times[0,\Delta]$ and thus is bounded for any given $x>0$.
Hence we can apply the Lebesgue's dominated convergence theorem to the integral $J_{2,1}$ as $\delta\to0$.
It is easy to see $J_{2,2}\to\frac{\varphi(\frac13+x)}{1+x}$ as $\delta\to0$ since the integrand is continuous on a compact set.
Now we consider
$$\frac{I_1(x+\delta)-I_1(x)}{\delta}
=\int_0^1\frac1\delta\Big(\varphi\Big(\frac{t^3}3+x+\delta\Big)-\varphi\Big(\frac{t^3}3+x\Big)\Big)\, dt
=\int_0^1\varphi'\Big(\frac{t^3}3+x+\theta_1\delta\Big)\, dt
$$
for some function $\theta_1:=\theta_1(\frac{t^3}3+x,\delta)\in[0,1]$.
As $\varphi\in C^{(1)}$ on a compact set, it is uniformly bounded with respect to $(t,\delta)$ for any given $x$. We can thus apply again Lebesgue's dominated convergence theorem to the above integral.
Summing up equation (1) is proved.
