Why is integration so much harder than differentiation?

Here is an extremely generic answer. Differentiation is a "local" operation: to compute the derivative of a function at a point you only have to know how it behaves in a neighborhood of that point. But integration is a "global" operation: to compute the definite integral of a function in an interval you have to know how it behaves on the entire interval (and to compute the indefinite integral you have to know how it behaves on all intervals). That is a lot of information to summarize. Generally, local things are much easier than global things.

On the other hand, if you can do the global things, they tend to be useful because of how much information goes into them. That's why theorems like the fundamental theorem of calculus, the full form of Stokes' theorem, and the main theorems of complex analysis are so powerful: they let us calculate global things in terms of slightly less global things.

The family of functions you generally consider (e.g., elementary functions) is closed under differentiation, that is, the derivative of such function is still in the family. However, the family is not in general closed under integration. For instance, even the family of rational functions is not closed under integration because you $\int 1/x = \log$.

I guess the OP asks about the symbolic integration. Other answers already dealt with the numeric case where integration is easy and differentiation is hard.

If you recall the definition of the differentiation, you can see it's just a subtraction and division by a constant. Even if you can't do any algebraic changes, it won't get any more complex than that. But usually you can do many simplifications due to the zero limit, as many terms fall out as being too small. From this definition it can be shown that if you know the derivative of $f(x)$ and $g(x)$, then you can use these derivatives to express the derivative of $f(x) \pm g(x)$, $f(x)g(x)$ and $f(g(x))$. This makes symbolic differentiation easy as you just need to apply the rules recursively.

Now about integration. Integration is basically an infinite sum of small quantities. So if you see an $\int f(x) \, dx$. You can imagine it as an infinite sum of $(f_1 + f_2 + ...) \, dx$ where $f_i$ are consecutive values of the function.

This means if you need to calculate integral of $\int (a f(x) + b g(x)) \,d x$. Then you can imagine the sum $((af_1 + bg_1) + (af_2 + bg_2) + ...) \,d x$. Using the associativity and distributivity, you can transform this into: $a(f_1 + f_2 +...)\,d x + b(g_1 + g_2 + ...)\,d x$. So this means $\int (a f(x) + b g(x)) \, d x = a \int f(x) \,d x + b \int g(x) \, dx$.

But if you have $\int f(x) g(x) \, d x$, you have the sum $(f_1 g_1 + f_2 g_2 + ...) \,d x$. From which you cannot factor out the sum of $f$s and $g$s. This means there is no recursive rule for multiplication.

Same goes for $\int f(g(x)) \,d x$. You cannot extract anything from the sum $(f(g_1) + f(g_2) + ...) \,d x$ in general.

So far, only linearity is the useful property. What about the analogues of the Differentiation rules? We have the product rule: $$\frac{d f(x)g(x) }{\, d x} = f(x) \frac{d g(x)}{\, d x} + g(x) \frac{d f(x)}{\, d x}.$$ Integrating both sides and rearranging the terms, we get the well-known integral by parts formula:

$$\int f(x) \frac{d g(x)}{\, d x} \, d x = f(x)g(x) - \int g(x) \frac{d f(x)}{\, d x} \, d x.$$

But this formula is only useful if $\frac{d f(x)}{dx} \int g(x) \, d x$ or $\frac{d g(x)}{dx} \int f(x) \, d x$ is easier to integrate than $f(x)g(x)$.

And it's often hard to see when this rule is useful. For example, when you try to integrate $\mathrm{ln}(x)$, it's not obvious to see that it's $1 \mathrm{ln}(x)$. The integral of $1$ is $x$ and the derivative of $\mathrm{ln}(x)$ is $\frac{1}{x}$, which lead to a very simple integral of $x\frac{1}{x} = 1$, whose integral is again $x$.

Another well-known differential rule is the chain rule $$\frac{d f(g(x))}{\, d x} = \frac{d f(g(x))}{d g(x)} \frac{d g(x)}{\, d x}.$$

Integrating both sides, you get the reverse chain rule:

$$f(g(x)) = \int \frac{d f(g(x))}{d g(x)} \frac{d g(x)}{\, d x} \, d x.$$

But again it's hard to see when it is useful. For example what about the integration of $\frac{x}{\sqrt{x^2 + c}}$? Is it obvious to you that $\frac{x}{\sqrt{x^2 + c}} = 2x \frac{1}{2\sqrt{x^2 + c}}$ and this is the derivative of $\sqrt{x^2 + c}$? I guess not, unless someone showed you the trick.

For differentiation, you can mechanically apply the rules. For integration, you need to recognize patterns and even need to introduce cancellations to bring the expression into the desired form and this requires lot of practice and intuition.

For example how would you integrate $\sqrt{x^2 + 1}$?

First you turn it into a fraction:

$$\frac{x^2 + 1}{\sqrt{x^2+1}}$$

Then multiply and divide by 2:

$$\frac{2x^2 + 2}{2\sqrt{x^2+1}}$$

Separate the terms like this:

$$\frac{1}{2}\left(\frac{1}{\sqrt{x^2+1}}+\frac{x^2+1}{\sqrt{x^2+1}}+\frac{x^2}{\sqrt{x^2+1}} \right)$$

Play with 2nd and 3rd term:

$$\frac{1}{2} \left( \frac{1}{\sqrt{x^2+1}}+ 1\sqrt{x^2+1}+ x2x\frac{1}{2\sqrt{x^2+1}} \right)$$

Now you can see the first bracketed term is the derivative of $\mathrm{arsinh(x)}$. The second and third term is the derivative of the $x\sqrt{x^2+1}$. Thus the integral will be:

$$\frac{\mathrm{arsinh}(x)}{2} + \frac{x\sqrt{x^2+1}}{2} + C$$

Were these transformations obvious to you? Probably not. That's why differentiation is just a mechanic while integration is an art.