L'Hopital is a curse here!

Let the integral $\displaystyle I = \int f(x) \ dx$ . Then:

$\begin{aligned} f(x) & = \frac {(x-\sin x)^\frac 32}{\sqrt x}\left(\frac {6x^2\sin^2 \frac x2}{x-\sin x} + 3x\right) \\ & = \frac {(x-\sin x)^\frac 32}{\sqrt x}\left(\frac {\frac {6x^2}2(1-\cos x)}{x-\sin x} + 3x\right) \\ & = \frac {3x(x-\sin x)^\frac 32}{\sqrt x}\left(\frac {x(1-\cos x)}{x-\sin x} + 1\right) \\ & = 3\sqrt x(x-\sin x)^\frac 32 \left(\frac {x-x\cos x + x-\sin x}{x-\sin x}\right) \\ & = 3 x^{\color{#D61F06}\frac 12} (x-\sin x)^{\color{#D61F06}\frac 12} \left(2x-x\cos x -\sin x\right) & ...(1) \end{aligned}$

We note that:

$\begin{aligned} f(x) & = \frac d{dx} \left(ax^b(x-\sin x)^c + C\right) \\ & = abx^{b-1}(x-\sin x)^c + acx^b(x-\sin x)^{c-1}(1-\cos x) \\ & = ax^{b-1}(x-\sin x)^{c-1}\left(b(x-\sin x) + cx(1-\cos x)\right) \\ & = ax^{\color{#D61F06}b-1}(x-\sin x)^{\color{#D61F06}c-1}\left((b+c)x-cx\cos x - b\sin x\right) & \small \color{#D61F06} \text{Comparing with (1): } b = c = \frac 32 \\ & = ax^{\color{#D61F06}\frac 12}(x-\sin x)^{\color{#D61F06}\frac 12}\left({\color{#D61F06}3}x-{\color{#D61F06}\frac 32}x\cos x - {\color{#D61F06}\frac 32}\sin x\right) \\ & = {\color{#D61F06}\frac {3a}2}x^\frac 12(x-\sin x)^\frac 12 \left(2x-x\cos x -\sin x\right) & \small \color{#D61F06} \implies a = 2 \end{aligned}$

$\implies a+2b+4c = 2 + 2\cdot \dfrac 32 + 4 \cdot \dfrac 32 = \boxed{11}$

Continue from what @Chew-Seong Cheong has done in the previous solution, we have $\begin{aligned} \int f(x)\ dx&=\int 3x^{\frac{1}{2}}(x-\sin x)^{\frac{1}{2}}(x-x\cos x+x-\sin x)\ dx\\ &={\color{#3D99F6}\int 3x^{\frac{3}{2}}(x-\sin x)^{\frac{1}{2}}(1-\cos x)\ dx}+{\color{magenta}\int 3x^{\frac{1}{2}}(x-\sin x)^{\frac{3}{2}}\ dx}.\\ \end{aligned}$ We shall put more focus on the term highlighted in blue by applying integration by parts on it. We let $\begin{aligned} u'=(x-\sin x)^{\frac{1}{2}}(1-\cos x)&\Rightarrow u=\frac{2}{3}(x-\sin x)^{\frac{3}{2}}\ \text{(See note)}\\ v=x^{\frac{3}{2}}&\Rightarrow v'=\frac{3}{2}x^{\frac{1}{2}}. \end{aligned}$ Thus, we have $\begin{aligned} \int 3x^{\frac{3}{2}}(x-\sin x)^{\frac{1}{2}}(1-\cos x)\ dx=2x^{\frac{3}{2}}(x-\sin x)^{\frac{3}{2}}-{\color{magenta}\int 3x^{\frac{1}{2}}(x-\sin x)^{\frac{3}{2}}\ dx}. \end{aligned}$ Observe that the terms highlighted in magenta cancels each other. Therefore, the integral evaluates to $\int f(x)\ dx=2x^{\frac{3}{2}}(x-\sin x)^{\frac{3}{2}}+C.$

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Note: $\begin{aligned} \int (x-\sin x)^{\frac{1}{2}}(1-\cos x)\ dx&=\int t^{\frac{1}{2}}\ dt &({\color{#3D99F6} t=x-\sin x\Rightarrow dt=(1-\cos x)\ dx})\\ &=\frac{2}{3}t^{\frac{3}{2}}\ &\text{(Arbitrary constants omitted)}\\ &=\frac{2}{3}(x-\sin x)^{\frac{3}{2}} \end{aligned}$