A Headache for sure!

Defining the function $F(a,\nu) \; =\; \int_0^\pi \sin^{\nu-1}x \sin ax\,dx \; =\; \frac{\pi \sin\frac12\pi a}{2^{\nu-1} \nu B\big(\tfrac{\nu+a+1}{2},\frac{\nu-a+1}{2}\big)}$ we see that the desired integral is $I \; = \; \int_0^\pi \frac{x \cos\frac13x}{\sqrt[3]{\sin x}}\,dx \; = \; \frac{\partial F}{\partial a}(\tfrac13,\tfrac23)\;.$ If we write $\begin{array}{rcl} G(a) \; = \; F(a,\tfrac23) & = & \displaystyle \frac{\pi \sin \frac12\pi a}{2^{-\frac13} \frac23 B(\frac56 + \frac12a,\frac56 - \frac12a)} \; = \; \frac{\pi 2^{\frac13}\sin \frac12\pi a \Gamma(\frac53)}{\frac23 \Gamma(\frac56+\frac12a)\Gamma(\frac56-\frac12a)} \\ & = & \displaystyle \frac{\pi 2^{\frac13}\sin\frac12\pi a \Gamma(\frac23)}{\Gamma(\frac56 + \frac12a)\Gamma(\frac56 - \frac12a)} \end{array}$ then $G'(a) \; =\; \frac{\pi 2^{\frac13}\Gamma(\frac23)}{\Gamma(\frac56 + \frac12a)\Gamma(\frac56 - \frac12a)} \left[ \begin{array}{c} \tfrac12\pi \cos\tfrac12\pi a - \tfrac12\sin \tfrac12\pi a \,\psi(\tfrac56 + \tfrac12a) \\ + \tfrac12\sin\tfrac12\pi a \,\psi(\tfrac56 - \tfrac12a) \end{array}\right]$ and so the integral is $\begin{array}{rcl} G'(\tfrac13) & = & \displaystyle \frac{\pi \sqrt[3]{2} \Gamma(\tfrac23)}{4\Gamma(1)\Gamma(\tfrac23)}\big[ \pi \sqrt{3} - \psi(1) + \psi(\tfrac23)\big] \\ & = & \displaystyle \frac{\pi \sqrt[3]{2}}{4}\big[\pi \sqrt{3} + \gamma - \gamma + \tfrac{1}{2\sqrt{3}}\pi - \tfrac32 \ln3\big] \\ & = & \displaystyle \frac{\pi \sqrt[3]{2}}{4}\big[\tfrac{7}{2\sqrt{3}}\pi - \tfrac32\ln3\big] \\ & = & \displaystyle \frac{\pi \sqrt[3]{2}}{24}\big[7\pi\sqrt{3} - 9\ln3\big] \end{array}$ This gives $a=3$ , $b=2$ , $c=24$ , $d=7$ and $e=9$ , making the answer $\boxed{45}$ .