Looks Are Very Deceptive (Fixed)

$\begin{aligned} \int_0^\frac \pi 2 \sqrt[5]{\tan x} \ dx & = \int_0^\frac \pi 2 \sin^\frac 15 x \cos^{-\frac 15} x \ dx & \small \blue{\text{Beta function B}\left(m,n\right) = 2\int_0^\frac \pi 2 \sin^{2m-1}x \cos^{2n-1}x\ dx} \\ & = \frac {\text B \left(\frac 35, \frac 25 \right)}2 & \small \blue{\text B(m,n) = \frac {\Gamma(m)\Gamma(n)}{\Gamma(m+n)}, \Gamma(\cdot) \text{ denotes gamma function.}} \\ & = \frac {\Gamma \left(\frac 35 \right)\Gamma \left(\frac 25 \right)}{2 \blue{\Gamma (1)}} & \small \blue{\Gamma(n) = (n-1)!} \\ & = \frac {\blue{\Gamma \left(1-\frac 25 \right)}\Gamma \left(\frac 25 \right)}2 & \small \blue{\Gamma(1-z) = -z\Gamma (-z)} \\ & = - \frac {\blue{\Gamma \left(-\frac 25 \right) \Gamma \left(\frac 25 \right)}}5 & \small \blue{\Gamma(z)\Gamma (-z)= - \frac \pi{z\sin z\pi}} \\ & = \frac \pi {2\sin \frac {2\pi}5} \end{aligned}$

Therefore $(a+b+c)^2 = (2+2+5)^2 = \boxed{81}$ .

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References :

• Beta function
• Gamma function