Can we show that $$ \int_{0}^{1}K^\prime(x)^5\text{d}x =\int_{0}^{1}(74x^2-68x+14)K(x)^5\text{d}x? $$ Where $K^\prime(x)=K\left(\sqrt{1-x^2}\right)$ and $K(x)$ is known as a complete elliptic integral of the first kind with the elliptic modulus $x$.
I have checked it numercally. The following equalities seem to be true as well. $$\begin{aligned} &\int_{0}^{1}xK^\prime(x)^5\text{d}x =\int_{0}^{1}(-56x^2+61x-8)K(x)^5\text{d}x\\ &\int_{0}^{1}xK(x)^2K^\prime(x)^3\text{d}x =\int_{0}^{1}\left(-\frac{61}{10}x^2+7x-\frac{5}{6}\right)K(x)^5\text{d}x\\ &\int_{0}^{1}x^3K(x)^4K^\prime(x)\text{d}x =\int_{0}^{1}\left(-\frac{58}{15}x^2+\frac{27}{5}x-\frac{6}{5}\right)K(x)^5\text{d}x\\ \end{aligned} $$