$\frac{d}{dx}\left(\ln\left(\sec\left(x^2+x^3\right)+\tan\left(x\right)\right)\right)=\frac{\left(2x+3x^{2}\right)\sec\left(x^2+x^3\right)\tan\left(x^2+x^3\right)+\sec\left(x\right)^2}{\sec\left(x^2+x^3\right)+\tan\left(x\right)}$
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