Integrand size = 17, antiderivative size = 87 \[ \int x^2 \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {4 e^{2 i a} x^3 \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (2-\frac {3 i}{b n}\right ),\frac {1}{2} \left (4-\frac {3 i}{b n}\right ),-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{3+2 i b n} \]
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Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {4605, 4601, 371} \[ \int x^2 \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {4 e^{2 i a} x^3 \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (2-\frac {3 i}{b n}\right ),\frac {1}{2} \left (4-\frac {3 i}{b n}\right ),-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{3+2 i b n} \]
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Rule 371
Rule 4601
Rule 4605
Rubi steps \begin{align*} \text {integral}& = \frac {\left (x^3 \left (c x^n\right )^{-3/n}\right ) \text {Subst}\left (\int x^{-1+\frac {3}{n}} \sec ^2(a+b \log (x)) \, dx,x,c x^n\right )}{n} \\ & = \frac {\left (4 e^{2 i a} x^3 \left (c x^n\right )^{-3/n}\right ) \text {Subst}\left (\int \frac {x^{-1+2 i b+\frac {3}{n}}}{\left (1+e^{2 i a} x^{2 i b}\right )^2} \, dx,x,c x^n\right )}{n} \\ & = \frac {4 e^{2 i a} x^3 \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (2-\frac {3 i}{b n}\right ),\frac {1}{2} \left (4-\frac {3 i}{b n}\right ),-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{3+2 i b n} \\ \end{align*}
Time = 4.22 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.84 \[ \int x^2 \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^3 \left (3 e^{2 i a} \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (1,1-\frac {3 i}{2 b n},2-\frac {3 i}{2 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )+(-3 i+2 b n) \left (-i \operatorname {Hypergeometric2F1}\left (1,-\frac {3 i}{2 b n},1-\frac {3 i}{2 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )+\tan \left (a+b \log \left (c x^n\right )\right )\right )\right )}{b n (-3 i+2 b n)} \]
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\[\int x^{2} {\sec \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}d x\]
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\[ \int x^2 \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x^{2} \sec \left (b \log \left (c x^{n}\right ) + a\right )^{2} \,d x } \]
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\[ \int x^2 \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^{2} \sec ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}\, dx \]
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\[ \int x^2 \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x^{2} \sec \left (b \log \left (c x^{n}\right ) + a\right )^{2} \,d x } \]
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\[ \int x^2 \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x^{2} \sec \left (b \log \left (c x^{n}\right ) + a\right )^{2} \,d x } \]
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Timed out. \[ \int x^2 \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int \frac {x^2}{{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}^2} \,d x \]
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