Integrand size = 15, antiderivative size = 109 \[ \int \frac {1}{\cos ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {2 x \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{4} \left (5-\frac {2 i}{b n}\right ),\frac {1}{4} \left (9-\frac {2 i}{b n}\right ),-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(2+5 i b n) \cos ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \]
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Time = 0.08 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4572, 4580, 371} \[ \int \frac {1}{\cos ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {2 x \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{4} \left (5-\frac {2 i}{b n}\right ),\frac {1}{4} \left (9-\frac {2 i}{b n}\right ),-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(2+5 i b n) \cos ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \]
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Rule 371
Rule 4572
Rule 4580
Rubi steps \begin{align*} \text {integral}& = \frac {\left (x \left (c x^n\right )^{-1/n}\right ) \text {Subst}\left (\int \frac {x^{-1+\frac {1}{n}}}{\cos ^{\frac {5}{2}}(a+b \log (x))} \, dx,x,c x^n\right )}{n} \\ & = \frac {\left (x \left (c x^n\right )^{-\frac {5 i b}{2}-\frac {1}{n}} \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2}\right ) \text {Subst}\left (\int \frac {x^{-1+\frac {5 i b}{2}+\frac {1}{n}}}{\left (1+e^{2 i a} x^{2 i b}\right )^{5/2}} \, dx,x,c x^n\right )}{n \cos ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \\ & = \frac {2 x \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{4} \left (5-\frac {2 i}{b n}\right ),\frac {1}{4} \left (9-\frac {2 i}{b n}\right ),-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(2+5 i b n) \cos ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \\ \end{align*}
Time = 2.01 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.72 \[ \int \frac {1}{\cos ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {2 x \left (\frac {(2-i b n) \sqrt {2+2 e^{2 i a} \left (c x^n\right )^{2 i b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4}-\frac {i}{2 b n},\frac {5}{4}-\frac {i}{2 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )}{\sqrt {e^{-i a} \left (c x^n\right )^{-i b}+e^{i a} \left (c x^n\right )^{i b}}}+\frac {-2 \cos \left (a+b \log \left (c x^n\right )\right )+b n \sin \left (a+b \log \left (c x^n\right )\right )}{\cos ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}\right )}{3 b^2 n^2} \]
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\[\int \frac {1}{{\cos \left (a +b \ln \left (c \,x^{n}\right )\right )}^{\frac {5}{2}}}d x\]
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Exception generated. \[ \int \frac {1}{\cos ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int \frac {1}{\cos ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {1}{\cos ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\int { \frac {1}{\cos \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {1}{\cos ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {1}{\cos ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\int \frac {1}{{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}^{5/2}} \,d x \]
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