Integrand size = 21, antiderivative size = 139 \[ \int (e x)^m \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(e x)^{1+m} \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \operatorname {Hypergeometric2F1}\left (p,-\frac {i+i m-b d n p}{2 b d n},\frac {1}{2} \left (2-\frac {i (1+m)}{b d n}+p\right ),-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right ) \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e (1+m+i b d n p)} \]
[Out]
Time = 0.15 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4605, 4603, 371} \[ \int (e x)^m \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(e x)^{m+1} \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \operatorname {Hypergeometric2F1}\left (p,\frac {1}{2} \left (p-\frac {i (m+1)}{b d n}\right ),\frac {1}{2} \left (-\frac {i (m+1)}{b d n}+p+2\right ),-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right ) \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e (i b d n p+m+1)} \]
[In]
[Out]
Rule 371
Rule 4603
Rule 4605
Rubi steps \begin{align*} \text {integral}& = \frac {\left ((e x)^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}}\right ) \text {Subst}\left (\int x^{-1+\frac {1+m}{n}} \sec ^p(d (a+b \log (x))) \, dx,x,c x^n\right )}{e n} \\ & = \frac {\left ((e x)^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}-i b d p} \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right )\right ) \text {Subst}\left (\int x^{-1+\frac {1+m}{n}+i b d p} \left (1+e^{2 i a d} x^{2 i b d}\right )^{-p} \, dx,x,c x^n\right )}{e n} \\ & = \frac {(e x)^{1+m} \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \operatorname {Hypergeometric2F1}\left (p,\frac {1}{2} \left (-\frac {i (1+m)}{b d n}+p\right ),\frac {1}{2} \left (2-\frac {i (1+m)}{b d n}+p\right ),-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right ) \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e (1+m+i b d n p)} \\ \end{align*}
Time = 1.18 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.22 \[ \int (e x)^m \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {2^p x (e x)^m \left (\frac {e^{i a d} \left (c x^n\right )^{i b d}}{1+e^{2 i a d} \left (c x^n\right )^{2 i b d}}\right )^p \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \operatorname {Hypergeometric2F1}\left (p,-\frac {i (1+m+i b d n p)}{2 b d n},\frac {1}{2} \left (2-\frac {i (1+m)}{b d n}+p\right ),-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{1+m+i b d n p} \]
[In]
[Out]
\[\int \left (e x \right )^{m} {\sec \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}^{p}d x\]
[In]
[Out]
\[ \int (e x)^m \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \sec \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{p} \,d x } \]
[In]
[Out]
\[ \int (e x)^m \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int \left (e x\right )^{m} \sec ^{p}{\left (a d + b d \log {\left (c x^{n} \right )} \right )}\, dx \]
[In]
[Out]
\[ \int (e x)^m \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \sec \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{p} \,d x } \]
[In]
[Out]
\[ \int (e x)^m \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \sec \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{p} \,d x } \]
[In]
[Out]
Timed out. \[ \int (e x)^m \sec ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int {\left (e\,x\right )}^m\,{\left (\frac {1}{\cos \left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}\right )}^p \,d x \]
[In]
[Out]