Integrand size = 21, antiderivative size = 139 \[ \int (e x)^m \csc ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(e x)^{1+m} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \csc ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (p,-\frac {i+i m-b d n p}{2 b d n},\frac {1}{2} \left (2-\frac {i (1+m)}{b d n}+p\right ),e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{e (1+m+i b d n p)} \]
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Time = 0.13 (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 = {4606, 4604, 371} \[ \int (e x)^m \csc ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(e x)^{m+1} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \operatorname {Hypergeometric2F1}\left (p,\frac {1}{2} \left (p-\frac {i (m+1)}{b d n}\right ),\frac {1}{2} \left (-\frac {i (m+1)}{b d n}+p+2\right ),e^{2 i a d} \left (c x^n\right )^{2 i b d}\right ) \csc ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e (i b d n p+m+1)} \]
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Rule 371
Rule 4604
Rule 4606
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}} \csc ^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 \csc ^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 \csc ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \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 )}{e (1+m+i b d n p)} \\ \end{align*}
Time = 1.28 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.22 \[ \int (e x)^m \csc ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {x (e x)^m \left (2-2 e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \left (\frac {i e^{i a d} \left (c x^n\right )^{i b d}}{-1+e^{2 i a d} \left (c x^n\right )^{2 i b d}}\right )^p \operatorname {Hypergeometric2F1}\left (p,-\frac {i (1+m+i b d n p)}{2 b d n},\frac {1}{2} \left (2-\frac {i (1+m)}{b d n}+p\right ),e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{1+m+i b d n p} \]
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\[\int \left (e x \right )^{m} {\csc \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}^{p}d x\]
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\[ \int (e x)^m \csc ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \csc \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{p} \,d x } \]
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\[ \int (e x)^m \csc ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int \left (e x\right )^{m} \csc ^{p}{\left (a d + b d \log {\left (c x^{n} \right )} \right )}\, dx \]
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\[ \int (e x)^m \csc ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \csc \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{p} \,d x } \]
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\[ \int (e x)^m \csc ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \csc \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{p} \,d x } \]
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Timed out. \[ \int (e x)^m \csc ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int {\left (e\,x\right )}^m\,{\left (\frac {1}{\sin \left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}\right )}^p \,d x \]
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