Integrand size = 21, antiderivative size = 210 \[ \int (e x)^m \tan ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(e x)^{1+m} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^{-p} \left (\frac {i \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{1+e^{2 i a d} \left (c x^n\right )^{2 i b d}}\right )^p \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \operatorname {AppellF1}\left (-\frac {i (1+m)}{2 b d n},-p,p,1-\frac {i (1+m)}{2 b d n},e^{2 i a d} \left (c x^n\right )^{2 i b d},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{e (1+m)} \]
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Time = 0.23 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {4593, 4591, 1986, 525, 524} \[ \int (e x)^m \tan ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(e x)^{m+1} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^{-p} \left (\frac {i \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{1+e^{2 i a d} \left (c x^n\right )^{2 i b d}}\right )^p \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \operatorname {AppellF1}\left (-\frac {i (m+1)}{2 b d n},-p,p,1-\frac {i (m+1)}{2 b d n},e^{2 i a d} \left (c x^n\right )^{2 i b d},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{e (m+1)} \]
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Rule 524
Rule 525
Rule 1986
Rule 4591
Rule 4593
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}} \tan ^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}}\right ) \text {Subst}\left (\int x^{-1+\frac {1+m}{n}} \left (\frac {i-i e^{2 i a d} x^{2 i b d}}{1+e^{2 i a d} x^{2 i b d}}\right )^p \, dx,x,c x^n\right )}{e n} \\ & = \frac {\left ((e x)^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}} \left (i-i e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^{-p} \left (\frac {i-i e^{2 i a d} \left (c x^n\right )^{2 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\right ) \text {Subst}\left (\int x^{-1+\frac {1+m}{n}} \left (i-i e^{2 i a d} x^{2 i b d}\right )^p \left (1+e^{2 i a d} x^{2 i b d}\right )^{-p} \, dx,x,c x^n\right )}{e n} \\ & = \frac {\left ((e x)^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}} \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^{-p} \left (\frac {i-i e^{2 i a d} \left (c x^n\right )^{2 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\right ) \text {Subst}\left (\int x^{-1+\frac {1+m}{n}} \left (1-e^{2 i a d} x^{2 i b d}\right )^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} \left (\frac {i \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{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 {AppellF1}\left (-\frac {i (1+m)}{2 b d n},-p,p,1-\frac {i (1+m)}{2 b d n},e^{2 i a d} \left (c x^n\right )^{2 i b d},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{e (1+m)} \\ \end{align*}
Time = 1.15 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.98 \[ \int (e x)^m \tan ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {x (e x)^m \left (1-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^{-p} \left (-\frac {i \left (-1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{1+e^{2 i a d} \left (c x^n\right )^{2 i b d}}\right )^p \left (1+e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )^p \operatorname {AppellF1}\left (-\frac {i (1+m)}{2 b d n},-p,p,1-\frac {i (1+m)}{2 b d n},e^{2 i a d} \left (c x^n\right )^{2 i b d},-e^{2 i a d} \left (c x^n\right )^{2 i b d}\right )}{1+m} \]
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\[\int \left (e x \right )^{m} {\tan \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}^{p}d x\]
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\[ \int (e x)^m \tan ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \tan \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{p} \,d x } \]
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\[ \int (e x)^m \tan ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int \left (e x\right )^{m} \tan ^{p}{\left (a d + b d \log {\left (c x^{n} \right )} \right )}\, dx \]
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\[ \int (e x)^m \tan ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \tan \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{p} \,d x } \]
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Timed out. \[ \int (e x)^m \tan ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {Timed out} \]
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Timed out. \[ \int (e x)^m \tan ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int {\mathrm {tan}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}^p\,{\left (e\,x\right )}^m \,d x \]
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