Integrand size = 15, antiderivative size = 162 \[ \int (e x)^m \tan ^p(a+b \log (x)) \, dx=\frac {(e x)^{1+m} \left (1-e^{2 i a} x^{2 i b}\right )^{-p} \left (\frac {i \left (1-e^{2 i a} x^{2 i b}\right )}{1+e^{2 i a} x^{2 i b}}\right )^p \left (1+e^{2 i a} x^{2 i b}\right )^p \operatorname {AppellF1}\left (-\frac {i (1+m)}{2 b},-p,p,1-\frac {i (1+m)}{2 b},e^{2 i a} x^{2 i b},-e^{2 i a} x^{2 i b}\right )}{e (1+m)} \]
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Time = 0.14 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {4591, 1986, 525, 524} \[ \int (e x)^m \tan ^p(a+b \log (x)) \, dx=\frac {(e x)^{m+1} \left (1-e^{2 i a} x^{2 i b}\right )^{-p} \left (\frac {i \left (1-e^{2 i a} x^{2 i b}\right )}{1+e^{2 i a} x^{2 i b}}\right )^p \left (1+e^{2 i a} x^{2 i b}\right )^p \operatorname {AppellF1}\left (-\frac {i (m+1)}{2 b},-p,p,1-\frac {i (m+1)}{2 b},e^{2 i a} x^{2 i b},-e^{2 i a} x^{2 i b}\right )}{e (m+1)} \]
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Rule 524
Rule 525
Rule 1986
Rule 4591
Rubi steps \begin{align*} \text {integral}& = \int (e x)^m \left (\frac {i-i e^{2 i a} x^{2 i b}}{1+e^{2 i a} x^{2 i b}}\right )^p \, dx \\ & = \left (\left (i-i e^{2 i a} x^{2 i b}\right )^{-p} \left (\frac {i-i e^{2 i a} x^{2 i b}}{1+e^{2 i a} x^{2 i b}}\right )^p \left (1+e^{2 i a} x^{2 i b}\right )^p\right ) \int (e x)^m \left (i-i e^{2 i a} x^{2 i b}\right )^p \left (1+e^{2 i a} x^{2 i b}\right )^{-p} \, dx \\ & = \left (\left (1-e^{2 i a} x^{2 i b}\right )^{-p} \left (\frac {i-i e^{2 i a} x^{2 i b}}{1+e^{2 i a} x^{2 i b}}\right )^p \left (1+e^{2 i a} x^{2 i b}\right )^p\right ) \int (e x)^m \left (1-e^{2 i a} x^{2 i b}\right )^p \left (1+e^{2 i a} x^{2 i b}\right )^{-p} \, dx \\ & = \frac {(e x)^{1+m} \left (1-e^{2 i a} x^{2 i b}\right )^{-p} \left (\frac {i \left (1-e^{2 i a} x^{2 i b}\right )}{1+e^{2 i a} x^{2 i b}}\right )^p \left (1+e^{2 i a} x^{2 i b}\right )^p \operatorname {AppellF1}\left (-\frac {i (1+m)}{2 b},-p,p,1-\frac {i (1+m)}{2 b},e^{2 i a} x^{2 i b},-e^{2 i a} x^{2 i b}\right )}{e (1+m)} \\ \end{align*}
Time = 0.70 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.97 \[ \int (e x)^m \tan ^p(a+b \log (x)) \, dx=\frac {x (e x)^m \left (1-e^{2 i a} x^{2 i b}\right )^{-p} \left (-\frac {i \left (-1+e^{2 i a} x^{2 i b}\right )}{1+e^{2 i a} x^{2 i b}}\right )^p \left (1+e^{2 i a} x^{2 i b}\right )^p \operatorname {AppellF1}\left (-\frac {i (1+m)}{2 b},-p,p,1-\frac {i (1+m)}{2 b},e^{2 i a} x^{2 i b},-e^{2 i a} x^{2 i b}\right )}{1+m} \]
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\[\int \left (e x \right )^{m} \tan \left (a +b \ln \left (x \right )\right )^{p}d x\]
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\[ \int (e x)^m \tan ^p(a+b \log (x)) \, dx=\int { \left (e x\right )^{m} \tan \left (b \log \left (x\right ) + a\right )^{p} \,d x } \]
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\[ \int (e x)^m \tan ^p(a+b \log (x)) \, dx=\int \left (e x\right )^{m} \tan ^{p}{\left (a + b \log {\left (x \right )} \right )}\, dx \]
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\[ \int (e x)^m \tan ^p(a+b \log (x)) \, dx=\int { \left (e x\right )^{m} \tan \left (b \log \left (x\right ) + a\right )^{p} \,d x } \]
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\[ \int (e x)^m \tan ^p(a+b \log (x)) \, dx=\int { \left (e x\right )^{m} \tan \left (b \log \left (x\right ) + a\right )^{p} \,d x } \]
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Timed out. \[ \int (e x)^m \tan ^p(a+b \log (x)) \, dx=\int {\mathrm {tan}\left (a+b\,\ln \left (x\right )\right )}^p\,{\left (e\,x\right )}^m \,d x \]
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