Integrand size = 9, antiderivative size = 79 \[ \int \tanh ^p(a+b \log (x)) \, dx=x \left (1-e^{2 a} x^{2 b}\right )^{-p} \left (-1+e^{2 a} x^{2 b}\right )^p \operatorname {AppellF1}\left (\frac {1}{2 b},-p,p,\frac {1}{2} \left (2+\frac {1}{b}\right ),e^{2 a} x^{2 b},-e^{2 a} x^{2 b}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5652, 441, 440} \[ \int \tanh ^p(a+b \log (x)) \, dx=x \left (1-e^{2 a} x^{2 b}\right )^{-p} \left (e^{2 a} x^{2 b}-1\right )^p \operatorname {AppellF1}\left (\frac {1}{2 b},-p,p,\frac {1}{2} \left (2+\frac {1}{b}\right ),e^{2 a} x^{2 b},-e^{2 a} x^{2 b}\right ) \]
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Rule 440
Rule 441
Rule 5652
Rubi steps \begin{align*} \text {integral}& = \int \left (-1+e^{2 a} x^{2 b}\right )^p \left (1+e^{2 a} x^{2 b}\right )^{-p} \, dx \\ & = \left (\left (1-e^{2 a} x^{2 b}\right )^{-p} \left (-1+e^{2 a} x^{2 b}\right )^p\right ) \int \left (1-e^{2 a} x^{2 b}\right )^p \left (1+e^{2 a} x^{2 b}\right )^{-p} \, dx \\ & = x \left (1-e^{2 a} x^{2 b}\right )^{-p} \left (-1+e^{2 a} x^{2 b}\right )^p \operatorname {AppellF1}\left (\frac {1}{2 b},-p,p,\frac {1}{2} \left (2+\frac {1}{b}\right ),e^{2 a} x^{2 b},-e^{2 a} x^{2 b}\right ) \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(259\) vs. \(2(79)=158\).
Time = 0.48 (sec) , antiderivative size = 259, normalized size of antiderivative = 3.28 \[ \int \tanh ^p(a+b \log (x)) \, dx=\frac {(1+2 b) x \left (\frac {-1+e^{2 a} x^{2 b}}{1+e^{2 a} x^{2 b}}\right )^p \operatorname {AppellF1}\left (\frac {1}{2 b},-p,p,1+\frac {1}{2 b},e^{2 a} x^{2 b},-e^{2 a} x^{2 b}\right )}{-2 b e^{2 a} p x^{2 b} \operatorname {AppellF1}\left (1+\frac {1}{2 b},1-p,p,2+\frac {1}{2 b},e^{2 a} x^{2 b},-e^{2 a} x^{2 b}\right )-2 b e^{2 a} p x^{2 b} \operatorname {AppellF1}\left (1+\frac {1}{2 b},-p,1+p,2+\frac {1}{2 b},e^{2 a} x^{2 b},-e^{2 a} x^{2 b}\right )+(1+2 b) \operatorname {AppellF1}\left (\frac {1}{2 b},-p,p,1+\frac {1}{2 b},e^{2 a} x^{2 b},-e^{2 a} x^{2 b}\right )} \]
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\[\int \tanh \left (a +b \ln \left (x \right )\right )^{p}d x\]
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\[ \int \tanh ^p(a+b \log (x)) \, dx=\int { \tanh \left (b \log \left (x\right ) + a\right )^{p} \,d x } \]
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\[ \int \tanh ^p(a+b \log (x)) \, dx=\int \tanh ^{p}{\left (a + b \log {\left (x \right )} \right )}\, dx \]
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\[ \int \tanh ^p(a+b \log (x)) \, dx=\int { \tanh \left (b \log \left (x\right ) + a\right )^{p} \,d x } \]
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\[ \int \tanh ^p(a+b \log (x)) \, dx=\int { \tanh \left (b \log \left (x\right ) + a\right )^{p} \,d x } \]
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Timed out. \[ \int \tanh ^p(a+b \log (x)) \, dx=\int {\mathrm {tanh}\left (a+b\,\ln \left (x\right )\right )}^p \,d x \]
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