Integrand size = 11, antiderivative size = 162 \[ \int \coth ^p\left (a+\frac {\log (x)}{6}\right ) \, dx=e^{-6 a} p \left (-1-e^{2 a} \sqrt [3]{x}\right )^{1+p} \left (1-e^{2 a} \sqrt [3]{x}\right )^{1-p}+e^{-4 a} \left (-1-e^{2 a} \sqrt [3]{x}\right )^{1+p} \left (1-e^{2 a} \sqrt [3]{x}\right )^{1-p} \sqrt [3]{x}-\frac {2^{-p} e^{-6 a} \left (1+2 p^2\right ) \left (-1-e^{2 a} \sqrt [3]{x}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (p,1+p,2+p,\frac {1}{2} \left (1+e^{2 a} \sqrt [3]{x}\right )\right )}{1+p} \]
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Time = 0.09 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {5653, 383, 92, 81, 71} \[ \int \coth ^p\left (a+\frac {\log (x)}{6}\right ) \, dx=-\frac {e^{-6 a} 2^{-p} \left (2 p^2+1\right ) \left (-e^{2 a} \sqrt [3]{x}-1\right )^{p+1} \operatorname {Hypergeometric2F1}\left (p,p+1,p+2,\frac {1}{2} \left (e^{2 a} \sqrt [3]{x}+1\right )\right )}{p+1}+e^{-6 a} p \left (-e^{2 a} \sqrt [3]{x}-1\right )^{p+1} \left (1-e^{2 a} \sqrt [3]{x}\right )^{1-p}+e^{-4 a} \sqrt [3]{x} \left (-e^{2 a} \sqrt [3]{x}-1\right )^{p+1} \left (1-e^{2 a} \sqrt [3]{x}\right )^{1-p} \]
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Rule 71
Rule 81
Rule 92
Rule 383
Rule 5653
Rubi steps \begin{align*} \text {integral}& = \int \left (-1-e^{2 a} \sqrt [3]{x}\right )^p \left (1-e^{2 a} \sqrt [3]{x}\right )^{-p} \, dx \\ & = 3 \text {Subst}\left (\int x^2 \left (-1-e^{2 a} x\right )^p \left (1-e^{2 a} x\right )^{-p} \, dx,x,\sqrt [3]{x}\right ) \\ & = e^{-4 a} \left (-1-e^{2 a} \sqrt [3]{x}\right )^{1+p} \left (1-e^{2 a} \sqrt [3]{x}\right )^{1-p} \sqrt [3]{x}+e^{-4 a} \text {Subst}\left (\int \left (-1-e^{2 a} x\right )^p \left (1-e^{2 a} x\right )^{-p} \left (1+2 e^{2 a} p x\right ) \, dx,x,\sqrt [3]{x}\right ) \\ & = e^{-6 a} p \left (-1-e^{2 a} \sqrt [3]{x}\right )^{1+p} \left (1-e^{2 a} \sqrt [3]{x}\right )^{1-p}+e^{-4 a} \left (-1-e^{2 a} \sqrt [3]{x}\right )^{1+p} \left (1-e^{2 a} \sqrt [3]{x}\right )^{1-p} \sqrt [3]{x}+\left (e^{-4 a} \left (1+2 p^2\right )\right ) \text {Subst}\left (\int \left (-1-e^{2 a} x\right )^p \left (1-e^{2 a} x\right )^{-p} \, dx,x,\sqrt [3]{x}\right ) \\ & = e^{-6 a} p \left (-1-e^{2 a} \sqrt [3]{x}\right )^{1+p} \left (1-e^{2 a} \sqrt [3]{x}\right )^{1-p}+e^{-4 a} \left (-1-e^{2 a} \sqrt [3]{x}\right )^{1+p} \left (1-e^{2 a} \sqrt [3]{x}\right )^{1-p} \sqrt [3]{x}-\frac {2^{-p} e^{-6 a} \left (1+2 p^2\right ) \left (-1-e^{2 a} \sqrt [3]{x}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (p,1+p,2+p,\frac {1}{2} \left (1+e^{2 a} \sqrt [3]{x}\right )\right )}{1+p} \\ \end{align*}
Time = 0.88 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.88 \[ \int \coth ^p\left (a+\frac {\log (x)}{6}\right ) \, dx=\frac {e^{-6 a} \left (1+e^{2 a} \sqrt [3]{x}\right )^{1-p} \left (\frac {1+e^{2 a} \sqrt [3]{x}}{-1+e^{2 a} \sqrt [3]{x}}\right )^{-1+p} \left ((-1+p) \left (1+e^{2 a} \sqrt [3]{x}\right )^{1+p} \left (p+e^{2 a} \sqrt [3]{x}\right )-2^p \left (1+2 p^2\right ) \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,\frac {1}{2}-\frac {1}{2} e^{2 a} \sqrt [3]{x}\right )\right )}{-1+p} \]
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\[\int \coth \left (a +\frac {\ln \left (x \right )}{6}\right )^{p}d x\]
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\[ \int \coth ^p\left (a+\frac {\log (x)}{6}\right ) \, dx=\int { \coth \left (a + \frac {1}{6} \, \log \left (x\right )\right )^{p} \,d x } \]
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\[ \int \coth ^p\left (a+\frac {\log (x)}{6}\right ) \, dx=\int \coth ^{p}{\left (a + \frac {\log {\left (x \right )}}{6} \right )}\, dx \]
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\[ \int \coth ^p\left (a+\frac {\log (x)}{6}\right ) \, dx=\int { \coth \left (a + \frac {1}{6} \, \log \left (x\right )\right )^{p} \,d x } \]
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\[ \int \coth ^p\left (a+\frac {\log (x)}{6}\right ) \, dx=\int { \coth \left (a + \frac {1}{6} \, \log \left (x\right )\right )^{p} \,d x } \]
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Timed out. \[ \int \coth ^p\left (a+\frac {\log (x)}{6}\right ) \, dx=\int {\mathrm {coth}\left (a+\frac {\ln \left (x\right )}{6}\right )}^p \,d x \]
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