Integrand size = 8, antiderivative size = 69 \[ \int \sqrt [3]{\tanh (8 x)} \, dx=-\frac {1}{16} \sqrt {3} \arctan \left (\frac {1+2 \tanh ^{\frac {2}{3}}(8 x)}{\sqrt {3}}\right )-\frac {1}{16} \log \left (1-\tanh ^{\frac {2}{3}}(8 x)\right )+\frac {1}{32} \log \left (1+\tanh ^{\frac {2}{3}}(8 x)+\tanh ^{\frac {4}{3}}(8 x)\right ) \]
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Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.125, Rules used = {3557, 335, 281, 298, 31, 648, 632, 210, 642} \[ \int \sqrt [3]{\tanh (8 x)} \, dx=-\frac {1}{16} \sqrt {3} \arctan \left (\frac {2 \tanh ^{\frac {2}{3}}(8 x)+1}{\sqrt {3}}\right )-\frac {1}{16} \log \left (1-\tanh ^{\frac {2}{3}}(8 x)\right )+\frac {1}{32} \log \left (\tanh ^{\frac {4}{3}}(8 x)+\tanh ^{\frac {2}{3}}(8 x)+1\right ) \]
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Rule 31
Rule 210
Rule 281
Rule 298
Rule 335
Rule 632
Rule 642
Rule 648
Rule 3557
Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{8} \text {Subst}\left (\int \frac {\sqrt [3]{x}}{-1+x^2} \, dx,x,\tanh (8 x)\right )\right ) \\ & = -\left (\frac {3}{8} \text {Subst}\left (\int \frac {x^3}{-1+x^6} \, dx,x,\sqrt [3]{\tanh (8 x)}\right )\right ) \\ & = -\left (\frac {3}{16} \text {Subst}\left (\int \frac {x}{-1+x^3} \, dx,x,\tanh ^{\frac {2}{3}}(8 x)\right )\right ) \\ & = -\left (\frac {1}{16} \text {Subst}\left (\int \frac {1}{-1+x} \, dx,x,\tanh ^{\frac {2}{3}}(8 x)\right )\right )+\frac {1}{16} \text {Subst}\left (\int \frac {-1+x}{1+x+x^2} \, dx,x,\tanh ^{\frac {2}{3}}(8 x)\right ) \\ & = -\frac {1}{16} \log \left (1-\tanh ^{\frac {2}{3}}(8 x)\right )+\frac {1}{32} \text {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,\tanh ^{\frac {2}{3}}(8 x)\right )-\frac {3}{32} \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\tanh ^{\frac {2}{3}}(8 x)\right ) \\ & = -\frac {1}{16} \log \left (1-\tanh ^{\frac {2}{3}}(8 x)\right )+\frac {1}{32} \log \left (1+\tanh ^{\frac {2}{3}}(8 x)+\tanh ^{\frac {4}{3}}(8 x)\right )+\frac {3}{16} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \tanh ^{\frac {2}{3}}(8 x)\right ) \\ & = -\frac {1}{16} \sqrt {3} \arctan \left (\frac {1+2 \tanh ^{\frac {2}{3}}(8 x)}{\sqrt {3}}\right )-\frac {1}{16} \log \left (1-\tanh ^{\frac {2}{3}}(8 x)\right )+\frac {1}{32} \log \left (1+\tanh ^{\frac {2}{3}}(8 x)+\tanh ^{\frac {4}{3}}(8 x)\right ) \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.30 \[ \int \sqrt [3]{\tanh (8 x)} \, dx=-\frac {\left (\log \left (1-\sqrt [3]{\tanh ^2(8 x)}\right )-\sqrt [3]{-1} \log \left (1+\sqrt [3]{-1} \sqrt [3]{\tanh ^2(8 x)}\right )+(-1)^{2/3} \log \left (1-(-1)^{2/3} \sqrt [3]{\tanh ^2(8 x)}\right )\right ) \tanh ^{\frac {4}{3}}(8 x)}{16 \tanh ^2(8 x)^{2/3}} \]
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Time = 0.05 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.48
method | result | size |
derivativedivides | \(-\frac {\ln \left (\tanh \left (8 x \right )^{\frac {1}{3}}-1\right )}{16}+\frac {\ln \left (\tanh \left (8 x \right )^{\frac {2}{3}}+\tanh \left (8 x \right )^{\frac {1}{3}}+1\right )}{32}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 \tanh \left (8 x \right )^{\frac {1}{3}}+1\right ) \sqrt {3}}{3}\right )}{16}-\frac {\ln \left (\tanh \left (8 x \right )^{\frac {1}{3}}+1\right )}{16}+\frac {\ln \left (\tanh \left (8 x \right )^{\frac {2}{3}}-\tanh \left (8 x \right )^{\frac {1}{3}}+1\right )}{32}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 \tanh \left (8 x \right )^{\frac {1}{3}}-1\right ) \sqrt {3}}{3}\right )}{16}\) | \(102\) |
default | \(-\frac {\ln \left (\tanh \left (8 x \right )^{\frac {1}{3}}-1\right )}{16}+\frac {\ln \left (\tanh \left (8 x \right )^{\frac {2}{3}}+\tanh \left (8 x \right )^{\frac {1}{3}}+1\right )}{32}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 \tanh \left (8 x \right )^{\frac {1}{3}}+1\right ) \sqrt {3}}{3}\right )}{16}-\frac {\ln \left (\tanh \left (8 x \right )^{\frac {1}{3}}+1\right )}{16}+\frac {\ln \left (\tanh \left (8 x \right )^{\frac {2}{3}}-\tanh \left (8 x \right )^{\frac {1}{3}}+1\right )}{32}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 \tanh \left (8 x \right )^{\frac {1}{3}}-1\right ) \sqrt {3}}{3}\right )}{16}\) | \(102\) |
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Leaf count of result is larger than twice the leaf count of optimal. 179 vs. \(2 (52) = 104\).
Time = 0.25 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.59 \[ \int \sqrt [3]{\tanh (8 x)} \, dx=-\frac {1}{16} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} \left (\frac {\sinh \left (8 \, x\right )}{\cosh \left (8 \, x\right )}\right )^{\frac {2}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - \frac {1}{16} \, \log \left (\left (\frac {\sinh \left (8 \, x\right )}{\cosh \left (8 \, x\right )}\right )^{\frac {2}{3}} - 1\right ) + \frac {1}{32} \, \log \left (\frac {\cosh \left (8 \, x\right )^{2} + 2 \, \cosh \left (8 \, x\right ) \sinh \left (8 \, x\right ) + \sinh \left (8 \, x\right )^{2} + {\left (\cosh \left (8 \, x\right )^{2} + 2 \, \cosh \left (8 \, x\right ) \sinh \left (8 \, x\right ) + \sinh \left (8 \, x\right )^{2} + 1\right )} \left (\frac {\sinh \left (8 \, x\right )}{\cosh \left (8 \, x\right )}\right )^{\frac {2}{3}} + {\left (\cosh \left (8 \, x\right )^{2} + 2 \, \cosh \left (8 \, x\right ) \sinh \left (8 \, x\right ) + \sinh \left (8 \, x\right )^{2} - 1\right )} \left (\frac {\sinh \left (8 \, x\right )}{\cosh \left (8 \, x\right )}\right )^{\frac {1}{3}} + 1}{\cosh \left (8 \, x\right )^{2} + 2 \, \cosh \left (8 \, x\right ) \sinh \left (8 \, x\right ) + \sinh \left (8 \, x\right )^{2} + 1}\right ) \]
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Time = 1.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91 \[ \int \sqrt [3]{\tanh (8 x)} \, dx=- \frac {\log {\left (\tanh ^{\frac {2}{3}}{\left (8 x \right )} - 1 \right )}}{16} + \frac {\log {\left (\tanh ^{\frac {4}{3}}{\left (8 x \right )} + \tanh ^{\frac {2}{3}}{\left (8 x \right )} + 1 \right )}}{32} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} \left (\tanh ^{\frac {2}{3}}{\left (8 x \right )} + \frac {1}{2}\right )}{3} \right )}}{16} \]
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\[ \int \sqrt [3]{\tanh (8 x)} \, dx=\int { \tanh \left (8 \, x\right )^{\frac {1}{3}} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (52) = 104\).
Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.59 \[ \int \sqrt [3]{\tanh (8 x)} \, dx=-\frac {1}{16} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \left (\frac {e^{\left (16 \, x\right )} - 1}{e^{\left (16 \, x\right )} + 1}\right )^{\frac {2}{3}} + 1\right )}\right ) + \frac {1}{32} \, \log \left (\left (\frac {e^{\left (16 \, x\right )} - 1}{e^{\left (16 \, x\right )} + 1}\right )^{\frac {2}{3}} + \frac {\left (\frac {e^{\left (16 \, x\right )} - 1}{e^{\left (16 \, x\right )} + 1}\right )^{\frac {1}{3}} {\left (e^{\left (16 \, x\right )} - 1\right )}}{e^{\left (16 \, x\right )} + 1} + 1\right ) - \frac {1}{16} \, \log \left ({\left | \left (\frac {e^{\left (16 \, x\right )} - 1}{e^{\left (16 \, x\right )} + 1}\right )^{\frac {2}{3}} - 1 \right |}\right ) \]
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Time = 2.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03 \[ \int \sqrt [3]{\tanh (8 x)} \, dx=-\frac {\ln \left (81\,{\mathrm {tanh}\left (8\,x\right )}^{2/3}-81\right )}{16}-\ln \left (162\,{\mathrm {tanh}\left (8\,x\right )}^{2/3}\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )-81\right )\,\left (-\frac {1}{32}+\frac {\sqrt {3}\,1{}\mathrm {i}}{32}\right )+\ln \left (-162\,{\mathrm {tanh}\left (8\,x\right )}^{2/3}\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )-81\right )\,\left (\frac {1}{32}+\frac {\sqrt {3}\,1{}\mathrm {i}}{32}\right ) \]
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