Integrand size = 11, antiderivative size = 43 \[ \int \frac {\tanh ^4(x)}{1+\coth (x)} \, dx=\frac {5 x}{2}-2 \log (\cosh (x))-\frac {5 \tanh (x)}{2}+\tanh ^2(x)-\frac {5 \tanh ^3(x)}{6}+\frac {\tanh ^3(x)}{2 (1+\coth (x))} \]
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Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3633, 3610, 3612, 3556} \[ \int \frac {\tanh ^4(x)}{1+\coth (x)} \, dx=\frac {5 x}{2}-\frac {5 \tanh ^3(x)}{6}+\tanh ^2(x)-\frac {5 \tanh (x)}{2}-2 \log (\cosh (x))+\frac {\tanh ^3(x)}{2 (\coth (x)+1)} \]
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Rule 3556
Rule 3610
Rule 3612
Rule 3633
Rubi steps \begin{align*} \text {integral}& = \frac {\tanh ^3(x)}{2 (1+\coth (x))}-\frac {1}{2} \int (-5+4 \coth (x)) \tanh ^4(x) \, dx \\ & = -\frac {5}{6} \tanh ^3(x)+\frac {\tanh ^3(x)}{2 (1+\coth (x))}-\frac {1}{2} i \int (-4 i+5 i \coth (x)) \tanh ^3(x) \, dx \\ & = \tanh ^2(x)-\frac {5 \tanh ^3(x)}{6}+\frac {\tanh ^3(x)}{2 (1+\coth (x))}+\frac {1}{2} \int (5-4 \coth (x)) \tanh ^2(x) \, dx \\ & = -\frac {5 \tanh (x)}{2}+\tanh ^2(x)-\frac {5 \tanh ^3(x)}{6}+\frac {\tanh ^3(x)}{2 (1+\coth (x))}+\frac {1}{2} i \int (4 i-5 i \coth (x)) \tanh (x) \, dx \\ & = \frac {5 x}{2}-\frac {5 \tanh (x)}{2}+\tanh ^2(x)-\frac {5 \tanh ^3(x)}{6}+\frac {\tanh ^3(x)}{2 (1+\coth (x))}-2 \int \tanh (x) \, dx \\ & = \frac {5 x}{2}-2 \log (\cosh (x))-\frac {5 \tanh (x)}{2}+\tanh ^2(x)-\frac {5 \tanh ^3(x)}{6}+\frac {\tanh ^3(x)}{2 (1+\coth (x))} \\ \end{align*}
Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.93 \[ \int \frac {\tanh ^4(x)}{1+\coth (x)} \, dx=\frac {5}{2} \text {arctanh}(\tanh (x))-2 \log (\cosh (x))-\frac {5 \tanh (x)}{2}+\tanh ^2(x)+\left (-\frac {5}{6}+\frac {1}{2+2 \coth (x)}\right ) \tanh ^3(x) \]
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Time = 0.16 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.02
method | result | size |
risch | \(\frac {9 x}{2}+\frac {{\mathrm e}^{-2 x}}{4}+\frac {4 \,{\mathrm e}^{4 x}+6 \,{\mathrm e}^{2 x}+\frac {14}{3}}{\left (1+{\mathrm e}^{2 x}\right )^{3}}-2 \ln \left (1+{\mathrm e}^{2 x}\right )\) | \(44\) |
parallelrisch | \(\frac {\left (12 \tanh \left (x \right )+12\right ) \ln \left (1-\tanh \left (x \right )\right )-2 \tanh \left (x \right )^{4}+\tanh \left (x \right )^{3}+27 \tanh \left (x \right ) x -9 \tanh \left (x \right )^{2}+27 x +15}{6+6 \tanh \left (x \right )}\) | \(50\) |
default | \(\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {1}{\tanh \left (\frac {x}{2}\right )+1}+\frac {9 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2}-\frac {4 \left (\tanh \left (\frac {x}{2}\right )^{5}-\frac {\tanh \left (\frac {x}{2}\right )^{4}}{2}+\frac {8 \tanh \left (\frac {x}{2}\right )^{3}}{3}-\frac {\tanh \left (\frac {x}{2}\right )^{2}}{2}+\tanh \left (\frac {x}{2}\right )\right )}{\left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )^{3}}-2 \ln \left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2}\) | \(96\) |
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Leaf count of result is larger than twice the leaf count of optimal. 571 vs. \(2 (35) = 70\).
Time = 0.26 (sec) , antiderivative size = 571, normalized size of antiderivative = 13.28 \[ \int \frac {\tanh ^4(x)}{1+\coth (x)} \, dx=\text {Too large to display} \]
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\[ \int \frac {\tanh ^4(x)}{1+\coth (x)} \, dx=\int \frac {\tanh ^{4}{\left (x \right )}}{\coth {\left (x \right )} + 1}\, dx \]
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Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \frac {\tanh ^4(x)}{1+\coth (x)} \, dx=\frac {1}{2} \, x - \frac {2 \, {\left (15 \, e^{\left (-2 \, x\right )} + 12 \, e^{\left (-4 \, x\right )} + 7\right )}}{3 \, {\left (3 \, e^{\left (-2 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} + 1\right )}} + \frac {1}{4} \, e^{\left (-2 \, x\right )} - 2 \, \log \left (e^{\left (-2 \, x\right )} + 1\right ) \]
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Time = 0.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.09 \[ \int \frac {\tanh ^4(x)}{1+\coth (x)} \, dx=\frac {9}{2} \, x + \frac {{\left (51 \, e^{\left (6 \, x\right )} + 81 \, e^{\left (4 \, x\right )} + 65 \, e^{\left (2 \, x\right )} + 3\right )} e^{\left (-2 \, x\right )}}{12 \, {\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} - 2 \, \log \left (e^{\left (2 \, x\right )} + 1\right ) \]
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Time = 1.96 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.60 \[ \int \frac {\tanh ^4(x)}{1+\coth (x)} \, dx=\frac {9\,x}{2}-2\,\ln \left ({\mathrm {e}}^{2\,x}+1\right )+\frac {{\mathrm {e}}^{-2\,x}}{4}+\frac {8}{3\,\left (3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1\right )}-\frac {2}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}+\frac {4}{{\mathrm {e}}^{2\,x}+1} \]
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