Integrand size = 11, antiderivative size = 43 \[ \int \frac {\coth ^4(x)}{1+\tanh (x)} \, dx=\frac {5 x}{2}-\frac {5 \coth (x)}{2}+\coth ^2(x)-\frac {5 \coth ^3(x)}{6}-2 \log (\sinh (x))+\frac {\coth ^3(x)}{2 (1+\tanh (x))} \]
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Time = 0.08 (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} \[ \int \frac {\coth ^4(x)}{1+\tanh (x)} \, dx=\frac {5 x}{2}-\frac {5 \coth ^3(x)}{6}+\coth ^2(x)-\frac {5 \coth (x)}{2}-2 \log (\sinh (x))+\frac {\coth ^3(x)}{2 (\tanh (x)+1)} \]
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Rule 3556
Rule 3610
Rule 3612
Rule 3633
Rubi steps \begin{align*} \text {integral}& = \frac {\coth ^3(x)}{2 (1+\tanh (x))}-\frac {1}{2} \int \coth ^4(x) (-5+4 \tanh (x)) \, dx \\ & = -\frac {5}{6} \coth ^3(x)+\frac {\coth ^3(x)}{2 (1+\tanh (x))}-\frac {1}{2} i \int \coth ^3(x) (-4 i+5 i \tanh (x)) \, dx \\ & = \coth ^2(x)-\frac {5 \coth ^3(x)}{6}+\frac {\coth ^3(x)}{2 (1+\tanh (x))}+\frac {1}{2} \int \coth ^2(x) (5-4 \tanh (x)) \, dx \\ & = -\frac {5 \coth (x)}{2}+\coth ^2(x)-\frac {5 \coth ^3(x)}{6}+\frac {\coth ^3(x)}{2 (1+\tanh (x))}+\frac {1}{2} i \int \coth (x) (4 i-5 i \tanh (x)) \, dx \\ & = \frac {5 x}{2}-\frac {5 \coth (x)}{2}+\coth ^2(x)-\frac {5 \coth ^3(x)}{6}+\frac {\coth ^3(x)}{2 (1+\tanh (x))}-2 \int \coth (x) \, dx \\ & = \frac {5 x}{2}-\frac {5 \coth (x)}{2}+\coth ^2(x)-\frac {5 \coth ^3(x)}{6}-2 \log (\sinh (x))+\frac {\coth ^3(x)}{2 (1+\tanh (x))} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.23 \[ \int \frac {\coth ^4(x)}{1+\tanh (x)} \, dx=\frac {1}{2} \left (2 \coth ^2(x)+\coth ^4(x)+\frac {\coth ^6(x)}{1+\coth (x)}-\coth ^5(x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},\tanh ^2(x)\right )-4 (\log (\cosh (x))+\log (\tanh (x)))\right ) \]
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Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.02
method | result | size |
risch | \(\frac {9 x}{2}-\frac {{\mathrm e}^{-2 x}}{4}-\frac {2 \left (6 \,{\mathrm e}^{4 x}-9 \,{\mathrm e}^{2 x}+7\right )}{3 \left ({\mathrm e}^{2 x}-1\right )^{3}}-2 \ln \left ({\mathrm e}^{2 x}-1\right )\) | \(44\) |
parallelrisch | \(\frac {\left (12 \tanh \left (x \right )+12\right ) \ln \left (1-\tanh \left (x \right )\right )+\left (-12 \tanh \left (x \right )-12\right ) \ln \left (\tanh \left (x \right )\right )-2 \coth \left (x \right )^{3}+27 \tanh \left (x \right ) x +\coth \left (x \right )^{2}+27 x -9 \coth \left (x \right )-15}{6+6 \tanh \left (x \right )}\) | \(58\) |
default | \(-\frac {\tanh \left (\frac {x}{2}\right )^{3}}{24}+\frac {\tanh \left (\frac {x}{2}\right )^{2}}{8}-\frac {9 \tanh \left (\frac {x}{2}\right )}{8}-\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {1}{\tanh \left (\frac {x}{2}\right )+1}+\frac {9 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2}-\frac {1}{24 \tanh \left (\frac {x}{2}\right )^{3}}+\frac {1}{8 \tanh \left (\frac {x}{2}\right )^{2}}-\frac {9}{8 \tanh \left (\frac {x}{2}\right )}-2 \ln \left (\tanh \left (\frac {x}{2}\right )\right )-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2}\) | \(91\) |
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Leaf count of result is larger than twice the leaf count of optimal. 582 vs. \(2 (35) = 70\).
Time = 0.25 (sec) , antiderivative size = 582, normalized size of antiderivative = 13.53 \[ \int \frac {\coth ^4(x)}{1+\tanh (x)} \, dx=\text {Too large to display} \]
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\[ \int \frac {\coth ^4(x)}{1+\tanh (x)} \, dx=\int \frac {\coth ^{4}{\left (x \right )}}{\tanh {\left (x \right )} + 1}\, dx \]
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none
Time = 0.20 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.49 \[ \int \frac {\coth ^4(x)}{1+\tanh (x)} \, dx=\frac {1}{2} \, x - \frac {2 \, {\left (15 \, e^{\left (-2 \, x\right )} - 12 \, e^{\left (-4 \, x\right )} - 7\right )}}{3 \, {\left (3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} - 1\right )}} - \frac {1}{4} \, e^{\left (-2 \, x\right )} - 2 \, \log \left (e^{\left (-x\right )} + 1\right ) - 2 \, \log \left (e^{\left (-x\right )} - 1\right ) \]
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Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.12 \[ \int \frac {\coth ^4(x)}{1+\tanh (x)} \, dx=\frac {9}{2} \, x - \frac {{\left (51 \, e^{\left (6 \, x\right )} - 81 \, e^{\left (4 \, x\right )} + 65 \, e^{\left (2 \, x\right )} - 3\right )} e^{\left (-2 \, x\right )}}{12 \, {\left (e^{\left (2 \, x\right )} - 1\right )}^{3}} - 2 \, \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right ) \]
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Time = 1.68 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.60 \[ \int \frac {\coth ^4(x)}{1+\tanh (x)} \, dx=\frac {9\,x}{2}-2\,\ln \left ({\mathrm {e}}^{2\,x}-1\right )-\frac {{\mathrm {e}}^{-2\,x}}{4}-\frac {8}{3\,\left (3\,{\mathrm {e}}^{2\,x}-3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}-1\right )}-\frac {2}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1}-\frac {4}{{\mathrm {e}}^{2\,x}-1} \]
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