Integrand size = 13, antiderivative size = 46 \[ \int \frac {\tanh ^6(x)}{a+a \cosh (x)} \, dx=\frac {3 \arctan (\sinh (x))}{8 a}-\frac {3 \text {sech}(x) \tanh (x)}{8 a}-\frac {\text {sech}(x) \tanh ^3(x)}{4 a}-\frac {\tanh ^5(x)}{5 a} \]
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Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2785, 2687, 30, 2691, 3855} \[ \int \frac {\tanh ^6(x)}{a+a \cosh (x)} \, dx=\frac {3 \arctan (\sinh (x))}{8 a}-\frac {\tanh ^5(x)}{5 a}-\frac {\tanh ^3(x) \text {sech}(x)}{4 a}-\frac {3 \tanh (x) \text {sech}(x)}{8 a} \]
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Rule 30
Rule 2687
Rule 2691
Rule 2785
Rule 3855
Rubi steps \begin{align*} \text {integral}& = \frac {\int \text {sech}(x) \tanh ^4(x) \, dx}{a}-\frac {\int \text {sech}^2(x) \tanh ^4(x) \, dx}{a} \\ & = -\frac {\text {sech}(x) \tanh ^3(x)}{4 a}+\frac {i \text {Subst}\left (\int x^4 \, dx,x,i \tanh (x)\right )}{a}+\frac {3 \int \text {sech}(x) \tanh ^2(x) \, dx}{4 a} \\ & = -\frac {3 \text {sech}(x) \tanh (x)}{8 a}-\frac {\text {sech}(x) \tanh ^3(x)}{4 a}-\frac {\tanh ^5(x)}{5 a}+\frac {3 \int \text {sech}(x) \, dx}{8 a} \\ & = \frac {3 \arctan (\sinh (x))}{8 a}-\frac {3 \text {sech}(x) \tanh (x)}{8 a}-\frac {\text {sech}(x) \tanh ^3(x)}{4 a}-\frac {\tanh ^5(x)}{5 a} \\ \end{align*}
Time = 0.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.26 \[ \int \frac {\tanh ^6(x)}{a+a \cosh (x)} \, dx=\frac {\cosh ^2\left (\frac {x}{2}\right ) \left (30 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )+\left (-8-25 \text {sech}(x)+16 \text {sech}^2(x)+10 \text {sech}^3(x)-8 \text {sech}^4(x)\right ) \tanh (x)\right )}{20 a (1+\cosh (x))} \]
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Time = 0.60 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.39
method | result | size |
default | \(\frac {\frac {64 \left (\frac {3 \tanh \left (\frac {x}{2}\right )^{9}}{256}+\frac {7 \tanh \left (\frac {x}{2}\right )^{7}}{128}-\frac {\tanh \left (\frac {x}{2}\right )^{5}}{10}-\frac {7 \tanh \left (\frac {x}{2}\right )^{3}}{128}-\frac {3 \tanh \left (\frac {x}{2}\right )}{256}\right )}{\left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )^{5}}+\frac {3 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{4}}{a}\) | \(64\) |
risch | \(-\frac {25 \,{\mathrm e}^{9 x}-40 \,{\mathrm e}^{8 x}+10 \,{\mathrm e}^{7 x}-80 \,{\mathrm e}^{4 x}-10 \,{\mathrm e}^{3 x}-25 \,{\mathrm e}^{x}-8}{20 \left (1+{\mathrm e}^{2 x}\right )^{5} a}+\frac {3 i \ln \left ({\mathrm e}^{x}+i\right )}{8 a}-\frac {3 i \ln \left ({\mathrm e}^{x}-i\right )}{8 a}\) | \(75\) |
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Leaf count of result is larger than twice the leaf count of optimal. 750 vs. \(2 (38) = 76\).
Time = 0.25 (sec) , antiderivative size = 750, normalized size of antiderivative = 16.30 \[ \int \frac {\tanh ^6(x)}{a+a \cosh (x)} \, dx=\text {Too large to display} \]
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\[ \int \frac {\tanh ^6(x)}{a+a \cosh (x)} \, dx=\frac {\int \frac {\tanh ^{6}{\left (x \right )}}{\cosh {\left (x \right )} + 1}\, dx}{a} \]
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Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (38) = 76\).
Time = 0.29 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.93 \[ \int \frac {\tanh ^6(x)}{a+a \cosh (x)} \, dx=-\frac {25 \, e^{\left (-x\right )} + 10 \, e^{\left (-3 \, x\right )} + 80 \, e^{\left (-4 \, x\right )} - 10 \, e^{\left (-7 \, x\right )} + 40 \, e^{\left (-8 \, x\right )} - 25 \, e^{\left (-9 \, x\right )} + 8}{20 \, {\left (5 \, a e^{\left (-2 \, x\right )} + 10 \, a e^{\left (-4 \, x\right )} + 10 \, a e^{\left (-6 \, x\right )} + 5 \, a e^{\left (-8 \, x\right )} + a e^{\left (-10 \, x\right )} + a\right )}} - \frac {3 \, \arctan \left (e^{\left (-x\right )}\right )}{4 \, a} \]
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none
Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.26 \[ \int \frac {\tanh ^6(x)}{a+a \cosh (x)} \, dx=\frac {3 \, \arctan \left (e^{x}\right )}{4 \, a} - \frac {25 \, e^{\left (9 \, x\right )} - 40 \, e^{\left (8 \, x\right )} + 10 \, e^{\left (7 \, x\right )} - 80 \, e^{\left (4 \, x\right )} - 10 \, e^{\left (3 \, x\right )} - 25 \, e^{x} - 8}{20 \, a {\left (e^{\left (2 \, x\right )} + 1\right )}^{5}} \]
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Time = 1.77 (sec) , antiderivative size = 183, normalized size of antiderivative = 3.98 \[ \int \frac {\tanh ^6(x)}{a+a \cosh (x)} \, dx=\frac {\frac {16}{a}-\frac {6\,{\mathrm {e}}^x}{a}}{3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1}-\frac {\frac {8}{a}-\frac {9\,{\mathrm {e}}^x}{2\,a}}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}+\frac {32}{5\,a\,\left (5\,{\mathrm {e}}^{2\,x}+10\,{\mathrm {e}}^{4\,x}+10\,{\mathrm {e}}^{6\,x}+5\,{\mathrm {e}}^{8\,x}+{\mathrm {e}}^{10\,x}+1\right )}-\frac {\frac {16}{a}-\frac {4\,{\mathrm {e}}^x}{a}}{4\,{\mathrm {e}}^{2\,x}+6\,{\mathrm {e}}^{4\,x}+4\,{\mathrm {e}}^{6\,x}+{\mathrm {e}}^{8\,x}+1}+\frac {\frac {2}{a}-\frac {5\,{\mathrm {e}}^x}{4\,a}}{{\mathrm {e}}^{2\,x}+1}+\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,\sqrt {a^2}}{a}\right )}{4\,\sqrt {a^2}} \]
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