Integrand size = 13, antiderivative size = 48 \[ \int \frac {\tanh ^6(x)}{a+a \text {sech}(x)} \, dx=\frac {x}{a}-\frac {3 \arctan (\sinh (x))}{8 a}-\frac {(8-3 \text {sech}(x)) \tanh (x)}{8 a}-\frac {(4-3 \text {sech}(x)) \tanh ^3(x)}{12 a} \]
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Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3973, 3966, 3855} \[ \int \frac {\tanh ^6(x)}{a+a \text {sech}(x)} \, dx=-\frac {3 \arctan (\sinh (x))}{8 a}+\frac {x}{a}-\frac {\tanh ^3(x) (4-3 \text {sech}(x))}{12 a}-\frac {\tanh (x) (8-3 \text {sech}(x))}{8 a} \]
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Rule 3855
Rule 3966
Rule 3973
Rubi steps \begin{align*} \text {integral}& = -\frac {\int (-a+a \text {sech}(x)) \tanh ^4(x) \, dx}{a^2} \\ & = -\frac {(4-3 \text {sech}(x)) \tanh ^3(x)}{12 a}-\frac {\int (-4 a+3 a \text {sech}(x)) \tanh ^2(x) \, dx}{4 a^2} \\ & = -\frac {(8-3 \text {sech}(x)) \tanh (x)}{8 a}-\frac {(4-3 \text {sech}(x)) \tanh ^3(x)}{12 a}-\frac {\int (-8 a+3 a \text {sech}(x)) \, dx}{8 a^2} \\ & = \frac {x}{a}-\frac {(8-3 \text {sech}(x)) \tanh (x)}{8 a}-\frac {(4-3 \text {sech}(x)) \tanh ^3(x)}{12 a}-\frac {3 \int \text {sech}(x) \, dx}{8 a} \\ & = \frac {x}{a}-\frac {3 \arctan (\sinh (x))}{8 a}-\frac {(8-3 \text {sech}(x)) \tanh (x)}{8 a}-\frac {(4-3 \text {sech}(x)) \tanh ^3(x)}{12 a} \\ \end{align*}
Time = 0.39 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.25 \[ \int \frac {\tanh ^6(x)}{a+a \text {sech}(x)} \, dx=\frac {\cosh ^2\left (\frac {x}{2}\right ) \text {sech}(x) \left (6 \left (4 x-3 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )\right )+\left (-32+15 \text {sech}(x)+8 \text {sech}^2(x)-6 \text {sech}^3(x)\right ) \tanh (x)\right )}{12 a (1+\text {sech}(x))} \]
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Time = 0.67 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.56
method | result | size |
default | \(\frac {\frac {2 \left (-\frac {11 \tanh \left (\frac {x}{2}\right )^{7}}{8}-\frac {137 \tanh \left (\frac {x}{2}\right )^{5}}{24}-\frac {71 \tanh \left (\frac {x}{2}\right )^{3}}{24}-\frac {5 \tanh \left (\frac {x}{2}\right )}{8}\right )}{\left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )^{4}}-\frac {3 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{4}+\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )-\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a}\) | \(75\) |
risch | \(\frac {x}{a}+\frac {15 \,{\mathrm e}^{7 x}+48 \,{\mathrm e}^{6 x}-9 \,{\mathrm e}^{5 x}+96 \,{\mathrm e}^{4 x}+9 \,{\mathrm e}^{3 x}+80 \,{\mathrm e}^{2 x}-15 \,{\mathrm e}^{x}+32}{12 \left (1+{\mathrm e}^{2 x}\right )^{4} a}+\frac {3 i \ln \left ({\mathrm e}^{x}-i\right )}{8 a}-\frac {3 i \ln \left ({\mathrm e}^{x}+i\right )}{8 a}\) | \(86\) |
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Leaf count of result is larger than twice the leaf count of optimal. 686 vs. \(2 (42) = 84\).
Time = 0.27 (sec) , antiderivative size = 686, normalized size of antiderivative = 14.29 \[ \int \frac {\tanh ^6(x)}{a+a \text {sech}(x)} \, dx=\text {Too large to display} \]
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\[ \int \frac {\tanh ^6(x)}{a+a \text {sech}(x)} \, dx=\frac {\int \frac {\tanh ^{6}{\left (x \right )}}{\operatorname {sech}{\left (x \right )} + 1}\, dx}{a} \]
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Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (42) = 84\).
Time = 0.28 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.94 \[ \int \frac {\tanh ^6(x)}{a+a \text {sech}(x)} \, dx=\frac {x}{a} + \frac {15 \, e^{\left (-x\right )} - 80 \, e^{\left (-2 \, x\right )} - 9 \, e^{\left (-3 \, x\right )} - 96 \, e^{\left (-4 \, x\right )} + 9 \, e^{\left (-5 \, x\right )} - 48 \, e^{\left (-6 \, x\right )} - 15 \, e^{\left (-7 \, x\right )} - 32}{12 \, {\left (4 \, a e^{\left (-2 \, x\right )} + 6 \, a e^{\left (-4 \, x\right )} + 4 \, a e^{\left (-6 \, x\right )} + a e^{\left (-8 \, x\right )} + a\right )}} + \frac {3 \, \arctan \left (e^{\left (-x\right )}\right )}{4 \, a} \]
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none
Time = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.44 \[ \int \frac {\tanh ^6(x)}{a+a \text {sech}(x)} \, dx=\frac {x}{a} - \frac {3 \, \arctan \left (e^{x}\right )}{4 \, a} + \frac {15 \, e^{\left (7 \, x\right )} + 48 \, e^{\left (6 \, x\right )} - 9 \, e^{\left (5 \, x\right )} + 96 \, e^{\left (4 \, x\right )} + 9 \, e^{\left (3 \, x\right )} + 80 \, e^{\left (2 \, x\right )} - 15 \, e^{x} + 32}{12 \, a {\left (e^{\left (2 \, x\right )} + 1\right )}^{4}} \]
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Time = 2.11 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.98 \[ \int \frac {\tanh ^6(x)}{a+a \text {sech}(x)} \, dx=\frac {\frac {8}{3\,a}+\frac {6\,{\mathrm {e}}^x}{a}}{3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1}-\frac {\frac {4}{a}+\frac {9\,{\mathrm {e}}^x}{2\,a}}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}+\frac {x}{a}+\frac {\frac {4}{a}+\frac {5\,{\mathrm {e}}^x}{4\,a}}{{\mathrm {e}}^{2\,x}+1}-\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,\sqrt {a^2}}{a}\right )}{4\,\sqrt {a^2}}-\frac {4\,{\mathrm {e}}^x}{a\,\left (4\,{\mathrm {e}}^{2\,x}+6\,{\mathrm {e}}^{4\,x}+4\,{\mathrm {e}}^{6\,x}+{\mathrm {e}}^{8\,x}+1\right )} \]
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