Integrand size = 12, antiderivative size = 90 \[ \int x \text {sech}^7\left (a+b x^2\right ) \, dx=\frac {5 \arctan \left (\sinh \left (a+b x^2\right )\right )}{32 b}+\frac {5 \text {sech}\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{32 b}+\frac {5 \text {sech}^3\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{48 b}+\frac {\text {sech}^5\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{12 b} \]
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Time = 0.06 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5544, 3853, 3855} \[ \int x \text {sech}^7\left (a+b x^2\right ) \, dx=\frac {5 \arctan \left (\sinh \left (a+b x^2\right )\right )}{32 b}+\frac {\tanh \left (a+b x^2\right ) \text {sech}^5\left (a+b x^2\right )}{12 b}+\frac {5 \tanh \left (a+b x^2\right ) \text {sech}^3\left (a+b x^2\right )}{48 b}+\frac {5 \tanh \left (a+b x^2\right ) \text {sech}\left (a+b x^2\right )}{32 b} \]
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Rule 3853
Rule 3855
Rule 5544
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \text {sech}^7(a+b x) \, dx,x,x^2\right ) \\ & = \frac {\text {sech}^5\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{12 b}+\frac {5}{12} \text {Subst}\left (\int \text {sech}^5(a+b x) \, dx,x,x^2\right ) \\ & = \frac {5 \text {sech}^3\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{48 b}+\frac {\text {sech}^5\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{12 b}+\frac {5}{16} \text {Subst}\left (\int \text {sech}^3(a+b x) \, dx,x,x^2\right ) \\ & = \frac {5 \text {sech}\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{32 b}+\frac {5 \text {sech}^3\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{48 b}+\frac {\text {sech}^5\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{12 b}+\frac {5}{32} \text {Subst}\left (\int \text {sech}(a+b x) \, dx,x,x^2\right ) \\ & = \frac {5 \arctan \left (\sinh \left (a+b x^2\right )\right )}{32 b}+\frac {5 \text {sech}\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{32 b}+\frac {5 \text {sech}^3\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{48 b}+\frac {\text {sech}^5\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{12 b} \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00 \[ \int x \text {sech}^7\left (a+b x^2\right ) \, dx=\frac {5 \arctan \left (\sinh \left (a+b x^2\right )\right )}{32 b}+\frac {5 \text {sech}\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{32 b}+\frac {5 \text {sech}^3\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{48 b}+\frac {\text {sech}^5\left (a+b x^2\right ) \tanh \left (a+b x^2\right )}{12 b} \]
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Time = 1.38 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.69
method | result | size |
derivativedivides | \(\frac {\left (\frac {\operatorname {sech}\left (b \,x^{2}+a \right )^{5}}{6}+\frac {5 \operatorname {sech}\left (b \,x^{2}+a \right )^{3}}{24}+\frac {5 \,\operatorname {sech}\left (b \,x^{2}+a \right )}{16}\right ) \tanh \left (b \,x^{2}+a \right )+\frac {5 \arctan \left ({\mathrm e}^{b \,x^{2}+a}\right )}{8}}{2 b}\) | \(62\) |
default | \(\frac {\left (\frac {\operatorname {sech}\left (b \,x^{2}+a \right )^{5}}{6}+\frac {5 \operatorname {sech}\left (b \,x^{2}+a \right )^{3}}{24}+\frac {5 \,\operatorname {sech}\left (b \,x^{2}+a \right )}{16}\right ) \tanh \left (b \,x^{2}+a \right )+\frac {5 \arctan \left ({\mathrm e}^{b \,x^{2}+a}\right )}{8}}{2 b}\) | \(62\) |
risch | \(\frac {{\mathrm e}^{b \,x^{2}+a} \left (15 \,{\mathrm e}^{10 b \,x^{2}+10 a}+85 \,{\mathrm e}^{8 b \,x^{2}+8 a}+198 \,{\mathrm e}^{6 b \,x^{2}+6 a}-198 \,{\mathrm e}^{4 b \,x^{2}+4 a}-85 \,{\mathrm e}^{2 b \,x^{2}+2 a}-15\right )}{48 b \left ({\mathrm e}^{2 b \,x^{2}+2 a}+1\right )^{6}}+\frac {5 i \ln \left ({\mathrm e}^{b \,x^{2}+a}+i\right )}{32 b}-\frac {5 i \ln \left ({\mathrm e}^{b \,x^{2}+a}-i\right )}{32 b}\) | \(133\) |
parallelrisch | \(\frac {15 i \left (-10-\cosh \left (6 b \,x^{2}+6 a \right )-6 \cosh \left (4 b \,x^{2}+4 a \right )-15 \cosh \left (2 b \,x^{2}+2 a \right )\right ) \ln \left (\tanh \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right )-i\right )+15 i \left (10+\cosh \left (6 b \,x^{2}+6 a \right )+6 \cosh \left (4 b \,x^{2}+4 a \right )+15 \cosh \left (2 b \,x^{2}+2 a \right )\right ) \ln \left (\tanh \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right )+i\right )+396 \sinh \left (b \,x^{2}+a \right )+170 \sinh \left (3 b \,x^{2}+3 a \right )+30 \sinh \left (5 b \,x^{2}+5 a \right )}{96 b \left (10+\cosh \left (6 b \,x^{2}+6 a \right )+6 \cosh \left (4 b \,x^{2}+4 a \right )+15 \cosh \left (2 b \,x^{2}+2 a \right )\right )}\) | \(200\) |
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Leaf count of result is larger than twice the leaf count of optimal. 1918 vs. \(2 (82) = 164\).
Time = 0.26 (sec) , antiderivative size = 1918, normalized size of antiderivative = 21.31 \[ \int x \text {sech}^7\left (a+b x^2\right ) \, dx=\text {Too large to display} \]
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\[ \int x \text {sech}^7\left (a+b x^2\right ) \, dx=\int x \operatorname {sech}^{7}{\left (a + b x^{2} \right )}\, dx \]
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Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (82) = 164\).
Time = 0.29 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.02 \[ \int x \text {sech}^7\left (a+b x^2\right ) \, dx=-\frac {5 \, \arctan \left (e^{\left (-b x^{2} - a\right )}\right )}{16 \, b} + \frac {15 \, e^{\left (-b x^{2} - a\right )} + 85 \, e^{\left (-3 \, b x^{2} - 3 \, a\right )} + 198 \, e^{\left (-5 \, b x^{2} - 5 \, a\right )} - 198 \, e^{\left (-7 \, b x^{2} - 7 \, a\right )} - 85 \, e^{\left (-9 \, b x^{2} - 9 \, a\right )} - 15 \, e^{\left (-11 \, b x^{2} - 11 \, a\right )}}{48 \, b {\left (6 \, e^{\left (-2 \, b x^{2} - 2 \, a\right )} + 15 \, e^{\left (-4 \, b x^{2} - 4 \, a\right )} + 20 \, e^{\left (-6 \, b x^{2} - 6 \, a\right )} + 15 \, e^{\left (-8 \, b x^{2} - 8 \, a\right )} + 6 \, e^{\left (-10 \, b x^{2} - 10 \, a\right )} + e^{\left (-12 \, b x^{2} - 12 \, a\right )} + 1\right )}} \]
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Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.62 \[ \int x \text {sech}^7\left (a+b x^2\right ) \, dx=\frac {5 \, {\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x^{2} + 2 \, a\right )} - 1\right )} e^{\left (-b x^{2} - a\right )}\right )\right )}}{64 \, b} + \frac {15 \, {\left (e^{\left (b x^{2} + a\right )} - e^{\left (-b x^{2} - a\right )}\right )}^{5} + 160 \, {\left (e^{\left (b x^{2} + a\right )} - e^{\left (-b x^{2} - a\right )}\right )}^{3} + 528 \, e^{\left (b x^{2} + a\right )} - 528 \, e^{\left (-b x^{2} - a\right )}}{48 \, {\left ({\left (e^{\left (b x^{2} + a\right )} - e^{\left (-b x^{2} - a\right )}\right )}^{2} + 4\right )}^{3} b} \]
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Time = 0.11 (sec) , antiderivative size = 395, normalized size of antiderivative = 4.39 \[ \int x \text {sech}^7\left (a+b x^2\right ) \, dx=\frac {5\,\mathrm {atan}\left (\frac {{\mathrm {e}}^a\,{\mathrm {e}}^{b\,x^2}\,\sqrt {b^2}}{b}\right )}{16\,\sqrt {b^2}}-\frac {8\,{\mathrm {e}}^{3\,b\,x^2+3\,a}}{3\,b\,\left (5\,{\mathrm {e}}^{2\,b\,x^2+2\,a}+10\,{\mathrm {e}}^{4\,b\,x^2+4\,a}+10\,{\mathrm {e}}^{6\,b\,x^2+6\,a}+5\,{\mathrm {e}}^{8\,b\,x^2+8\,a}+{\mathrm {e}}^{10\,b\,x^2+10\,a}+1\right )}-\frac {{\mathrm {e}}^{b\,x^2+a}}{b\,\left (4\,{\mathrm {e}}^{2\,b\,x^2+2\,a}+6\,{\mathrm {e}}^{4\,b\,x^2+4\,a}+4\,{\mathrm {e}}^{6\,b\,x^2+6\,a}+{\mathrm {e}}^{8\,b\,x^2+8\,a}+1\right )}+\frac {5\,{\mathrm {e}}^{b\,x^2+a}}{24\,b\,\left (2\,{\mathrm {e}}^{2\,b\,x^2+2\,a}+{\mathrm {e}}^{4\,b\,x^2+4\,a}+1\right )}-\frac {16\,{\mathrm {e}}^{5\,b\,x^2+5\,a}}{3\,b\,\left (6\,{\mathrm {e}}^{2\,b\,x^2+2\,a}+15\,{\mathrm {e}}^{4\,b\,x^2+4\,a}+20\,{\mathrm {e}}^{6\,b\,x^2+6\,a}+15\,{\mathrm {e}}^{8\,b\,x^2+8\,a}+6\,{\mathrm {e}}^{10\,b\,x^2+10\,a}+{\mathrm {e}}^{12\,b\,x^2+12\,a}+1\right )}+\frac {{\mathrm {e}}^{b\,x^2+a}}{6\,b\,\left (3\,{\mathrm {e}}^{2\,b\,x^2+2\,a}+3\,{\mathrm {e}}^{4\,b\,x^2+4\,a}+{\mathrm {e}}^{6\,b\,x^2+6\,a}+1\right )}+\frac {5\,{\mathrm {e}}^{b\,x^2+a}}{16\,b\,\left ({\mathrm {e}}^{2\,b\,x^2+2\,a}+1\right )} \]
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