Integrand size = 19, antiderivative size = 36 \[ \int \text {sech}^{3+n}(a+b x) \sinh ^3(a+b x) \, dx=-\frac {\text {sech}^n(a+b x)}{b n}+\frac {\text {sech}^{2+n}(a+b x)}{b (2+n)} \]
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Time = 0.03 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2702, 14} \[ \int \text {sech}^{3+n}(a+b x) \sinh ^3(a+b x) \, dx=\frac {\text {sech}^{n+2}(a+b x)}{b (n+2)}-\frac {\text {sech}^n(a+b x)}{b n} \]
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Rule 14
Rule 2702
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int x^{-1+n} \left (-1+x^2\right ) \, dx,x,\text {sech}(a+b x)\right )}{b} \\ & = \frac {\text {Subst}\left (\int \left (-x^{-1+n}+x^{1+n}\right ) \, dx,x,\text {sech}(a+b x)\right )}{b} \\ & = -\frac {\text {sech}^n(a+b x)}{b n}+\frac {\text {sech}^{2+n}(a+b x)}{b (2+n)} \\ \end{align*}
Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.89 \[ \int \text {sech}^{3+n}(a+b x) \sinh ^3(a+b x) \, dx=\frac {\text {sech}^n(a+b x) \left (-\frac {1}{n}+\frac {\text {sech}^2(a+b x)}{2+n}\right )}{b} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 6.18 (sec) , antiderivative size = 209, normalized size of antiderivative = 5.81
method | result | size |
risch | \(-\frac {\left (n \,{\mathrm e}^{4 b x +4 a}+2 \,{\mathrm e}^{4 b x +4 a}-2 n \,{\mathrm e}^{2 b x +2 a}+4 \,{\mathrm e}^{2 b x +2 a}+n +2\right ) 2^{n} \left ({\mathrm e}^{b x +a}\right )^{n} \left (1+{\mathrm e}^{2 b x +2 a}\right )^{-n} {\mathrm e}^{-\frac {i \operatorname {csgn}\left (\frac {i {\mathrm e}^{b x +a}}{1+{\mathrm e}^{2 b x +2 a}}\right ) \pi n \left (-\operatorname {csgn}\left (\frac {i {\mathrm e}^{b x +a}}{1+{\mathrm e}^{2 b x +2 a}}\right )+\operatorname {csgn}\left (\frac {i}{1+{\mathrm e}^{2 b x +2 a}}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i {\mathrm e}^{b x +a}}{1+{\mathrm e}^{2 b x +2 a}}\right )+\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )\right )}{2}}}{b n \left (n +2\right ) \left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}\) | \(209\) |
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Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (36) = 72\).
Time = 0.24 (sec) , antiderivative size = 219, normalized size of antiderivative = 6.08 \[ \int \text {sech}^{3+n}(a+b x) \sinh ^3(a+b x) \, dx=-\frac {{\left ({\left (n + 2\right )} \cosh \left (b x + a\right )^{2} + {\left (n + 2\right )} \sinh \left (b x + a\right )^{2} - n + 2\right )} \cosh \left (n \log \left (\frac {2 \, {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1}\right )\right ) + {\left ({\left (n + 2\right )} \cosh \left (b x + a\right )^{2} + {\left (n + 2\right )} \sinh \left (b x + a\right )^{2} - n + 2\right )} \sinh \left (n \log \left (\frac {2 \, {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1}\right )\right )}{b n^{2} + {\left (b n^{2} + 2 \, b n\right )} \cosh \left (b x + a\right )^{2} + {\left (b n^{2} + 2 \, b n\right )} \sinh \left (b x + a\right )^{2} + 2 \, b n} \]
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\[ \int \text {sech}^{3+n}(a+b x) \sinh ^3(a+b x) \, dx=\begin {cases} x \tanh ^{3}{\left (a \right )} \operatorname {sech}^{n}{\left (a \right )} & \text {for}\: b = 0 \\\int \frac {\tanh ^{3}{\left (a + b x \right )}}{\operatorname {sech}^{2}{\left (a + b x \right )}}\, dx & \text {for}\: n = -2 \\x - \frac {\log {\left (\tanh {\left (a + b x \right )} + 1 \right )}}{b} - \frac {\tanh ^{2}{\left (a + b x \right )}}{2 b} & \text {for}\: n = 0 \\- \frac {n \tanh ^{2}{\left (a + b x \right )} \operatorname {sech}^{n}{\left (a + b x \right )}}{b n^{2} + 2 b n} - \frac {2 \operatorname {sech}^{n}{\left (a + b x \right )}}{b n^{2} + 2 b n} & \text {otherwise} \end {cases} \]
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Leaf count of result is larger than twice the leaf count of optimal. 345 vs. \(2 (36) = 72\).
Time = 0.34 (sec) , antiderivative size = 345, normalized size of antiderivative = 9.58 \[ \int \text {sech}^{3+n}(a+b x) \sinh ^3(a+b x) \, dx=-\frac {2^{n} n e^{\left (-{\left (b x + a\right )} n - n \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )\right )}}{{\left (n^{2} + 2 \, {\left (n^{2} + 2 \, n\right )} e^{\left (-2 \, b x - 2 \, a\right )} + {\left (n^{2} + 2 \, n\right )} e^{\left (-4 \, b x - 4 \, a\right )} + 2 \, n\right )} b} + \frac {{\left (2^{n + 1} n - 2^{n + 2}\right )} e^{\left (-{\left (b x + a\right )} n - 2 \, b x - n \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) - 2 \, a\right )}}{{\left (n^{2} + 2 \, {\left (n^{2} + 2 \, n\right )} e^{\left (-2 \, b x - 2 \, a\right )} + {\left (n^{2} + 2 \, n\right )} e^{\left (-4 \, b x - 4 \, a\right )} + 2 \, n\right )} b} - \frac {{\left (2^{n} n + 2^{n + 1}\right )} e^{\left (-{\left (b x + a\right )} n - 4 \, b x - n \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) - 4 \, a\right )}}{{\left (n^{2} + 2 \, {\left (n^{2} + 2 \, n\right )} e^{\left (-2 \, b x - 2 \, a\right )} + {\left (n^{2} + 2 \, n\right )} e^{\left (-4 \, b x - 4 \, a\right )} + 2 \, n\right )} b} - \frac {2^{n + 1} e^{\left (-{\left (b x + a\right )} n - n \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )\right )}}{{\left (n^{2} + 2 \, {\left (n^{2} + 2 \, n\right )} e^{\left (-2 \, b x - 2 \, a\right )} + {\left (n^{2} + 2 \, n\right )} e^{\left (-4 \, b x - 4 \, a\right )} + 2 \, n\right )} b} \]
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\[ \int \text {sech}^{3+n}(a+b x) \sinh ^3(a+b x) \, dx=\int { \operatorname {sech}\left (b x + a\right )^{n} \tanh \left (b x + a\right )^{3} \,d x } \]
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Time = 2.38 (sec) , antiderivative size = 101, normalized size of antiderivative = 2.81 \[ \int \text {sech}^{3+n}(a+b x) \sinh ^3(a+b x) \, dx=-\frac {{\left (\frac {1}{\frac {{\mathrm {e}}^{a+b\,x}}{2}+\frac {{\mathrm {e}}^{-a-b\,x}}{2}}\right )}^n\,\left (\frac {1}{b\,n}+\frac {{\mathrm {e}}^{4\,a+4\,b\,x}}{b\,n}-\frac {{\mathrm {e}}^{2\,a+2\,b\,x}\,\left (2\,n-4\right )}{b\,n\,\left (n+2\right )}\right )}{2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1} \]
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