Integrand size = 14, antiderivative size = 67 \[ \int (e x)^m \sinh \left (a+\frac {b}{x}\right ) \, dx=-\frac {1}{2} b e^a \left (-\frac {b}{x}\right )^m (e x)^m \Gamma \left (-1-m,-\frac {b}{x}\right )-\frac {1}{2} b e^{-a} \left (\frac {b}{x}\right )^m (e x)^m \Gamma \left (-1-m,\frac {b}{x}\right ) \]
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Time = 0.06 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5458, 3389, 2212} \[ \int (e x)^m \sinh \left (a+\frac {b}{x}\right ) \, dx=-\frac {1}{2} e^a b \left (-\frac {b}{x}\right )^m (e x)^m \Gamma \left (-m-1,-\frac {b}{x}\right )-\frac {1}{2} e^{-a} b \left (\frac {b}{x}\right )^m (e x)^m \Gamma \left (-m-1,\frac {b}{x}\right ) \]
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Rule 2212
Rule 3389
Rule 5458
Rubi steps \begin{align*} \text {integral}& = -\left (\left (\left (\frac {1}{x}\right )^m (e x)^m\right ) \text {Subst}\left (\int x^{-2-m} \sinh (a+b x) \, dx,x,\frac {1}{x}\right )\right ) \\ & = -\left (\frac {1}{2} \left (\left (\frac {1}{x}\right )^m (e x)^m\right ) \text {Subst}\left (\int e^{-i (i a+i b x)} x^{-2-m} \, dx,x,\frac {1}{x}\right )\right )+\frac {1}{2} \left (\left (\frac {1}{x}\right )^m (e x)^m\right ) \text {Subst}\left (\int e^{i (i a+i b x)} x^{-2-m} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {1}{2} b e^a \left (-\frac {b}{x}\right )^m (e x)^m \Gamma \left (-1-m,-\frac {b}{x}\right )-\frac {1}{2} b e^{-a} \left (\frac {b}{x}\right )^m (e x)^m \Gamma \left (-1-m,\frac {b}{x}\right ) \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.91 \[ \int (e x)^m \sinh \left (a+\frac {b}{x}\right ) \, dx=-\frac {1}{2} b e^{-a} (e x)^m \left (e^{2 a} \left (-\frac {b}{x}\right )^m \Gamma \left (-1-m,-\frac {b}{x}\right )+\left (\frac {b}{x}\right )^m \Gamma \left (-1-m,\frac {b}{x}\right )\right ) \]
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Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 0.66 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.04
method | result | size |
meijerg | \(\frac {\left (e x \right )^{m} b \operatorname {hypergeom}\left (\left [-\frac {m}{2}\right ], \left [\frac {3}{2}, 1-\frac {m}{2}\right ], \frac {b^{2}}{4 x^{2}}\right ) \cosh \left (a \right )}{m}+\frac {\left (e x \right )^{m} x \operatorname {hypergeom}\left (\left [-\frac {1}{2}-\frac {m}{2}\right ], \left [\frac {1}{2}, \frac {1}{2}-\frac {m}{2}\right ], \frac {b^{2}}{4 x^{2}}\right ) \sinh \left (a \right )}{1+m}\) | \(70\) |
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\[ \int (e x)^m \sinh \left (a+\frac {b}{x}\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (a + \frac {b}{x}\right ) \,d x } \]
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\[ \int (e x)^m \sinh \left (a+\frac {b}{x}\right ) \, dx=\int \left (e x\right )^{m} \sinh {\left (a + \frac {b}{x} \right )}\, dx \]
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\[ \int (e x)^m \sinh \left (a+\frac {b}{x}\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (a + \frac {b}{x}\right ) \,d x } \]
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\[ \int (e x)^m \sinh \left (a+\frac {b}{x}\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (a + \frac {b}{x}\right ) \,d x } \]
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Timed out. \[ \int (e x)^m \sinh \left (a+\frac {b}{x}\right ) \, dx=\int \mathrm {sinh}\left (a+\frac {b}{x}\right )\,{\left (e\,x\right )}^m \,d x \]
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