Integrand size = 8, antiderivative size = 76 \[ \int x^m \text {Shi}(b x) \, dx=-\frac {x^m (-b x)^{-m} \Gamma (1+m,-b x)}{2 b (1+m)}-\frac {x^m (b x)^{-m} \Gamma (1+m,b x)}{2 b (1+m)}+\frac {x^{1+m} \text {Shi}(b x)}{1+m} \]
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Time = 0.05 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6667, 12, 3389, 2212} \[ \int x^m \text {Shi}(b x) \, dx=\frac {x^{m+1} \text {Shi}(b x)}{m+1}-\frac {x^m (-b x)^{-m} \Gamma (m+1,-b x)}{2 b (m+1)}-\frac {x^m (b x)^{-m} \Gamma (m+1,b x)}{2 b (m+1)} \]
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Rule 12
Rule 2212
Rule 3389
Rule 6667
Rubi steps \begin{align*} \text {integral}& = \frac {x^{1+m} \text {Shi}(b x)}{1+m}-\frac {b \int \frac {x^m \sinh (b x)}{b} \, dx}{1+m} \\ & = \frac {x^{1+m} \text {Shi}(b x)}{1+m}-\frac {\int x^m \sinh (b x) \, dx}{1+m} \\ & = \frac {x^{1+m} \text {Shi}(b x)}{1+m}+\frac {\int e^{-b x} x^m \, dx}{2 (1+m)}-\frac {\int e^{b x} x^m \, dx}{2 (1+m)} \\ & = -\frac {x^m (-b x)^{-m} \Gamma (1+m,-b x)}{2 b (1+m)}-\frac {x^m (b x)^{-m} \Gamma (1+m,b x)}{2 b (1+m)}+\frac {x^{1+m} \text {Shi}(b x)}{1+m} \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.74 \[ \int x^m \text {Shi}(b x) \, dx=-\frac {x^m \left ((-b x)^{-m} \Gamma (1+m,-b x)+(b x)^{-m} \Gamma (1+m,b x)-2 b x \text {Shi}(b x)\right )}{2 b (1+m)} \]
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Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 0.60 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.49
method | result | size |
meijerg | \(\frac {b \,x^{2+m} \operatorname {hypergeom}\left (\left [\frac {1}{2}, 1+\frac {m}{2}\right ], \left [\frac {3}{2}, \frac {3}{2}, 2+\frac {m}{2}\right ], \frac {b^{2} x^{2}}{4}\right )}{2+m}\) | \(37\) |
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\[ \int x^m \text {Shi}(b x) \, dx=\int { x^{m} {\rm Shi}\left (b x\right ) \,d x } \]
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Time = 0.57 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.55 \[ \int x^m \text {Shi}(b x) \, dx=\frac {b x^{m + 2} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{3}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + 1 \\ \frac {3}{2}, \frac {3}{2}, \frac {m}{2} + 2 \end {matrix}\middle | {\frac {b^{2} x^{2}}{4}} \right )}}{2 \Gamma \left (\frac {m}{2} + 2\right )} \]
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\[ \int x^m \text {Shi}(b x) \, dx=\int { x^{m} {\rm Shi}\left (b x\right ) \,d x } \]
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\[ \int x^m \text {Shi}(b x) \, dx=\int { x^{m} {\rm Shi}\left (b x\right ) \,d x } \]
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Timed out. \[ \int x^m \text {Shi}(b x) \, dx=\int x^m\,\mathrm {sinhint}\left (b\,x\right ) \,d x \]
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