Integrand size = 8, antiderivative size = 63 \[ \int x^3 \text {Shi}(b x) \, dx=-\frac {3 x \cosh (b x)}{2 b^3}-\frac {x^3 \cosh (b x)}{4 b}+\frac {3 \sinh (b x)}{2 b^4}+\frac {3 x^2 \sinh (b x)}{4 b^2}+\frac {1}{4} x^4 \text {Shi}(b x) \]
[Out]
Time = 0.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6667, 12, 3377, 2717} \[ \int x^3 \text {Shi}(b x) \, dx=\frac {3 \sinh (b x)}{2 b^4}-\frac {3 x \cosh (b x)}{2 b^3}+\frac {3 x^2 \sinh (b x)}{4 b^2}+\frac {1}{4} x^4 \text {Shi}(b x)-\frac {x^3 \cosh (b x)}{4 b} \]
[In]
[Out]
Rule 12
Rule 2717
Rule 3377
Rule 6667
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} x^4 \text {Shi}(b x)-\frac {1}{4} b \int \frac {x^3 \sinh (b x)}{b} \, dx \\ & = \frac {1}{4} x^4 \text {Shi}(b x)-\frac {1}{4} \int x^3 \sinh (b x) \, dx \\ & = -\frac {x^3 \cosh (b x)}{4 b}+\frac {1}{4} x^4 \text {Shi}(b x)+\frac {3 \int x^2 \cosh (b x) \, dx}{4 b} \\ & = -\frac {x^3 \cosh (b x)}{4 b}+\frac {3 x^2 \sinh (b x)}{4 b^2}+\frac {1}{4} x^4 \text {Shi}(b x)-\frac {3 \int x \sinh (b x) \, dx}{2 b^2} \\ & = -\frac {3 x \cosh (b x)}{2 b^3}-\frac {x^3 \cosh (b x)}{4 b}+\frac {3 x^2 \sinh (b x)}{4 b^2}+\frac {1}{4} x^4 \text {Shi}(b x)+\frac {3 \int \cosh (b x) \, dx}{2 b^3} \\ & = -\frac {3 x \cosh (b x)}{2 b^3}-\frac {x^3 \cosh (b x)}{4 b}+\frac {3 \sinh (b x)}{2 b^4}+\frac {3 x^2 \sinh (b x)}{4 b^2}+\frac {1}{4} x^4 \text {Shi}(b x) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.84 \[ \int x^3 \text {Shi}(b x) \, dx=-\frac {x \left (6+b^2 x^2\right ) \cosh (b x)}{4 b^3}+\frac {3 \left (2+b^2 x^2\right ) \sinh (b x)}{4 b^4}+\frac {1}{4} x^4 \text {Shi}(b x) \]
[In]
[Out]
Time = 0.32 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.86
method | result | size |
parts | \(\frac {x^{4} \operatorname {Shi}\left (b x \right )}{4}-\frac {b^{3} x^{3} \cosh \left (b x \right )-3 b^{2} x^{2} \sinh \left (b x \right )+6 b x \cosh \left (b x \right )-6 \sinh \left (b x \right )}{4 b^{4}}\) | \(54\) |
derivativedivides | \(\frac {\frac {b^{4} x^{4} \operatorname {Shi}\left (b x \right )}{4}-\frac {b^{3} x^{3} \cosh \left (b x \right )}{4}+\frac {3 b^{2} x^{2} \sinh \left (b x \right )}{4}-\frac {3 b x \cosh \left (b x \right )}{2}+\frac {3 \sinh \left (b x \right )}{2}}{b^{4}}\) | \(56\) |
default | \(\frac {\frac {b^{4} x^{4} \operatorname {Shi}\left (b x \right )}{4}-\frac {b^{3} x^{3} \cosh \left (b x \right )}{4}+\frac {3 b^{2} x^{2} \sinh \left (b x \right )}{4}-\frac {3 b x \cosh \left (b x \right )}{2}+\frac {3 \sinh \left (b x \right )}{2}}{b^{4}}\) | \(56\) |
meijerg | \(-\frac {4 i \sqrt {\pi }\, \left (-\frac {i x b \left (\frac {5 b^{2} x^{2}}{2}+15\right ) \cosh \left (b x \right )}{40 \sqrt {\pi }}+\frac {i \left (\frac {15 b^{2} x^{2}}{2}+15\right ) \sinh \left (b x \right )}{40 \sqrt {\pi }}+\frac {i x^{4} b^{4} \operatorname {Shi}\left (b x \right )}{16 \sqrt {\pi }}\right )}{b^{4}}\) | \(69\) |
[In]
[Out]
\[ \int x^3 \text {Shi}(b x) \, dx=\int { x^{3} {\rm Shi}\left (b x\right ) \,d x } \]
[In]
[Out]
Time = 0.65 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.97 \[ \int x^3 \text {Shi}(b x) \, dx=\frac {x^{4} \operatorname {Shi}{\left (b x \right )}}{4} - \frac {x^{3} \cosh {\left (b x \right )}}{4 b} + \frac {3 x^{2} \sinh {\left (b x \right )}}{4 b^{2}} - \frac {3 x \cosh {\left (b x \right )}}{2 b^{3}} + \frac {3 \sinh {\left (b x \right )}}{2 b^{4}} \]
[In]
[Out]
\[ \int x^3 \text {Shi}(b x) \, dx=\int { x^{3} {\rm Shi}\left (b x\right ) \,d x } \]
[In]
[Out]
\[ \int x^3 \text {Shi}(b x) \, dx=\int { x^{3} {\rm Shi}\left (b x\right ) \,d x } \]
[In]
[Out]
Timed out. \[ \int x^3 \text {Shi}(b x) \, dx=\int x^3\,\mathrm {sinhint}\left (b\,x\right ) \,d x \]
[In]
[Out]