Integrand size = 8, antiderivative size = 63 \[ \int x^3 \text {Shi}(b x) \, dx=-\frac {3 x \cosh (b x)}{2 b^3}-\frac {x^3 \cosh (b x)}{4 b}+\frac {3 \sinh (b x)}{2 b^4}+\frac {3 x^2 \sinh (b x)}{4 b^2}+\frac {1}{4} x^4 \text {Shi}(b x) \]
-3/2*x*cosh(b*x)/b^3-1/4*x^3*cosh(b*x)/b+1/4*x^4*Shi(b*x)+3/2*sinh(b*x)/b^ 4+3/4*x^2*sinh(b*x)/b^2
Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.84 \[ \int x^3 \text {Shi}(b x) \, dx=-\frac {x \left (6+b^2 x^2\right ) \cosh (b x)}{4 b^3}+\frac {3 \left (2+b^2 x^2\right ) \sinh (b x)}{4 b^4}+\frac {1}{4} x^4 \text {Shi}(b x) \]
-1/4*(x*(6 + b^2*x^2)*Cosh[b*x])/b^3 + (3*(2 + b^2*x^2)*Sinh[b*x])/(4*b^4) + (x^4*SinhIntegral[b*x])/4
Result contains complex when optimal does not.
Time = 0.48 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.32, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.625, Rules used = {7086, 27, 3042, 26, 3777, 3042, 3777, 26, 3042, 26, 3777, 3042, 3117}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^3 \text {Shi}(b x) \, dx\) |
\(\Big \downarrow \) 7086 |
\(\displaystyle \frac {1}{4} x^4 \text {Shi}(b x)-\frac {1}{4} b \int \frac {x^3 \sinh (b x)}{b}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} x^4 \text {Shi}(b x)-\frac {1}{4} \int x^3 \sinh (b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} x^4 \text {Shi}(b x)-\frac {1}{4} \int -i x^3 \sin (i b x)dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {1}{4} x^4 \text {Shi}(b x)+\frac {1}{4} i \int x^3 \sin (i b x)dx\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{4} x^4 \text {Shi}(b x)+\frac {1}{4} i \left (\frac {i x^3 \cosh (b x)}{b}-\frac {3 i \int x^2 \cosh (b x)dx}{b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} x^4 \text {Shi}(b x)+\frac {1}{4} i \left (\frac {i x^3 \cosh (b x)}{b}-\frac {3 i \int x^2 \sin \left (i b x+\frac {\pi }{2}\right )dx}{b}\right )\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{4} x^4 \text {Shi}(b x)+\frac {1}{4} i \left (\frac {i x^3 \cosh (b x)}{b}-\frac {3 i \left (\frac {x^2 \sinh (b x)}{b}-\frac {2 i \int -i x \sinh (b x)dx}{b}\right )}{b}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {1}{4} x^4 \text {Shi}(b x)+\frac {1}{4} i \left (\frac {i x^3 \cosh (b x)}{b}-\frac {3 i \left (\frac {x^2 \sinh (b x)}{b}-\frac {2 \int x \sinh (b x)dx}{b}\right )}{b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} x^4 \text {Shi}(b x)+\frac {1}{4} i \left (\frac {i x^3 \cosh (b x)}{b}-\frac {3 i \left (\frac {x^2 \sinh (b x)}{b}-\frac {2 \int -i x \sin (i b x)dx}{b}\right )}{b}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {1}{4} x^4 \text {Shi}(b x)+\frac {1}{4} i \left (\frac {i x^3 \cosh (b x)}{b}-\frac {3 i \left (\frac {x^2 \sinh (b x)}{b}+\frac {2 i \int x \sin (i b x)dx}{b}\right )}{b}\right )\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{4} x^4 \text {Shi}(b x)+\frac {1}{4} i \left (\frac {i x^3 \cosh (b x)}{b}-\frac {3 i \left (\frac {x^2 \sinh (b x)}{b}+\frac {2 i \left (\frac {i x \cosh (b x)}{b}-\frac {i \int \cosh (b x)dx}{b}\right )}{b}\right )}{b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} x^4 \text {Shi}(b x)+\frac {1}{4} i \left (\frac {i x^3 \cosh (b x)}{b}-\frac {3 i \left (\frac {x^2 \sinh (b x)}{b}+\frac {2 i \left (\frac {i x \cosh (b x)}{b}-\frac {i \int \sin \left (i b x+\frac {\pi }{2}\right )dx}{b}\right )}{b}\right )}{b}\right )\) |
\(\Big \downarrow \) 3117 |
\(\displaystyle \frac {1}{4} x^4 \text {Shi}(b x)+\frac {1}{4} i \left (\frac {i x^3 \cosh (b x)}{b}-\frac {3 i \left (\frac {x^2 \sinh (b x)}{b}+\frac {2 i \left (\frac {i x \cosh (b x)}{b}-\frac {i \sinh (b x)}{b^2}\right )}{b}\right )}{b}\right )\) |
(I/4)*((I*x^3*Cosh[b*x])/b - ((3*I)*((x^2*Sinh[b*x])/b + ((2*I)*((I*x*Cosh [b*x])/b - (I*Sinh[b*x])/b^2))/b))/b) + (x^4*SinhIntegral[b*x])/4
3.1.2.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( -(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*C os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
Int[((c_.) + (d_.)*(x_))^(m_.)*SinhIntegral[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[(c + d*x)^(m + 1)*(SinhIntegral[a + b*x]/(d*(m + 1))), x] - Simp[b/ (d*(m + 1)) Int[(c + d*x)^(m + 1)*(Sinh[a + b*x]/(a + b*x)), x], x] /; Fr eeQ[{a, b, c, d, m}, x] && NeQ[m, -1]
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 | \(-\frac {4 i \sqrt {\pi }\, \left (-\frac {i x b \left (\frac {5 b^{2} x^{2}}{2}+15\right ) \cosh \left (b x \right )}{40 \sqrt {\pi }}+\frac {i \left (\frac {15 b^{2} x^{2}}{2}+15\right ) \sinh \left (b x \right )}{40 \sqrt {\pi }}+\frac {i x^{4} b^{4} \operatorname {Shi}\left (b x \right )}{16 \sqrt {\pi }}\right )}{b^{4}}\) | \(69\) |
1/4*x^4*Shi(b*x)-1/4/b^4*(b^3*x^3*cosh(b*x)-3*b^2*x^2*sinh(b*x)+6*b*x*cosh (b*x)-6*sinh(b*x))
\[ \int x^3 \text {Shi}(b x) \, dx=\int { x^{3} {\rm Shi}\left (b x\right ) \,d x } \]
Time = 0.65 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.97 \[ \int x^3 \text {Shi}(b x) \, dx=\frac {x^{4} \operatorname {Shi}{\left (b x \right )}}{4} - \frac {x^{3} \cosh {\left (b x \right )}}{4 b} + \frac {3 x^{2} \sinh {\left (b x \right )}}{4 b^{2}} - \frac {3 x \cosh {\left (b x \right )}}{2 b^{3}} + \frac {3 \sinh {\left (b x \right )}}{2 b^{4}} \]
x**4*Shi(b*x)/4 - x**3*cosh(b*x)/(4*b) + 3*x**2*sinh(b*x)/(4*b**2) - 3*x*c osh(b*x)/(2*b**3) + 3*sinh(b*x)/(2*b**4)
\[ \int x^3 \text {Shi}(b x) \, dx=\int { x^{3} {\rm Shi}\left (b x\right ) \,d x } \]
\[ \int x^3 \text {Shi}(b x) \, dx=\int { x^{3} {\rm Shi}\left (b x\right ) \,d x } \]
Timed out. \[ \int x^3 \text {Shi}(b x) \, dx=\int x^3\,\mathrm {sinhint}\left (b\,x\right ) \,d x \]