Integrand size = 8, antiderivative size = 63 \[ \int x^3 \text {Si}(b x) \, dx=-\frac {3 x \cos (b x)}{2 b^3}+\frac {x^3 \cos (b x)}{4 b}+\frac {3 \sin (b x)}{2 b^4}-\frac {3 x^2 \sin (b x)}{4 b^2}+\frac {1}{4} x^4 \text {Si}(b x) \] Output:
-3/2*x*cos(b*x)/b^3+1/4*x^3*cos(b*x)/b+3/2*sin(b*x)/b^4-3/4*x^2*sin(b*x)/b ^2+1/4*x^4*Si(b*x)
Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.79 \[ \int x^3 \text {Si}(b x) \, dx=\frac {b x \left (-6+b^2 x^2\right ) \cos (b x)-3 \left (-2+b^2 x^2\right ) \sin (b x)+b^4 x^4 \text {Si}(b x)}{4 b^4} \] Input:
Integrate[x^3*SinIntegral[b*x],x]
Output:
(b*x*(-6 + b^2*x^2)*Cos[b*x] - 3*(-2 + b^2*x^2)*Sin[b*x] + b^4*x^4*SinInte gral[b*x])/(4*b^4)
Time = 0.49 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.375, Rules used = {7057, 27, 3042, 3777, 3042, 3777, 25, 3042, 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.
\(\displaystyle \int x^3 \text {Si}(b x) \, dx\) |
\(\Big \downarrow \) 7057 |
\(\displaystyle \frac {1}{4} x^4 \text {Si}(b x)-\frac {1}{4} b \int \frac {x^3 \sin (b x)}{b}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} x^4 \text {Si}(b x)-\frac {1}{4} \int x^3 \sin (b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} x^4 \text {Si}(b x)-\frac {1}{4} \int x^3 \sin (b x)dx\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{4} \left (\frac {x^3 \cos (b x)}{b}-\frac {3 \int x^2 \cos (b x)dx}{b}\right )+\frac {1}{4} x^4 \text {Si}(b x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \left (\frac {x^3 \cos (b x)}{b}-\frac {3 \int x^2 \sin \left (b x+\frac {\pi }{2}\right )dx}{b}\right )+\frac {1}{4} x^4 \text {Si}(b x)\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{4} \left (\frac {x^3 \cos (b x)}{b}-\frac {3 \left (\frac {2 \int -x \sin (b x)dx}{b}+\frac {x^2 \sin (b x)}{b}\right )}{b}\right )+\frac {1}{4} x^4 \text {Si}(b x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \left (\frac {x^3 \cos (b x)}{b}-\frac {3 \left (\frac {x^2 \sin (b x)}{b}-\frac {2 \int x \sin (b x)dx}{b}\right )}{b}\right )+\frac {1}{4} x^4 \text {Si}(b x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \left (\frac {x^3 \cos (b x)}{b}-\frac {3 \left (\frac {x^2 \sin (b x)}{b}-\frac {2 \int x \sin (b x)dx}{b}\right )}{b}\right )+\frac {1}{4} x^4 \text {Si}(b x)\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{4} \left (\frac {x^3 \cos (b x)}{b}-\frac {3 \left (\frac {x^2 \sin (b x)}{b}-\frac {2 \left (\frac {\int \cos (b x)dx}{b}-\frac {x \cos (b x)}{b}\right )}{b}\right )}{b}\right )+\frac {1}{4} x^4 \text {Si}(b x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \left (\frac {x^3 \cos (b x)}{b}-\frac {3 \left (\frac {x^2 \sin (b x)}{b}-\frac {2 \left (\frac {\int \sin \left (b x+\frac {\pi }{2}\right )dx}{b}-\frac {x \cos (b x)}{b}\right )}{b}\right )}{b}\right )+\frac {1}{4} x^4 \text {Si}(b x)\) |
\(\Big \downarrow \) 3117 |
\(\displaystyle \frac {1}{4} \left (\frac {x^3 \cos (b x)}{b}-\frac {3 \left (\frac {x^2 \sin (b x)}{b}-\frac {2 \left (\frac {\sin (b x)}{b^2}-\frac {x \cos (b x)}{b}\right )}{b}\right )}{b}\right )+\frac {1}{4} x^4 \text {Si}(b x)\) |
Input:
Int[x^3*SinIntegral[b*x],x]
Output:
((x^3*Cos[b*x])/b - (3*((x^2*Sin[b*x])/b - (2*(-((x*Cos[b*x])/b) + Sin[b*x ]/b^2))/b))/b)/4 + (x^4*SinIntegral[b*x])/4
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_.)*SinIntegral[(a_.) + (b_.)*(x_)], x_Symbol] : > Simp[(c + d*x)^(m + 1)*(SinIntegral[a + b*x]/(d*(m + 1))), x] - Simp[b/(d *(m + 1)) Int[(c + d*x)^(m + 1)*(Sin[a + b*x]/(a + b*x)), x], x] /; FreeQ [{a, b, c, d, m}, x] && NeQ[m, -1]
Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 0.69 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.37
method | result | size |
meijerg | \(\frac {b \,x^{5} \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {5}{2}\right ], \left [\frac {3}{2}, \frac {3}{2}, \frac {7}{2}\right ], -\frac {b^{2} x^{2}}{4}\right )}{5}\) | \(23\) |
parts | \(\frac {x^{4} \operatorname {Si}\left (b x \right )}{4}-\frac {-b^{3} x^{3} \cos \left (b x \right )+3 b^{2} x^{2} \sin \left (b x \right )-6 \sin \left (b x \right )+6 b x \cos \left (b x \right )}{4 b^{4}}\) | \(55\) |
derivativedivides | \(\frac {\frac {b^{4} x^{4} \operatorname {Si}\left (b x \right )}{4}+\frac {b^{3} x^{3} \cos \left (b x \right )}{4}-\frac {3 b^{2} x^{2} \sin \left (b x \right )}{4}+\frac {3 \sin \left (b x \right )}{2}-\frac {3 b x \cos \left (b x \right )}{2}}{b^{4}}\) | \(56\) |
default | \(\frac {\frac {b^{4} x^{4} \operatorname {Si}\left (b x \right )}{4}+\frac {b^{3} x^{3} \cos \left (b x \right )}{4}-\frac {3 b^{2} x^{2} \sin \left (b x \right )}{4}+\frac {3 \sin \left (b x \right )}{2}-\frac {3 b x \cos \left (b x \right )}{2}}{b^{4}}\) | \(56\) |
orering | \(\frac {\left (b^{4} x^{4}+18 b^{2} x^{2}-72\right ) \operatorname {Si}\left (b x \right )}{4 b^{4}}-\frac {\left (2 b^{2} x^{2}-9\right ) \left (3 x^{2} \operatorname {Si}\left (b x \right )+x^{2} \sin \left (b x \right )\right )}{x^{2} b^{4}}+\frac {\left (b^{2} x^{2}-6\right ) \left (6 x \,\operatorname {Si}\left (b x \right )+5 x \sin \left (b x \right )+x^{2} b \cos \left (b x \right )\right )}{4 x \,b^{4}}\) | \(105\) |
Input:
int(x^3*Si(b*x),x,method=_RETURNVERBOSE)
Output:
1/5*b*x^5*hypergeom([1/2,5/2],[3/2,3/2,7/2],-1/4*b^2*x^2)
Time = 0.09 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.78 \[ \int x^3 \text {Si}(b x) \, dx=\frac {b^{4} x^{4} \operatorname {Si}\left (b x\right ) + {\left (b^{3} x^{3} - 6 \, b x\right )} \cos \left (b x\right ) - 3 \, {\left (b^{2} x^{2} - 2\right )} \sin \left (b x\right )}{4 \, b^{4}} \] Input:
integrate(x^3*sin_integral(b*x),x, algorithm="fricas")
Output:
1/4*(b^4*x^4*sin_integral(b*x) + (b^3*x^3 - 6*b*x)*cos(b*x) - 3*(b^2*x^2 - 2)*sin(b*x))/b^4
Time = 0.71 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.97 \[ \int x^3 \text {Si}(b x) \, dx=\frac {x^{4} \operatorname {Si}{\left (b x \right )}}{4} + \frac {x^{3} \cos {\left (b x \right )}}{4 b} - \frac {3 x^{2} \sin {\left (b x \right )}}{4 b^{2}} - \frac {3 x \cos {\left (b x \right )}}{2 b^{3}} + \frac {3 \sin {\left (b x \right )}}{2 b^{4}} \] Input:
integrate(x**3*Si(b*x),x)
Output:
x**4*Si(b*x)/4 + x**3*cos(b*x)/(4*b) - 3*x**2*sin(b*x)/(4*b**2) - 3*x*cos( b*x)/(2*b**3) + 3*sin(b*x)/(2*b**4)
Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.76 \[ \int x^3 \text {Si}(b x) \, dx=\frac {1}{4} \, x^{4} \operatorname {Si}\left (b x\right ) + \frac {{\left (b^{3} x^{3} - 6 \, b x\right )} \cos \left (b x\right ) - 3 \, {\left (b^{2} x^{2} - 2\right )} \sin \left (b x\right )}{4 \, b^{4}} \] Input:
integrate(x^3*sin_integral(b*x),x, algorithm="maxima")
Output:
1/4*x^4*sin_integral(b*x) + 1/4*((b^3*x^3 - 6*b*x)*cos(b*x) - 3*(b^2*x^2 - 2)*sin(b*x))/b^4
Time = 0.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.78 \[ \int x^3 \text {Si}(b x) \, dx=\frac {1}{4} \, x^{4} \operatorname {Si}\left (b x\right ) + \frac {{\left (b^{3} x^{3} - 6 \, b x\right )} \cos \left (b x\right )}{4 \, b^{4}} - \frac {3 \, {\left (b^{2} x^{2} - 2\right )} \sin \left (b x\right )}{4 \, b^{4}} \] Input:
integrate(x^3*sin_integral(b*x),x, algorithm="giac")
Output:
1/4*x^4*sin_integral(b*x) + 1/4*(b^3*x^3 - 6*b*x)*cos(b*x)/b^4 - 3/4*(b^2* x^2 - 2)*sin(b*x)/b^4
Timed out. \[ \int x^3 \text {Si}(b x) \, dx=\frac {\sin \left (b\,x\right )\,\left (\frac {6}{b^4}-\frac {3\,x^2}{b^2}\right )}{4}+\frac {x^4\,\mathrm {sinint}\left (b\,x\right )}{4}-\frac {\cos \left (b\,x\right )\,\left (\frac {6\,x}{b^3}-\frac {x^3}{b}\right )}{4} \] Input:
int(x^3*sinint(b*x),x)
Output:
(sin(b*x)*(6/b^4 - (3*x^2)/b^2))/4 + (x^4*sinint(b*x))/4 - (cos(b*x)*((6*x )/b^3 - x^3/b))/4
Time = 0.22 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.86 \[ \int x^3 \text {Si}(b x) \, dx=\frac {\cos \left (b x \right ) b^{3} x^{3}-6 \cos \left (b x \right ) b x +\mathit {si} \left (b x \right ) b^{4} x^{4}-3 \sin \left (b x \right ) b^{2} x^{2}+6 \sin \left (b x \right )}{4 b^{4}} \] Input:
int(x^3*Si(b*x),x)
Output:
(cos(b*x)*b**3*x**3 - 6*cos(b*x)*b*x + si(b*x)*b**4*x**4 - 3*sin(b*x)*b**2 *x**2 + 6*sin(b*x))/(4*b**4)