Integrand size = 8, antiderivative size = 63 \[ \int x^3 \operatorname {CosIntegral}(b x) \, dx=\frac {3 \cos (b x)}{2 b^4}-\frac {3 x^2 \cos (b x)}{4 b^2}+\frac {1}{4} x^4 \operatorname {CosIntegral}(b x)+\frac {3 x \sin (b x)}{2 b^3}-\frac {x^3 \sin (b x)}{4 b} \] Output:
3/2*cos(b*x)/b^4-3/4*x^2*cos(b*x)/b^2+1/4*x^4*Ci(b*x)+3/2*x*sin(b*x)/b^3-1 /4*x^3*sin(b*x)/b
Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.84 \[ \int x^3 \operatorname {CosIntegral}(b x) \, dx=-\frac {3 \left (-2+b^2 x^2\right ) \cos (b x)}{4 b^4}+\frac {1}{4} x^4 \operatorname {CosIntegral}(b x)-\frac {x \left (-6+b^2 x^2\right ) \sin (b x)}{4 b^3} \] Input:
Integrate[x^3*CosIntegral[b*x],x]
Output:
(-3*(-2 + b^2*x^2)*Cos[b*x])/(4*b^4) + (x^4*CosIntegral[b*x])/4 - (x*(-6 + b^2*x^2)*Sin[b*x])/(4*b^3)
Time = 0.48 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.11, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.500, Rules used = {7058, 27, 3042, 3777, 25, 3042, 3777, 3042, 3777, 25, 3042, 3118}
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.
\(\displaystyle \int x^3 \operatorname {CosIntegral}(b x) \, dx\) |
\(\Big \downarrow \) 7058 |
\(\displaystyle \frac {1}{4} x^4 \operatorname {CosIntegral}(b x)-\frac {1}{4} b \int \frac {x^3 \cos (b x)}{b}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} x^4 \operatorname {CosIntegral}(b x)-\frac {1}{4} \int x^3 \cos (b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} x^4 \operatorname {CosIntegral}(b x)-\frac {1}{4} \int x^3 \sin \left (b x+\frac {\pi }{2}\right )dx\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 \int -x^2 \sin (b x)dx}{b}-\frac {x^3 \sin (b x)}{b}\right )+\frac {1}{4} x^4 \operatorname {CosIntegral}(b x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \left (\frac {3 \int x^2 \sin (b x)dx}{b}-\frac {x^3 \sin (b x)}{b}\right )+\frac {1}{4} x^4 \operatorname {CosIntegral}(b x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \left (\frac {3 \int x^2 \sin (b x)dx}{b}-\frac {x^3 \sin (b x)}{b}\right )+\frac {1}{4} x^4 \operatorname {CosIntegral}(b x)\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{4} \left (\frac {3 \left (\frac {2 \int x \cos (b x)dx}{b}-\frac {x^2 \cos (b x)}{b}\right )}{b}-\frac {x^3 \sin (b x)}{b}\right )+\frac {1}{4} x^4 \operatorname {CosIntegral}(b x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \left (\frac {3 \left (\frac {2 \int x \sin \left (b x+\frac {\pi }{2}\right )dx}{b}-\frac {x^2 \cos (b x)}{b}\right )}{b}-\frac {x^3 \sin (b x)}{b}\right )+\frac {1}{4} x^4 \operatorname {CosIntegral}(b x)\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{4} \left (\frac {3 \left (\frac {2 \left (\frac {\int -\sin (b x)dx}{b}+\frac {x \sin (b x)}{b}\right )}{b}-\frac {x^2 \cos (b x)}{b}\right )}{b}-\frac {x^3 \sin (b x)}{b}\right )+\frac {1}{4} x^4 \operatorname {CosIntegral}(b x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \left (\frac {3 \left (\frac {2 \left (\frac {x \sin (b x)}{b}-\frac {\int \sin (b x)dx}{b}\right )}{b}-\frac {x^2 \cos (b x)}{b}\right )}{b}-\frac {x^3 \sin (b x)}{b}\right )+\frac {1}{4} x^4 \operatorname {CosIntegral}(b x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \left (\frac {3 \left (\frac {2 \left (\frac {x \sin (b x)}{b}-\frac {\int \sin (b x)dx}{b}\right )}{b}-\frac {x^2 \cos (b x)}{b}\right )}{b}-\frac {x^3 \sin (b x)}{b}\right )+\frac {1}{4} x^4 \operatorname {CosIntegral}(b x)\) |
\(\Big \downarrow \) 3118 |
\(\displaystyle \frac {1}{4} \left (\frac {3 \left (\frac {2 \left (\frac {\cos (b x)}{b^2}+\frac {x \sin (b x)}{b}\right )}{b}-\frac {x^2 \cos (b x)}{b}\right )}{b}-\frac {x^3 \sin (b x)}{b}\right )+\frac {1}{4} x^4 \operatorname {CosIntegral}(b x)\) |
Input:
Int[x^3*CosIntegral[b*x],x]
Output:
(x^4*CosIntegral[b*x])/4 + (-((x^3*Sin[b*x])/b) + (3*(-((x^2*Cos[b*x])/b) + (2*(Cos[b*x]/b^2 + (x*Sin[b*x])/b))/b))/b)/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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[CosIntegral[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] : > Simp[(c + d*x)^(m + 1)*(CosIntegral[a + b*x]/(d*(m + 1))), x] - Simp[b/(d *(m + 1)) Int[(c + d*x)^(m + 1)*(Cos[a + b*x]/(a + b*x)), x], x] /; FreeQ [{a, b, c, d, m}, x] && NeQ[m, -1]
Time = 0.82 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.86
method | result | size |
parts | \(\frac {x^{4} \operatorname {Ci}\left (b x \right )}{4}-\frac {b^{3} x^{3} \sin \left (b x \right )+3 b^{2} x^{2} \cos \left (b x \right )-6 \cos \left (b x \right )-6 b x \sin \left (b x \right )}{4 b^{4}}\) | \(54\) |
derivativedivides | \(\frac {\frac {b^{4} x^{4} \operatorname {Ci}\left (b x \right )}{4}-\frac {b^{3} x^{3} \sin \left (b x \right )}{4}-\frac {3 b^{2} x^{2} \cos \left (b x \right )}{4}+\frac {3 \cos \left (b x \right )}{2}+\frac {3 b x \sin \left (b x \right )}{2}}{b^{4}}\) | \(56\) |
default | \(\frac {\frac {b^{4} x^{4} \operatorname {Ci}\left (b x \right )}{4}-\frac {b^{3} x^{3} \sin \left (b x \right )}{4}-\frac {3 b^{2} x^{2} \cos \left (b x \right )}{4}+\frac {3 \cos \left (b x \right )}{2}+\frac {3 b x \sin \left (b x \right )}{2}}{b^{4}}\) | \(56\) |
meijerg | \(\frac {4 \sqrt {\pi }\, \left (\frac {\left (-\frac {1}{2}+2 \gamma +2 \ln \left (x \right )+2 \ln \left (b \right )\right ) x^{4} b^{4}}{32 \sqrt {\pi }}-\frac {b^{6} x^{6} \operatorname {hypergeom}\left (\left [1, 1, 3\right ], \left [\frac {3}{2}, 2, 2, 4\right ], -\frac {b^{2} x^{2}}{4}\right )}{96 \sqrt {\pi }}\right )}{b^{4}}\) | \(63\) |
orering | \(\frac {\left (b^{4} x^{4}+18 b^{2} x^{2}-72\right ) \operatorname {Ci}\left (b x \right )}{4 b^{4}}-\frac {\left (2 b^{2} x^{2}-9\right ) \left (3 x^{2} \operatorname {Ci}\left (b x \right )+x^{2} \cos \left (b x \right )\right )}{x^{2} b^{4}}+\frac {\left (b^{2} x^{2}-6\right ) \left (6 x \,\operatorname {Ci}\left (b x \right )+5 x \cos \left (b x \right )-x^{2} b \sin \left (b x \right )\right )}{4 x \,b^{4}}\) | \(106\) |
Input:
int(x^3*Ci(b*x),x,method=_RETURNVERBOSE)
Output:
1/4*x^4*Ci(b*x)-1/4/b^4*(b^3*x^3*sin(b*x)+3*b^2*x^2*cos(b*x)-6*cos(b*x)-6* b*x*sin(b*x))
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.94 \[ \int x^3 \operatorname {CosIntegral}(b x) \, dx=-\frac {\pi b^{3} x^{3} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 3 \, b x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - {\left (\pi ^{2} b^{4} x^{4} + 3\right )} \operatorname {C}\left (b x\right )}{4 \, \pi ^{2} b^{4}} \] Input:
integrate(x^3*fresnel_cos(b*x),x, algorithm="fricas")
Output:
-1/4*(pi*b^3*x^3*sin(1/2*pi*b^2*x^2) + 3*b*x*cos(1/2*pi*b^2*x^2) - (pi^2*b ^4*x^4 + 3)*fresnel_cos(b*x))/(pi^2*b^4)
Time = 1.55 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.35 \[ \int x^3 \operatorname {CosIntegral}(b x) \, dx=- \frac {x^{4} \log {\left (b x \right )}}{4} + \frac {x^{4} \log {\left (b^{2} x^{2} \right )}}{8} + \frac {x^{4} \operatorname {Ci}{\left (b x \right )}}{4} - \frac {x^{3} \sin {\left (b x \right )}}{4 b} - \frac {3 x^{2} \cos {\left (b x \right )}}{4 b^{2}} + \frac {3 x \sin {\left (b x \right )}}{2 b^{3}} + \frac {3 \cos {\left (b x \right )}}{2 b^{4}} \] Input:
integrate(x**3*Ci(b*x),x)
Output:
-x**4*log(b*x)/4 + x**4*log(b**2*x**2)/8 + x**4*Ci(b*x)/4 - x**3*sin(b*x)/ (4*b) - 3*x**2*cos(b*x)/(4*b**2) + 3*x*sin(b*x)/(2*b**3) + 3*cos(b*x)/(2*b **4)
Result contains complex when optimal does not.
Time = 0.14 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.49 \[ \int x^3 \operatorname {CosIntegral}(b x) \, dx=\frac {1}{4} \, x^{4} \operatorname {C}\left (b x\right ) - \frac {\sqrt {\frac {1}{2}} {\left (4 \, \sqrt {\frac {1}{2}} \pi ^{2} b^{3} x^{3} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 12 \, \sqrt {\frac {1}{2}} \pi b x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + \left (3 i - 3\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {\frac {1}{2} i \, \pi } b x\right ) - \left (3 i + 3\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {-\frac {1}{2} i \, \pi } b x\right )\right )}}{8 \, \pi ^{3} b^{4}} \] Input:
integrate(x^3*fresnel_cos(b*x),x, algorithm="maxima")
Output:
1/4*x^4*fresnel_cos(b*x) - 1/8*sqrt(1/2)*(4*sqrt(1/2)*pi^2*b^3*x^3*sin(1/2 *pi*b^2*x^2) + 12*sqrt(1/2)*pi*b*x*cos(1/2*pi*b^2*x^2) + (3*I - 3)*(1/4)^( 1/4)*pi*erf(sqrt(1/2*I*pi)*b*x) - (3*I + 3)*(1/4)^(1/4)*pi*erf(sqrt(-1/2*I *pi)*b*x))/(pi^3*b^4)
\[ \int x^3 \operatorname {CosIntegral}(b x) \, dx=\int { x^{3} \operatorname {C}\left (b x\right ) \,d x } \] Input:
integrate(x^3*fresnel_cos(b*x),x, algorithm="giac")
Output:
integrate(x^3*fresnel_cos(b*x), x)
Timed out. \[ \int x^3 \operatorname {CosIntegral}(b x) \, dx=\frac {6\,\cos \left (b\,x\right )-3\,b^2\,x^2\,\cos \left (b\,x\right )-b^3\,x^3\,\sin \left (b\,x\right )+6\,b\,x\,\sin \left (b\,x\right )}{4\,b^4}+\frac {x^4\,\mathrm {cosint}\left (b\,x\right )}{4} \] Input:
int(x^3*cosint(b*x),x)
Output:
(6*cos(b*x) - 3*b^2*x^2*cos(b*x) - b^3*x^3*sin(b*x) + 6*b*x*sin(b*x))/(4*b ^4) + (x^4*cosint(b*x))/4
Time = 0.19 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.87 \[ \int x^3 \operatorname {CosIntegral}(b x) \, dx=\frac {\mathit {ci} \left (b x \right ) b^{4} x^{4}-3 \cos \left (b x \right ) b^{2} x^{2}+6 \cos \left (b x \right )-\sin \left (b x \right ) b^{3} x^{3}+6 \sin \left (b x \right ) b x}{4 b^{4}} \] Input:
int(x^3*Ci(b*x),x)
Output:
(ci(b*x)*b**4*x**4 - 3*cos(b*x)*b**2*x**2 + 6*cos(b*x) - sin(b*x)*b**3*x** 3 + 6*sin(b*x)*b*x)/(4*b**4)