Integrand size = 8, antiderivative size = 49 \[ \int x^2 \operatorname {CosIntegral}(b x) \, dx=-\frac {2 x \cos (b x)}{3 b^2}+\frac {1}{3} x^3 \operatorname {CosIntegral}(b x)+\frac {2 \sin (b x)}{3 b^3}-\frac {x^2 \sin (b x)}{3 b} \] Output:
-2/3*x*cos(b*x)/b^2+1/3*x^3*Ci(b*x)+2/3*sin(b*x)/b^3-1/3*x^2*sin(b*x)/b
Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90 \[ \int x^2 \operatorname {CosIntegral}(b x) \, dx=-\frac {2 x \cos (b x)}{3 b^2}+\frac {1}{3} x^3 \operatorname {CosIntegral}(b x)-\frac {\left (-2+b^2 x^2\right ) \sin (b x)}{3 b^3} \] Input:
Integrate[x^2*CosIntegral[b*x],x]
Output:
(-2*x*Cos[b*x])/(3*b^2) + (x^3*CosIntegral[b*x])/3 - ((-2 + b^2*x^2)*Sin[b *x])/(3*b^3)
Time = 0.38 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.125, Rules used = {7058, 27, 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.
\(\displaystyle \int x^2 \operatorname {CosIntegral}(b x) \, dx\) |
\(\Big \downarrow \) 7058 |
\(\displaystyle \frac {1}{3} x^3 \operatorname {CosIntegral}(b x)-\frac {1}{3} b \int \frac {x^2 \cos (b x)}{b}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} x^3 \operatorname {CosIntegral}(b x)-\frac {1}{3} \int x^2 \cos (b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} x^3 \operatorname {CosIntegral}(b x)-\frac {1}{3} \int x^2 \sin \left (b x+\frac {\pi }{2}\right )dx\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{3} \left (-\frac {2 \int -x \sin (b x)dx}{b}-\frac {x^2 \sin (b x)}{b}\right )+\frac {1}{3} x^3 \operatorname {CosIntegral}(b x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} \left (\frac {2 \int x \sin (b x)dx}{b}-\frac {x^2 \sin (b x)}{b}\right )+\frac {1}{3} x^3 \operatorname {CosIntegral}(b x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \left (\frac {2 \int x \sin (b x)dx}{b}-\frac {x^2 \sin (b x)}{b}\right )+\frac {1}{3} x^3 \operatorname {CosIntegral}(b x)\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{3} \left (\frac {2 \left (\frac {\int \cos (b x)dx}{b}-\frac {x \cos (b x)}{b}\right )}{b}-\frac {x^2 \sin (b x)}{b}\right )+\frac {1}{3} x^3 \operatorname {CosIntegral}(b x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \left (\frac {2 \left (\frac {\int \sin \left (b x+\frac {\pi }{2}\right )dx}{b}-\frac {x \cos (b x)}{b}\right )}{b}-\frac {x^2 \sin (b x)}{b}\right )+\frac {1}{3} x^3 \operatorname {CosIntegral}(b x)\) |
\(\Big \downarrow \) 3117 |
\(\displaystyle \frac {1}{3} \left (\frac {2 \left (\frac {\sin (b x)}{b^2}-\frac {x \cos (b x)}{b}\right )}{b}-\frac {x^2 \sin (b x)}{b}\right )+\frac {1}{3} x^3 \operatorname {CosIntegral}(b x)\) |
Input:
Int[x^2*CosIntegral[b*x],x]
Output:
(x^3*CosIntegral[b*x])/3 + (-((x^2*Sin[b*x])/b) + (2*(-((x*Cos[b*x])/b) + Sin[b*x]/b^2))/b)/3
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[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 = 0.81 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.86
method | result | size |
parts | \(\frac {x^{3} \operatorname {Ci}\left (b x \right )}{3}-\frac {b^{2} x^{2} \sin \left (b x \right )-2 \sin \left (b x \right )+2 b x \cos \left (b x \right )}{3 b^{3}}\) | \(42\) |
derivativedivides | \(\frac {\frac {b^{3} x^{3} \operatorname {Ci}\left (b x \right )}{3}-\frac {b^{2} x^{2} \sin \left (b x \right )}{3}+\frac {2 \sin \left (b x \right )}{3}-\frac {2 b x \cos \left (b x \right )}{3}}{b^{3}}\) | \(44\) |
default | \(\frac {\frac {b^{3} x^{3} \operatorname {Ci}\left (b x \right )}{3}-\frac {b^{2} x^{2} \sin \left (b x \right )}{3}+\frac {2 \sin \left (b x \right )}{3}-\frac {2 b x \cos \left (b x \right )}{3}}{b^{3}}\) | \(44\) |
meijerg | \(\frac {2 \sqrt {\pi }\, \left (\frac {\left (-\frac {2}{3}+2 \gamma +2 \ln \left (x \right )+2 \ln \left (b \right )\right ) x^{3} b^{3}}{12 \sqrt {\pi }}-\frac {b^{5} x^{5} \operatorname {hypergeom}\left (\left [1, 1, \frac {5}{2}\right ], \left [\frac {3}{2}, 2, 2, \frac {7}{2}\right ], -\frac {b^{2} x^{2}}{4}\right )}{40 \sqrt {\pi }}\right )}{b^{3}}\) | \(63\) |
orering | \(\frac {\left (b^{4} x^{4}+8 b^{2} x^{2}-8\right ) \operatorname {Ci}\left (b x \right )}{3 b^{4} x}-\frac {\left (5 b^{2} x^{2}-6\right ) \left (2 x \,\operatorname {Ci}\left (b x \right )+x \cos \left (b x \right )\right )}{3 b^{4} x^{2}}+\frac {\left (b^{2} x^{2}-2\right ) \left (2 \,\operatorname {Ci}\left (b x \right )+3 \cos \left (b x \right )-b x \sin \left (b x \right )\right )}{3 b^{4} x}\) | \(101\) |
Input:
int(x^2*Ci(b*x),x,method=_RETURNVERBOSE)
Output:
1/3*x^3*Ci(b*x)-1/3/b^3*(b^2*x^2*sin(b*x)-2*sin(b*x)+2*b*x*cos(b*x))
Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.10 \[ \int x^2 \operatorname {CosIntegral}(b x) \, dx=\frac {\pi ^{2} b^{3} x^{3} \operatorname {C}\left (b x\right ) - \pi b^{2} x^{2} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 2 \, \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{3 \, \pi ^{2} b^{3}} \] Input:
integrate(x^2*fresnel_cos(b*x),x, algorithm="fricas")
Output:
1/3*(pi^2*b^3*x^3*fresnel_cos(b*x) - pi*b^2*x^2*sin(1/2*pi*b^2*x^2) - 2*co s(1/2*pi*b^2*x^2))/(pi^2*b^3)
Time = 1.33 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.43 \[ \int x^2 \operatorname {CosIntegral}(b x) \, dx=- \frac {x^{3} \log {\left (b x \right )}}{3} + \frac {x^{3} \log {\left (b^{2} x^{2} \right )}}{6} + \frac {x^{3} \operatorname {Ci}{\left (b x \right )}}{3} - \frac {x^{2} \sin {\left (b x \right )}}{3 b} - \frac {2 x \cos {\left (b x \right )}}{3 b^{2}} + \frac {2 \sin {\left (b x \right )}}{3 b^{3}} \] Input:
integrate(x**2*Ci(b*x),x)
Output:
-x**3*log(b*x)/3 + x**3*log(b**2*x**2)/6 + x**3*Ci(b*x)/3 - x**2*sin(b*x)/ (3*b) - 2*x*cos(b*x)/(3*b**2) + 2*sin(b*x)/(3*b**3)
Time = 0.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int x^2 \operatorname {CosIntegral}(b x) \, dx=\frac {1}{3} \, x^{3} \operatorname {C}\left (b x\right ) - \frac {\pi b^{2} x^{2} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 2 \, \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{3 \, \pi ^{2} b^{3}} \] Input:
integrate(x^2*fresnel_cos(b*x),x, algorithm="maxima")
Output:
1/3*x^3*fresnel_cos(b*x) - 1/3*(pi*b^2*x^2*sin(1/2*pi*b^2*x^2) + 2*cos(1/2 *pi*b^2*x^2))/(pi^2*b^3)
\[ \int x^2 \operatorname {CosIntegral}(b x) \, dx=\int { x^{2} \operatorname {C}\left (b x\right ) \,d x } \] Input:
integrate(x^2*fresnel_cos(b*x),x, algorithm="giac")
Output:
integrate(x^2*fresnel_cos(b*x), x)
Timed out. \[ \int x^2 \operatorname {CosIntegral}(b x) \, dx=\frac {x^3\,\mathrm {cosint}\left (b\,x\right )}{3}-\frac {b^2\,x^2\,\sin \left (b\,x\right )-2\,\sin \left (b\,x\right )+2\,b\,x\,\cos \left (b\,x\right )}{3\,b^3} \] Input:
int(x^2*cosint(b*x),x)
Output:
(x^3*cosint(b*x))/3 - (b^2*x^2*sin(b*x) - 2*sin(b*x) + 2*b*x*cos(b*x))/(3* b^3)
Time = 0.18 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.88 \[ \int x^2 \operatorname {CosIntegral}(b x) \, dx=\frac {\mathit {ci} \left (b x \right ) b^{3} x^{3}-2 \cos \left (b x \right ) b x -\sin \left (b x \right ) b^{2} x^{2}+2 \sin \left (b x \right )}{3 b^{3}} \] Input:
int(x^2*Ci(b*x),x)
Output:
(ci(b*x)*b**3*x**3 - 2*cos(b*x)*b*x - sin(b*x)*b**2*x**2 + 2*sin(b*x))/(3* b**3)