Integrand size = 6, antiderivative size = 35 \[ \int x \operatorname {CosIntegral}(b x) \, dx=-\frac {\cos (b x)}{2 b^2}+\frac {1}{2} x^2 \operatorname {CosIntegral}(b x)-\frac {x \sin (b x)}{2 b} \] Output:
-1/2*cos(b*x)/b^2+1/2*x^2*Ci(b*x)-1/2*x*sin(b*x)/b
Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int x \operatorname {CosIntegral}(b x) \, dx=-\frac {\cos (b x)}{2 b^2}+\frac {1}{2} x^2 \operatorname {CosIntegral}(b x)-\frac {x \sin (b x)}{2 b} \] Input:
Integrate[x*CosIntegral[b*x],x]
Output:
-1/2*Cos[b*x]/b^2 + (x^2*CosIntegral[b*x])/2 - (x*Sin[b*x])/(2*b)
Time = 0.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.167, Rules used = {7058, 27, 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.
\(\displaystyle \int x \operatorname {CosIntegral}(b x) \, dx\) |
\(\Big \downarrow \) 7058 |
\(\displaystyle \frac {1}{2} x^2 \operatorname {CosIntegral}(b x)-\frac {1}{2} b \int \frac {x \cos (b x)}{b}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} x^2 \operatorname {CosIntegral}(b x)-\frac {1}{2} \int x \cos (b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} x^2 \operatorname {CosIntegral}(b x)-\frac {1}{2} \int x \sin \left (b x+\frac {\pi }{2}\right )dx\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{2} \left (-\frac {\int -\sin (b x)dx}{b}-\frac {x \sin (b x)}{b}\right )+\frac {1}{2} x^2 \operatorname {CosIntegral}(b x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \sin (b x)dx}{b}-\frac {x \sin (b x)}{b}\right )+\frac {1}{2} x^2 \operatorname {CosIntegral}(b x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \sin (b x)dx}{b}-\frac {x \sin (b x)}{b}\right )+\frac {1}{2} x^2 \operatorname {CosIntegral}(b x)\) |
\(\Big \downarrow \) 3118 |
\(\displaystyle \frac {1}{2} \left (-\frac {\cos (b x)}{b^2}-\frac {x \sin (b x)}{b}\right )+\frac {1}{2} x^2 \operatorname {CosIntegral}(b x)\) |
Input:
Int[x*CosIntegral[b*x],x]
Output:
(x^2*CosIntegral[b*x])/2 + (-(Cos[b*x]/b^2) - (x*Sin[b*x])/b)/2
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 = 0.78 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80
method | result | size |
parts | \(\frac {x^{2} \operatorname {Ci}\left (b x \right )}{2}-\frac {\cos \left (b x \right )+b x \sin \left (b x \right )}{2 b^{2}}\) | \(28\) |
derivativedivides | \(\frac {\frac {b^{2} x^{2} \operatorname {Ci}\left (b x \right )}{2}-\frac {\cos \left (b x \right )}{2}-\frac {b x \sin \left (b x \right )}{2}}{b^{2}}\) | \(32\) |
default | \(\frac {\frac {b^{2} x^{2} \operatorname {Ci}\left (b x \right )}{2}-\frac {\cos \left (b x \right )}{2}-\frac {b x \sin \left (b x \right )}{2}}{b^{2}}\) | \(32\) |
orering | \(\frac {\left (b^{2} x^{2}+2\right ) \operatorname {Ci}\left (b x \right )}{2 b^{2}}-\frac {\operatorname {Ci}\left (b x \right )+\cos \left (b x \right )}{b^{2}}+\frac {x \left (\frac {\cos \left (b x \right )}{x}-b \sin \left (b x \right )\right )}{2 b^{2}}\) | \(56\) |
meijerg | \(\frac {\sqrt {\pi }\, \left (\frac {\left (2 \gamma -1+2 \ln \left (x \right )+2 \ln \left (b \right )\right ) x^{2} b^{2}}{4 \sqrt {\pi }}+\frac {\frac {b^{2} x^{2}}{2}+1}{2 \sqrt {\pi }}-\frac {b^{2} x^{2} \gamma }{2 \sqrt {\pi }}-\frac {b^{2} x^{2} \ln \left (2\right )}{2 \sqrt {\pi }}-\frac {b^{2} x^{2} \ln \left (\frac {b x}{2}\right )}{2 \sqrt {\pi }}-\frac {\cos \left (b x \right )}{2 \sqrt {\pi }}-\frac {b x \sin \left (b x \right )}{2 \sqrt {\pi }}+\frac {b^{2} x^{2} \operatorname {Ci}\left (b x \right )}{2 \sqrt {\pi }}\right )}{b^{2}}\) | \(124\) |
Input:
int(x*Ci(b*x),x,method=_RETURNVERBOSE)
Output:
1/2*x^2*Ci(b*x)-1/2/b^2*(cos(b*x)+b*x*sin(b*x))
Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.46 \[ \int x \operatorname {CosIntegral}(b x) \, dx=\frac {\pi b^{3} x^{2} \operatorname {C}\left (b x\right ) - b^{2} x \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + \sqrt {b^{2}} \operatorname {S}\left (\sqrt {b^{2}} x\right )}{2 \, \pi b^{3}} \] Input:
integrate(x*fresnel_cos(b*x),x, algorithm="fricas")
Output:
1/2*(pi*b^3*x^2*fresnel_cos(b*x) - b^2*x*sin(1/2*pi*b^2*x^2) + sqrt(b^2)*f resnel_sin(sqrt(b^2)*x))/(pi*b^3)
Time = 1.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.51 \[ \int x \operatorname {CosIntegral}(b x) \, dx=- \frac {x^{2} \log {\left (b x \right )}}{2} + \frac {x^{2} \log {\left (b^{2} x^{2} \right )}}{4} + \frac {x^{2} \operatorname {Ci}{\left (b x \right )}}{2} - \frac {x \sin {\left (b x \right )}}{2 b} - \frac {\cos {\left (b x \right )}}{2 b^{2}} \] Input:
integrate(x*Ci(b*x),x)
Output:
-x**2*log(b*x)/2 + x**2*log(b**2*x**2)/4 + x**2*Ci(b*x)/2 - x*sin(b*x)/(2* b) - cos(b*x)/(2*b**2)
Result contains complex when optimal does not.
Time = 0.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.00 \[ \int x \operatorname {CosIntegral}(b x) \, dx=\frac {1}{2} \, x^{2} \operatorname {C}\left (b x\right ) - \frac {\sqrt {\frac {1}{2}} {\left (4 \, \sqrt {\frac {1}{2}} \pi b x \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - \left (i + 1\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {\frac {1}{2} i \, \pi } b x\right ) + \left (i - 1\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {-\frac {1}{2} i \, \pi } b x\right )\right )}}{4 \, \pi ^{2} b^{2}} \] Input:
integrate(x*fresnel_cos(b*x),x, algorithm="maxima")
Output:
1/2*x^2*fresnel_cos(b*x) - 1/4*sqrt(1/2)*(4*sqrt(1/2)*pi*b*x*sin(1/2*pi*b^ 2*x^2) - (I + 1)*(1/4)^(1/4)*pi*erf(sqrt(1/2*I*pi)*b*x) + (I - 1)*(1/4)^(1 /4)*pi*erf(sqrt(-1/2*I*pi)*b*x))/(pi^2*b^2)
\[ \int x \operatorname {CosIntegral}(b x) \, dx=\int { x \operatorname {C}\left (b x\right ) \,d x } \] Input:
integrate(x*fresnel_cos(b*x),x, algorithm="giac")
Output:
integrate(x*fresnel_cos(b*x), x)
Timed out. \[ \int x \operatorname {CosIntegral}(b x) \, dx=\frac {x^2\,\mathrm {cosint}\left (b\,x\right )}{2}-\frac {\cos \left (b\,x\right )+b\,x\,\sin \left (b\,x\right )}{2\,b^2} \] Input:
int(x*cosint(b*x),x)
Output:
(x^2*cosint(b*x))/2 - (cos(b*x) + b*x*sin(b*x))/(2*b^2)
Time = 0.22 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int x \operatorname {CosIntegral}(b x) \, dx=\frac {\mathit {ci} \left (b x \right ) b^{2} x^{2}-\cos \left (b x \right )-\sin \left (b x \right ) b x}{2 b^{2}} \] Input:
int(x*Ci(b*x),x)
Output:
(ci(b*x)*b**2*x**2 - cos(b*x) - sin(b*x)*b*x)/(2*b**2)