\(\int x \cos (b x) \operatorname {CosIntegral}(b x) \, dx\) [118]

Optimal result

Integrand size = 10, antiderivative size = 60 \[ \int x \cos (b x) \operatorname {CosIntegral}(b x) \, dx=\frac {\cos (b x) \operatorname {CosIntegral}(b x)}{b^2}-\frac {\operatorname {CosIntegral}(2 b x)}{2 b^2}-\frac {\log (x)}{2 b^2}+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}-\frac {\sin ^2(b x)}{2 b^2} \] Output:

cos(b*x)*Ci(b*x)/b^2-1/2*Ci(2*b*x)/b^2-1/2*ln(x)/b^2+x*Ci(b*x)*sin(b*x)/b- 
1/2*sin(b*x)^2/b^2
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.70 \[ \int x \cos (b x) \operatorname {CosIntegral}(b x) \, dx=\frac {\cos (2 b x)-2 \operatorname {CosIntegral}(2 b x)-2 \log (x)+4 \operatorname {CosIntegral}(b x) (\cos (b x)+b x \sin (b x))}{4 b^2} \] Input:

Integrate[x*Cos[b*x]*CosIntegral[b*x],x]
 

Output:

(Cos[2*b*x] - 2*CosIntegral[2*b*x] - 2*Log[x] + 4*CosIntegral[b*x]*(Cos[b* 
x] + b*x*Sin[b*x]))/(4*b^2)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {7068, 27, 3042, 3044, 15, 7072, 27, 3042, 3793, 2009}

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.

\(\displaystyle \int x \operatorname {CosIntegral}(b x) \cos (b x) \, dx\)

\(\Big \downarrow \) 7068

\(\displaystyle -\frac {\int \operatorname {CosIntegral}(b x) \sin (b x)dx}{b}-\int \frac {\cos (b x) \sin (b x)}{b}dx+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \operatorname {CosIntegral}(b x) \sin (b x)dx}{b}-\frac {\int \cos (b x) \sin (b x)dx}{b}+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \operatorname {CosIntegral}(b x) \sin (b x)dx}{b}-\frac {\int \cos (b x) \sin (b x)dx}{b}+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}\)

\(\Big \downarrow \) 3044

\(\displaystyle -\frac {\int \sin (b x)d\sin (b x)}{b^2}-\frac {\int \operatorname {CosIntegral}(b x) \sin (b x)dx}{b}+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}\)

\(\Big \downarrow \) 15

\(\displaystyle -\frac {\int \operatorname {CosIntegral}(b x) \sin (b x)dx}{b}-\frac {\sin ^2(b x)}{2 b^2}+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}\)

\(\Big \downarrow \) 7072

\(\displaystyle -\frac {\int \frac {\cos ^2(b x)}{b x}dx-\frac {\operatorname {CosIntegral}(b x) \cos (b x)}{b}}{b}-\frac {\sin ^2(b x)}{2 b^2}+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\int \frac {\cos ^2(b x)}{x}dx}{b}-\frac {\operatorname {CosIntegral}(b x) \cos (b x)}{b}}{b}-\frac {\sin ^2(b x)}{2 b^2}+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\int \frac {\sin \left (b x+\frac {\pi }{2}\right )^2}{x}dx}{b}-\frac {\operatorname {CosIntegral}(b x) \cos (b x)}{b}}{b}-\frac {\sin ^2(b x)}{2 b^2}+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}\)

\(\Big \downarrow \) 3793

\(\displaystyle -\frac {\frac {\int \left (\frac {\cos (2 b x)}{2 x}+\frac {1}{2 x}\right )dx}{b}-\frac {\operatorname {CosIntegral}(b x) \cos (b x)}{b}}{b}-\frac {\sin ^2(b x)}{2 b^2}+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\sin ^2(b x)}{2 b^2}+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}-\frac {\frac {\frac {\operatorname {CosIntegral}(2 b x)}{2}+\frac {\log (x)}{2}}{b}-\frac {\operatorname {CosIntegral}(b x) \cos (b x)}{b}}{b}\)

Input:

Int[x*Cos[b*x]*CosIntegral[b*x],x]
 

Output:

-((-((Cos[b*x]*CosIntegral[b*x])/b) + (CosIntegral[2*b*x]/2 + Log[x]/2)/b) 
/b) + (x*CosIntegral[b*x]*Sin[b*x])/b - Sin[b*x]^2/(2*b^2)
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3044

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ 
Symbol] :> Simp[1/(a*f)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a 
*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&  !(I 
ntegerQ[(m - 1)/2] && LtQ[0, m, n])
 

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
 

rule 7068

Int[Cos[(a_.) + (b_.)*(x_)]*CosIntegral[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)* 
(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^m*Sin[a + b*x]*(CosIntegral[c + d* 
x]/b), x] + (-Simp[d/b   Int[(e + f*x)^m*Sin[a + b*x]*(Cos[c + d*x]/(c + d* 
x)), x], x] - Simp[f*(m/b)   Int[(e + f*x)^(m - 1)*Sin[a + b*x]*CosIntegral 
[c + d*x], x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0]
 

rule 7072

Int[CosIntegral[(c_.) + (d_.)*(x_)]*Sin[(a_.) + (b_.)*(x_)], x_Symbol] :> S 
imp[(-Cos[a + b*x])*(CosIntegral[c + d*x]/b), x] + Simp[d/b   Int[Cos[a + b 
*x]*(Cos[c + d*x]/(c + d*x)), x], x] /; FreeQ[{a, b, c, d}, x]
 

Maple [A] (verified)

Time = 5.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {\operatorname {Ci}\left (b x \right ) \left (\cos \left (b x \right )+b x \sin \left (b x \right )\right )-\frac {\ln \left (b x \right )}{2}-\frac {\operatorname {Ci}\left (2 b x \right )}{2}+\frac {\cos \left (b x \right )^{2}}{2}}{b^{2}}\)	\(44\)
default	\(\frac {\operatorname {Ci}\left (b x \right ) \left (\cos \left (b x \right )+b x \sin \left (b x \right )\right )-\frac {\ln \left (b x \right )}{2}-\frac {\operatorname {Ci}\left (2 b x \right )}{2}+\frac {\cos \left (b x \right )^{2}}{2}}{b^{2}}\)	\(44\)

Input:

int(x*cos(b*x)*Ci(b*x),x,method=_RETURNVERBOSE)
 

Output:

1/b^2*(Ci(b*x)*(cos(b*x)+b*x*sin(b*x))-1/2*ln(b*x)-1/2*Ci(2*b*x)+1/2*cos(b 
*x)^2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (54) = 108\).

Time = 0.10 (sec) , antiderivative size = 221, normalized size of antiderivative = 3.68 \[ \int x \cos (b x) \operatorname {CosIntegral}(b x) \, dx=\frac {2 \, \pi b^{2} x \operatorname {C}\left (b x\right ) \sin \left (b x\right ) + 2 \, \pi b \cos \left (b x\right ) \operatorname {C}\left (b x\right ) - 2 \, b \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \sin \left (b x\right ) - \sqrt {b^{2}} {\left (\pi \cos \left (\frac {1}{2 \, \pi }\right ) + \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {C}\left (\frac {{\left (\pi b x + 1\right )} \sqrt {b^{2}}}{\pi b}\right ) - \sqrt {b^{2}} {\left (\pi \cos \left (\frac {1}{2 \, \pi }\right ) + \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {C}\left (\frac {{\left (\pi b x - 1\right )} \sqrt {b^{2}}}{\pi b}\right ) - \sqrt {b^{2}} {\left (\pi \sin \left (\frac {1}{2 \, \pi }\right ) - \cos \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {S}\left (\frac {{\left (\pi b x + 1\right )} \sqrt {b^{2}}}{\pi b}\right ) - \sqrt {b^{2}} {\left (\pi \sin \left (\frac {1}{2 \, \pi }\right ) - \cos \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {S}\left (\frac {{\left (\pi b x - 1\right )} \sqrt {b^{2}}}{\pi b}\right )}{2 \, \pi b^{3}} \] Input:

integrate(x*cos(b*x)*fresnel_cos(b*x),x, algorithm="fricas")
 

Output:

1/2*(2*pi*b^2*x*fresnel_cos(b*x)*sin(b*x) + 2*pi*b*cos(b*x)*fresnel_cos(b* 
x) - 2*b*sin(1/2*pi*b^2*x^2)*sin(b*x) - sqrt(b^2)*(pi*cos(1/2/pi) + sin(1/ 
2/pi))*fresnel_cos((pi*b*x + 1)*sqrt(b^2)/(pi*b)) - sqrt(b^2)*(pi*cos(1/2/ 
pi) + sin(1/2/pi))*fresnel_cos((pi*b*x - 1)*sqrt(b^2)/(pi*b)) - sqrt(b^2)* 
(pi*sin(1/2/pi) - cos(1/2/pi))*fresnel_sin((pi*b*x + 1)*sqrt(b^2)/(pi*b)) 
- sqrt(b^2)*(pi*sin(1/2/pi) - cos(1/2/pi))*fresnel_sin((pi*b*x - 1)*sqrt(b 
^2)/(pi*b)))/(pi*b^3)
 

Sympy [F]

\[ \int x \cos (b x) \operatorname {CosIntegral}(b x) \, dx=\int x \cos {\left (b x \right )} \operatorname {Ci}{\left (b x \right )}\, dx \] Input:

integrate(x*cos(b*x)*Ci(b*x),x)
 

Output:

Integral(x*cos(b*x)*Ci(b*x), x)
 

Maxima [F]

\[ \int x \cos (b x) \operatorname {CosIntegral}(b x) \, dx=\int { x \cos \left (b x\right ) \operatorname {C}\left (b x\right ) \,d x } \] Input:

integrate(x*cos(b*x)*fresnel_cos(b*x),x, algorithm="maxima")
 

Output:

integrate(x*cos(b*x)*fresnel_cos(b*x), x)
 

Giac [F]

\[ \int x \cos (b x) \operatorname {CosIntegral}(b x) \, dx=\int { x \cos \left (b x\right ) \operatorname {C}\left (b x\right ) \,d x } \] Input:

integrate(x*cos(b*x)*fresnel_cos(b*x),x, algorithm="giac")
 

Output:

integrate(x*cos(b*x)*fresnel_cos(b*x), x)
 

Mupad [F(-1)]

Timed out. \[ \int x \cos (b x) \operatorname {CosIntegral}(b x) \, dx=\int x\,\mathrm {cosint}\left (b\,x\right )\,\cos \left (b\,x\right ) \,d x \] Input:

int(x*cosint(b*x)*cos(b*x),x)
 

Output:

int(x*cosint(b*x)*cos(b*x), x)
 

Reduce [F]

\[ \int x \cos (b x) \operatorname {CosIntegral}(b x) \, dx=\int \mathit {ci} \left (b x \right ) \cos \left (b x \right ) x d x \] Input:

int(x*cos(b*x)*Ci(b*x),x)
 

Output:

int(ci(b*x)*cos(b*x)*x,x)