\(\int x \cos (a+b x) \operatorname {CosIntegral}(a+b x) \, dx\) [128]

Optimal result

Integrand size = 14, antiderivative size = 96 \[ \int x \cos (a+b x) \operatorname {CosIntegral}(a+b x) \, dx=\frac {\cos (2 a+2 b x)}{4 b^2}+\frac {\cos (a+b x) \operatorname {CosIntegral}(a+b x)}{b^2}-\frac {\operatorname {CosIntegral}(2 a+2 b x)}{2 b^2}-\frac {\log (a+b x)}{2 b^2}+\frac {x \operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b}+\frac {a \text {Si}(2 a+2 b x)}{2 b^2} \] Output:

1/4*cos(2*b*x+2*a)/b^2+cos(b*x+a)*Ci(b*x+a)/b^2-1/2*Ci(2*b*x+2*a)/b^2-1/2* 
ln(b*x+a)/b^2+x*Ci(b*x+a)*sin(b*x+a)/b+1/2*a*Si(2*b*x+2*a)/b^2
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.72 \[ \int x \cos (a+b x) \operatorname {CosIntegral}(a+b x) \, dx=\frac {\cos (2 (a+b x))-2 \operatorname {CosIntegral}(2 (a+b x))-2 \log (a+b x)+4 \operatorname {CosIntegral}(a+b x) (\cos (a+b x)+b x \sin (a+b x))+2 a \text {Si}(2 (a+b x))}{4 b^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {7068, 5084, 7072, 3042, 3793, 2009, 7292, 7293, 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.

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

\(\Big \downarrow \) 7068

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

\(\Big \downarrow \) 5084

\(\displaystyle -\frac {\int \operatorname {CosIntegral}(a+b x) \sin (a+b x)dx}{b}-\frac {1}{2} \int \frac {x \sin (2 (a+b x))}{a+b x}dx+\frac {x \operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b}\)

\(\Big \downarrow \) 7072

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3793

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {1}{2} \int \frac {x \sin (2 (a+b x))}{a+b x}dx+\frac {x \operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b}-\frac {\frac {\operatorname {CosIntegral}(2 a+2 b x)}{2 b}-\frac {\operatorname {CosIntegral}(a+b x) \cos (a+b x)}{b}+\frac {\log (a+b x)}{2 b}}{b}\)

\(\Big \downarrow \) 7292

\(\displaystyle -\frac {1}{2} \int \frac {x \sin (2 a+2 b x)}{a+b x}dx+\frac {x \operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b}-\frac {\frac {\operatorname {CosIntegral}(2 a+2 b x)}{2 b}-\frac {\operatorname {CosIntegral}(a+b x) \cos (a+b x)}{b}+\frac {\log (a+b x)}{2 b}}{b}\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {a \text {Si}(2 a+2 b x)}{b^2}+\frac {\cos (2 a+2 b x)}{2 b^2}\right )+\frac {x \operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b}-\frac {\frac {\operatorname {CosIntegral}(2 a+2 b x)}{2 b}-\frac {\operatorname {CosIntegral}(a+b x) \cos (a+b x)}{b}+\frac {\log (a+b x)}{2 b}}{b}\)

Input:

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

Output:

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

Defintions of rubi rules used

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 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 5084

Int[Cos[w_]^(p_.)*(u_.)*Sin[v_]^(p_.), x_Symbol] :> Simp[1/2^p   Int[u*Sin[ 
2*v]^p, x], x] /; EqQ[w, v] && IntegerQ[p]
 

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]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 10.38 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {\operatorname {Ci}\left (b x +a \right ) \left (\cos \left (b x +a \right )+\left (b x +a \right ) \sin \left (b x +a \right )-a \sin \left (b x +a \right )\right )-\frac {\ln \left (b x +a \right )}{2}-\frac {\operatorname {Ci}\left (2 b x +2 a \right )}{2}+\frac {\cos \left (b x +a \right )^{2}}{2}+\frac {a \,\operatorname {Si}\left (2 b x +2 a \right )}{2}}{b^{2}}\)	\(82\)
default	\(\frac {\operatorname {Ci}\left (b x +a \right ) \left (\cos \left (b x +a \right )+\left (b x +a \right ) \sin \left (b x +a \right )-a \sin \left (b x +a \right )\right )-\frac {\ln \left (b x +a \right )}{2}-\frac {\operatorname {Ci}\left (2 b x +2 a \right )}{2}+\frac {\cos \left (b x +a \right )^{2}}{2}+\frac {a \,\operatorname {Si}\left (2 b x +2 a \right )}{2}}{b^{2}}\)	\(82\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 276 vs. \(2 (88) = 176\).

Time = 0.10 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.88 \[ \int x \cos (a+b x) \operatorname {CosIntegral}(a+b x) \, dx=\frac {2 \, \pi b^{2} x \operatorname {C}\left (b x + a\right ) \sin \left (b x + a\right ) + 2 \, \pi b \cos \left (b x + a\right ) \operatorname {C}\left (b x + a\right ) - 2 \, b \sin \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) \sin \left (b x + a\right ) - \sqrt {b^{2}} {\left (\pi \cos \left (\frac {1}{2 \, \pi }\right ) + {\left (\pi a + 1\right )} \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {C}\left (\frac {{\left (\pi b x + \pi a + 1\right )} \sqrt {b^{2}}}{\pi b}\right ) - \sqrt {b^{2}} {\left (\pi \cos \left (\frac {1}{2 \, \pi }\right ) - {\left (\pi a - 1\right )} \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {C}\left (\frac {{\left (\pi b x + \pi a - 1\right )} \sqrt {b^{2}}}{\pi b}\right ) + \sqrt {b^{2}} {\left ({\left (\pi a + 1\right )} \cos \left (\frac {1}{2 \, \pi }\right ) - \pi \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {S}\left (\frac {{\left (\pi b x + \pi a + 1\right )} \sqrt {b^{2}}}{\pi b}\right ) - \sqrt {b^{2}} {\left ({\left (\pi a - 1\right )} \cos \left (\frac {1}{2 \, \pi }\right ) + \pi \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {S}\left (\frac {{\left (\pi b x + \pi a - 1\right )} \sqrt {b^{2}}}{\pi b}\right )}{2 \, \pi b^{3}} \] Input:

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

Output:

1/2*(2*pi*b^2*x*fresnel_cos(b*x + a)*sin(b*x + a) + 2*pi*b*cos(b*x + a)*fr 
esnel_cos(b*x + a) - 2*b*sin(1/2*pi*b^2*x^2 + pi*a*b*x + 1/2*pi*a^2)*sin(b 
*x + a) - sqrt(b^2)*(pi*cos(1/2/pi) + (pi*a + 1)*sin(1/2/pi))*fresnel_cos( 
(pi*b*x + pi*a + 1)*sqrt(b^2)/(pi*b)) - sqrt(b^2)*(pi*cos(1/2/pi) - (pi*a 
- 1)*sin(1/2/pi))*fresnel_cos((pi*b*x + pi*a - 1)*sqrt(b^2)/(pi*b)) + sqrt 
(b^2)*((pi*a + 1)*cos(1/2/pi) - pi*sin(1/2/pi))*fresnel_sin((pi*b*x + pi*a 
 + 1)*sqrt(b^2)/(pi*b)) - sqrt(b^2)*((pi*a - 1)*cos(1/2/pi) + pi*sin(1/2/p 
i))*fresnel_sin((pi*b*x + pi*a - 1)*sqrt(b^2)/(pi*b)))/(pi*b^3)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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