\(\int x \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx\) [124]

Optimal result

Integrand size = 14, antiderivative size = 109 \[ \int x \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx=\frac {x}{2 b}-\frac {x \cos (a+b x) \operatorname {CosIntegral}(a+b x)}{b}-\frac {a \operatorname {CosIntegral}(2 a+2 b x)}{2 b^2}-\frac {a \log (a+b x)}{2 b^2}+\frac {\cos (a+b x) \sin (a+b x)}{2 b^2}+\frac {\operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b^2}-\frac {\text {Si}(2 a+2 b x)}{2 b^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.70 \[ \int x \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx=\frac {2 b x-2 a \operatorname {CosIntegral}(2 (a+b x))-2 a \log (a+b x)+\operatorname {CosIntegral}(a+b x) (-4 b x \cos (a+b x)+4 \sin (a+b x))+\sin (2 (a+b x))-2 \text {Si}(2 (a+b x))}{4 b^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {7074, 7066, 4906, 27, 3042, 3780, 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.

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

\(\Big \downarrow \) 7074

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

\(\Big \downarrow \) 7066

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

\(\Big \downarrow \) 4906

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3780

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

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 3780

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinInte 
gral[e + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*e - c*f, 0]
 

rule 4906

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b 
_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x 
]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG 
tQ[p, 0]
 

rule 7066

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

rule 7074

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

rule 7293

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

Maple [A] (verified)

Time = 9.48 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {\operatorname {Ci}\left (b x +a \right ) \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )+a \cos \left (b x +a \right )\right )-\frac {\operatorname {Si}\left (2 b x +2 a \right )}{2}+\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}-a \left (\frac {\ln \left (b x +a \right )}{2}+\frac {\operatorname {Ci}\left (2 b x +2 a \right )}{2}\right )}{b^{2}}\)	\(96\)
default	\(\frac {\operatorname {Ci}\left (b x +a \right ) \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )+a \cos \left (b x +a \right )\right )-\frac {\operatorname {Si}\left (2 b x +2 a \right )}{2}+\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}-a \left (\frac {\ln \left (b x +a \right )}{2}+\frac {\operatorname {Ci}\left (2 b x +2 a \right )}{2}\right )}{b^{2}}\)	\(96\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 274 vs. \(2 (99) = 198\).

Time = 0.12 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.51 \[ \int x \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx=-\frac {2 \, \pi b^{2} x \cos \left (b x + a\right ) \operatorname {C}\left (b x + a\right ) - 2 \, \pi b \operatorname {C}\left (b x + a\right ) \sin \left (b x + a\right ) - 2 \, b \cos \left (b x + a\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) + \sqrt {b^{2}} {\left ({\left (\pi a + 1\right )} \cos \left (\frac {1}{2 \, \pi }\right ) - \pi \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {C}\left (\frac {{\left (\pi b x + \pi a + 1\right )} \sqrt {b^{2}}}{\pi b}\right ) + \sqrt {b^{2}} {\left ({\left (\pi a - 1\right )} \cos \left (\frac {1}{2 \, \pi }\right ) + \pi \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {C}\left (\frac {{\left (\pi b x + \pi a - 1\right )} \sqrt {b^{2}}}{\pi b}\right ) + \sqrt {b^{2}} {\left (\pi \cos \left (\frac {1}{2 \, \pi }\right ) + {\left (\pi a + 1\right )} \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {S}\left (\frac {{\left (\pi b x + \pi a + 1\right )} \sqrt {b^{2}}}{\pi b}\right ) - \sqrt {b^{2}} {\left (\pi \cos \left (\frac {1}{2 \, \pi }\right ) - {\left (\pi a - 1\right )} \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {S}\left (\frac {{\left (\pi b x + \pi a - 1\right )} \sqrt {b^{2}}}{\pi b}\right )}{2 \, \pi b^{3}} \] Input:

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

Output:

-1/2*(2*pi*b^2*x*cos(b*x + a)*fresnel_cos(b*x + a) - 2*pi*b*fresnel_cos(b* 
x + a)*sin(b*x + a) - 2*b*cos(b*x + a)*sin(1/2*pi*b^2*x^2 + pi*a*b*x + 1/2 
*pi*a^2) + sqrt(b^2)*((pi*a + 1)*cos(1/2/pi) - pi*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_cos((pi*b*x + pi*a - 1)*sqrt(b^2)/(pi*b)) + sqr 
t(b^2)*(pi*cos(1/2/pi) + (pi*a + 1)*sin(1/2/pi))*fresnel_sin((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_sin((pi*b*x + pi*a - 1)*sqrt(b^2)/(pi*b)))/(pi*b^3)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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