\(\int x^2 \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx\) [123]

Optimal result

Integrand size = 16, antiderivative size = 201 \[ \int x^2 \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx=\frac {(a-b x)^2}{4 b^3}+\frac {\cos ^2(a+b x)}{4 b^3}+\frac {\cos (2 a+2 b x)}{2 b^3}+\frac {2 \cos (a+b x) \operatorname {CosIntegral}(a+b x)}{b^3}-\frac {x^2 \cos (a+b x) \operatorname {CosIntegral}(a+b x)}{b}-\frac {\operatorname {CosIntegral}(2 a+2 b x)}{b^3}+\frac {a^2 \operatorname {CosIntegral}(2 a+2 b x)}{2 b^3}-\frac {\log (a+b x)}{b^3}+\frac {a^2 \log (a+b x)}{2 b^3}-\frac {(a-b x) \cos (a+b x) \sin (a+b x)}{2 b^3}+\frac {2 x \operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b^2}+\frac {a \text {Si}(2 a+2 b x)}{b^3} \] Output:

1/4*(-b*x+a)^2/b^3+1/4*cos(b*x+a)^2/b^3+1/2*cos(2*b*x+2*a)/b^3+2*cos(b*x+a 
)*Ci(b*x+a)/b^3-x^2*cos(b*x+a)*Ci(b*x+a)/b-Ci(2*b*x+2*a)/b^3+1/2*a^2*Ci(2* 
b*x+2*a)/b^3-ln(b*x+a)/b^3+1/2*a^2*ln(b*x+a)/b^3-1/2*(-b*x+a)*cos(b*x+a)*s 
in(b*x+a)/b^3+2*x*Ci(b*x+a)*sin(b*x+a)/b^2+a*Si(2*b*x+2*a)/b^3
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.67 \[ \int x^2 \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx=\frac {-4 a b x+2 b^2 x^2+5 \cos (2 (a+b x))+4 \left (-2+a^2\right ) \operatorname {CosIntegral}(2 (a+b x))-8 \log (a+b x)+4 a^2 \log (a+b x)-8 \operatorname {CosIntegral}(a+b x) \left (\left (-2+b^2 x^2\right ) \cos (a+b x)-2 b x \sin (a+b x)\right )-2 a \sin (2 (a+b x))+2 b x \sin (2 (a+b x))+8 a \text {Si}(2 (a+b x))}{8 b^3} \] Input:

Integrate[x^2*CosIntegral[a + b*x]*Sin[a + b*x],x]
 

Output:

(-4*a*b*x + 2*b^2*x^2 + 5*Cos[2*(a + b*x)] + 4*(-2 + a^2)*CosIntegral[2*(a 
 + b*x)] - 8*Log[a + b*x] + 4*a^2*Log[a + b*x] - 8*CosIntegral[a + b*x]*(( 
-2 + b^2*x^2)*Cos[a + b*x] - 2*b*x*Sin[a + b*x]) - 2*a*Sin[2*(a + b*x)] + 
2*b*x*Sin[2*(a + b*x)] + 8*a*SinIntegral[2*(a + b*x)])/(8*b^3)
 

Rubi [A] (verified)

Time = 1.95 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.19, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {7074, 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.

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

\(\Big \downarrow \) 7074

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

\(\Big \downarrow \) 7068

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

\(\Big \downarrow \) 5084

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

\(\Big \downarrow \) 7072

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3793

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^2 \operatorname {CosIntegral}(2 a+2 b x)}{2 b^3}+\frac {a^2 \log (a+b x)}{2 b^3}+\frac {\cos ^2(a+b x)}{4 b^3}-\frac {a \sin (a+b x) \cos (a+b x)}{2 b^3}+\frac {2 \left (\frac {1}{2} \left (\frac {a \text {Si}(2 a+2 b x)}{b^2}+\frac {\cos (2 a+2 b x)}{2 b^2}\right )+\frac {x \operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b}-\frac {\frac {\operatorname {CosIntegral}(2 a+2 b x)}{2 b}-\frac {\operatorname {CosIntegral}(a+b x) \cos (a+b x)}{b}+\frac {\log (a+b x)}{2 b}}{b}\right )}{b}-\frac {a x}{2 b^2}+\frac {x \sin (a+b x) \cos (a+b x)}{2 b^2}-\frac {x^2 \operatorname {CosIntegral}(a+b x) \cos (a+b x)}{b}+\frac {x^2}{4 b}\)

Input:

Int[x^2*CosIntegral[a + b*x]*Sin[a + b*x],x]
 

Output:

-1/2*(a*x)/b^2 + x^2/(4*b) + Cos[a + b*x]^2/(4*b^3) - (x^2*Cos[a + b*x]*Co 
sIntegral[a + b*x])/b + (a^2*CosIntegral[2*a + 2*b*x])/(2*b^3) + (a^2*Log[ 
a + b*x])/(2*b^3) - (a*Cos[a + b*x]*Sin[a + b*x])/(2*b^3) + (x*Cos[a + b*x 
]*Sin[a + b*x])/(2*b^2) + (2*(-((-((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 + (Cos[2*a + 2*b*x]/(2*b^2) + (a*SinIntegral[2*a 
 + 2*b*x])/b^2)/2))/b
 

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 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 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 = 10.63 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(\frac {\operatorname {Ci}\left (b x +a \right ) \left (-a^{2} \cos \left (b x +a \right )-2 a \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )-\left (b x +a \right )^{2} \cos \left (b x +a \right )+2 \cos \left (b x +a \right )+2 \left (b x +a \right ) \sin \left (b x +a \right )\right )+\frac {a^{2} \ln \left (b x +a \right )}{2}+\frac {a^{2} \operatorname {Ci}\left (2 b x +2 a \right )}{2}-\sin \left (b x +a \right ) \cos \left (b x +a \right ) a -\left (b x +a \right ) a +\left (b x +a \right ) \left (\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )-\frac {\left (b x +a \right )^{2}}{4}-\frac {\sin \left (b x +a \right )^{2}}{4}+a \,\operatorname {Si}\left (2 b x +2 a \right )+\cos \left (b x +a \right )^{2}-\ln \left (b x +a \right )-\operatorname {Ci}\left (2 b x +2 a \right )}{b^{3}}\)	\(217\)
default	\(\frac {\operatorname {Ci}\left (b x +a \right ) \left (-a^{2} \cos \left (b x +a \right )-2 a \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )-\left (b x +a \right )^{2} \cos \left (b x +a \right )+2 \cos \left (b x +a \right )+2 \left (b x +a \right ) \sin \left (b x +a \right )\right )+\frac {a^{2} \ln \left (b x +a \right )}{2}+\frac {a^{2} \operatorname {Ci}\left (2 b x +2 a \right )}{2}-\sin \left (b x +a \right ) \cos \left (b x +a \right ) a -\left (b x +a \right ) a +\left (b x +a \right ) \left (\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )-\frac {\left (b x +a \right )^{2}}{4}-\frac {\sin \left (b x +a \right )^{2}}{4}+a \,\operatorname {Si}\left (2 b x +2 a \right )+\cos \left (b x +a \right )^{2}-\ln \left (b x +a \right )-\operatorname {Ci}\left (2 b x +2 a \right )}{b^{3}}\)	\(217\)

Input:

int(x^2*Ci(b*x+a)*sin(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

1/b^3*(Ci(b*x+a)*(-a^2*cos(b*x+a)-2*a*(sin(b*x+a)-(b*x+a)*cos(b*x+a))-(b*x 
+a)^2*cos(b*x+a)+2*cos(b*x+a)+2*(b*x+a)*sin(b*x+a))+1/2*a^2*ln(b*x+a)+1/2* 
a^2*Ci(2*b*x+2*a)-sin(b*x+a)*cos(b*x+a)*a-(b*x+a)*a+(b*x+a)*(1/2*sin(b*x+a 
)*cos(b*x+a)+1/2*b*x+1/2*a)-1/4*(b*x+a)^2-1/4*sin(b*x+a)^2+a*Si(2*b*x+2*a) 
+cos(b*x+a)^2-ln(b*x+a)-Ci(2*b*x+2*a))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 414 vs. \(2 (191) = 382\).

Time = 0.13 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.06 \[ \int x^2 \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx=-\frac {2 \, {\left (\pi ^{2} b^{3} x^{2} - 2 \, \pi ^{2} b\right )} \cos \left (b x + a\right ) \operatorname {C}\left (b x + a\right ) - \sqrt {b^{2}} {\left ({\left (\pi ^{2} {\left (a^{2} - 2\right )} + 2 \, \pi a + 1\right )} \cos \left (\frac {1}{2 \, \pi }\right ) - {\left (\pi + 2 \, \pi ^{2} a\right )} \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {C}\left (\frac {{\left (\pi b x + \pi a + 1\right )} \sqrt {b^{2}}}{\pi b}\right ) - \sqrt {b^{2}} {\left ({\left (\pi ^{2} {\left (a^{2} - 2\right )} - 2 \, \pi a + 1\right )} \cos \left (\frac {1}{2 \, \pi }\right ) - {\left (\pi - 2 \, \pi ^{2} a\right )} \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {C}\left (\frac {{\left (\pi b x + \pi a - 1\right )} \sqrt {b^{2}}}{\pi b}\right ) - \sqrt {b^{2}} {\left ({\left (\pi + 2 \, \pi ^{2} a\right )} \cos \left (\frac {1}{2 \, \pi }\right ) + {\left (\pi ^{2} {\left (a^{2} - 2\right )} + 2 \, \pi a + 1\right )} \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {S}\left (\frac {{\left (\pi b x + \pi a + 1\right )} \sqrt {b^{2}}}{\pi b}\right ) - \sqrt {b^{2}} {\left ({\left (\pi - 2 \, \pi ^{2} a\right )} \cos \left (\frac {1}{2 \, \pi }\right ) + {\left (\pi ^{2} {\left (a^{2} - 2\right )} - 2 \, \pi a + 1\right )} \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {S}\left (\frac {{\left (\pi b x + \pi a - 1\right )} \sqrt {b^{2}}}{\pi b}\right ) + 2 \, {\left (2 \, \pi b \sin \left (b x + a\right ) - {\left (\pi b^{2} x - \pi a b\right )} \cos \left (b x + a\right )\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) - 2 \, {\left (2 \, \pi ^{2} b^{2} x \operatorname {C}\left (b x + a\right ) - b \cos \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right )\right )} \sin \left (b x + a\right )}{2 \, \pi ^{2} b^{4}} \] Input:

integrate(x^2*fresnel_cos(b*x+a)*sin(b*x+a),x, algorithm="fricas")
 

Output:

-1/2*(2*(pi^2*b^3*x^2 - 2*pi^2*b)*cos(b*x + a)*fresnel_cos(b*x + a) - sqrt 
(b^2)*((pi^2*(a^2 - 2) + 2*pi*a + 1)*cos(1/2/pi) - (pi + 2*pi^2*a)*sin(1/2 
/pi))*fresnel_cos((pi*b*x + pi*a + 1)*sqrt(b^2)/(pi*b)) - sqrt(b^2)*((pi^2 
*(a^2 - 2) - 2*pi*a + 1)*cos(1/2/pi) - (pi - 2*pi^2*a)*sin(1/2/pi))*fresne 
l_cos((pi*b*x + pi*a - 1)*sqrt(b^2)/(pi*b)) - sqrt(b^2)*((pi + 2*pi^2*a)*c 
os(1/2/pi) + (pi^2*(a^2 - 2) + 2*pi*a + 1)*sin(1/2/pi))*fresnel_sin((pi*b* 
x + pi*a + 1)*sqrt(b^2)/(pi*b)) - sqrt(b^2)*((pi - 2*pi^2*a)*cos(1/2/pi) + 
 (pi^2*(a^2 - 2) - 2*pi*a + 1)*sin(1/2/pi))*fresnel_sin((pi*b*x + pi*a - 1 
)*sqrt(b^2)/(pi*b)) + 2*(2*pi*b*sin(b*x + a) - (pi*b^2*x - pi*a*b)*cos(b*x 
 + a))*sin(1/2*pi*b^2*x^2 + pi*a*b*x + 1/2*pi*a^2) - 2*(2*pi^2*b^2*x*fresn 
el_cos(b*x + a) - b*cos(1/2*pi*b^2*x^2 + pi*a*b*x + 1/2*pi*a^2))*sin(b*x + 
 a))/(pi^2*b^4)
 

Sympy [F]

\[ \int x^2 \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx=\int x^{2} \sin {\left (a + b x \right )} \operatorname {Ci}{\left (a + b x \right )}\, dx \] Input:

integrate(x**2*Ci(b*x+a)*sin(b*x+a),x)
 

Output:

Integral(x**2*sin(a + b*x)*Ci(a + b*x), x)
 

Maxima [F]

\[ \int x^2 \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx=\int { x^{2} \operatorname {C}\left (b x + a\right ) \sin \left (b x + a\right ) \,d x } \] Input:

integrate(x^2*fresnel_cos(b*x+a)*sin(b*x+a),x, algorithm="maxima")
 

Output:

integrate(x^2*fresnel_cos(b*x + a)*sin(b*x + a), x)
 

Giac [F]

\[ \int x^2 \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx=\int { x^{2} \operatorname {C}\left (b x + a\right ) \sin \left (b x + a\right ) \,d x } \] Input:

integrate(x^2*fresnel_cos(b*x+a)*sin(b*x+a),x, algorithm="giac")
 

Output:

integrate(x^2*fresnel_cos(b*x + a)*sin(b*x + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int x^2 \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx=\int x^2\,\mathrm {cosint}\left (a+b\,x\right )\,\sin \left (a+b\,x\right ) \,d x \] Input:

int(x^2*cosint(a + b*x)*sin(a + b*x),x)
 

Output:

int(x^2*cosint(a + b*x)*sin(a + b*x), x)
 

Reduce [F]

\[ \int x^2 \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx=\int \mathit {ci} \left (b x +a \right ) \sin \left (b x +a \right ) x^{2}d x \] Input:

int(x^2*Ci(b*x+a)*sin(b*x+a),x)
 

Output:

int(ci(a + b*x)*sin(a + b*x)*x**2,x)