\(\int x^2 \sin (a+b x) \text {Si}(a+b x) \, dx\) [55]

Optimal result

Integrand size = 16, antiderivative size = 175 \[ \int x^2 \sin (a+b x) \text {Si}(a+b x) \, dx=-\frac {x}{b^2}+\frac {(a-b x) \cos (2 a+2 b x)}{4 b^3}-\frac {a \operatorname {CosIntegral}(2 a+2 b x)}{b^3}+\frac {a \log (a+b x)}{b^3}+\frac {\cos (a+b x) \sin (a+b x)}{b^3}+\frac {\sin (2 a+2 b x)}{8 b^3}+\frac {2 \cos (a+b x) \text {Si}(a+b x)}{b^3}-\frac {x^2 \cos (a+b x) \text {Si}(a+b x)}{b}+\frac {2 x \sin (a+b x) \text {Si}(a+b x)}{b^2}-\frac {\text {Si}(2 a+2 b x)}{b^3}+\frac {a^2 \text {Si}(2 a+2 b x)}{2 b^3} \] Output:

-x/b^2+1/4*(-b*x+a)*cos(2*b*x+2*a)/b^3-a*Ci(2*b*x+2*a)/b^3+a*ln(b*x+a)/b^3 
+cos(b*x+a)*sin(b*x+a)/b^3+1/8*sin(2*b*x+2*a)/b^3+2*cos(b*x+a)*Si(b*x+a)/b 
^3-x^2*cos(b*x+a)*Si(b*x+a)/b+2*x*sin(b*x+a)*Si(b*x+a)/b^2-Si(2*b*x+2*a)/b 
^3+1/2*a^2*Si(2*b*x+2*a)/b^3
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.70 \[ \int x^2 \sin (a+b x) \text {Si}(a+b x) \, dx=\frac {-8 b x+2 a \cos (2 (a+b x))-2 b x \cos (2 (a+b x))-8 a \operatorname {CosIntegral}(2 (a+b x))+8 a \log (a+b x)+5 \sin (2 (a+b x))-8 \left (\left (-2+b^2 x^2\right ) \cos (a+b x)-2 b x \sin (a+b x)\right ) \text {Si}(a+b x)-8 \text {Si}(2 (a+b x))+4 a^2 \text {Si}(2 (a+b x))}{8 b^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.81 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.21, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.688, Rules used = {7067, 5084, 7073, 7065, 4906, 27, 3042, 3780, 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.

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

\(\Big \downarrow \) 7067

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

\(\Big \downarrow \) 5084

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

\(\Big \downarrow \) 7073

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

\(\Big \downarrow \) 7065

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

\(\Big \downarrow \) 4906

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3780

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-((x^2*Cos[a + b*x]*SinIntegral[a + b*x])/b) + ((a*Cos[2*a + 2*b*x])/(2*b^ 
3) - (x*Cos[2*a + 2*b*x])/(2*b^2) + Sin[2*a + 2*b*x]/(4*b^3) + (a^2*SinInt 
egral[2*a + 2*b*x])/b^3)/2 + (2*(-1/2*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) + 
(x*Sin[a + b*x]*SinIntegral[a + b*x])/b - (-((Cos[a + b*x]*SinIntegral[a + 
 b*x])/b) + SinIntegral[2*a + 2*b*x]/(2*b))/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 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 7065

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

rule 7067

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

rule 7073

Int[Cos[(a_.) + (b_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.)*SinIntegral[(c_.) + 
(d_.)*(x_)], x_Symbol] :> Simp[(e + f*x)^m*Sin[a + b*x]*(SinIntegral[c + d* 
x]/b), x] + (-Simp[d/b   Int[(e + f*x)^m*Sin[a + b*x]*(Sin[c + d*x]/(c + d* 
x)), x], x] - Simp[f*(m/b)   Int[(e + f*x)^(m - 1)*Sin[a + b*x]*SinIntegral 
[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 = 9.98 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {\operatorname {Si}\left (b x +a \right ) \left (-a^{2} \cos \left (b x +a \right )-2 a \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )-\left (b x +a \right )^{2} \cos \left (b x +a \right )+2 \cos \left (b x +a \right )+2 \left (b x +a \right ) \sin \left (b x +a \right )\right )+\frac {a^{2} \operatorname {Si}\left (2 b x +2 a \right )}{2}+a \cos \left (b x +a \right )^{2}-\frac {\left (b x +a \right ) \cos \left (b x +a \right )^{2}}{2}+\frac {5 \sin \left (b x +a \right ) \cos \left (b x +a \right )}{4}-\frac {3 b x}{4}-\frac {3 a}{4}-a \,\operatorname {Ci}\left (2 b x +2 a \right )+a \ln \left (b x +a \right )-\operatorname {Si}\left (2 b x +2 a \right )}{b^{3}}\)	\(175\)
default	\(\frac {\operatorname {Si}\left (b x +a \right ) \left (-a^{2} \cos \left (b x +a \right )-2 a \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )-\left (b x +a \right )^{2} \cos \left (b x +a \right )+2 \cos \left (b x +a \right )+2 \left (b x +a \right ) \sin \left (b x +a \right )\right )+\frac {a^{2} \operatorname {Si}\left (2 b x +2 a \right )}{2}+a \cos \left (b x +a \right )^{2}-\frac {\left (b x +a \right ) \cos \left (b x +a \right )^{2}}{2}+\frac {5 \sin \left (b x +a \right ) \cos \left (b x +a \right )}{4}-\frac {3 b x}{4}-\frac {3 a}{4}-a \,\operatorname {Ci}\left (2 b x +2 a \right )+a \ln \left (b x +a \right )-\operatorname {Si}\left (2 b x +2 a \right )}{b^{3}}\)	\(175\)

Input:

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

Output:

1/b^3*(Si(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*Si(2*b*x+2*a)+ 
a*cos(b*x+a)^2-1/2*(b*x+a)*cos(b*x+a)^2+5/4*sin(b*x+a)*cos(b*x+a)-3/4*b*x- 
3/4*a-a*Ci(2*b*x+2*a)+a*ln(b*x+a)-Si(2*b*x+2*a))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.65 \[ \int x^2 \sin (a+b x) \text {Si}(a+b x) \, dx=-\frac {2 \, {\left (b x - a\right )} \cos \left (b x + a\right )^{2} + 4 \, {\left (b^{2} x^{2} - 2\right )} \cos \left (b x + a\right ) \operatorname {Si}\left (b x + a\right ) + 3 \, b x + 4 \, a \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) - 4 \, a \log \left (b x + a\right ) - {\left (8 \, b x \operatorname {Si}\left (b x + a\right ) + 5 \, \cos \left (b x + a\right )\right )} \sin \left (b x + a\right ) - 2 \, {\left (a^{2} - 2\right )} \operatorname {Si}\left (2 \, b x + 2 \, a\right )}{4 \, b^{3}} \] Input:

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

Output:

-1/4*(2*(b*x - a)*cos(b*x + a)^2 + 4*(b^2*x^2 - 2)*cos(b*x + a)*sin_integr 
al(b*x + a) + 3*b*x + 4*a*cos_integral(2*b*x + 2*a) - 4*a*log(b*x + a) - ( 
8*b*x*sin_integral(b*x + a) + 5*cos(b*x + a))*sin(b*x + a) - 2*(a^2 - 2)*s 
in_integral(2*b*x + 2*a))/b^3
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.16 (sec) , antiderivative size = 398, normalized size of antiderivative = 2.27 \[ \int x^2 \sin (a+b x) \text {Si}(a+b x) \, dx={\left (\frac {2 \, x \sin \left (b x + a\right )}{b^{2}} - \frac {{\left (b^{2} x^{2} - 2\right )} \cos \left (b x + a\right )}{b^{3}}\right )} \operatorname {Si}\left (b x + a\right ) + \frac {a^{2} \Im \left ( \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x + a\right )^{2} - a^{2} \Im \left ( \operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x + a\right )^{2} + 2 \, a^{2} \operatorname {Si}\left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )^{2} - 3 \, b x \tan \left (b x + a\right )^{2} + 4 \, a \log \left ({\left | b x + a \right |}\right ) \tan \left (b x + a\right )^{2} - 2 \, a \Re \left ( \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x + a\right )^{2} - 2 \, a \Re \left ( \operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x + a\right )^{2} + a^{2} \Im \left ( \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) - a^{2} \Im \left ( \operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) + 2 \, a^{2} \operatorname {Si}\left (2 \, b x + 2 \, a\right ) - a \tan \left (b x + a\right )^{2} - 2 \, \Im \left ( \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x + a\right )^{2} + 2 \, \Im \left ( \operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x + a\right )^{2} - 4 \, \operatorname {Si}\left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )^{2} - 5 \, b x + 4 \, a \log \left ({\left | b x + a \right |}\right ) - 2 \, a \Re \left ( \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) - 2 \, a \Re \left ( \operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) + a - 2 \, \Im \left ( \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) + 2 \, \Im \left ( \operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) - 4 \, \operatorname {Si}\left (2 \, b x + 2 \, a\right ) + 5 \, \tan \left (b x + a\right )}{4 \, {\left (b^{3} \tan \left (b x + a\right )^{2} + b^{3}\right )}} \] Input:

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

Output:

(2*x*sin(b*x + a)/b^2 - (b^2*x^2 - 2)*cos(b*x + a)/b^3)*sin_integral(b*x + 
 a) + 1/4*(a^2*imag_part(cos_integral(2*b*x + 2*a))*tan(b*x + a)^2 - a^2*i 
mag_part(cos_integral(-2*b*x - 2*a))*tan(b*x + a)^2 + 2*a^2*sin_integral(2 
*b*x + 2*a)*tan(b*x + a)^2 - 3*b*x*tan(b*x + a)^2 + 4*a*log(abs(b*x + a))* 
tan(b*x + a)^2 - 2*a*real_part(cos_integral(2*b*x + 2*a))*tan(b*x + a)^2 - 
 2*a*real_part(cos_integral(-2*b*x - 2*a))*tan(b*x + a)^2 + a^2*imag_part( 
cos_integral(2*b*x + 2*a)) - a^2*imag_part(cos_integral(-2*b*x - 2*a)) + 2 
*a^2*sin_integral(2*b*x + 2*a) - a*tan(b*x + a)^2 - 2*imag_part(cos_integr 
al(2*b*x + 2*a))*tan(b*x + a)^2 + 2*imag_part(cos_integral(-2*b*x - 2*a))* 
tan(b*x + a)^2 - 4*sin_integral(2*b*x + 2*a)*tan(b*x + a)^2 - 5*b*x + 4*a* 
log(abs(b*x + a)) - 2*a*real_part(cos_integral(2*b*x + 2*a)) - 2*a*real_pa 
rt(cos_integral(-2*b*x - 2*a)) + a - 2*imag_part(cos_integral(2*b*x + 2*a) 
) + 2*imag_part(cos_integral(-2*b*x - 2*a)) - 4*sin_integral(2*b*x + 2*a) 
+ 5*tan(b*x + a))/(b^3*tan(b*x + a)^2 + b^3)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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