\(\int x^2 \sin (b x) \text {Si}(b x) \, dx\) [44]

Optimal result

Integrand size = 12, antiderivative size = 91 \[ \int x^2 \sin (b x) \text {Si}(b x) \, dx=-\frac {5 x}{4 b^2}+\frac {5 \cos (b x) \sin (b x)}{4 b^3}+\frac {x \sin ^2(b x)}{2 b^2}+\frac {2 \cos (b x) \text {Si}(b x)}{b^3}-\frac {x^2 \cos (b x) \text {Si}(b x)}{b}+\frac {2 x \sin (b x) \text {Si}(b x)}{b^2}-\frac {\text {Si}(2 b x)}{b^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.70 \[ \int x^2 \sin (b x) \text {Si}(b x) \, dx=-\frac {8 b x+2 b x \cos (2 b x)-5 \sin (2 b x)+8 \left (\left (-2+b^2 x^2\right ) \cos (b x)-2 b x \sin (b x)\right ) \text {Si}(b x)+8 \text {Si}(2 b x)}{8 b^3} \] Input:

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

Output:

-1/8*(8*b*x + 2*b*x*Cos[2*b*x] - 5*Sin[2*b*x] + 8*((-2 + b^2*x^2)*Cos[b*x] 
 - 2*b*x*Sin[b*x])*SinIntegral[b*x] + 8*SinIntegral[2*b*x])/b^3
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.54, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.417, Rules used = {7067, 27, 3924, 3042, 3115, 24, 7073, 27, 3042, 3115, 24, 7065, 27, 4906, 27, 3042, 3780}

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.

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

\(\Big \downarrow \) 7067

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3924

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 7073

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 7065

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 4906

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3780

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

Input:

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

Output:

((x*Sin[b*x]^2)/(2*b) - (x/2 - (Cos[b*x]*Sin[b*x])/(2*b))/(2*b))/b - (x^2* 
Cos[b*x]*SinIntegral[b*x])/b + (2*(-((x/2 - (Cos[b*x]*Sin[b*x])/(2*b))/b) 
+ (x*Sin[b*x]*SinIntegral[b*x])/b - (-((Cos[b*x]*SinIntegral[b*x])/b) + Si 
nIntegral[2*b*x]/(2*b))/b))/b
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

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 3042

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

rule 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

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 3924

Int[Cos[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^(n_.)]^ 
(p_.), x_Symbol] :> Simp[x^(m - n + 1)*(Sin[a + b*x^n]^(p + 1)/(b*n*(p + 1) 
)), x] - Simp[(m - n + 1)/(b*n*(p + 1))   Int[x^(m - n)*Sin[a + b*x^n]^(p + 
 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]
 

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 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]
 

Maple [A] (verified)

Time = 4.89 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\frac {\operatorname {Si}\left (b x \right ) \left (-b^{2} x^{2} \cos \left (b x \right )+2 \cos \left (b x \right )+2 b x \sin \left (b x \right )\right )-\frac {\cos \left (b x \right )^{2} b x}{2}+\frac {5 \sin \left (b x \right ) \cos \left (b x \right )}{4}-\frac {3 b x}{4}-\operatorname {Si}\left (2 b x \right )}{b^{3}}\)	\(69\)
default	\(\frac {\operatorname {Si}\left (b x \right ) \left (-b^{2} x^{2} \cos \left (b x \right )+2 \cos \left (b x \right )+2 b x \sin \left (b x \right )\right )-\frac {\cos \left (b x \right )^{2} b x}{2}+\frac {5 \sin \left (b x \right ) \cos \left (b x \right )}{4}-\frac {3 b x}{4}-\operatorname {Si}\left (2 b x \right )}{b^{3}}\)	\(69\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74 \[ \int x^2 \sin (b x) \text {Si}(b x) \, dx=-\frac {2 \, b x \cos \left (b x\right )^{2} + 4 \, {\left (b^{2} x^{2} - 2\right )} \cos \left (b x\right ) \operatorname {Si}\left (b x\right ) + 3 \, b x - {\left (8 \, b x \operatorname {Si}\left (b x\right ) + 5 \, \cos \left (b x\right )\right )} \sin \left (b x\right ) + 4 \, \operatorname {Si}\left (2 \, b x\right )}{4 \, b^{3}} \] Input:

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

Output:

-1/4*(2*b*x*cos(b*x)^2 + 4*(b^2*x^2 - 2)*cos(b*x)*sin_integral(b*x) + 3*b* 
x - (8*b*x*sin_integral(b*x) + 5*cos(b*x))*sin(b*x) + 4*sin_integral(2*b*x 
))/b^3
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [C] (verification not implemented)

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

Time = 0.13 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.52 \[ \int x^2 \sin (b x) \text {Si}(b x) \, dx={\left (\frac {2 \, x \sin \left (b x\right )}{b^{2}} - \frac {{\left (b^{2} x^{2} - 2\right )} \cos \left (b x\right )}{b^{3}}\right )} \operatorname {Si}\left (b x\right ) - \frac {3 \, b x \tan \left (b x\right )^{2} + 2 \, \Im \left ( \operatorname {Ci}\left (2 \, b x\right ) \right ) \tan \left (b x\right )^{2} - 2 \, \Im \left ( \operatorname {Ci}\left (-2 \, b x\right ) \right ) \tan \left (b x\right )^{2} + 4 \, \operatorname {Si}\left (2 \, b x\right ) \tan \left (b x\right )^{2} + 5 \, b x + 2 \, \Im \left ( \operatorname {Ci}\left (2 \, b x\right ) \right ) - 2 \, \Im \left ( \operatorname {Ci}\left (-2 \, b x\right ) \right ) + 4 \, \operatorname {Si}\left (2 \, b x\right ) - 5 \, \tan \left (b x\right )}{4 \, {\left (b^{3} \tan \left (b x\right )^{2} + b^{3}\right )}} \] Input:

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

Output:

(2*x*sin(b*x)/b^2 - (b^2*x^2 - 2)*cos(b*x)/b^3)*sin_integral(b*x) - 1/4*(3 
*b*x*tan(b*x)^2 + 2*imag_part(cos_integral(2*b*x))*tan(b*x)^2 - 2*imag_par 
t(cos_integral(-2*b*x))*tan(b*x)^2 + 4*sin_integral(2*b*x)*tan(b*x)^2 + 5* 
b*x + 2*imag_part(cos_integral(2*b*x)) - 2*imag_part(cos_integral(-2*b*x)) 
 + 4*sin_integral(2*b*x) - 5*tan(b*x))/(b^3*tan(b*x)^2 + b^3)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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