\(\int x \sin (a+b x) \text {Si}(a+b x) \, dx\) [56]

Optimal result

Integrand size = 14, antiderivative size = 97 \[ \int x \sin (a+b x) \text {Si}(a+b x) \, dx=-\frac {\cos (2 a+2 b x)}{4 b^2}+\frac {\operatorname {CosIntegral}(2 a+2 b x)}{2 b^2}-\frac {\log (a+b x)}{2 b^2}-\frac {x \cos (a+b x) \text {Si}(a+b x)}{b}+\frac {\sin (a+b x) \text {Si}(a+b x)}{b^2}-\frac {a \text {Si}(2 a+2 b x)}{2 b^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.73 \[ \int x \sin (a+b x) \text {Si}(a+b x) \, dx=-\frac {\cos (2 (a+b x))-2 \operatorname {CosIntegral}(2 (a+b x))+2 \log (a+b x)+4 (b x \cos (a+b x)-\sin (a+b x)) \text {Si}(a+b x)+2 a \text {Si}(2 (a+b x))}{4 b^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {7067, 5084, 7071, 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.

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

\(\Big \downarrow \) 7067

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

\(\Big \downarrow \) 5084

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

\(\Big \downarrow \) 7071

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3793

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-((x*Cos[a + b*x]*SinIntegral[a + b*x])/b) + (CosIntegral[2*a + 2*b*x]/(2* 
b) - Log[a + b*x]/(2*b) + (Sin[a + b*x]*SinIntegral[a + b*x])/b)/b + (-1/2 
*Cos[2*a + 2*b*x]/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 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 7071

Int[Cos[(a_.) + (b_.)*(x_)]*SinIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> S 
imp[Sin[a + b*x]*(SinIntegral[c + d*x]/b), x] - Simp[d/b   Int[Sin[a + b*x] 
*(Sin[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 = 8.83 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85

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

Input:

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

Output:

1/b^2*(Si(b*x+a)*(sin(b*x+a)-(b*x+a)*cos(b*x+a)+a*cos(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 [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74 \[ \int x \sin (a+b x) \text {Si}(a+b x) \, dx=-\frac {2 \, b x \cos \left (b x + a\right ) \operatorname {Si}\left (b x + a\right ) + \cos \left (b x + a\right )^{2} + a \operatorname {Si}\left (2 \, b x + 2 \, a\right ) - 2 \, \sin \left (b x + a\right ) \operatorname {Si}\left (b x + a\right ) - \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) + \log \left (b x + a\right )}{2 \, b^{2}} \] Input:

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

Output:

-1/2*(2*b*x*cos(b*x + a)*sin_integral(b*x + a) + cos(b*x + a)^2 + a*sin_in 
tegral(2*b*x + 2*a) - 2*sin(b*x + a)*sin_integral(b*x + a) - cos_integral( 
2*b*x + 2*a) + log(b*x + a))/b^2
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate(x*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 = 0.14 (sec) , antiderivative size = 507, normalized size of antiderivative = 5.23 \[ \int x \sin (a+b x) \text {Si}(a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

-(x*cos(b*x + a)/b - sin(b*x + a)/b^2)*sin_integral(b*x + a) - 1/4*(a*imag 
_part(cos_integral(2*b*x + 2*a))*tan(b*x)^2*tan(a)^2 - a*imag_part(cos_int 
egral(-2*b*x - 2*a))*tan(b*x)^2*tan(a)^2 + 2*a*sin_integral(2*b*x + 2*a)*t 
an(b*x)^2*tan(a)^2 + 2*log(abs(b*x + a))*tan(b*x)^2*tan(a)^2 - real_part(c 
os_integral(2*b*x + 2*a))*tan(b*x)^2*tan(a)^2 - real_part(cos_integral(-2* 
b*x - 2*a))*tan(b*x)^2*tan(a)^2 + a*imag_part(cos_integral(2*b*x + 2*a))*t 
an(b*x)^2 - a*imag_part(cos_integral(-2*b*x - 2*a))*tan(b*x)^2 + 2*a*sin_i 
ntegral(2*b*x + 2*a)*tan(b*x)^2 + a*imag_part(cos_integral(2*b*x + 2*a))*t 
an(a)^2 - a*imag_part(cos_integral(-2*b*x - 2*a))*tan(a)^2 + 2*a*sin_integ 
ral(2*b*x + 2*a)*tan(a)^2 + tan(b*x)^2*tan(a)^2 + 2*log(abs(b*x + a))*tan( 
b*x)^2 - real_part(cos_integral(2*b*x + 2*a))*tan(b*x)^2 - real_part(cos_i 
ntegral(-2*b*x - 2*a))*tan(b*x)^2 + 2*log(abs(b*x + a))*tan(a)^2 - real_pa 
rt(cos_integral(2*b*x + 2*a))*tan(a)^2 - real_part(cos_integral(-2*b*x - 2 
*a))*tan(a)^2 + a*imag_part(cos_integral(2*b*x + 2*a)) - a*imag_part(cos_i 
ntegral(-2*b*x - 2*a)) + 2*a*sin_integral(2*b*x + 2*a) - tan(b*x)^2 - 4*ta 
n(b*x)*tan(a) - tan(a)^2 + 2*log(abs(b*x + a)) - real_part(cos_integral(2* 
b*x + 2*a)) - real_part(cos_integral(-2*b*x - 2*a)) + 1)/(b^2*tan(b*x)^2*t 
an(a)^2 + b^2*tan(b*x)^2 + b^2*tan(a)^2 + b^2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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