\(\int \sin (a+b x) \text {Si}(c+d x) \, dx\) [64]

Optimal result

Integrand size = 13, antiderivative size = 154 \[ \int \sin (a+b x) \text {Si}(c+d x) \, dx=-\frac {\operatorname {CosIntegral}\left (\frac {c (b-d)}{d}+(b-d) x\right ) \sin \left (a-\frac {b c}{d}\right )}{2 b}+\frac {\operatorname {CosIntegral}\left (\frac {c (b+d)}{d}+(b+d) x\right ) \sin \left (a-\frac {b c}{d}\right )}{2 b}-\frac {\cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {c (b-d)}{d}+(b-d) x\right )}{2 b}-\frac {\cos (a+b x) \text {Si}(c+d x)}{b}+\frac {\cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {c (b+d)}{d}+(b+d) x\right )}{2 b} \] Output:

-1/2*Ci(c*(b-d)/d+(b-d)*x)*sin(a-b*c/d)/b+1/2*Ci(c*(b+d)/d+(b+d)*x)*sin(a- 
b*c/d)/b-1/2*cos(a-b*c/d)*Si(c*(b-d)/d+(b-d)*x)/b-cos(b*x+a)*Si(d*x+c)/b+1 
/2*cos(a-b*c/d)*Si(c*(b+d)/d+(b+d)*x)/b
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.98 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.09 \[ \int \sin (a+b x) \text {Si}(c+d x) \, dx=\frac {i e^{-\frac {i (b c+a d)}{d}} \left (-e^{\frac {2 i b c}{d}} \operatorname {ExpIntegralEi}\left (-\frac {i (b-d) (c+d x)}{d}\right )+e^{2 i a} \operatorname {ExpIntegralEi}\left (\frac {i (b-d) (c+d x)}{d}\right )+e^{\frac {2 i b c}{d}} \operatorname {ExpIntegralEi}\left (-\frac {i (b+d) (c+d x)}{d}\right )-e^{2 i a} \operatorname {ExpIntegralEi}\left (\frac {i (b+d) (c+d x)}{d}\right )+4 i e^{\frac {i (b c+a d)}{d}} \cos (a+b x) \text {Si}(c+d x)\right )}{4 b} \] Input:

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

Output:

((I/4)*(-(E^(((2*I)*b*c)/d)*ExpIntegralEi[((-I)*(b - d)*(c + d*x))/d]) + E 
^((2*I)*a)*ExpIntegralEi[(I*(b - d)*(c + d*x))/d] + E^(((2*I)*b*c)/d)*ExpI 
ntegralEi[((-I)*(b + d)*(c + d*x))/d] - E^((2*I)*a)*ExpIntegralEi[(I*(b + 
d)*(c + d*x))/d] + (4*I)*E^((I*(b*c + a*d))/d)*Cos[a + b*x]*SinIntegral[c 
+ d*x]))/(b*E^((I*(b*c + a*d))/d))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {7065, 4930, 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.

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

\(\Big \downarrow \) 7065

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

\(\Big \downarrow \) 4930

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4930

Int[Cos[(c_.) + (d_.)*(x_)]^(q_.)*((e_.) + (f_.)*(x_))^(m_.)*Sin[(a_.) + (b 
_.)*(x_)]^(p_.), x_Symbol] :> Int[ExpandTrigReduce[(e + f*x)^m, Sin[a + b*x 
]^p*Cos[c + d*x]^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p, 0] 
 && IGtQ[q, 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]
 

Maple [A] (verified)

Time = 1.07 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.73

\[\frac {-\frac {\operatorname {Si}\left (d x +c \right ) d \cos \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -c b}{d}\right )}{b}+\frac {d \left (-\frac {d \left (\frac {\operatorname {Si}\left (\left (-1+\frac {b}{d}\right ) \left (d x +c \right )+a -\frac {c b}{d}+\frac {-a d +c b}{d}\right ) \cos \left (\frac {-a d +c b}{d}\right )}{d}-\frac {\operatorname {Ci}\left (\left (-1+\frac {b}{d}\right ) \left (d x +c \right )+a -\frac {c b}{d}+\frac {-a d +c b}{d}\right ) \sin \left (\frac {-a d +c b}{d}\right )}{d}\right )}{2}+\frac {d \left (\frac {\operatorname {Si}\left (\left (1+\frac {b}{d}\right ) \left (d x +c \right )+a -\frac {c b}{d}+\frac {-a d +c b}{d}\right ) \cos \left (\frac {-a d +c b}{d}\right )}{d}-\frac {\operatorname {Ci}\left (\left (1+\frac {b}{d}\right ) \left (d x +c \right )+a -\frac {c b}{d}+\frac {-a d +c b}{d}\right ) \sin \left (\frac {-a d +c b}{d}\right )}{d}\right )}{2}\right )}{b}}{d}\]

Input:

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

Output:

(-Si(d*x+c)/b*d*cos(b*(d*x+c)/d+(a*d-b*c)/d)+1/b*d*(-1/2*d*(Si((-1+b/d)*(d 
*x+c)+a-c*b/d+(-a*d+b*c)/d)*cos((-a*d+b*c)/d)/d-Ci((-1+b/d)*(d*x+c)+a-c*b/ 
d+(-a*d+b*c)/d)*sin((-a*d+b*c)/d)/d)+1/2*d*(Si((1+b/d)*(d*x+c)+a-c*b/d+(-a 
*d+b*c)/d)*cos((-a*d+b*c)/d)/d-Ci((1+b/d)*(d*x+c)+a-c*b/d+(-a*d+b*c)/d)*si 
n((-a*d+b*c)/d)/d)))/d
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.95 \[ \int \sin (a+b x) \text {Si}(c+d x) \, dx=\frac {{\left (\operatorname {Si}\left (\frac {b c + c d + {\left (b d + d^{2}\right )} x}{d}\right ) + \operatorname {Si}\left (-\frac {b c - c d + {\left (b d - d^{2}\right )} x}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + {\left (\operatorname {Ci}\left (\frac {b c + c d + {\left (b d + d^{2}\right )} x}{d}\right ) - \operatorname {Ci}\left (-\frac {b c - c d + {\left (b d - d^{2}\right )} x}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right ) - 2 \, \cos \left (b x + a\right ) \operatorname {Si}\left (d x + c\right )}{2 \, b} \] Input:

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

Output:

1/2*((sin_integral((b*c + c*d + (b*d + d^2)*x)/d) + sin_integral(-(b*c - c 
*d + (b*d - d^2)*x)/d))*cos(-(b*c - a*d)/d) + (cos_integral((b*c + c*d + ( 
b*d + d^2)*x)/d) - cos_integral(-(b*c - c*d + (b*d - d^2)*x)/d))*sin(-(b*c 
 - a*d)/d) - 2*cos(b*x + a)*sin_integral(d*x + c))/b
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate(sin(b*x + a)*sin_integral(d*x + c), x)
 

Giac [C] (verification not implemented)

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

Time = 0.39 (sec) , antiderivative size = 9541, normalized size of antiderivative = 61.95 \[ \int \sin (a+b x) \text {Si}(c+d x) \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(imag_part(cos_integral(b*x + d*x + c + b*c/d))*tan(1/2*a + 1/2*c)^2*t 
an(1/2*a - 1/2*c)^2*tan(1/2*(b*c + c*d)/d)^2*tan(1/2*(b*c - c*d)/d)^2 - im 
ag_part(cos_integral(b*x - d*x - c + b*c/d))*tan(1/2*a + 1/2*c)^2*tan(1/2* 
a - 1/2*c)^2*tan(1/2*(b*c + c*d)/d)^2*tan(1/2*(b*c - c*d)/d)^2 + imag_part 
(cos_integral(-b*x + d*x + c - b*c/d))*tan(1/2*a + 1/2*c)^2*tan(1/2*a - 1/ 
2*c)^2*tan(1/2*(b*c + c*d)/d)^2*tan(1/2*(b*c - c*d)/d)^2 - imag_part(cos_i 
ntegral(-b*x - d*x - c - b*c/d))*tan(1/2*a + 1/2*c)^2*tan(1/2*a - 1/2*c)^2 
*tan(1/2*(b*c + c*d)/d)^2*tan(1/2*(b*c - c*d)/d)^2 + 2*sin_integral((b*d*x 
 + d^2*x + b*c + c*d)/d)*tan(1/2*a + 1/2*c)^2*tan(1/2*a - 1/2*c)^2*tan(1/2 
*(b*c + c*d)/d)^2*tan(1/2*(b*c - c*d)/d)^2 - 2*sin_integral((b*d*x - d^2*x 
 + b*c - c*d)/d)*tan(1/2*a + 1/2*c)^2*tan(1/2*a - 1/2*c)^2*tan(1/2*(b*c + 
c*d)/d)^2*tan(1/2*(b*c - c*d)/d)^2 - 2*real_part(cos_integral(b*x - d*x - 
c + b*c/d))*tan(1/2*a + 1/2*c)^2*tan(1/2*a - 1/2*c)^2*tan(1/2*(b*c + c*d)/ 
d)^2*tan(1/2*(b*c - c*d)/d) - 2*real_part(cos_integral(-b*x + d*x + c - b* 
c/d))*tan(1/2*a + 1/2*c)^2*tan(1/2*a - 1/2*c)^2*tan(1/2*(b*c + c*d)/d)^2*t 
an(1/2*(b*c - c*d)/d) + 2*real_part(cos_integral(b*x + d*x + c + b*c/d))*t 
an(1/2*a + 1/2*c)^2*tan(1/2*a - 1/2*c)^2*tan(1/2*(b*c + c*d)/d)*tan(1/2*(b 
*c - c*d)/d)^2 + 2*real_part(cos_integral(-b*x - d*x - c - b*c/d))*tan(1/2 
*a + 1/2*c)^2*tan(1/2*a - 1/2*c)^2*tan(1/2*(b*c + c*d)/d)*tan(1/2*(b*c - c 
*d)/d)^2 + 2*real_part(cos_integral(b*x - d*x - c + b*c/d))*tan(1/2*a +...
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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