Integrand size = 13, antiderivative size = 153 \[ \int \cos (a+b x) \text {Si}(c+d x) \, dx=-\frac {\cos \left (a-\frac {b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {c (b-d)}{d}+(b-d) x\right )}{2 b}+\frac {\cos \left (a-\frac {b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {c (b+d)}{d}+(b+d) x\right )}{2 b}+\frac {\sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {c (b-d)}{d}+(b-d) x\right )}{2 b}+\frac {\sin (a+b x) \text {Si}(c+d x)}{b}-\frac {\sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {c (b+d)}{d}+(b+d) x\right )}{2 b} \] Output:
-1/2*cos(a-b*c/d)*Ci(c*(b-d)/d+(b-d)*x)/b+1/2*cos(a-b*c/d)*Ci(c*(b+d)/d+(b +d)*x)/b+1/2*sin(a-b*c/d)*Si(c*(b-d)/d+(b-d)*x)/b+sin(b*x+a)*Si(d*x+c)/b-1 /2*sin(a-b*c/d)*Si(c*(b+d)/d+(b+d)*x)/b
Result contains complex when optimal does not.
Time = 0.99 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.07 \[ \int \cos (a+b x) \text {Si}(c+d x) \, dx=\frac {e^{-\frac {i (b c+a d)}{d}} \left (-e^{\frac {2 i b c}{d}} \operatorname {ExpIntegralEi}\left (-\frac {i (b-d) (c+d x)}{d}\right )-e^{2 i a} \operatorname {ExpIntegralEi}\left (\frac {i (b-d) (c+d x)}{d}\right )+e^{\frac {2 i b c}{d}} \operatorname {ExpIntegralEi}\left (-\frac {i (b+d) (c+d x)}{d}\right )+e^{2 i a} \operatorname {ExpIntegralEi}\left (\frac {i (b+d) (c+d x)}{d}\right )+4 e^{\frac {i (b c+a d)}{d}} \sin (a+b x) \text {Si}(c+d x)\right )}{4 b} \] Input:
Integrate[Cos[a + b*x]*SinIntegral[c + d*x],x]
Output:
(-(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)*ExpIntegral Ei[((-I)*(b + d)*(c + d*x))/d] + E^((2*I)*a)*ExpIntegralEi[(I*(b + d)*(c + d*x))/d] + 4*E^((I*(b*c + a*d))/d)*Sin[a + b*x]*SinIntegral[c + d*x])/(4* b*E^((I*(b*c + a*d))/d))
Time = 0.49 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {7071, 4928, 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.
\(\displaystyle \int \cos (a+b x) \text {Si}(c+d x) \, dx\) |
\(\Big \downarrow \) 7071 |
\(\displaystyle \frac {\sin (a+b x) \text {Si}(c+d x)}{b}-\frac {d \int \frac {\sin (a+b x) \sin (c+d x)}{c+d x}dx}{b}\) |
\(\Big \downarrow \) 4928 |
\(\displaystyle \frac {\sin (a+b x) \text {Si}(c+d x)}{b}-\frac {d \int \left (\frac {\cos (a-c+(b-d) x)}{2 (c+d x)}-\frac {\cos (a+c+(b+d) x)}{2 (c+d x)}\right )dx}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sin (a+b x) \text {Si}(c+d x)}{b}-\frac {d \left (\frac {\cos \left (a-\frac {b c}{d}\right ) \operatorname {CosIntegral}\left (x (b-d)+\frac {c (b-d)}{d}\right )}{2 d}-\frac {\cos \left (a-\frac {b c}{d}\right ) \operatorname {CosIntegral}\left (x (b+d)+\frac {c (b+d)}{d}\right )}{2 d}-\frac {\sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (x (b-d)+\frac {c (b-d)}{d}\right )}{2 d}+\frac {\sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (x (b+d)+\frac {c (b+d)}{d}\right )}{2 d}\right )}{b}\) |
Input:
Int[Cos[a + b*x]*SinIntegral[c + d*x],x]
Output:
(Sin[a + b*x]*SinIntegral[c + d*x])/b - (d*((Cos[a - (b*c)/d]*CosIntegral[ (c*(b - d))/d + (b - d)*x])/(2*d) - (Cos[a - (b*c)/d]*CosIntegral[(c*(b + d))/d + (b + d)*x])/(2*d) - (Sin[a - (b*c)/d]*SinIntegral[(c*(b - d))/d + (b - d)*x])/(2*d) + (Sin[a - (b*c)/d]*SinIntegral[(c*(b + d))/d + (b + d)* x])/(2*d)))/b
Int[((e_.) + (f_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(p_.)*Sin[(c_.) + (d _.)*(x_)]^(q_.), x_Symbol] :> Int[ExpandTrigReduce[(e + f*x)^m, Sin[a + b*x ]^p*Sin[c + d*x]^q, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && IGtQ[q, 0] && IntegerQ[m]
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]
Time = 0.96 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.73
\[\frac {\frac {\operatorname {Si}\left (d x +c \right ) d \sin \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -c b}{d}\right )}{b}-\frac {d \left (\frac {d \left (\frac {\operatorname {Si}\left (\left (-1+\frac {b}{d}\right ) \left (d x +c \right )+a -\frac {c b}{d}+\frac {-a d +c b}{d}\right ) \sin \left (\frac {-a d +c b}{d}\right )}{d}+\frac {\operatorname {Ci}\left (\left (-1+\frac {b}{d}\right ) \left (d x +c \right )+a -\frac {c b}{d}+\frac {-a d +c b}{d}\right ) \cos \left (\frac {-a d +c b}{d}\right )}{d}\right )}{2}-\frac {d \left (\frac {\operatorname {Si}\left (\left (1+\frac {b}{d}\right ) \left (d x +c \right )+a -\frac {c b}{d}+\frac {-a d +c b}{d}\right ) \sin \left (\frac {-a d +c b}{d}\right )}{d}+\frac {\operatorname {Ci}\left (\left (1+\frac {b}{d}\right ) \left (d x +c \right )+a -\frac {c b}{d}+\frac {-a d +c b}{d}\right ) \cos \left (\frac {-a d +c b}{d}\right )}{d}\right )}{2}\right )}{b}}{d}\]
Input:
int(cos(b*x+a)*Si(d*x+c),x)
Output:
(Si(d*x+c)/b*d*sin(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)*sin((-a*d+b*c)/d)/d+Ci((-1+b/d)*(d*x+c)+a-c*b/d+ (-a*d+b*c)/d)*cos((-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)*sin((-a*d+b*c)/d)/d+Ci((1+b/d)*(d*x+c)+a-c*b/d+(-a*d+b*c)/d)*cos( (-a*d+b*c)/d)/d)))/d
Time = 0.09 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.96 \[ \int \cos (a+b x) \text {Si}(c+d x) \, dx=\frac {{\left (\operatorname {Ci}\left (\frac {b c + c d + {\left (b d + d^{2}\right )} x}{d}\right ) - \operatorname {Ci}\left (-\frac {b c - c d + {\left (b d - d^{2}\right )} x}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) - {\left (\operatorname {Si}\left (\frac {b c + c d + {\left (b d + d^{2}\right )} x}{d}\right ) + \operatorname {Si}\left (-\frac {b c - c d + {\left (b d - d^{2}\right )} x}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right ) + 2 \, \sin \left (b x + a\right ) \operatorname {Si}\left (d x + c\right )}{2 \, b} \] Input:
integrate(cos(b*x+a)*sin_integral(d*x+c),x, algorithm="fricas")
Output:
1/2*((cos_integral((b*c + c*d + (b*d + d^2)*x)/d) - cos_integral(-(b*c - c *d + (b*d - d^2)*x)/d))*cos(-(b*c - a*d)/d) - (sin_integral((b*c + c*d + ( b*d + d^2)*x)/d) + sin_integral(-(b*c - c*d + (b*d - d^2)*x)/d))*sin(-(b*c - a*d)/d) + 2*sin(b*x + a)*sin_integral(d*x + c))/b
\[ \int \cos (a+b x) \text {Si}(c+d x) \, dx=\int \cos {\left (a + b x \right )} \operatorname {Si}{\left (c + d x \right )}\, dx \] Input:
integrate(cos(b*x+a)*Si(d*x+c),x)
Output:
Integral(cos(a + b*x)*Si(c + d*x), x)
\[ \int \cos (a+b x) \text {Si}(c+d x) \, dx=\int { \cos \left (b x + a\right ) \operatorname {Si}\left (d x + c\right ) \,d x } \] Input:
integrate(cos(b*x+a)*sin_integral(d*x+c),x, algorithm="maxima")
Output:
integrate(cos(b*x + a)*sin_integral(d*x + c), x)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.40 (sec) , antiderivative size = 9214, normalized size of antiderivative = 60.22 \[ \int \cos (a+b x) \text {Si}(c+d x) \, dx=\text {Too large to display} \] Input:
integrate(cos(b*x+a)*sin_integral(d*x+c),x, algorithm="giac")
Output:
1/4*(real_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 - re al_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*tan(1/2*(b*c - c*d)/d)^2 + real_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*imag_part(cos_integ ral(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) + 4*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*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)*tan( 1/2*(b*c - c*d)/d)^2 + 2*imag_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 - 4*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)*tan(1/2*(b*c - c*d)/ d)^2 - 2*imag_part(cos_integral(b*x - d*x - c + b*c/d))*tan(1/2*a + 1/2...
Timed out. \[ \int \cos (a+b x) \text {Si}(c+d x) \, dx=\int \mathrm {sinint}\left (c+d\,x\right )\,\cos \left (a+b\,x\right ) \,d x \] Input:
int(sinint(c + d*x)*cos(a + b*x),x)
Output:
int(sinint(c + d*x)*cos(a + b*x), x)
\[ \int \cos (a+b x) \text {Si}(c+d x) \, dx=\int \cos \left (b x +a \right ) \mathit {si} \left (d x +c \right )d x \] Input:
int(cos(b*x+a)*Si(d*x+c),x)
Output:
int(cos(a + b*x)*si(c + d*x),x)