3.60 \(\int x \cos (a+b x) \text {Si}(a+b x) \, dx\)

Optimal. Leaf size=108 \[ -\frac {a \text {Ci}(2 a+2 b x)}{2 b^2}-\frac {\text {Si}(2 a+2 b x)}{2 b^2}+\frac {\text {Si}(a+b x) \cos (a+b x)}{b^2}+\frac {a \log (a+b x)}{2 b^2}+\frac {\sin (a+b x) \cos (a+b x)}{2 b^2}+\frac {x \text {Si}(a+b x) \sin (a+b x)}{b}-\frac {x}{2 b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {6519, 6742, 2635, 8, 3312, 3302, 6511, 4406, 12, 3299} \[ -\frac {a \text {CosIntegral}(2 a+2 b x)}{2 b^2}-\frac {\text {Si}(2 a+2 b x)}{2 b^2}+\frac {\text {Si}(a+b x) \cos (a+b x)}{b^2}+\frac {a \log (a+b x)}{2 b^2}+\frac {\sin (a+b x) \cos (a+b x)}{2 b^2}+\frac {x \text {Si}(a+b x) \sin (a+b x)}{b}-\frac {x}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2635

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] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
 e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[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 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(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]
&& IGtQ[p, 0]

Rule 6511

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

Rule 6519

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] + (-Dist[d/b, Int[((e + f*x)^m*Sin[a + b*x]*Sin[c + d*x])/(c
 + d*x), x], x] - Dist[(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 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int x \cos (a+b x) \text {Si}(a+b x) \, dx &=\frac {x \sin (a+b x) \text {Si}(a+b x)}{b}-\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 {\cos (a+b x) \text {Si}(a+b x)}{b^2}+\frac {x \sin (a+b x) \text {Si}(a+b x)}{b}-\frac {\int \frac {\cos (a+b x) \sin (a+b x)}{a+b x} \, dx}{b}-\int \left (\frac {\sin ^2(a+b x)}{b}-\frac {a \sin ^2(a+b x)}{b (a+b x)}\right ) \, dx\\ &=\frac {\cos (a+b x) \text {Si}(a+b x)}{b^2}+\frac {x \sin (a+b x) \text {Si}(a+b x)}{b}-\frac {\int \sin ^2(a+b x) \, dx}{b}-\frac {\int \frac {\sin (2 a+2 b x)}{2 (a+b x)} \, dx}{b}+\frac {a \int \frac {\sin ^2(a+b x)}{a+b x} \, dx}{b}\\ &=\frac {\cos (a+b x) \sin (a+b x)}{2 b^2}+\frac {\cos (a+b x) \text {Si}(a+b x)}{b^2}+\frac {x \sin (a+b x) \text {Si}(a+b x)}{b}-\frac {\int 1 \, dx}{2 b}-\frac {\int \frac {\sin (2 a+2 b x)}{a+b x} \, dx}{2 b}+\frac {a \int \left (\frac {1}{2 (a+b x)}-\frac {\cos (2 a+2 b x)}{2 (a+b x)}\right ) \, dx}{b}\\ &=-\frac {x}{2 b}+\frac {a \log (a+b x)}{2 b^2}+\frac {\cos (a+b x) \sin (a+b x)}{2 b^2}+\frac {\cos (a+b x) \text {Si}(a+b x)}{b^2}+\frac {x \sin (a+b x) \text {Si}(a+b x)}{b}-\frac {\text {Si}(2 a+2 b x)}{2 b^2}-\frac {a \int \frac {\cos (2 a+2 b x)}{a+b x} \, dx}{2 b}\\ &=-\frac {x}{2 b}-\frac {a \text {Ci}(2 a+2 b x)}{2 b^2}+\frac {a \log (a+b x)}{2 b^2}+\frac {\cos (a+b x) \sin (a+b x)}{2 b^2}+\frac {\cos (a+b x) \text {Si}(a+b x)}{b^2}+\frac {x \sin (a+b x) \text {Si}(a+b x)}{b}-\frac {\text {Si}(2 a+2 b x)}{2 b^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.23, size = 74, normalized size = 0.69 \[ \frac {-2 a \text {Ci}(2 (a+b x))-2 \text {Si}(2 (a+b x))+4 \text {Si}(a+b x) (b x \sin (a+b x)+\cos (a+b x))+2 a \log (a+b x)+\sin (2 (a+b x))-2 b x}{4 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x \cos \left (b x + a\right ) \operatorname {Si}\left (b x + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cos(b*x+a)*Si(b*x+a),x, algorithm="fricas")

[Out]

integral(x*cos(b*x + a)*sin_integral(b*x + a), x)

________________________________________________________________________________________

giac [C]  time = 0.26, size = 528, normalized size = 4.89 \[ {\left (\frac {x \sin \left (b x + a\right )}{b} + \frac {\cos \left (b x + a\right )}{b^{2}}\right )} \operatorname {Si}\left (b x + a\right ) - \frac {2 \, b x \tan \left (b x\right )^{2} \tan \relax (a)^{2} - 2 \, a \log \left ({\left | b x + a \right |}\right ) \tan \left (b x\right )^{2} \tan \relax (a)^{2} + a \Re \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x\right )^{2} \tan \relax (a)^{2} + a \Re \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x\right )^{2} \tan \relax (a)^{2} + \Im \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x\right )^{2} \tan \relax (a)^{2} - \Im \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x\right )^{2} \tan \relax (a)^{2} + 2 \, \operatorname {Si}\left (2 \, b x + 2 \, a\right ) \tan \left (b x\right )^{2} \tan \relax (a)^{2} + 2 \, b x \tan \left (b x\right )^{2} - 2 \, a \log \left ({\left | b x + a \right |}\right ) \tan \left (b x\right )^{2} + a \Re \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x\right )^{2} + a \Re \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x\right )^{2} + 2 \, b x \tan \relax (a)^{2} - 2 \, a \log \left ({\left | b x + a \right |}\right ) \tan \relax (a)^{2} + a \Re \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \relax (a)^{2} + a \Re \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \relax (a)^{2} + \Im \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x\right )^{2} - \Im \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x\right )^{2} + 2 \, \operatorname {Si}\left (2 \, b x + 2 \, a\right ) \tan \left (b x\right )^{2} + 2 \, \tan \left (b x\right )^{2} \tan \relax (a) + \Im \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \relax (a)^{2} - \Im \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \relax (a)^{2} + 2 \, \operatorname {Si}\left (2 \, b x + 2 \, a\right ) \tan \relax (a)^{2} + 2 \, \tan \left (b x\right ) \tan \relax (a)^{2} + 2 \, b x - 2 \, a \log \left ({\left | b x + a \right |}\right ) + a \Re \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) + a \Re \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) + \Im \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) - \Im \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) + 2 \, \operatorname {Si}\left (2 \, b x + 2 \, a\right ) - 2 \, \tan \left (b x\right ) - 2 \, \tan \relax (a)}{4 \, {\left (b^{2} \tan \left (b x\right )^{2} \tan \relax (a)^{2} + b^{2} \tan \left (b x\right )^{2} + b^{2} \tan \relax (a)^{2} + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cos(b*x+a)*Si(b*x+a),x, algorithm="giac")

[Out]

(x*sin(b*x + a)/b + cos(b*x + a)/b^2)*sin_integral(b*x + a) - 1/4*(2*b*x*tan(b*x)^2*tan(a)^2 - 2*a*log(abs(b*x
 + a))*tan(b*x)^2*tan(a)^2 + a*real_part(cos_integral(2*b*x + 2*a))*tan(b*x)^2*tan(a)^2 + a*real_part(cos_inte
gral(-2*b*x - 2*a))*tan(b*x)^2*tan(a)^2 + imag_part(cos_integral(2*b*x + 2*a))*tan(b*x)^2*tan(a)^2 - imag_part
(cos_integral(-2*b*x - 2*a))*tan(b*x)^2*tan(a)^2 + 2*sin_integral(2*b*x + 2*a)*tan(b*x)^2*tan(a)^2 + 2*b*x*tan
(b*x)^2 - 2*a*log(abs(b*x + a))*tan(b*x)^2 + a*real_part(cos_integral(2*b*x + 2*a))*tan(b*x)^2 + a*real_part(c
os_integral(-2*b*x - 2*a))*tan(b*x)^2 + 2*b*x*tan(a)^2 - 2*a*log(abs(b*x + a))*tan(a)^2 + a*real_part(cos_inte
gral(2*b*x + 2*a))*tan(a)^2 + a*real_part(cos_integral(-2*b*x - 2*a))*tan(a)^2 + imag_part(cos_integral(2*b*x
+ 2*a))*tan(b*x)^2 - imag_part(cos_integral(-2*b*x - 2*a))*tan(b*x)^2 + 2*sin_integral(2*b*x + 2*a)*tan(b*x)^2
 + 2*tan(b*x)^2*tan(a) + imag_part(cos_integral(2*b*x + 2*a))*tan(a)^2 - imag_part(cos_integral(-2*b*x - 2*a))
*tan(a)^2 + 2*sin_integral(2*b*x + 2*a)*tan(a)^2 + 2*tan(b*x)*tan(a)^2 + 2*b*x - 2*a*log(abs(b*x + a)) + a*rea
l_part(cos_integral(2*b*x + 2*a)) + a*real_part(cos_integral(-2*b*x - 2*a)) + imag_part(cos_integral(2*b*x + 2
*a)) - imag_part(cos_integral(-2*b*x - 2*a)) + 2*sin_integral(2*b*x + 2*a) - 2*tan(b*x) - 2*tan(a))/(b^2*tan(b
*x)^2*tan(a)^2 + b^2*tan(b*x)^2 + b^2*tan(a)^2 + b^2)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 105, normalized size = 0.97 \[ \frac {x \Si \left (b x +a \right ) \sin \left (b x +a \right )}{b}+\frac {\cos \left (b x +a \right ) \Si \left (b x +a \right )}{b^{2}}-\frac {\Si \left (2 b x +2 a \right )}{2 b^{2}}+\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2 b^{2}}-\frac {x}{2 b}-\frac {a}{2 b^{2}}+\frac {a \ln \left (b x +a \right )}{2 b^{2}}-\frac {a \Ci \left (2 b x +2 a \right )}{2 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x {\rm Si}\left (b x + a\right ) \cos \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cos(b*x+a)*Si(b*x+a),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\mathrm {sinint}\left (a+b\,x\right )\,\cos \left (a+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sinint(a + b*x)*cos(a + b*x),x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \cos {\left (a + b x \right )} \operatorname {Si}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________