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

Optimal. Leaf size=97 \[ \frac {\text {Ci}(2 a+2 b x)}{2 b^2}-\frac {a \text {Si}(2 a+2 b x)}{2 b^2}+\frac {\text {Si}(a+b x) \sin (a+b x)}{b^2}-\frac {\log (a+b x)}{2 b^2}-\frac {\cos (2 a+2 b x)}{4 b^2}-\frac {x \text {Si}(a+b x) \cos (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {6513, 4573, 6741, 6742, 2638, 3299, 6517, 3312, 3302} \[ \frac {\text {CosIntegral}(2 a+2 b x)}{2 b^2}-\frac {a \text {Si}(2 a+2 b x)}{2 b^2}+\frac {\text {Si}(a+b x) \sin (a+b x)}{b^2}-\frac {\log (a+b x)}{2 b^2}-\frac {\cos (2 a+2 b x)}{4 b^2}-\frac {x \text {Si}(a+b x) \cos (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 4573

Int[Cos[w_]^(p_.)*(u_.)*Sin[v_]^(p_.), x_Symbol] :> Dist[1/2^p, Int[u*Sin[2*v]^p, x], x] /; EqQ[w, v] && Integ
erQ[p]

Rule 6513

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] + (Dist[d/b, Int[((e + f*x)^m*Cos[a + b*x]*Sin[c + d*x])/(c
+ d*x), x], x] + Dist[(f*m)/b, Int[(e + f*x)^(m - 1)*Cos[a + b*x]*SinIntegral[c + d*x], x], x]) /; FreeQ[{a,
b, c, d, e, f}, x] && IGtQ[m, 0]

Rule 6517

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

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {align*} \int x \sin (a+b x) \text {Si}(a+b x) \, dx &=-\frac {x \cos (a+b x) \text {Si}(a+b x)}{b}+\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 \cos (a+b x) \text {Si}(a+b x)}{b}+\frac {\sin (a+b x) \text {Si}(a+b x)}{b^2}+\frac {1}{2} \int \frac {x \sin (2 (a+b x))}{a+b x} \, dx-\frac {\int \frac {\sin ^2(a+b x)}{a+b x} \, dx}{b}\\ &=-\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 {1}{2} \int \frac {x \sin (2 a+2 b x)}{a+b x} \, dx-\frac {\int \left (\frac {1}{2 (a+b x)}-\frac {\cos (2 a+2 b x)}{2 (a+b x)}\right ) \, dx}{b}\\ &=-\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 {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 {\int \frac {\cos (2 a+2 b x)}{a+b x} \, dx}{2 b}\\ &=\frac {\text {Ci}(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 {\int \sin (2 a+2 b x) \, dx}{2 b}+\frac {a \int \frac {\sin (2 a+2 b x)}{-a-b x} \, dx}{2 b}\\ &=-\frac {\cos (2 a+2 b x)}{4 b^2}+\frac {\text {Ci}(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}\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 71, normalized size = 0.73 \[ -\frac {-2 \text {Ci}(2 (a+b x))+2 a \text {Si}(2 (a+b x))+4 \text {Si}(a+b x) (b x \cos (a+b x)-\sin (a+b x))+2 \log (a+b x)+\cos (2 (a+b x))}{4 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x \sin \left (b x + a\right ) \operatorname {Si}\left (b x + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.32, size = 507, normalized size = 5.23 \[ -{\left (\frac {x \cos \left (b x + a\right )}{b} - \frac {\sin \left (b x + a\right )}{b^{2}}\right )} \operatorname {Si}\left (b x + a\right ) - \frac {a \Im \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x\right )^{2} \tan \relax (a)^{2} - a \Im \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x\right )^{2} \tan \relax (a)^{2} + 2 \, a \operatorname {Si}\left (2 \, b x + 2 \, a\right ) \tan \left (b x\right )^{2} \tan \relax (a)^{2} + 2 \, \log \left ({\left | b x + a \right |}\right ) \tan \left (b x\right )^{2} \tan \relax (a)^{2} - \Re \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x\right )^{2} \tan \relax (a)^{2} - \Re \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x\right )^{2} \tan \relax (a)^{2} + a \Im \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x\right )^{2} - a \Im \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x\right )^{2} + 2 \, a \operatorname {Si}\left (2 \, b x + 2 \, a\right ) \tan \left (b x\right )^{2} + a \Im \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \relax (a)^{2} - a \Im \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \relax (a)^{2} + 2 \, a \operatorname {Si}\left (2 \, b x + 2 \, a\right ) \tan \relax (a)^{2} + \tan \left (b x\right )^{2} \tan \relax (a)^{2} + 2 \, \log \left ({\left | b x + a \right |}\right ) \tan \left (b x\right )^{2} - \Re \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x\right )^{2} - \Re \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x\right )^{2} + 2 \, \log \left ({\left | b x + a \right |}\right ) \tan \relax (a)^{2} - \Re \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \relax (a)^{2} - \Re \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \relax (a)^{2} + a \Im \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) - a \Im \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) + 2 \, a \operatorname {Si}\left (2 \, b x + 2 \, a\right ) - \tan \left (b x\right )^{2} - 4 \, \tan \left (b x\right ) \tan \relax (a) - \tan \relax (a)^{2} + 2 \, \log \left ({\left | b x + a \right |}\right ) - \Re \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) - \Re \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) + 1}{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*Si(b*x+a)*sin(b*x+a),x, algorithm="giac")

[Out]

-(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_integral(-2*b*x - 2*a))*tan(b*x)^2*tan(a)^2 + 2*a*sin_integral(2*b*x + 2*a)
*tan(b*x)^2*tan(a)^2 + 2*log(abs(b*x + a))*tan(b*x)^2*tan(a)^2 - real_part(cos_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
))*tan(b*x)^2 - a*imag_part(cos_integral(-2*b*x - 2*a))*tan(b*x)^2 + 2*a*sin_integral(2*b*x + 2*a)*tan(b*x)^2
+ a*imag_part(cos_integral(2*b*x + 2*a))*tan(a)^2 - a*imag_part(cos_integral(-2*b*x - 2*a))*tan(a)^2 + 2*a*sin
_integral(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_integra
l(2*b*x + 2*a))*tan(b*x)^2 - real_part(cos_integral(-2*b*x - 2*a))*tan(b*x)^2 + 2*log(abs(b*x + a))*tan(a)^2 -
real_part(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_integral(-2*b*x - 2*a)) + 2*a*sin_integral(2*b*x + 2*a) - tan(b*x
)^2 - 4*tan(b*x)*tan(a) - tan(a)^2 + 2*log(abs(b*x + a)) - real_part(cos_integral(2*b*x + 2*a)) - real_part(co
s_integral(-2*b*x - 2*a)) + 1)/(b^2*tan(b*x)^2*tan(a)^2 + b^2*tan(b*x)^2 + b^2*tan(a)^2 + b^2)

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maple [A]  time = 0.03, size = 89, normalized size = 0.92 \[ -\frac {x \cos \left (b x +a \right ) \Si \left (b x +a \right )}{b}+\frac {\Si \left (b x +a \right ) \sin \left (b x +a \right )}{b^{2}}-\frac {\cos ^{2}\left (b x +a \right )}{2 b^{2}}-\frac {a \Si \left (2 b x +2 a \right )}{2 b^{2}}-\frac {\ln \left (b x +a \right )}{2 b^{2}}+\frac {\Ci \left (2 b x +2 a \right )}{2 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sinint(a + b*x)*sin(a + b*x),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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