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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.274699, antiderivative size = 97, normalized size of antiderivative = 1., 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 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 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 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]]

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 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 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 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]

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*}

Mathematica [A]  time = 0.237806, size = 71, normalized size = 0.73 \[ -\frac{-2 \text{CosIntegral}(2 (a+b x))+2 a \text{Si}(2 (a+b x))+4 \text{Si}(a+b x) (b x \cos (a+b x)-\sin (a+b x))+2 \log (a+b x)+\cos (2 (a+b x))}{4 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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])*SinInte
gral[a + b*x] + 2*a*SinIntegral[2*(a + b*x)])/(4*b^2)

________________________________________________________________________________________

Maple [A]  time = 0.057, size = 89, normalized size = 0.9 \begin{align*} -{\frac{x\cos \left ( bx+a \right ){\it Si} \left ( bx+a \right ) }{b}}+{\frac{{\it Si} \left ( bx+a \right ) \sin \left ( bx+a \right ) }{{b}^{2}}}-{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{2}}{2\,{b}^{2}}}-{\frac{a{\it Si} \left ( 2\,bx+2\,a \right ) }{2\,{b}^{2}}}-{\frac{\ln \left ( bx+a \right ) }{2\,{b}^{2}}}+{\frac{{\it Ci} \left ( 2\,bx+2\,a \right ) }{2\,{b}^{2}}} \end{align*}

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x{\rm Si}\left (b x + a\right ) \sin \left (b x + a\right )\,{d x} \end{align*}

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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x \sin \left (b x + a\right ) \operatorname{Si}\left (b x + a\right ), x\right ) \end{align*}

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)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \sin{\left (a + b x \right )} \operatorname{Si}{\left (a + b x \right )}\, dx \end{align*}

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)

________________________________________________________________________________________

Giac [C]  time = 1.18991, size = 146, normalized size = 1.51 \begin{align*} -{\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 ) - a \Im \left ( \operatorname{Ci}\left (-2 \, b x - 2 \, a\right ) \right ) + 2 \, a \operatorname{Si}\left (2 \, b x + 2 \, a\right ) + 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 )}{4 \, b^{2}} \end{align*}

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)) - a
*imag_part(cos_integral(-2*b*x - 2*a)) + 2*a*sin_integral(2*b*x + 2*a) + 2*log(abs(b*x + a)) - real_part(cos_i
ntegral(2*b*x + 2*a)) - real_part(cos_integral(-2*b*x - 2*a)))/b^2