3.43 \(\int x \sin (b x) \text{Si}(b x) \, dx\)

Optimal. Leaf size=61 \[ \frac{\text{CosIntegral}(2 b x)}{2 b^2}+\frac{\text{Si}(b x) \sin (b x)}{b^2}-\frac{\log (x)}{2 b^2}+\frac{\sin ^2(b x)}{2 b^2}-\frac{x \text{Si}(b x) \cos (b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.0812445, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {6513, 12, 2564, 30, 6517, 3312, 3302} \[ \frac{\text{CosIntegral}(2 b x)}{2 b^2}+\frac{\text{Si}(b x) \sin (b x)}{b^2}-\frac{\log (x)}{2 b^2}+\frac{\sin ^2(b x)}{2 b^2}-\frac{x \text{Si}(b x) \cos (b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

CosIntegral[2*b*x]/(2*b^2) - Log[x]/(2*b^2) + Sin[b*x]^2/(2*b^2) - (x*Cos[b*x]*SinIntegral[b*x])/b + (Sin[b*x]
*SinIntegral[b*x])/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 12

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

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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

Mathematica [A]  time = 0.0789173, size = 44, normalized size = 0.72 \[ -\frac{-2 \text{CosIntegral}(2 b x)+4 \text{Si}(b x) (b x \cos (b x)-\sin (b x))+\cos (2 b x)+2 \log (x)}{4 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.062, size = 58, normalized size = 1. \begin{align*} -{\frac{x\cos \left ( bx \right ){\it Si} \left ( bx \right ) }{b}}+{\frac{{\it Si} \left ( bx \right ) \sin \left ( bx \right ) }{{b}^{2}}}-{\frac{ \left ( \cos \left ( bx \right ) \right ) ^{2}}{2\,{b}^{2}}}-{\frac{\ln \left ( bx \right ) }{2\,{b}^{2}}}+{\frac{{\it Ci} \left ( 2\,bx \right ) }{2\,{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*Si(b*x)*sin(b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sin(b*x)*Si(b*x), x)

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Giac [A]  time = 1.16351, size = 62, normalized size = 1.02 \begin{align*} -{\left (\frac{x \cos \left (b x\right )}{b} - \frac{\sin \left (b x\right )}{b^{2}}\right )} \operatorname{Si}\left (b x\right ) + \frac{\operatorname{Ci}\left (2 \, b x\right ) + \operatorname{Ci}\left (-2 \, b x\right ) - 2 \, \log \left (x\right )}{4 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*Si(b*x)*sin(b*x),x, algorithm="giac")

[Out]

-(x*cos(b*x)/b - sin(b*x)/b^2)*sin_integral(b*x) + 1/4*(cos_integral(2*b*x) + cos_integral(-2*b*x) - 2*log(x))
/b^2