3.111 \(\int x \text{CosIntegral}(b x) \sin (b x) \, dx\)

Optimal. Leaf size=62 \[ \frac{\text{CosIntegral}(b x) \sin (b x)}{b^2}-\frac{\text{Si}(2 b x)}{2 b^2}+\frac{\sin (b x) \cos (b x)}{2 b^2}-\frac{x \text{CosIntegral}(b x) \cos (b x)}{b}+\frac{x}{2 b} \]

[Out]

x/(2*b) - (x*Cos[b*x]*CosIntegral[b*x])/b + (Cos[b*x]*Sin[b*x])/(2*b^2) + (CosIntegral[b*x]*Sin[b*x])/b^2 - Si
nIntegral[2*b*x]/(2*b^2)

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Rubi [A]  time = 0.0691263, antiderivative size = 62, 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 = {6520, 12, 2635, 8, 6512, 4406, 3299} \[ \frac{\text{CosIntegral}(b x) \sin (b x)}{b^2}-\frac{\text{Si}(2 b x)}{2 b^2}+\frac{\sin (b x) \cos (b x)}{2 b^2}-\frac{x \text{CosIntegral}(b x) \cos (b x)}{b}+\frac{x}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/(2*b) - (x*Cos[b*x]*CosIntegral[b*x])/b + (Cos[b*x]*Sin[b*x])/(2*b^2) + (CosIntegral[b*x]*Sin[b*x])/b^2 - Si
nIntegral[2*b*x]/(2*b^2)

Rule 6520

Int[CosIntegral[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)], x_Symbol] :> -Simp[((e
+ f*x)^m*Cos[a + b*x]*CosIntegral[c + d*x])/b, x] + (Dist[d/b, Int[((e + f*x)^m*Cos[a + b*x]*Cos[c + d*x])/(c
+ d*x), x], x] + Dist[(f*m)/b, Int[(e + f*x)^(m - 1)*Cos[a + b*x]*CosIntegral[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 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 8

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

Rule 6512

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0504018, size = 44, normalized size = 0.71 \[ \frac{\text{CosIntegral}(b x) (4 \sin (b x)-4 b x \cos (b x))-2 \text{Si}(2 b x)+2 b x+\sin (2 b x)}{4 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.057, size = 45, normalized size = 0.7 \begin{align*}{\frac{1}{{b}^{2}} \left ({\it Ci} \left ( bx \right ) \left ( -bx\cos \left ( bx \right ) +\sin \left ( bx \right ) \right ) +{\frac{\sin \left ( bx \right ) \cos \left ( bx \right ) }{2}}+{\frac{bx}{2}}-{\frac{{\it Si} \left ( 2\,bx \right ) }{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x{\rm Ci}\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*Ci(b*x)*sin(b*x),x, algorithm="maxima")

[Out]

integrate(x*Ci(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 \operatorname{Ci}\left (b x\right ) \sin \left (b x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*cos_integral(b*x)*sin(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{Ci}{\left (b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sin(b*x)*Ci(b*x), x)

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Giac [C]  time = 1.17779, size = 77, normalized size = 1.24 \begin{align*} -{\left (\frac{x \cos \left (b x\right )}{b} - \frac{\sin \left (b x\right )}{b^{2}}\right )} \operatorname{Ci}\left (b x\right ) + \frac{2 \, b x - \Im \left ( \operatorname{Ci}\left (2 \, b x\right ) \right ) + \Im \left ( \operatorname{Ci}\left (-2 \, b x\right ) \right ) - 2 \, \operatorname{Si}\left (2 \, b x\right )}{4 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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