3.112 \(\int x^2 \text{CosIntegral}(b x) \sin (b x) \, dx\)

Optimal. Leaf size=111 \[ -\frac{\text{CosIntegral}(2 b x)}{b^3}+\frac{2 x \text{CosIntegral}(b x) \sin (b x)}{b^2}+\frac{2 \text{CosIntegral}(b x) \cos (b x)}{b^3}-\frac{\log (x)}{b^3}-\frac{\sin ^2(b x)}{b^3}+\frac{\cos ^2(b x)}{4 b^3}+\frac{x \sin (b x) \cos (b x)}{2 b^2}-\frac{x^2 \text{CosIntegral}(b x) \cos (b x)}{b}+\frac{x^2}{4 b} \]

[Out]

x^2/(4*b) + Cos[b*x]^2/(4*b^3) + (2*Cos[b*x]*CosIntegral[b*x])/b^3 - (x^2*Cos[b*x]*CosIntegral[b*x])/b - CosIn
tegral[2*b*x]/b^3 - Log[x]/b^3 + (x*Cos[b*x]*Sin[b*x])/(2*b^2) + (2*x*CosIntegral[b*x]*Sin[b*x])/b^2 - Sin[b*x
]^2/b^3

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Rubi [A]  time = 0.12765, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {6520, 12, 3310, 30, 6514, 2564, 6518, 3312, 3302} \[ -\frac{\text{CosIntegral}(2 b x)}{b^3}+\frac{2 x \text{CosIntegral}(b x) \sin (b x)}{b^2}+\frac{2 \text{CosIntegral}(b x) \cos (b x)}{b^3}-\frac{\log (x)}{b^3}-\frac{\sin ^2(b x)}{b^3}+\frac{\cos ^2(b x)}{4 b^3}+\frac{x \sin (b x) \cos (b x)}{2 b^2}-\frac{x^2 \text{CosIntegral}(b x) \cos (b x)}{b}+\frac{x^2}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*CosIntegral[b*x]*Sin[b*x],x]

[Out]

x^2/(4*b) + Cos[b*x]^2/(4*b^3) + (2*Cos[b*x]*CosIntegral[b*x])/b^3 - (x^2*Cos[b*x]*CosIntegral[b*x])/b - CosIn
tegral[2*b*x]/b^3 - Log[x]/b^3 + (x*Cos[b*x]*Sin[b*x])/(2*b^2) + (2*x*CosIntegral[b*x]*Sin[b*x])/b^2 - Sin[b*x
]^2/b^3

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 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rule 30

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

Rule 6514

Int[Cos[(a_.) + (b_.)*(x_)]*CosIntegral[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e
+ f*x)^m*Sin[a + b*x]*CosIntegral[c + d*x])/b, x] + (-Dist[d/b, Int[((e + f*x)^m*Sin[a + b*x]*Cos[c + d*x])/(c
+ d*x), x], x] - Dist[(f*m)/b, Int[(e + f*x)^(m - 1)*Sin[a + b*x]*CosIntegral[c + d*x], x], x]) /; FreeQ[{a,
b, c, d, e, f}, x] && IGtQ[m, 0]

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 6518

Int[CosIntegral[(c_.) + (d_.)*(x_)]*Sin[(a_.) + (b_.)*(x_)], x_Symbol] :> -Simp[(Cos[a + b*x]*CosIntegral[c +
d*x])/b, x] + Dist[d/b, Int[(Cos[a + b*x]*Cos[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^2 \text{Ci}(b x) \sin (b x) \, dx &=-\frac{x^2 \cos (b x) \text{Ci}(b x)}{b}+\frac{2 \int x \cos (b x) \text{Ci}(b x) \, dx}{b}+\int \frac{x \cos ^2(b x)}{b} \, dx\\ &=-\frac{x^2 \cos (b x) \text{Ci}(b x)}{b}+\frac{2 x \text{Ci}(b x) \sin (b x)}{b^2}-\frac{2 \int \text{Ci}(b x) \sin (b x) \, dx}{b^2}+\frac{\int x \cos ^2(b x) \, dx}{b}-\frac{2 \int \frac{\cos (b x) \sin (b x)}{b} \, dx}{b}\\ &=\frac{\cos ^2(b x)}{4 b^3}+\frac{2 \cos (b x) \text{Ci}(b x)}{b^3}-\frac{x^2 \cos (b x) \text{Ci}(b x)}{b}+\frac{x \cos (b x) \sin (b x)}{2 b^2}+\frac{2 x \text{Ci}(b x) \sin (b x)}{b^2}-\frac{2 \int \frac{\cos ^2(b x)}{b x} \, dx}{b^2}-\frac{2 \int \cos (b x) \sin (b x) \, dx}{b^2}+\frac{\int x \, dx}{2 b}\\ &=\frac{x^2}{4 b}+\frac{\cos ^2(b x)}{4 b^3}+\frac{2 \cos (b x) \text{Ci}(b x)}{b^3}-\frac{x^2 \cos (b x) \text{Ci}(b x)}{b}+\frac{x \cos (b x) \sin (b x)}{2 b^2}+\frac{2 x \text{Ci}(b x) \sin (b x)}{b^2}-\frac{2 \int \frac{\cos ^2(b x)}{x} \, dx}{b^3}-\frac{2 \operatorname{Subst}(\int x \, dx,x,\sin (b x))}{b^3}\\ &=\frac{x^2}{4 b}+\frac{\cos ^2(b x)}{4 b^3}+\frac{2 \cos (b x) \text{Ci}(b x)}{b^3}-\frac{x^2 \cos (b x) \text{Ci}(b x)}{b}+\frac{x \cos (b x) \sin (b x)}{2 b^2}+\frac{2 x \text{Ci}(b x) \sin (b x)}{b^2}-\frac{\sin ^2(b x)}{b^3}-\frac{2 \int \left (\frac{1}{2 x}+\frac{\cos (2 b x)}{2 x}\right ) \, dx}{b^3}\\ &=\frac{x^2}{4 b}+\frac{\cos ^2(b x)}{4 b^3}+\frac{2 \cos (b x) \text{Ci}(b x)}{b^3}-\frac{x^2 \cos (b x) \text{Ci}(b x)}{b}-\frac{\log (x)}{b^3}+\frac{x \cos (b x) \sin (b x)}{2 b^2}+\frac{2 x \text{Ci}(b x) \sin (b x)}{b^2}-\frac{\sin ^2(b x)}{b^3}-\frac{\int \frac{\cos (2 b x)}{x} \, dx}{b^3}\\ &=\frac{x^2}{4 b}+\frac{\cos ^2(b x)}{4 b^3}+\frac{2 \cos (b x) \text{Ci}(b x)}{b^3}-\frac{x^2 \cos (b x) \text{Ci}(b x)}{b}-\frac{\text{Ci}(2 b x)}{b^3}-\frac{\log (x)}{b^3}+\frac{x \cos (b x) \sin (b x)}{2 b^2}+\frac{2 x \text{Ci}(b x) \sin (b x)}{b^2}-\frac{\sin ^2(b x)}{b^3}\\ \end{align*}

Mathematica [A]  time = 0.0859148, size = 72, normalized size = 0.65 \[ \frac{-8 \text{CosIntegral}(b x) \left (\left (b^2 x^2-2\right ) \cos (b x)-2 b x \sin (b x)\right )+2 b^2 x^2-8 \text{CosIntegral}(2 b x)+2 b x \sin (2 b x)+5 \cos (2 b x)-8 \log (x)}{8 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*CosIntegral[b*x]*Sin[b*x],x]

[Out]

(2*b^2*x^2 + 5*Cos[2*b*x] - 8*CosIntegral[2*b*x] - 8*Log[x] - 8*CosIntegral[b*x]*((-2 + b^2*x^2)*Cos[b*x] - 2*
b*x*Sin[b*x]) + 2*b*x*Sin[2*b*x])/(8*b^3)

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Maple [A]  time = 0.062, size = 107, normalized size = 1. \begin{align*} -{\frac{{x}^{2}{\it Ci} \left ( bx \right ) \cos \left ( bx \right ) }{b}}+2\,{\frac{x{\it Ci} \left ( bx \right ) \sin \left ( bx \right ) }{{b}^{2}}}+2\,{\frac{{\it Ci} \left ( bx \right ) \cos \left ( bx \right ) }{{b}^{3}}}+{\frac{x\cos \left ( bx \right ) \sin \left ( bx \right ) }{2\,{b}^{2}}}+{\frac{{x}^{2}}{4\,b}}-{\frac{ \left ( \sin \left ( bx \right ) \right ) ^{2}}{4\,{b}^{3}}}+{\frac{ \left ( \cos \left ( bx \right ) \right ) ^{2}}{{b}^{3}}}-{\frac{\ln \left ( bx \right ) }{{b}^{3}}}-{\frac{{\it Ci} \left ( 2\,bx \right ) }{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*Ci(b*x)*sin(b*x),x)

[Out]

-x^2*Ci(b*x)*cos(b*x)/b+2*x*Ci(b*x)*sin(b*x)/b^2+2*Ci(b*x)*cos(b*x)/b^3+1/2*x*cos(b*x)*sin(b*x)/b^2+1/4*x^2/b-
1/4*sin(b*x)^2/b^3+cos(b*x)^2/b^3-1/b^3*ln(b*x)-Ci(2*b*x)/b^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2}{\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^2*Ci(b*x)*sin(b*x),x, algorithm="maxima")

[Out]

integrate(x^2*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^{2} \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^2*Ci(b*x)*sin(b*x),x, algorithm="fricas")

[Out]

integral(x^2*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^{2} \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**2*Ci(b*x)*sin(b*x),x)

[Out]

Integral(x**2*sin(b*x)*Ci(b*x), x)

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Giac [A]  time = 1.15272, size = 89, normalized size = 0.8 \begin{align*}{\left (\frac{2 \, x \sin \left (b x\right )}{b^{2}} - \frac{{\left (b^{2} x^{2} - 2\right )} \cos \left (b x\right )}{b^{3}}\right )} \operatorname{Ci}\left (b x\right ) + \frac{b^{2} x^{2} - 2 \, \operatorname{Ci}\left (2 \, b x\right ) - 2 \, \operatorname{Ci}\left (-2 \, b x\right ) - 4 \, \log \left (x\right )}{4 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*Ci(b*x)*sin(b*x),x, algorithm="giac")

[Out]

(2*x*sin(b*x)/b^2 - (b^2*x^2 - 2)*cos(b*x)/b^3)*cos_integral(b*x) + 1/4*(b^2*x^2 - 2*cos_integral(2*b*x) - 2*c
os_integral(-2*b*x) - 4*log(x))/b^3