3.119 \(\int x^2 \cos (b x) \text{CosIntegral}(b x) \, dx\)

Optimal. Leaf size=89 \[ -\frac{2 \text{CosIntegral}(b x) \sin (b x)}{b^3}+\frac{2 x \text{CosIntegral}(b x) \cos (b x)}{b^2}+\frac{\text{Si}(2 b x)}{b^3}-\frac{3 x}{4 b^2}-\frac{x \sin ^2(b x)}{2 b^2}-\frac{5 \sin (b x) \cos (b x)}{4 b^3}+\frac{x^2 \text{CosIntegral}(b x) \sin (b x)}{b} \]

[Out]

(-3*x)/(4*b^2) + (2*x*Cos[b*x]*CosIntegral[b*x])/b^2 - (5*Cos[b*x]*Sin[b*x])/(4*b^3) - (2*CosIntegral[b*x]*Sin
[b*x])/b^3 + (x^2*CosIntegral[b*x]*Sin[b*x])/b - (x*Sin[b*x]^2)/(2*b^2) + SinIntegral[2*b*x]/b^3

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Rubi [A]  time = 0.112837, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {6514, 12, 3443, 2635, 8, 6520, 6512, 4406, 3299} \[ -\frac{2 \text{CosIntegral}(b x) \sin (b x)}{b^3}+\frac{2 x \text{CosIntegral}(b x) \cos (b x)}{b^2}+\frac{\text{Si}(2 b x)}{b^3}-\frac{3 x}{4 b^2}-\frac{x \sin ^2(b x)}{2 b^2}-\frac{5 \sin (b x) \cos (b x)}{4 b^3}+\frac{x^2 \text{CosIntegral}(b x) \sin (b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x)/(4*b^2) + (2*x*Cos[b*x]*CosIntegral[b*x])/b^2 - (5*Cos[b*x]*Sin[b*x])/(4*b^3) - (2*CosIntegral[b*x]*Sin
[b*x])/b^3 + (x^2*CosIntegral[b*x]*Sin[b*x])/b - (x*Sin[b*x]^2)/(2*b^2) + SinIntegral[2*b*x]/b^3

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 12

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

Rule 3443

Int[Cos[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[(x^(m - n
+ 1)*Sin[a + b*x^n]^(p + 1))/(b*n*(p + 1)), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Sin[a + b*x^n]^
(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

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 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 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^2 \cos (b x) \text{Ci}(b x) \, dx &=\frac{x^2 \text{Ci}(b x) \sin (b x)}{b}-\frac{2 \int x \text{Ci}(b x) \sin (b x) \, dx}{b}-\int \frac{x \cos (b x) \sin (b x)}{b} \, dx\\ &=\frac{2 x \cos (b x) \text{Ci}(b x)}{b^2}+\frac{x^2 \text{Ci}(b x) \sin (b x)}{b}-\frac{2 \int \cos (b x) \text{Ci}(b x) \, dx}{b^2}-\frac{\int x \cos (b x) \sin (b x) \, dx}{b}-\frac{2 \int \frac{\cos ^2(b x)}{b} \, dx}{b}\\ &=\frac{2 x \cos (b x) \text{Ci}(b x)}{b^2}-\frac{2 \text{Ci}(b x) \sin (b x)}{b^3}+\frac{x^2 \text{Ci}(b x) \sin (b x)}{b}-\frac{x \sin ^2(b x)}{2 b^2}+\frac{\int \sin ^2(b x) \, dx}{2 b^2}-\frac{2 \int \cos ^2(b x) \, dx}{b^2}+\frac{2 \int \frac{\cos (b x) \sin (b x)}{b x} \, dx}{b^2}\\ &=\frac{2 x \cos (b x) \text{Ci}(b x)}{b^2}-\frac{5 \cos (b x) \sin (b x)}{4 b^3}-\frac{2 \text{Ci}(b x) \sin (b x)}{b^3}+\frac{x^2 \text{Ci}(b x) \sin (b x)}{b}-\frac{x \sin ^2(b x)}{2 b^2}+\frac{2 \int \frac{\cos (b x) \sin (b x)}{x} \, dx}{b^3}+\frac{\int 1 \, dx}{4 b^2}-\frac{\int 1 \, dx}{b^2}\\ &=-\frac{3 x}{4 b^2}+\frac{2 x \cos (b x) \text{Ci}(b x)}{b^2}-\frac{5 \cos (b x) \sin (b x)}{4 b^3}-\frac{2 \text{Ci}(b x) \sin (b x)}{b^3}+\frac{x^2 \text{Ci}(b x) \sin (b x)}{b}-\frac{x \sin ^2(b x)}{2 b^2}+\frac{2 \int \frac{\sin (2 b x)}{2 x} \, dx}{b^3}\\ &=-\frac{3 x}{4 b^2}+\frac{2 x \cos (b x) \text{Ci}(b x)}{b^2}-\frac{5 \cos (b x) \sin (b x)}{4 b^3}-\frac{2 \text{Ci}(b x) \sin (b x)}{b^3}+\frac{x^2 \text{Ci}(b x) \sin (b x)}{b}-\frac{x \sin ^2(b x)}{2 b^2}+\frac{\int \frac{\sin (2 b x)}{x} \, dx}{b^3}\\ &=-\frac{3 x}{4 b^2}+\frac{2 x \cos (b x) \text{Ci}(b x)}{b^2}-\frac{5 \cos (b x) \sin (b x)}{4 b^3}-\frac{2 \text{Ci}(b x) \sin (b x)}{b^3}+\frac{x^2 \text{Ci}(b x) \sin (b x)}{b}-\frac{x \sin ^2(b x)}{2 b^2}+\frac{\text{Si}(2 b x)}{b^3}\\ \end{align*}

Mathematica [A]  time = 0.0809702, size = 64, normalized size = 0.72 \[ \frac{8 \text{CosIntegral}(b x) \left (\left (b^2 x^2-2\right ) \sin (b x)+2 b x \cos (b x)\right )+8 \text{Si}(2 b x)-8 b x-5 \sin (2 b x)+2 b x \cos (2 b x)}{8 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.073, size = 66, normalized size = 0.7 \begin{align*}{\frac{1}{{b}^{3}} \left ({\it Ci} \left ( bx \right ) \left ({b}^{2}{x}^{2}\sin \left ( bx \right ) -2\,\sin \left ( bx \right ) +2\,bx\cos \left ( bx \right ) \right ) +{\frac{bx \left ( \cos \left ( bx \right ) \right ) ^{2}}{2}}-{\frac{5\,\sin \left ( bx \right ) \cos \left ( bx \right ) }{4}}-{\frac{5\,bx}{4}}+{\it Si} \left ( 2\,bx \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^3*(Ci(b*x)*(b^2*x^2*sin(b*x)-2*sin(b*x)+2*b*x*cos(b*x))+1/2*b*x*cos(b*x)^2-5/4*sin(b*x)*cos(b*x)-5/4*b*x+S
i(2*b*x))

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

[Out]

integrate(x^2*Ci(b*x)*cos(b*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{2} \cos \left (b x\right ) \operatorname{Ci}\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)*cos(b*x),x, algorithm="fricas")

[Out]

integral(x^2*cos(b*x)*cos_integral(b*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \cos{\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)*cos(b*x),x)

[Out]

Integral(x**2*cos(b*x)*Ci(b*x), x)

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Giac [C]  time = 1.19961, size = 88, normalized size = 0.99 \begin{align*}{\left (\frac{2 \, x \cos \left (b x\right )}{b^{2}} + \frac{{\left (b^{2} x^{2} - 2\right )} \sin \left (b x\right )}{b^{3}}\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 )}{2 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*x*cos(b*x)/b^2 + (b^2*x^2 - 2)*sin(b*x)/b^3)*cos_integral(b*x) - 1/2*(2*b*x - imag_part(cos_integral(2*b*x)
) + imag_part(cos_integral(-2*b*x)) - 2*sin_integral(2*b*x))/b^3