3.71 \(\int x^2 \text {Ci}(b x) \, dx\)

Optimal. Leaf size=49 \[ \frac {2 \sin (b x)}{3 b^3}-\frac {2 x \cos (b x)}{3 b^2}+\frac {1}{3} x^3 \text {Ci}(b x)-\frac {x^2 \sin (b x)}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6504, 12, 3296, 2637} \[ \frac {2 \sin (b x)}{3 b^3}-\frac {2 x \cos (b x)}{3 b^2}+\frac {1}{3} x^3 \text {CosIntegral}(b x)-\frac {x^2 \sin (b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

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

Rule 6504

Int[CosIntegral[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*CosIntegr
al[a + b*x])/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[((c + d*x)^(m + 1)*Cos[a + b*x])/(a + b*x), x], x] /; F
reeQ[{a, b, c, d, m}, x] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 44, normalized size = 0.90 \[ -\frac {2 x \cos (b x)}{3 b^2}-\frac {\left (b^2 x^2-2\right ) \sin (b x)}{3 b^3}+\frac {1}{3} x^3 \text {Ci}(b x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \operatorname {Ci}\left (b x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*cos_integral(b*x), x)

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giac [A]  time = 0.17, size = 38, normalized size = 0.78 \[ \frac {1}{3} \, x^{3} \operatorname {Ci}\left (b x\right ) - \frac {2 \, x \cos \left (b x\right )}{3 \, b^{2}} - \frac {{\left (b^{2} x^{2} - 2\right )} \sin \left (b x\right )}{3 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 44, normalized size = 0.90 \[ \frac {\frac {b^{3} x^{3} \Ci \left (b x \right )}{3}-\frac {b^{2} x^{2} \sin \left (b x \right )}{3}+\frac {2 \sin \left (b x \right )}{3}-\frac {2 b x \cos \left (b x \right )}{3}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} {\rm Ci}\left (b x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*Ci(b*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \frac {x^3\,\mathrm {cosint}\left (b\,x\right )}{3}-\frac {b^2\,x^2\,\sin \left (b\,x\right )-2\,\sin \left (b\,x\right )+2\,b\,x\,\cos \left (b\,x\right )}{3\,b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cosint(b*x),x)

[Out]

(x^3*cosint(b*x))/3 - (b^2*x^2*sin(b*x) - 2*sin(b*x) + 2*b*x*cos(b*x))/(3*b^3)

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sympy [A]  time = 1.79, size = 70, normalized size = 1.43 \[ - \frac {x^{3} \log {\left (b x \right )}}{3} + \frac {x^{3} \log {\left (b^{2} x^{2} \right )}}{6} + \frac {x^{3} \operatorname {Ci}{\left (b x \right )}}{3} - \frac {x^{2} \sin {\left (b x \right )}}{3 b} - \frac {2 x \cos {\left (b x \right )}}{3 b^{2}} + \frac {2 \sin {\left (b x \right )}}{3 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*Ci(b*x),x)

[Out]

-x**3*log(b*x)/3 + x**3*log(b**2*x**2)/6 + x**3*Ci(b*x)/3 - x**2*sin(b*x)/(3*b) - 2*x*cos(b*x)/(3*b**2) + 2*si
n(b*x)/(3*b**3)

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