3.1.71 \(\int x^2 \text {CosIntegral}(b x) \, dx\) [71]

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

[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 = 0.03, 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 = {6639, 12, 3377, 2717} \begin {gather*} \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} \end {gather*}

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 2717

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

Rule 3377

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 6639

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 = 0.02, size = 44, normalized size = 0.90 \begin {gather*} -\frac {2 x \cos (b x)}{3 b^2}+\frac {1}{3} x^3 \text {CosIntegral}(b x)-\frac {\left (-2+b^2 x^2\right ) \sin (b x)}{3 b^3} \end {gather*}

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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Maple [A]
time = 0.25, size = 44, normalized size = 0.90

method	result	size
derivativedivides	\(\frac {\frac {b^{3} x^{3} \cosineIntegral \left (b x \right )}{3}-\frac {b^{2} x^{2} \sin \left (b x \right )}{3}+\frac {2 \sin \left (b x \right )}{3}-\frac {2 b x \cos \left (b x \right )}{3}}{b^{3}}\)	\(44\)
default	\(\frac {\frac {b^{3} x^{3} \cosineIntegral \left (b x \right )}{3}-\frac {b^{2} x^{2} \sin \left (b x \right )}{3}+\frac {2 \sin \left (b x \right )}{3}-\frac {2 b x \cos \left (b x \right )}{3}}{b^{3}}\)	\(44\)
meijerg	\(\frac {2 \sqrt {\pi }\, \left (-\frac {x^{5} b^{5} \hypergeom \left (\left [1, 1, \frac {5}{2}\right ], \left [\frac {3}{2}, 2, 2, \frac {7}{2}\right ], -\frac {b^{2} x^{2}}{4}\right )}{40 \sqrt {\pi }}+\frac {\left (-\frac {2}{3}+2 \gamma +2 \ln \left (x \right )+2 \ln \left (b \right )\right ) x^{3} b^{3}}{12 \sqrt {\pi }}\right )}{b^{3}}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*Ci(b*x),x,method=_RETURNVERBOSE)

[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 [A]
time = 0.27, size = 49, normalized size = 1.00 \begin {gather*} \frac {1}{3} \, x^{3} \operatorname {C}\left (b x\right ) - \frac {\pi b^{2} x^{2} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 2 \, \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{3 \, \pi ^{2} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 54, normalized size = 1.10 \begin {gather*} \frac {\pi ^{2} b^{3} x^{3} \operatorname {C}\left (b x\right ) - \pi b^{2} x^{2} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 2 \, \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{3 \, \pi ^{2} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 1.17, size = 70, normalized size = 1.43 \begin {gather*} - \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}} \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*fresnel_cos(b*x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \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} \end {gather*}

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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