3.1.70 \(\int x^3 \text {CosIntegral}(b x) \, dx\) [70]

Optimal. Leaf size=63 \[ \frac {3 \cos (b x)}{2 b^4}-\frac {3 x^2 \cos (b x)}{4 b^2}+\frac {1}{4} x^4 \text {CosIntegral}(b x)+\frac {3 x \sin (b x)}{2 b^3}-\frac {x^3 \sin (b x)}{4 b} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6639, 12, 3377, 2718} \begin {gather*} \frac {3 \cos (b x)}{2 b^4}+\frac {3 x \sin (b x)}{2 b^3}-\frac {3 x^2 \cos (b x)}{4 b^2}+\frac {1}{4} x^4 \text {CosIntegral}(b x)-\frac {x^3 \sin (b x)}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*CosIntegral[b*x],x]

[Out]

(3*Cos[b*x])/(2*b^4) - (3*x^2*Cos[b*x])/(4*b^2) + (x^4*CosIntegral[b*x])/4 + (3*x*Sin[b*x])/(2*b^3) - (x^3*Sin
[b*x])/(4*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 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[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^3 \text {Ci}(b x) \, dx &=\frac {1}{4} x^4 \text {Ci}(b x)-\frac {1}{4} b \int \frac {x^3 \cos (b x)}{b} \, dx\\ &=\frac {1}{4} x^4 \text {Ci}(b x)-\frac {1}{4} \int x^3 \cos (b x) \, dx\\ &=\frac {1}{4} x^4 \text {Ci}(b x)-\frac {x^3 \sin (b x)}{4 b}+\frac {3 \int x^2 \sin (b x) \, dx}{4 b}\\ &=-\frac {3 x^2 \cos (b x)}{4 b^2}+\frac {1}{4} x^4 \text {Ci}(b x)-\frac {x^3 \sin (b x)}{4 b}+\frac {3 \int x \cos (b x) \, dx}{2 b^2}\\ &=-\frac {3 x^2 \cos (b x)}{4 b^2}+\frac {1}{4} x^4 \text {Ci}(b x)+\frac {3 x \sin (b x)}{2 b^3}-\frac {x^3 \sin (b x)}{4 b}-\frac {3 \int \sin (b x) \, dx}{2 b^3}\\ &=\frac {3 \cos (b x)}{2 b^4}-\frac {3 x^2 \cos (b x)}{4 b^2}+\frac {1}{4} x^4 \text {Ci}(b x)+\frac {3 x \sin (b x)}{2 b^3}-\frac {x^3 \sin (b x)}{4 b}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 53, normalized size = 0.84 \begin {gather*} -\frac {3 \left (-2+b^2 x^2\right ) \cos (b x)}{4 b^4}+\frac {1}{4} x^4 \text {CosIntegral}(b x)-\frac {x \left (-6+b^2 x^2\right ) \sin (b x)}{4 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*CosIntegral[b*x],x]

[Out]

(-3*(-2 + b^2*x^2)*Cos[b*x])/(4*b^4) + (x^4*CosIntegral[b*x])/4 - (x*(-6 + b^2*x^2)*Sin[b*x])/(4*b^3)

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Maple [A]
time = 0.27, size = 56, normalized size = 0.89

method	result	size
derivativedivides	\(\frac {\frac {b^{4} x^{4} \cosineIntegral \left (b x \right )}{4}-\frac {b^{3} x^{3} \sin \left (b x \right )}{4}-\frac {3 b^{2} x^{2} \cos \left (b x \right )}{4}+\frac {3 \cos \left (b x \right )}{2}+\frac {3 b x \sin \left (b x \right )}{2}}{b^{4}}\)	\(56\)
default	\(\frac {\frac {b^{4} x^{4} \cosineIntegral \left (b x \right )}{4}-\frac {b^{3} x^{3} \sin \left (b x \right )}{4}-\frac {3 b^{2} x^{2} \cos \left (b x \right )}{4}+\frac {3 \cos \left (b x \right )}{2}+\frac {3 b x \sin \left (b x \right )}{2}}{b^{4}}\)	\(56\)
meijerg	\(\frac {4 \sqrt {\pi }\, \left (-\frac {b^{6} x^{6} \hypergeom \left (\left [1, 1, 3\right ], \left [\frac {3}{2}, 2, 2, 4\right ], -\frac {b^{2} x^{2}}{4}\right )}{96 \sqrt {\pi }}+\frac {\left (-\frac {1}{2}+2 \gamma +2 \ln \left (x \right )+2 \ln \left (b \right )\right ) x^{4} b^{4}}{32 \sqrt {\pi }}\right )}{b^{4}}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*Ci(b*x),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.48, size = 94, normalized size = 1.49 \begin {gather*} \frac {1}{4} \, x^{4} \operatorname {C}\left (b x\right ) - \frac {\sqrt {\frac {1}{2}} {\left (4 \, \sqrt {\frac {1}{2}} \pi ^{2} b^{3} x^{3} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 12 \, \sqrt {\frac {1}{2}} \pi b x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + \left (3 i - 3\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {\frac {1}{2} i \, \pi } b x\right ) - \left (3 i + 3\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {-\frac {1}{2} i \, \pi } b x\right )\right )}}{8 \, \pi ^{3} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*fresnel_cos(b*x),x, algorithm="maxima")

[Out]

1/4*x^4*fresnel_cos(b*x) - 1/8*sqrt(1/2)*(4*sqrt(1/2)*pi^2*b^3*x^3*sin(1/2*pi*b^2*x^2) + 12*sqrt(1/2)*pi*b*x*c
os(1/2*pi*b^2*x^2) + (3*I - 3)*(1/4)^(1/4)*pi*erf(sqrt(1/2*I*pi)*b*x) - (3*I + 3)*(1/4)^(1/4)*pi*erf(sqrt(-1/2
*I*pi)*b*x))/(pi^3*b^4)

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Fricas [A]
time = 0.35, size = 59, normalized size = 0.94 \begin {gather*} -\frac {\pi b^{3} x^{3} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 3 \, b x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - {\left (\pi ^{2} b^{4} x^{4} + 3\right )} \operatorname {C}\left (b x\right )}{4 \, \pi ^{2} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*fresnel_cos(b*x),x, algorithm="fricas")

[Out]

-1/4*(pi*b^3*x^3*sin(1/2*pi*b^2*x^2) + 3*b*x*cos(1/2*pi*b^2*x^2) - (pi^2*b^4*x^4 + 3)*fresnel_cos(b*x))/(pi^2*
b^4)

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Sympy [A]
time = 1.45, size = 85, normalized size = 1.35 \begin {gather*} - \frac {x^{4} \log {\left (b x \right )}}{4} + \frac {x^{4} \log {\left (b^{2} x^{2} \right )}}{8} + \frac {x^{4} \operatorname {Ci}{\left (b x \right )}}{4} - \frac {x^{3} \sin {\left (b x \right )}}{4 b} - \frac {3 x^{2} \cos {\left (b x \right )}}{4 b^{2}} + \frac {3 x \sin {\left (b x \right )}}{2 b^{3}} + \frac {3 \cos {\left (b x \right )}}{2 b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*Ci(b*x),x)

[Out]

-x**4*log(b*x)/4 + x**4*log(b**2*x**2)/8 + x**4*Ci(b*x)/4 - x**3*sin(b*x)/(4*b) - 3*x**2*cos(b*x)/(4*b**2) + 3
*x*sin(b*x)/(2*b**3) + 3*cos(b*x)/(2*b**4)

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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^3*fresnel_cos(b*x),x, algorithm="giac")

[Out]

integrate(x^3*fresnel_cos(b*x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \frac {6\,\cos \left (b\,x\right )-3\,b^2\,x^2\,\cos \left (b\,x\right )-b^3\,x^3\,\sin \left (b\,x\right )+6\,b\,x\,\sin \left (b\,x\right )}{4\,b^4}+\frac {x^4\,\mathrm {cosint}\left (b\,x\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*cosint(b*x),x)

[Out]

(6*cos(b*x) - 3*b^2*x^2*cos(b*x) - b^3*x^3*sin(b*x) + 6*b*x*sin(b*x))/(4*b^4) + (x^4*cosint(b*x))/4

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