3.97 \(\int x^2 \cos (\frac {1}{2} b^2 \pi x^2) S(b x) \, dx\)

Optimal. Leaf size=73 \[ -\frac {S(b x)^2}{2 \pi b^3}+\frac {x S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\sin \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}-\frac {x^2}{4 \pi b} \]

[Out]

-1/4*x^2/b/Pi-1/2*FresnelS(b*x)^2/b^3/Pi+x*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2)/b^2/Pi+1/4*sin(b^2*Pi*x^2)/b^3/Pi
^2

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Rubi [A]  time = 0.05, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6462, 3379, 2634, 6440, 30} \[ \frac {x S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {S(b x)^2}{2 \pi b^3}+\frac {\sin \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}-\frac {x^2}{4 \pi b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x],x]

[Out]

-x^2/(4*b*Pi) - FresnelS[b*x]^2/(2*b^3*Pi) + (x*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi) + Sin[b^2*Pi*x^2]/
(4*b^3*Pi^2)

Rule 30

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

Rule 2634

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

Rule 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 6440

Int[FresnelS[(b_.)*(x_)]^(n_.)*Sin[(d_.)*(x_)^2], x_Symbol] :> Dist[(Pi*b)/(2*d), Subst[Int[x^n, x], x, Fresne
lS[b*x]], x] /; FreeQ[{b, d, n}, x] && EqQ[d^2, (Pi^2*b^4)/4]

Rule 6462

Int[Cos[(d_.)*(x_)^2]*FresnelS[(b_.)*(x_)]*(x_)^(m_), x_Symbol] :> Simp[(x^(m - 1)*Sin[d*x^2]*FresnelS[b*x])/(
2*d), x] + (-Dist[1/(Pi*b), Int[x^(m - 1)*Sin[d*x^2]^2, x], x] - Dist[(m - 1)/(2*d), Int[x^(m - 2)*Sin[d*x^2]*
FresnelS[b*x], x], x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2*b^4)/4] && IGtQ[m, 1]

Rubi steps

\begin {align*} \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x) \, dx &=\frac {x S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {\int S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b^2 \pi }-\frac {\int x \sin ^2\left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b \pi }\\ &=\frac {x S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {\operatorname {Subst}(\int x \, dx,x,S(b x))}{b^3 \pi }-\frac {\operatorname {Subst}\left (\int \sin ^2\left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{2 b \pi }\\ &=-\frac {x^2}{4 b \pi }-\frac {S(b x)^2}{2 b^3 \pi }+\frac {x S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }+\frac {\sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 73, normalized size = 1.00 \[ -\frac {S(b x)^2}{2 \pi b^3}+\frac {x S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\sin \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}-\frac {x^2}{4 \pi b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x],x]

[Out]

-1/4*x^2/(b*Pi) - FresnelS[b*x]^2/(2*b^3*Pi) + (x*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi) + Sin[b^2*Pi*x^2
]/(4*b^3*Pi^2)

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) {\rm fresnels}\left (b x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cos(1/2*b^2*pi*x^2)*fresnels(b*x),x, algorithm="fricas")

[Out]

integral(x^2*cos(1/2*pi*b^2*x^2)*fresnels(b*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) {\rm fresnels}\left (b x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cos(1/2*b^2*pi*x^2)*fresnels(b*x),x, algorithm="giac")

[Out]

integrate(x^2*cos(1/2*pi*b^2*x^2)*fresnels(b*x), x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right ) \mathrm {S}\left (b x \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x),x)

[Out]

int(x^2*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) {\rm fresnels}\left (b x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cos(1/2*b^2*pi*x^2)*fresnels(b*x),x, algorithm="maxima")

[Out]

integrate(x^2*cos(1/2*pi*b^2*x^2)*fresnels(b*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\mathrm {FresnelS}\left (b\,x\right )\,\cos \left (\frac {\Pi \,b^2\,x^2}{2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*FresnelS(b*x)*cos((Pi*b^2*x^2)/2),x)

[Out]

int(x^2*FresnelS(b*x)*cos((Pi*b^2*x^2)/2), x)

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sympy [A]  time = 2.89, size = 114, normalized size = 1.56 \[ \begin {cases} - \frac {x^{2} \sin ^{2}{\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{4 \pi b} - \frac {x^{2} \cos ^{2}{\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{4 \pi b} + \frac {x \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )}{\pi b^{2}} + \frac {\sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{2 \pi ^{2} b^{3}} - \frac {S^{2}\left (b x\right )}{2 \pi b^{3}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*cos(1/2*b**2*pi*x**2)*fresnels(b*x),x)

[Out]

Piecewise((-x**2*sin(pi*b**2*x**2/2)**2/(4*pi*b) - x**2*cos(pi*b**2*x**2/2)**2/(4*pi*b) + x*sin(pi*b**2*x**2/2
)*fresnels(b*x)/(pi*b**2) + sin(pi*b**2*x**2/2)*cos(pi*b**2*x**2/2)/(2*pi**2*b**3) - fresnels(b*x)**2/(2*pi*b*
*3), Ne(b, 0)), (0, True))

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