3.1.96 \(\int x^3 \cos (\frac {1}{2} b^2 \pi x^2) S(b x) \, dx\) [96]

Optimal. Leaf size=108 \[ -\frac {x^3}{6 b \pi }+\frac {2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^4 \pi ^2}-\frac {5 S\left (\sqrt {2} b x\right )}{4 \sqrt {2} b^4 \pi ^2}+\frac {x^2 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }+\frac {x \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2} \]

[Out]

-1/6*x^3/b/Pi+2*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x)/b^4/Pi^2+x^2*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2)/b^2/Pi+1/4*x*
sin(b^2*Pi*x^2)/b^3/Pi^2-5/8*FresnelS(b*x*2^(1/2))/b^4/Pi^2*2^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6597, 3472, 30, 3467, 3432, 6587} \begin {gather*} -\frac {5 S\left (\sqrt {2} b x\right )}{4 \sqrt {2} \pi ^2 b^4}+\frac {x^2 S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {2 S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}+\frac {x \sin \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}-\frac {x^3}{6 \pi b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x],x]

[Out]

-1/6*x^3/(b*Pi) + (2*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(b^4*Pi^2) - (5*FresnelS[Sqrt[2]*b*x])/(4*Sqrt[2]*b^4*
Pi^2) + (x^2*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi) + (x*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 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3467

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[e^(n - 1)*(e*x)^(m - n + 1)*(Sin[c + d*
x^n]/(d*n)), x] - Dist[e^n*((m - n + 1)/(d*n)), Int[(e*x)^(m - n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x
] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3472

Int[(x_)^(m_.)*Sin[(a_.) + ((b_.)*(x_)^(n_))/2]^2, x_Symbol] :> Dist[1/2, Int[x^m, x], x] - Dist[1/2, Int[x^m*
Cos[2*a + b*x^n], x], x] /; FreeQ[{a, b, m, n}, x]

Rule 6587

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

Rule 6597

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^2/4)*b^4] && IGtQ[m, 1]

Rubi steps

\begin {align*} \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x) \, dx &=\frac {x^2 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {2 \int x S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b^2 \pi }-\frac {\int x^2 \sin ^2\left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b \pi }\\ &=\frac {2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^4 \pi ^2}+\frac {x^2 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {\int \sin \left (b^2 \pi x^2\right ) \, dx}{b^3 \pi ^2}-\frac {\int x^2 \, dx}{2 b \pi }+\frac {\int x^2 \cos \left (b^2 \pi x^2\right ) \, dx}{2 b \pi }\\ &=-\frac {x^3}{6 b \pi }+\frac {2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^4 \pi ^2}-\frac {S\left (\sqrt {2} b x\right )}{\sqrt {2} b^4 \pi ^2}+\frac {x^2 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }+\frac {x \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac {\int \sin \left (b^2 \pi x^2\right ) \, dx}{4 b^3 \pi ^2}\\ &=-\frac {x^3}{6 b \pi }+\frac {2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{b^4 \pi ^2}-\frac {5 S\left (\sqrt {2} b x\right )}{4 \sqrt {2} b^4 \pi ^2}+\frac {x^2 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }+\frac {x \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 90, normalized size = 0.83 \begin {gather*} \frac {-4 b^3 \pi x^3-15 \sqrt {2} S\left (\sqrt {2} b x\right )+24 S(b x) \left (2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )+b^2 \pi x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )\right )+6 b x \sin \left (b^2 \pi x^2\right )}{24 b^4 \pi ^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x],x]

[Out]

(-4*b^3*Pi*x^3 - 15*Sqrt[2]*FresnelS[Sqrt[2]*b*x] + 24*FresnelS[b*x]*(2*Cos[(b^2*Pi*x^2)/2] + b^2*Pi*x^2*Sin[(
b^2*Pi*x^2)/2]) + 6*b*x*Sin[b^2*Pi*x^2])/(24*b^4*Pi^2)

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Maple [A]
time = 0.86, size = 119, normalized size = 1.10

method	result	size
default	\(\frac {\frac {\mathrm {S}\left (b x \right ) \left (\frac {b^{2} x^{2} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {2 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}\right )}{b^{3}}-\frac {\frac {\sqrt {2}\, \mathrm {S}\left (b x \sqrt {2}\right )}{2 \pi ^{2}}+\frac {b^{3} x^{3}}{6 \pi }-\frac {\frac {b x \sin \left (b^{2} \pi \,x^{2}\right )}{2 \pi }-\frac {\sqrt {2}\, \mathrm {S}\left (b x \sqrt {2}\right )}{4 \pi }}{2 \pi }}{b^{3}}}{b}\)	\(119\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x),x,method=_RETURNVERBOSE)

[Out]

(FresnelS(b*x)/b^3*(1/Pi*b^2*x^2*sin(1/2*b^2*Pi*x^2)+2/Pi^2*cos(1/2*b^2*Pi*x^2))-1/b^3*(1/2/Pi^2*2^(1/2)*Fresn
elS(b*x*2^(1/2))+1/6*b^3*x^3/Pi-1/2/Pi*(1/2/Pi*b*x*sin(b^2*Pi*x^2)-1/4/Pi*2^(1/2)*FresnelS(b*x*2^(1/2)))))/b

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cos(1/2*b^2*pi*x^2)*fresnel_sin(b*x),x, algorithm="maxima")

[Out]

integrate(x^3*cos(1/2*pi*b^2*x^2)*fresnel_sin(b*x), x)

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Fricas [A]
time = 0.35, size = 97, normalized size = 0.90 \begin {gather*} -\frac {4 \, \pi b^{4} x^{3} - 48 \, b \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {S}\left (b x\right ) + 15 \, \sqrt {2} \sqrt {b^{2}} \operatorname {S}\left (\sqrt {2} \sqrt {b^{2}} x\right ) - 12 \, {\left (2 \, \pi b^{3} x^{2} \operatorname {S}\left (b x\right ) + b^{2} x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{24 \, \pi ^{2} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cos(1/2*b^2*pi*x^2)*fresnel_sin(b*x),x, algorithm="fricas")

[Out]

-1/24*(4*pi*b^4*x^3 - 48*b*cos(1/2*pi*b^2*x^2)*fresnel_sin(b*x) + 15*sqrt(2)*sqrt(b^2)*fresnel_sin(sqrt(2)*sqr
t(b^2)*x) - 12*(2*pi*b^3*x^2*fresnel_sin(b*x) + b^2*x*cos(1/2*pi*b^2*x^2))*sin(1/2*pi*b^2*x^2))/(pi^2*b^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*cos(1/2*b**2*pi*x**2)*fresnels(b*x),x)

[Out]

Integral(x**3*cos(pi*b**2*x**2/2)*fresnels(b*x), x)

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

[Out]

integrate(x^3*cos(1/2*pi*b^2*x^2)*fresnel_sin(b*x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,\mathrm {FresnelS}\left (b\,x\right )\,\cos \left (\frac {\Pi \,b^2\,x^2}{2}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*FresnelS(b*x)*cos((Pi*b^2*x^2)/2),x)

[Out]

int(x^3*FresnelS(b*x)*cos((Pi*b^2*x^2)/2), x)

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