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

Optimal. Leaf size=49 \[ \frac {S\left (\sqrt {2} b x\right )}{2 \sqrt {2} \pi b^2}-\frac {S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6452, 3351} \[ \frac {S\left (\sqrt {2} b x\right )}{2 \sqrt {2} \pi b^2}-\frac {S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(b^2*Pi)) + FresnelS[Sqrt[2]*b*x]/(2*Sqrt[2]*b^2*Pi)

Rule 3351

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

Rule 6452

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 44, normalized size = 0.90 \[ \frac {\sqrt {2} S\left (\sqrt {2} b x\right )-4 S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{4 \pi b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x] + Sqrt[2]*FresnelS[Sqrt[2]*b*x])/(4*b^2*Pi)

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x {\rm fresnels}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 46, normalized size = 0.94 \[ \frac {-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right ) \mathrm {S}\left (b x \right )}{\pi b}+\frac {\mathrm {S}\left (b x \sqrt {2}\right ) \sqrt {2}}{4 b \pi }}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x\,\mathrm {FresnelS}\left (b\,x\right )\,\sin \left (\frac {\Pi \,b^2\,x^2}{2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sin(pi*b**2*x**2/2)*fresnels(b*x), x)

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