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

Optimal. Leaf size=13 \[ \frac {S(b x)^2}{2 b} \]

[Out]

1/2*FresnelS(b*x)^2/b

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6440, 30} \[ \frac {S(b x)^2}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

FresnelS[b*x]^2/(2*b)

Rule 30

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 13, normalized size = 1.00 \[ \frac {S(b x)^2}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

FresnelS[b*x]^2/(2*b)

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\rm fresnels}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(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 \[ \int {\rm fresnels}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 12, normalized size = 0.92 \[ \frac {\mathrm {S}\left (b x \right )^{2}}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*FresnelS(b*x)^2/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(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 = -0.08 \[ \int \mathrm {FresnelS}\left (b\,x\right )\,\sin \left (\frac {\Pi \,b^2\,x^2}{2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.37, size = 10, normalized size = 0.77 \[ \begin {cases} \frac {S^{2}\left (b x\right )}{2 b} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((fresnels(b*x)**2/(2*b), Ne(b, 0)), (0, True))

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