3.1.79 \(\int S(b x) \sin (\frac {1}{2} b^2 \pi x^2) \, dx\) [79]

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 = {6575, 30} \begin {gather*} \frac {S(b x)^2}{2 b} \end {gather*}

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 6575

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^2/4)*b^4]

Rubi steps

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

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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Maple [A]
time = 0.07, size = 12, normalized size = 0.92

method result size
derivativedivides \(\frac {\mathrm {S}\left (b x \right )^{2}}{2 b}\) \(12\)
default \(\frac {\mathrm {S}\left (b x \right )^{2}}{2 b}\) \(12\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.25, size = 11, normalized size = 0.85 \begin {gather*} \frac {\operatorname {S}\left (b x\right )^{2}}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 11, normalized size = 0.85 \begin {gather*} \frac {\operatorname {S}\left (b x\right )^{2}}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.10, size = 10, normalized size = 0.77 \begin {gather*} \begin {cases} \frac {S^{2}\left (b x\right )}{2 b} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \end {gather*}

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

[Out]

integrate(fresnel_sin(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 \begin {gather*} \int \mathrm {FresnelS}\left (b\,x\right )\,\sin \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(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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