3.1.68 \(\int \frac {\sin (\frac {1}{2} b^2 \pi x^2)}{S(b x)^2} \, dx\) [68]

Optimal. Leaf size=11 \[ -\frac {1}{b S(b x)} \]

[Out]

-1/b/FresnelS(b*x)

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Rubi [A]
time = 0.01, antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {6575, 30} \begin {gather*} -\frac {1}{b S(b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(b*FresnelS[b*x]))

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

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Mathematica [A]
time = 0.00, size = 11, normalized size = 1.00 \begin {gather*} -\frac {1}{b S(b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(b*FresnelS[b*x]))

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Maple [A]
time = 0.06, size = 12, normalized size = 1.09

method	result	size
derivativedivides	\(-\frac {1}{b \,\mathrm {S}\left (b x \right )}\)	\(12\)
default	\(-\frac {1}{b \,\mathrm {S}\left (b x \right )}\)	\(12\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/b/FresnelS(b*x)

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Maxima [A]
time = 0.26, size = 11, normalized size = 1.00 \begin {gather*} -\frac {1}{b \operatorname {S}\left (b x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(b*fresnel_sin(b*x))

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Fricas [A]
time = 0.34, size = 11, normalized size = 1.00 \begin {gather*} -\frac {1}{b \operatorname {S}\left (b x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(b*fresnel_sin(b*x))

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Sympy [A]
time = 0.26, size = 10, normalized size = 0.91 \begin {gather*} \begin {cases} - \frac {1}{b S\left (b x\right )} & \text {for}\: b \neq 0 \\\text {NaN} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-1/(b*fresnels(b*x)), Ne(b, 0)), (nan, 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(sin(1/2*b^2*pi*x^2)/fresnel_sin(b*x)^2,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.09 \begin {gather*} \int \frac {\sin \left (\frac {\Pi \,b^2\,x^2}{2}\right )}{{\mathrm {FresnelS}\left (b\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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