3.69 \(\int \frac{\sin (\frac{1}{2} b^2 \pi x^2)}{S(b x)^3} \, dx\)

Optimal. Leaf size=13 \[ -\frac{1}{2 b S(b x)^2} \]

[Out]

-1/(2*b*FresnelS[b*x]^2)

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Rubi [A]  time = 0.0149547, antiderivative size = 13, normalized size of antiderivative = 1., 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 = {6440, 30} \[ -\frac{1}{2 b S(b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*b*FresnelS[b*x]^2)

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]

Rule 30

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

Rubi steps

\begin{align*} \int \frac{\sin \left (\frac{1}{2} b^2 \pi x^2\right )}{S(b x)^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^3} \, dx,x,S(b x)\right )}{b}\\ &=-\frac{1}{2 b S(b x)^2}\\ \end{align*}

Mathematica [A]  time = 0.0040066, size = 13, normalized size = 1. \[ -\frac{1}{2 b S(b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*b*FresnelS[b*x]^2)

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Maple [A]  time = 0.044, size = 12, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,b \left ({\it FresnelS} \left ( bx \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(1/2*b^2*Pi*x^2)/FresnelS(b*x)^3,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (\frac{1}{2} \, \pi b^{2} x^{2}\right )}{{\rm fresnels}\left (b x\right )^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sin \left (\frac{1}{2} \, \pi b^{2} x^{2}\right )}{{\rm fresnels}\left (b x\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sin(1/2*pi*b^2*x^2)/fresnels(b*x)^3, x)

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Sympy [A]  time = 1.93431, size = 14, normalized size = 1.08 \begin{align*} \begin{cases} - \frac{1}{2 b S^{2}\left (b x\right )} & \text{for}\: b \neq 0 \\\text{NaN} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-1/(2*b*fresnels(b*x)**2), Ne(b, 0)), (nan, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (\frac{1}{2} \, \pi b^{2} x^{2}\right )}{{\rm fresnels}\left (b x\right )^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(1/2*b^2*pi*x^2)/fresnels(b*x)^3,x, algorithm="giac")

[Out]

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