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

Optimal. Leaf size=17 \[ \frac{S(b x)^{n+1}}{b (n+1)} \]

[Out]

FresnelS[b*x]^(1 + n)/(b*(1 + n))

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Rubi [A]  time = 0.0176465, antiderivative size = 17, 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} \[ \frac{S(b x)^{n+1}}{b (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

FresnelS[b*x]^(1 + n)/(b*(1 + n))

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

Mathematica [A]  time = 0.0057795, size = 17, normalized size = 1. \[ \frac{S(b x)^{n+1}}{b (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

FresnelS[b*x]^(1 + n)/(b*(1 + n))

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Maple [A]  time = 0.046, size = 18, normalized size = 1.1 \begin{align*}{\frac{ \left ({\it FresnelS} \left ( bx \right ) \right ) ^{1+n}}{b \left ( 1+n \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

FresnelS(b*x)^(1+n)/b/(1+n)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 4.62388, size = 31, normalized size = 1.82 \begin{align*} \begin{cases} 0 & \text{for}\: b = 0 \wedge \left (b = 0 \vee n = -1\right ) \\\frac{\log{\left (S\left (b x\right ) \right )}}{b} & \text{for}\: n = -1 \\\frac{S\left (b x\right ) S^{n}\left (b x\right )}{b n + b} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((0, Eq(b, 0) & (Eq(b, 0) | Eq(n, -1))), (log(fresnels(b*x))/b, Eq(n, -1)), (fresnels(b*x)*fresnels(b
*x)**n/(b*n + b), True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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