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 = 0.02, antiderivative size = 17, 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 = {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 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)^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*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 1.00 \[ \frac {S(b x)^{n+1}}{b (n+1)} \]

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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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\rm fresnels}\left (b x\right )^{n} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)^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)

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maple [A]  time = 0.01, size = 18, normalized size = 1.06 \[ \frac {\mathrm {S}\left (b x \right )^{1+n}}{b \left (1+n \right )} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)^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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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \[ \int {\mathrm {FresnelS}\left (b\,x\right )}^n\,\sin \left (\frac {\Pi \,b^2\,x^2}{2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.98, size = 31, normalized size = 1.82 \[ \begin {cases} 0 & \text {for}\: b = 0 \wedge \left (b = 0 \vee n = -1\right ) \\\frac {\log {\left (S\left (b x\right ) \right )}}{b} & \text {for}\: n = -1 \\\frac {S\left (b x\right ) S^{n}\left (b x\right )}{b n + b} & \text {otherwise} \end {cases} \]

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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