3.63 \(\int S(b x) \sin (c+\frac{1}{2} b^2 \pi x^2) \, dx\)

Optimal. Leaf size=101 \[ -\frac{1}{8} i b x^2 \sin (c) \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},-\frac{1}{2} i \pi b^2 x^2\right )+\frac{1}{8} i b x^2 \sin (c) \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},\frac{1}{2} i \pi b^2 x^2\right )+\frac{\sin (c) \text{FresnelC}(b x) S(b x)}{2 b}+\frac{\cos (c) S(b x)^2}{2 b} \]

[Out]

(Cos[c]*FresnelS[b*x]^2)/(2*b) + (FresnelC[b*x]*FresnelS[b*x]*Sin[c])/(2*b) - (I/8)*b*x^2*HypergeometricPFQ[{1
, 1}, {3/2, 2}, (-I/2)*b^2*Pi*x^2]*Sin[c] + (I/8)*b*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (I/2)*b^2*Pi*x^2]*
Sin[c]

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Rubi [A]  time = 0.0566188, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {6442, 6446, 6440, 30} \[ -\frac{1}{8} i b x^2 \sin (c) \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i b^2 \pi x^2\right )+\frac{1}{8} i b x^2 \sin (c) \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i b^2 \pi x^2\right )+\frac{\sin (c) \text{FresnelC}(b x) S(b x)}{2 b}+\frac{\cos (c) S(b x)^2}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c]*FresnelS[b*x]^2)/(2*b) + (FresnelC[b*x]*FresnelS[b*x]*Sin[c])/(2*b) - (I/8)*b*x^2*HypergeometricPFQ[{1
, 1}, {3/2, 2}, (-I/2)*b^2*Pi*x^2]*Sin[c] + (I/8)*b*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (I/2)*b^2*Pi*x^2]*
Sin[c]

Rule 6442

Int[FresnelS[(b_.)*(x_)]*Sin[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sin[c], Int[Cos[d*x^2]*FresnelS[b*x], x],
x] + Dist[Cos[c], Int[Sin[d*x^2]*FresnelS[b*x], x], x] /; FreeQ[{b, c, d}, x] && EqQ[d^2, (Pi^2*b^4)/4]

Rule 6446

Int[Cos[(d_.)*(x_)^2]*FresnelS[(b_.)*(x_)], x_Symbol] :> Simp[(FresnelC[b*x]*FresnelS[b*x])/(2*b), x] + (-Simp
[(1*I*b*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, -((I*b^2*Pi*x^2)/2)])/8, x] + Simp[(1*I*b*x^2*HypergeometricPF
Q[{1, 1}, {3/2, 2}, (1*I*b^2*Pi*x^2)/2])/8, x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2*b^4)/4]

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) \sin \left (c+\frac{1}{2} b^2 \pi x^2\right ) \, dx &=\cos (c) \int S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx+\sin (c) \int \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x) \, dx\\ &=\frac{C(b x) S(b x) \sin (c)}{2 b}-\frac{1}{8} i b x^2 \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i b^2 \pi x^2\right ) \sin (c)+\frac{1}{8} i b x^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i b^2 \pi x^2\right ) \sin (c)+\frac{\cos (c) \operatorname{Subst}(\int x \, dx,x,S(b x))}{b}\\ &=\frac{\cos (c) S(b x)^2}{2 b}+\frac{C(b x) S(b x) \sin (c)}{2 b}-\frac{1}{8} i b x^2 \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i b^2 \pi x^2\right ) \sin (c)+\frac{1}{8} i b x^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i b^2 \pi x^2\right ) \sin (c)\\ \end{align*}

Mathematica [F]  time = 0.0490403, size = 0, normalized size = 0. \[ \int S(b x) \sin \left (c+\frac{1}{2} b^2 \pi x^2\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.085, size = 0, normalized size = 0. \begin{align*} \int{\it FresnelS} \left ( bx \right ) \sin \left ( c+{\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(fresnels(b*x)*sin(1/2*pi*b^2*x^2 + c), 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 ) \sin \left (\frac{1}{2} \, \pi b^{2} x^{2} + c\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin{\left (\frac{\pi b^{2} x^{2}}{2} + c \right )} S\left (b x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(pi*b**2*x**2/2 + c)*fresnels(b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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