3.98 \(\int x \cos (\frac {1}{2} b^2 \pi x^2) S(b x) \, dx\)

Optimal. Leaf size=59 \[ \frac {C\left (\sqrt {2} b x\right )}{2 \sqrt {2} \pi b^2}+\frac {S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {x}{2 \pi b} \]

[Out]

-1/2*x/b/Pi+FresnelS(b*x)*sin(1/2*b^2*Pi*x^2)/b^2/Pi+1/4*FresnelC(b*x*2^(1/2))/b^2/Pi*2^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6460, 3357, 3352} \[ \frac {\text {FresnelC}\left (\sqrt {2} b x\right )}{2 \sqrt {2} \pi b^2}+\frac {S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {x}{2 \pi b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x/(2*b*Pi) + FresnelC[Sqrt[2]*b*x]/(2*Sqrt[2]*b^2*Pi) + (FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi)

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3357

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(a +
b*Sin[c + d*(e + f*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 1] && IGtQ[n, 1]

Rule 6460

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

Rubi steps

\begin {align*} \int x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x) \, dx &=\frac {S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {\int \sin ^2\left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b \pi }\\ &=\frac {S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {\int \left (\frac {1}{2}-\frac {1}{2} \cos \left (b^2 \pi x^2\right )\right ) \, dx}{b \pi }\\ &=-\frac {x}{2 b \pi }+\frac {S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }+\frac {\int \cos \left (b^2 \pi x^2\right ) \, dx}{2 b \pi }\\ &=-\frac {x}{2 b \pi }+\frac {C\left (\sqrt {2} b x\right )}{2 \sqrt {2} b^2 \pi }+\frac {S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 48, normalized size = 0.81 \[ \frac {4 S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )+\sqrt {2} C\left (\sqrt {2} b x\right )-2 b x}{4 \pi b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*x + Sqrt[2]*FresnelC[Sqrt[2]*b*x] + 4*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(4*b^2*Pi)

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fricas [F]  time = 1.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) {\rm fresnels}\left (b x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 52, normalized size = 0.88 \[ \frac {\frac {\mathrm {S}\left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b \pi }-\frac {\frac {b x}{2}-\frac {\sqrt {2}\, \FresnelC \left (b x \sqrt {2}\right )}{4}}{b \pi }}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(FresnelS(b*x)/b*sin(1/2*b^2*Pi*x^2)/Pi-1/b/Pi*(1/2*b*x-1/4*2^(1/2)*FresnelC(b*x*2^(1/2))))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x\,\mathrm {FresnelS}\left (b\,x\right )\,\cos \left (\frac {\Pi \,b^2\,x^2}{2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*cos(pi*b**2*x**2/2)*fresnels(b*x), x)

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