3.1.98 \(\int x \cos (\frac {1}{2} b^2 \pi x^2) S(b x) \, dx\) [98]

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

[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 = 0.02, 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 = {6595, 3438, 3433} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*x/(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 3433

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

Rule 3438

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 6595

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^2/4)*b^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 = 0.02, size = 48, normalized size = 0.81 \begin {gather*} \frac {-2 b x+\sqrt {2} \text {FresnelC}\left (\sqrt {2} b x\right )+4 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{4 b^2 \pi } \end {gather*}

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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Maple [A]
time = 0.81, size = 52, normalized size = 0.88

method	result	size
default	\(\frac {\frac {\mathrm {S}\left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b \pi }-\frac {\frac {b x}{2}-\frac {\sqrt {2}\, \FresnelC \left (b x \sqrt {2}\right )}{4}}{b \pi }}{b}\)	\(52\)

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,method=_RETURNVERBOSE)

[Out]

(FresnelS(b*x)/b/Pi*sin(1/2*b^2*Pi*x^2)-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 53, normalized size = 0.90 \begin {gather*} -\frac {2 \, b^{2} x - 4 \, b \operatorname {S}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - \sqrt {2} \sqrt {b^{2}} \operatorname {C}\left (\sqrt {2} \sqrt {b^{2}} x\right )}{4 \, \pi b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*b^2*x - 4*b*fresnel_sin(b*x)*sin(1/2*pi*b^2*x^2) - sqrt(2)*sqrt(b^2)*fresnel_cos(sqrt(2)*sqrt(b^2)*x))
/(pi*b^3)

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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