3.96 \(\int x^3 \cos (\frac{1}{2} b^2 \pi x^2) S(b x) \, dx\)

Optimal. Leaf size=108 \[ \frac{x^2 S(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac{2 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac{5 S\left (\sqrt{2} b x\right )}{4 \sqrt{2} \pi ^2 b^4}+\frac{x \sin \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}-\frac{x^3}{6 \pi b} \]

[Out]

-x^3/(6*b*Pi) + (2*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(b^4*Pi^2) - (5*FresnelS[Sqrt[2]*b*x])/(4*Sqrt[2]*b^4*Pi
^2) + (x^2*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi) + (x*Sin[b^2*Pi*x^2])/(4*b^3*Pi^2)

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Rubi [A]  time = 0.0898809, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {6462, 3391, 30, 3386, 3351, 6452} \[ \frac{x^2 S(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac{2 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac{5 S\left (\sqrt{2} b x\right )}{4 \sqrt{2} \pi ^2 b^4}+\frac{x \sin \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}-\frac{x^3}{6 \pi b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x],x]

[Out]

-x^3/(6*b*Pi) + (2*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(b^4*Pi^2) - (5*FresnelS[Sqrt[2]*b*x])/(4*Sqrt[2]*b^4*Pi
^2) + (x^2*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi) + (x*Sin[b^2*Pi*x^2])/(4*b^3*Pi^2)

Rule 6462

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

Rule 3391

Int[(x_)^(m_.)*Sin[(a_.) + ((b_.)*(x_)^(n_))/2]^2, x_Symbol] :> Dist[1/2, Int[x^m, x], x] - Dist[1/2, Int[x^m*
Cos[2*a + b*x^n], x], x] /; FreeQ[{a, b, m, n}, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 3386

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Sin[c + d*
x^n])/(d*n), x] - Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x
] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3351

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

Rule 6452

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

Rubi steps

\begin{align*} \int x^3 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x) \, dx &=\frac{x^2 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac{2 \int x S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{b^2 \pi }-\frac{\int x^2 \sin ^2\left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{b \pi }\\ &=\frac{2 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^4 \pi ^2}+\frac{x^2 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac{\int \sin \left (b^2 \pi x^2\right ) \, dx}{b^3 \pi ^2}-\frac{\int x^2 \, dx}{2 b \pi }+\frac{\int x^2 \cos \left (b^2 \pi x^2\right ) \, dx}{2 b \pi }\\ &=-\frac{x^3}{6 b \pi }+\frac{2 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^4 \pi ^2}-\frac{S\left (\sqrt{2} b x\right )}{\sqrt{2} b^4 \pi ^2}+\frac{x^2 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^2 \pi }+\frac{x \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac{\int \sin \left (b^2 \pi x^2\right ) \, dx}{4 b^3 \pi ^2}\\ &=-\frac{x^3}{6 b \pi }+\frac{2 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{b^4 \pi ^2}-\frac{5 S\left (\sqrt{2} b x\right )}{4 \sqrt{2} b^4 \pi ^2}+\frac{x^2 S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^2 \pi }+\frac{x \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}\\ \end{align*}

Mathematica [A]  time = 0.081416, size = 90, normalized size = 0.83 \[ \frac{24 S(b x) \left (\pi b^2 x^2 \sin \left (\frac{1}{2} \pi b^2 x^2\right )+2 \cos \left (\frac{1}{2} \pi b^2 x^2\right )\right )-4 \pi b^3 x^3+6 b x \sin \left (\pi b^2 x^2\right )-15 \sqrt{2} S\left (\sqrt{2} b x\right )}{24 \pi ^2 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x],x]

[Out]

(-4*b^3*Pi*x^3 - 15*Sqrt[2]*FresnelS[Sqrt[2]*b*x] + 24*FresnelS[b*x]*(2*Cos[(b^2*Pi*x^2)/2] + b^2*Pi*x^2*Sin[(
b^2*Pi*x^2)/2]) + 6*b*x*Sin[b^2*Pi*x^2])/(24*b^4*Pi^2)

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Maple [A]  time = 0.075, size = 119, normalized size = 1.1 \begin{align*}{\frac{1}{b} \left ({\frac{{\it FresnelS} \left ( bx \right ) }{{b}^{3}} \left ({\frac{{b}^{2}{x}^{2}}{\pi }\sin \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+2\,{\frac{\cos \left ( 1/2\,{b}^{2}\pi \,{x}^{2} \right ) }{{\pi }^{2}}} \right ) }-{\frac{1}{{b}^{3}} \left ({\frac{\sqrt{2}{\it FresnelS} \left ( bx\sqrt{2} \right ) }{2\,{\pi }^{2}}}+{\frac{{x}^{3}{b}^{3}}{6\,\pi }}-{\frac{1}{2\,\pi } \left ({\frac{bx\sin \left ({b}^{2}\pi \,{x}^{2} \right ) }{2\,\pi }}-{\frac{\sqrt{2}{\it FresnelS} \left ( bx\sqrt{2} \right ) }{4\,\pi }} \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x),x)

[Out]

(FresnelS(b*x)/b^3*(1/Pi*b^2*x^2*sin(1/2*b^2*Pi*x^2)+2/Pi^2*cos(1/2*b^2*Pi*x^2))-1/b^3*(1/2/Pi^2*2^(1/2)*Fresn
elS(b*x*2^(1/2))+1/6/Pi*b^3*x^3-1/2/Pi*(1/2/Pi*b*x*sin(b^2*Pi*x^2)-1/4/Pi*2^(1/2)*FresnelS(b*x*2^(1/2)))))/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \cos \left (\frac{1}{2} \, \pi b^{2} x^{2}\right ){\rm fresnels}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cos(1/2*b^2*pi*x^2)*fresnels(b*x),x, algorithm="maxima")

[Out]

integrate(x^3*cos(1/2*pi*b^2*x^2)*fresnels(b*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{3} \cos \left (\frac{1}{2} \, \pi b^{2} x^{2}\right ){\rm fresnels}\left (b x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cos(1/2*b^2*pi*x^2)*fresnels(b*x),x, algorithm="fricas")

[Out]

integral(x^3*cos(1/2*pi*b^2*x^2)*fresnels(b*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \cos{\left (\frac{\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*cos(1/2*b**2*pi*x**2)*fresnels(b*x),x)

[Out]

Integral(x**3*cos(pi*b**2*x**2/2)*fresnels(b*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \cos \left (\frac{1}{2} \, \pi b^{2} x^{2}\right ){\rm fresnels}\left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cos(1/2*b^2*pi*x^2)*fresnels(b*x),x, algorithm="giac")

[Out]

integrate(x^3*cos(1/2*pi*b^2*x^2)*fresnels(b*x), x)