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

Optimal. Leaf size=105 \[ \frac{5 \text{FresnelC}\left (\sqrt{2} b x\right )}{4 \sqrt{2} \pi ^2 b^4}+\frac{2 S(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac{x^2 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac{x \cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}-\frac{x}{\pi ^2 b^3} \]

[Out]

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

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Rubi [A]  time = 0.081234, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6454, 6460, 3357, 3352, 3385} \[ \frac{5 \text{FresnelC}\left (\sqrt{2} b x\right )}{4 \sqrt{2} \pi ^2 b^4}+\frac{2 S(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac{x^2 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac{x \cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}-\frac{x}{\pi ^2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6454

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

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 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 3385

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

Rubi steps

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

Mathematica [A]  time = 0.102146, size = 83, normalized size = 0.79 \[ \frac{-8 S(b x) \left (\pi b^2 x^2 \cos \left (\frac{1}{2} \pi b^2 x^2\right )-2 \sin \left (\frac{1}{2} \pi b^2 x^2\right )\right )-2 b x \left (\cos \left (\pi b^2 x^2\right )+4\right )+5 \sqrt{2} \text{FresnelC}\left (\sqrt{2} b x\right )}{8 \pi ^2 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*x*(4 + Cos[b^2*Pi*x^2]) + 5*Sqrt[2]*FresnelC[Sqrt[2]*b*x] - 8*FresnelS[b*x]*(b^2*Pi*x^2*Cos[(b^2*Pi*x^2)
/2] - 2*Sin[(b^2*Pi*x^2)/2]))/(8*b^4*Pi^2)

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Maple [A]  time = 0.083, size = 115, 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 }\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }+2\,{\frac{\sin \left ( 1/2\,{b}^{2}\pi \,{x}^{2} \right ) }{{\pi }^{2}}} \right ) }-{\frac{1}{{b}^{3}} \left ({\frac{bx}{{\pi }^{2}}}-{\frac{\sqrt{2}{\it FresnelC} \left ( bx\sqrt{2} \right ) }{2\,{\pi }^{2}}}-{\frac{1}{2\,\pi } \left ( -{\frac{bx\cos \left ({b}^{2}\pi \,{x}^{2} \right ) }{2\,\pi }}+{\frac{\sqrt{2}{\it FresnelC} \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*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \sin{\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*fresnels(b*x)*sin(1/2*b**2*pi*x**2),x)

[Out]

Integral(x**3*sin(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}{\rm fresnels}\left (b x\right ) \sin \left (\frac{1}{2} \, \pi b^{2} x^{2}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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