\(\int x^3 \cos (\frac {1}{2} b^2 \pi x^2) \operatorname {FresnelS}(b x) \, dx\) [96]

Optimal result

Integrand size = 20, antiderivative size = 108 \[ \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x) \, dx=-\frac {x^3}{6 b \pi }+\frac {2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^4 \pi ^2}-\frac {5 \operatorname {FresnelS}\left (\sqrt {2} b x\right )}{4 \sqrt {2} b^4 \pi ^2}+\frac {x^2 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }+\frac {x \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2} \]

[Out]

-1/6*x^3/b/Pi+2*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x)/b^4/Pi^2+x^2*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2)/b^2/Pi+1/4*x*
sin(b^2*Pi*x^2)/b^3/Pi^2-5/8*FresnelS(b*x*2^(1/2))/b^4/Pi^2*2^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6597, 3472, 30, 3467, 3432, 6587} \[ \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x) \, dx=-\frac {5 \operatorname {FresnelS}\left (\sqrt {2} b x\right )}{4 \sqrt {2} \pi ^2 b^4}+\frac {x^2 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {2 \operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}+\frac {x \sin \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}-\frac {x^3}{6 \pi b} \]

[In]

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

[Out]

-1/6*x^3/(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 30

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

Rule 3432

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

Rule 3467

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 3472

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 6587

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

Rule 6597

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^2/4)*b^4] && IGtQ[m, 1]

Rubi steps \begin{align*} \text {integral}& = \frac {x^2 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {2 \int x \operatorname {FresnelS}(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 ) \operatorname {FresnelS}(b x)}{b^4 \pi ^2}+\frac {x^2 \operatorname {FresnelS}(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 ) \operatorname {FresnelS}(b x)}{b^4 \pi ^2}-\frac {\operatorname {FresnelS}\left (\sqrt {2} b x\right )}{\sqrt {2} b^4 \pi ^2}+\frac {x^2 \operatorname {FresnelS}(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 ) \operatorname {FresnelS}(b x)}{b^4 \pi ^2}-\frac {5 \operatorname {FresnelS}\left (\sqrt {2} b x\right )}{4 \sqrt {2} b^4 \pi ^2}+\frac {x^2 \operatorname {FresnelS}(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] (verified)

Time = 0.06 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x) \, dx=\frac {-4 b^3 \pi x^3-15 \sqrt {2} \operatorname {FresnelS}\left (\sqrt {2} b x\right )+24 \operatorname {FresnelS}(b x) \left (2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )+b^2 \pi x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )\right )+6 b x \sin \left (b^2 \pi x^2\right )}{24 b^4 \pi ^2} \]

[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)

Maple [A] (verified)

Time = 1.13 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.10

method	result	size
default	\(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) \left (\frac {b^{2} x^{2} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {2 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}\right )}{b^{3}}-\frac {\frac {\sqrt {2}\, \operatorname {FresnelS}\left (b x \sqrt {2}\right )}{2 \pi ^{2}}+\frac {b^{3} x^{3}}{6 \pi }-\frac {\frac {b x \sin \left (b^{2} \pi \,x^{2}\right )}{2 \pi }-\frac {\sqrt {2}\, \operatorname {FresnelS}\left (b x \sqrt {2}\right )}{4 \pi }}{2 \pi }}{b^{3}}}{b}\)	\(119\)

[In]

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

[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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.90 \[ \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x) \, dx=-\frac {4 \, \pi b^{4} x^{3} - 48 \, b \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {S}\left (b x\right ) + 15 \, \sqrt {2} \sqrt {b^{2}} \operatorname {S}\left (\sqrt {2} \sqrt {b^{2}} x\right ) - 12 \, {\left (2 \, \pi b^{3} x^{2} \operatorname {S}\left (b x\right ) + b^{2} x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{24 \, \pi ^{2} b^{5}} \]

[In]

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

[Out]

-1/24*(4*pi*b^4*x^3 - 48*b*cos(1/2*pi*b^2*x^2)*fresnel_sin(b*x) + 15*sqrt(2)*sqrt(b^2)*fresnel_sin(sqrt(2)*sqr
t(b^2)*x) - 12*(2*pi*b^3*x^2*fresnel_sin(b*x) + b^2*x*cos(1/2*pi*b^2*x^2))*sin(1/2*pi*b^2*x^2))/(pi^2*b^5)

Sympy [F]

\[ \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x) \, dx=\int x^{3} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )\, dx \]

[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)

Maxima [F]

\[ \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x) \, dx=\int { x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {S}\left (b x\right ) \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x) \, dx=\int { x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {S}\left (b x\right ) \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x) \, dx=\int x^3\,\mathrm {FresnelS}\left (b\,x\right )\,\cos \left (\frac {\Pi \,b^2\,x^2}{2}\right ) \,d x \]

[In]

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

[Out]

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