\(\int x^3 \operatorname {FresnelS}(b x) \sin (\frac {1}{2} b^2 \pi x^2) \, dx\) [76]

Optimal result

Integrand size = 20, antiderivative size = 105 \[ \int x^3 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=-\frac {x}{b^3 \pi ^2}-\frac {x \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {5 \operatorname {FresnelC}\left (\sqrt {2} b x\right )}{4 \sqrt {2} b^4 \pi ^2}-\frac {x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^2 \pi }+\frac {2 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6589, 6595, 3438, 3433, 3466} \[ \int x^3 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\frac {5 \operatorname {FresnelC}\left (\sqrt {2} b x\right )}{4 \sqrt {2} \pi ^2 b^4}-\frac {x}{\pi ^2 b^3}-\frac {x^2 \operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {2 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac {x \cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3} \]

[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 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 3466

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]

Rule 6589

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^2/4)*b^4] && IGtQ[m, 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*} \text {integral}& = -\frac {x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^2 \pi }+\frac {2 \int x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(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 ) \operatorname {FresnelS}(b x)}{b^2 \pi }+\frac {2 \operatorname {FresnelS}(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 {\operatorname {FresnelC}\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 ) \operatorname {FresnelS}(b x)}{b^2 \pi }+\frac {2 \operatorname {FresnelS}(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 {\operatorname {FresnelC}\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 ) \operatorname {FresnelS}(b x)}{b^2 \pi }+\frac {2 \operatorname {FresnelS}(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 \operatorname {FresnelC}\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 ) \operatorname {FresnelS}(b x)}{b^2 \pi }+\frac {2 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.79 \[ \int x^3 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\frac {-2 b x \left (4+\cos \left (b^2 \pi x^2\right )\right )+5 \sqrt {2} \operatorname {FresnelC}\left (\sqrt {2} b x\right )-8 \operatorname {FresnelS}(b x) \left (b^2 \pi x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )-2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )\right )}{8 b^4 \pi ^2} \]

[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)

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.10

method	result	size
default	\(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) \left (-\frac {b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}\right )}{b^{3}}-\frac {\frac {b x}{\pi ^{2}}-\frac {\sqrt {2}\, \operatorname {FresnelC}\left (b x \sqrt {2}\right )}{2 \pi ^{2}}-\frac {-\frac {b x \cos \left (b^{2} \pi \,x^{2}\right )}{2 \pi }+\frac {\sqrt {2}\, \operatorname {FresnelC}\left (b x \sqrt {2}\right )}{4 \pi }}{2 \pi }}{b^{3}}}{b}\)	\(115\)

[In]

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

[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*(b*x/Pi^2-1/2/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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90 \[ \int x^3 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=-\frac {8 \, \pi b^{3} x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {S}\left (b x\right ) + 4 \, b^{2} x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )^{2} + 6 \, b^{2} x - 16 \, b \operatorname {S}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 5 \, \sqrt {2} \sqrt {b^{2}} \operatorname {C}\left (\sqrt {2} \sqrt {b^{2}} x\right )}{8 \, \pi ^{2} b^{5}} \]

[In]

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

[Out]

-1/8*(8*pi*b^3*x^2*cos(1/2*pi*b^2*x^2)*fresnel_sin(b*x) + 4*b^2*x*cos(1/2*pi*b^2*x^2)^2 + 6*b^2*x - 16*b*fresn
el_sin(b*x)*sin(1/2*pi*b^2*x^2) - 5*sqrt(2)*sqrt(b^2)*fresnel_cos(sqrt(2)*sqrt(b^2)*x))/(pi^2*b^5)

Sympy [F]

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

[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)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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