3.1.36 \(\int x^2 S(b x)^2 \, dx\) [36]

Optimal. Leaf size=124 \[ \frac {2 x}{3 b^2 \pi ^2}+\frac {x \cos \left (b^2 \pi x^2\right )}{6 b^2 \pi ^2}-\frac {5 \text {FresnelC}\left (\sqrt {2} b x\right )}{6 \sqrt {2} b^3 \pi ^2}+\frac {2 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{3 b \pi }+\frac {1}{3} x^3 S(b x)^2-\frac {4 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2} \]

[Out]

2/3*x/b^2/Pi^2+1/6*x*cos(b^2*Pi*x^2)/b^2/Pi^2+2/3*x^2*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x)/b/Pi+1/3*x^3*FresnelS(
b*x)^2-4/3*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2)/b^3/Pi^2-5/12*FresnelC(b*x*2^(1/2))/b^3/Pi^2*2^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6565, 6589, 6595, 3438, 3433, 3466} \begin {gather*} -\frac {5 \text {FresnelC}\left (\sqrt {2} b x\right )}{6 \sqrt {2} \pi ^2 b^3}+\frac {2 x^2 S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{3 \pi b}+\frac {x \cos \left (\pi b^2 x^2\right )}{6 \pi ^2 b^2}+\frac {2 x}{3 \pi ^2 b^2}-\frac {4 S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{3 \pi ^2 b^3}+\frac {1}{3} x^3 S(b x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*FresnelS[b*x]^2,x]

[Out]

(2*x)/(3*b^2*Pi^2) + (x*Cos[b^2*Pi*x^2])/(6*b^2*Pi^2) - (5*FresnelC[Sqrt[2]*b*x])/(6*Sqrt[2]*b^3*Pi^2) + (2*x^
2*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(3*b*Pi) + (x^3*FresnelS[b*x]^2)/3 - (4*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2]
)/(3*b^3*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 6565

Int[FresnelS[(b_.)*(x_)]^2*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*(FresnelS[b*x]^2/(m + 1)), x] - Dist[2*(b/(
m + 1)), Int[x^(m + 1)*Sin[(Pi/2)*b^2*x^2]*FresnelS[b*x], x], x] /; FreeQ[b, x] && IntegerQ[m] && NeQ[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*} \int x^2 S(b x)^2 \, dx &=\frac {1}{3} x^3 S(b x)^2-\frac {1}{3} (2 b) \int x^3 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx\\ &=\frac {2 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{3 b \pi }+\frac {1}{3} x^3 S(b x)^2-\frac {\int x^2 \sin \left (b^2 \pi x^2\right ) \, dx}{3 \pi }-\frac {4 \int x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x) \, dx}{3 b \pi }\\ &=\frac {x \cos \left (b^2 \pi x^2\right )}{6 b^2 \pi ^2}+\frac {2 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{3 b \pi }+\frac {1}{3} x^3 S(b x)^2-\frac {4 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2}-\frac {\int \cos \left (b^2 \pi x^2\right ) \, dx}{6 b^2 \pi ^2}+\frac {4 \int \sin ^2\left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{3 b^2 \pi ^2}\\ &=\frac {x \cos \left (b^2 \pi x^2\right )}{6 b^2 \pi ^2}-\frac {C\left (\sqrt {2} b x\right )}{6 \sqrt {2} b^3 \pi ^2}+\frac {2 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{3 b \pi }+\frac {1}{3} x^3 S(b x)^2-\frac {4 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2}+\frac {4 \int \left (\frac {1}{2}-\frac {1}{2} \cos \left (b^2 \pi x^2\right )\right ) \, dx}{3 b^2 \pi ^2}\\ &=\frac {2 x}{3 b^2 \pi ^2}+\frac {x \cos \left (b^2 \pi x^2\right )}{6 b^2 \pi ^2}-\frac {C\left (\sqrt {2} b x\right )}{6 \sqrt {2} b^3 \pi ^2}+\frac {2 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{3 b \pi }+\frac {1}{3} x^3 S(b x)^2-\frac {4 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2}-\frac {2 \int \cos \left (b^2 \pi x^2\right ) \, dx}{3 b^2 \pi ^2}\\ &=\frac {2 x}{3 b^2 \pi ^2}+\frac {x \cos \left (b^2 \pi x^2\right )}{6 b^2 \pi ^2}-\frac {C\left (\sqrt {2} b x\right )}{6 \sqrt {2} b^3 \pi ^2}-\frac {\sqrt {2} C\left (\sqrt {2} b x\right )}{3 b^3 \pi ^2}+\frac {2 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{3 b \pi }+\frac {1}{3} x^3 S(b x)^2-\frac {4 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.09, size = 100, normalized size = 0.81 \begin {gather*} \frac {2 b x \left (4+\cos \left (b^2 \pi x^2\right )\right )-5 \sqrt {2} \text {FresnelC}\left (\sqrt {2} b x\right )+4 b^3 \pi ^2 x^3 S(b x)^2+8 S(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 )}{12 b^3 \pi ^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*FresnelS[b*x]^2,x]

[Out]

(2*b*x*(4 + Cos[b^2*Pi*x^2]) - 5*Sqrt[2]*FresnelC[Sqrt[2]*b*x] + 4*b^3*Pi^2*x^3*FresnelS[b*x]^2 + 8*FresnelS[b
*x]*(b^2*Pi*x^2*Cos[(b^2*Pi*x^2)/2] - 2*Sin[(b^2*Pi*x^2)/2]))/(12*b^3*Pi^2)

________________________________________________________________________________________

Maple [A]
time = 0.66, size = 122, normalized size = 0.98

method	result	size
derivativedivides	\(\frac {\frac {\mathrm {S}\left (b x \right )^{2} b^{3} x^{3}}{3}-2 \,\mathrm {S}\left (b x \right ) \left (-\frac {b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi }+\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi ^{2}}\right )+\frac {2 b x}{3 \pi ^{2}}-\frac {\sqrt {2}\, \FresnelC \left (b x \sqrt {2}\right )}{3 \pi ^{2}}-\frac {-\frac {b x \cos \left (b^{2} \pi \,x^{2}\right )}{2 \pi }+\frac {\sqrt {2}\, \FresnelC \left (b x \sqrt {2}\right )}{4 \pi }}{3 \pi }}{b^{3}}\)	\(122\)
default	\(\frac {\frac {\mathrm {S}\left (b x \right )^{2} b^{3} x^{3}}{3}-2 \,\mathrm {S}\left (b x \right ) \left (-\frac {b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi }+\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi ^{2}}\right )+\frac {2 b x}{3 \pi ^{2}}-\frac {\sqrt {2}\, \FresnelC \left (b x \sqrt {2}\right )}{3 \pi ^{2}}-\frac {-\frac {b x \cos \left (b^{2} \pi \,x^{2}\right )}{2 \pi }+\frac {\sqrt {2}\, \FresnelC \left (b x \sqrt {2}\right )}{4 \pi }}{3 \pi }}{b^{3}}\)	\(122\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*FresnelS(b*x)^2,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

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^2*fresnel_sin(b*x)^2,x, algorithm="maxima")

[Out]

integrate(x^2*fresnel_sin(b*x)^2, x)

________________________________________________________________________________________

Fricas [A]
time = 0.35, size = 111, normalized size = 0.90 \begin {gather*} \frac {4 \, \pi ^{2} b^{4} x^{3} \operatorname {S}\left (b x\right )^{2} + 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 )}{12 \, \pi ^{2} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*fresnel_sin(b*x)^2,x, algorithm="fricas")

[Out]

1/12*(4*pi^2*b^4*x^3*fresnel_sin(b*x)^2 + 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*fresnel_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^4)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*fresnels(b*x)**2,x)

[Out]

Integral(x**2*fresnels(b*x)**2, x)

________________________________________________________________________________________

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^2*fresnel_sin(b*x)^2,x, algorithm="giac")

[Out]

integrate(x^2*fresnel_sin(b*x)^2, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,{\mathrm {FresnelS}\left (b\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*FresnelS(b*x)^2,x)

[Out]

int(x^2*FresnelS(b*x)^2, x)

________________________________________________________________________________________