3.1.35 \(\int x^3 S(b x)^2 \, dx\) [35]

Optimal. Leaf size=140 \[ \frac {3 x^2}{8 b^2 \pi ^2}+\frac {x^2 \cos \left (b^2 \pi x^2\right )}{8 b^2 \pi ^2}+\frac {x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{2 b \pi }+\frac {3 S(b x)^2}{4 b^4 \pi ^2}+\frac {1}{4} x^4 S(b x)^2-\frac {3 x S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{2 b^3 \pi ^2}-\frac {\sin \left (b^2 \pi x^2\right )}{2 b^4 \pi ^3} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {6565, 6589, 6597, 3460, 2714, 6575, 30, 3377, 2717} \begin {gather*} \frac {3 S(b x)^2}{4 \pi ^2 b^4}+\frac {x^3 S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{2 \pi b}+\frac {3 x^2}{8 \pi ^2 b^2}+\frac {x^2 \cos \left (\pi b^2 x^2\right )}{8 \pi ^2 b^2}-\frac {\sin \left (\pi b^2 x^2\right )}{2 \pi ^3 b^4}-\frac {3 x S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{2 \pi ^2 b^3}+\frac {1}{4} x^4 S(b x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*FresnelS[b*x]^2,x]

[Out]

(3*x^2)/(8*b^2*Pi^2) + (x^2*Cos[b^2*Pi*x^2])/(8*b^2*Pi^2) + (x^3*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(2*b*Pi) +
(3*FresnelS[b*x]^2)/(4*b^4*Pi^2) + (x^4*FresnelS[b*x]^2)/4 - (3*x*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(2*b^3*P
i^2) - Sin[b^2*Pi*x^2]/(2*b^4*Pi^3)

Rule 30

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

Rule 2714

Int[sin[(c_.) + ((d_.)*(x_))/2]^2, x_Symbol] :> Simp[x/2, x] - Simp[Sin[2*c + d*x]/(2*d), x] /; FreeQ[{c, d},
x]

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3460

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

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 6575

Int[FresnelS[(b_.)*(x_)]^(n_.)*Sin[(d_.)*(x_)^2], x_Symbol] :> Dist[Pi*(b/(2*d)), Subst[Int[x^n, x], x, Fresne
lS[b*x]], x] /; FreeQ[{b, d, n}, x] && EqQ[d^2, (Pi^2/4)*b^4]

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 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*} \int x^3 S(b x)^2 \, dx &=\frac {1}{4} x^4 S(b x)^2-\frac {1}{2} b \int x^4 S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx\\ &=\frac {x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{2 b \pi }+\frac {1}{4} x^4 S(b x)^2-\frac {\int x^3 \sin \left (b^2 \pi x^2\right ) \, dx}{4 \pi }-\frac {3 \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x) \, dx}{2 b \pi }\\ &=\frac {x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{2 b \pi }+\frac {1}{4} x^4 S(b x)^2-\frac {3 x S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{2 b^3 \pi ^2}+\frac {3 \int S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{2 b^3 \pi ^2}+\frac {3 \int x \sin ^2\left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{2 b^2 \pi ^2}-\frac {\text {Subst}\left (\int x \sin \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{8 \pi }\\ &=\frac {x^2 \cos \left (b^2 \pi x^2\right )}{8 b^2 \pi ^2}+\frac {x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{2 b \pi }+\frac {1}{4} x^4 S(b x)^2-\frac {3 x S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{2 b^3 \pi ^2}+\frac {3 \text {Subst}(\int x \, dx,x,S(b x))}{2 b^4 \pi ^2}-\frac {\text {Subst}\left (\int \cos \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{8 b^2 \pi ^2}+\frac {3 \text {Subst}\left (\int \sin ^2\left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b^2 \pi ^2}\\ &=\frac {3 x^2}{8 b^2 \pi ^2}+\frac {x^2 \cos \left (b^2 \pi x^2\right )}{8 b^2 \pi ^2}+\frac {x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{2 b \pi }+\frac {3 S(b x)^2}{4 b^4 \pi ^2}+\frac {1}{4} x^4 S(b x)^2-\frac {3 x S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{2 b^3 \pi ^2}-\frac {\sin \left (b^2 \pi x^2\right )}{2 b^4 \pi ^3}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 140, normalized size = 1.00 \begin {gather*} \frac {3 x^2}{8 b^2 \pi ^2}+\frac {x^2 \cos \left (b^2 \pi x^2\right )}{8 b^2 \pi ^2}+\frac {x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{2 b \pi }+\frac {3 S(b x)^2}{4 b^4 \pi ^2}+\frac {1}{4} x^4 S(b x)^2-\frac {3 x S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{2 b^3 \pi ^2}-\frac {\sin \left (b^2 \pi x^2\right )}{2 b^4 \pi ^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*FresnelS[b*x]^2,x]

[Out]

(3*x^2)/(8*b^2*Pi^2) + (x^2*Cos[b^2*Pi*x^2])/(8*b^2*Pi^2) + (x^3*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(2*b*Pi) +
(3*FresnelS[b*x]^2)/(4*b^4*Pi^2) + (x^4*FresnelS[b*x]^2)/4 - (3*x*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(2*b^3*P
i^2) - Sin[b^2*Pi*x^2]/(2*b^4*Pi^3)

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Maple [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int x^{3} \mathrm {S}\left (b x \right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*FresnelS(b*x)^2,x)

[Out]

int(x^3*FresnelS(b*x)^2,x)

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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^3*fresnel_sin(b*x)^2,x, algorithm="maxima")

[Out]

integrate(x^3*fresnel_sin(b*x)^2, x)

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Fricas [A]
time = 0.37, size = 117, normalized size = 0.84 \begin {gather*} \frac {2 \, \pi ^{2} b^{3} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {S}\left (b x\right ) + \pi b^{2} x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )^{2} + \pi b^{2} x^{2} + {\left (3 \, \pi + \pi ^{3} b^{4} x^{4}\right )} \operatorname {S}\left (b x\right )^{2} - 2 \, {\left (3 \, \pi b x \operatorname {S}\left (b x\right ) + 2 \, \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{4 \, \pi ^{3} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*fresnel_sin(b*x)^2,x, algorithm="fricas")

[Out]

1/4*(2*pi^2*b^3*x^3*cos(1/2*pi*b^2*x^2)*fresnel_sin(b*x) + pi*b^2*x^2*cos(1/2*pi*b^2*x^2)^2 + pi*b^2*x^2 + (3*
pi + pi^3*b^4*x^4)*fresnel_sin(b*x)^2 - 2*(3*pi*b*x*fresnel_sin(b*x) + 2*cos(1/2*pi*b^2*x^2))*sin(1/2*pi*b^2*x
^2))/(pi^3*b^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} S^{2}\left (b x\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*fresnels(b*x)**2,x)

[Out]

Integral(x**3*fresnels(b*x)**2, x)

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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^3*fresnel_sin(b*x)^2,x, algorithm="giac")

[Out]

integrate(x^3*fresnel_sin(b*x)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,{\mathrm {FresnelS}\left (b\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*FresnelS(b*x)^2,x)

[Out]

int(x^3*FresnelS(b*x)^2, x)

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