\(\int x^6 \cos (\frac {1}{2} b^2 \pi x^2) \operatorname {FresnelS}(b x) \, dx\) [93]

Optimal result

Integrand size = 20, antiderivative size = 184 \[ \int x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x) \, dx=\frac {15 x^2}{4 b^5 \pi ^3}-\frac {x^6}{12 b \pi }+\frac {7 x^2 \cos \left (b^2 \pi x^2\right )}{4 b^5 \pi ^3}+\frac {5 x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^4 \pi ^2}+\frac {15 \operatorname {FresnelS}(b x)^2}{2 b^7 \pi ^3}-\frac {15 x \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^6 \pi ^3}+\frac {x^5 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {11 \sin \left (b^2 \pi x^2\right )}{2 b^7 \pi ^4}+\frac {x^4 \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2} \]

[Out]

15/4*x^2/b^5/Pi^3-1/12*x^6/b/Pi+7/4*x^2*cos(b^2*Pi*x^2)/b^5/Pi^3+5*x^3*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x)/b^4/P
i^2+15/2*FresnelS(b*x)^2/b^7/Pi^3-15*x*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2)/b^6/Pi^3+x^5*FresnelS(b*x)*sin(1/2*b^
2*Pi*x^2)/b^2/Pi-11/2*sin(b^2*Pi*x^2)/b^7/Pi^4+1/4*x^4*sin(b^2*Pi*x^2)/b^3/Pi^2

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {6597, 3460, 3390, 30, 3377, 2717, 6589, 2714, 6575} \[ \int x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x) \, dx=\frac {15 \operatorname {FresnelS}(b x)^2}{2 \pi ^3 b^7}+\frac {15 x^2}{4 \pi ^3 b^5}+\frac {x^5 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {11 \sin \left (\pi b^2 x^2\right )}{2 \pi ^4 b^7}-\frac {15 x \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^3 b^6}+\frac {7 x^2 \cos \left (\pi b^2 x^2\right )}{4 \pi ^3 b^5}+\frac {5 x^3 \operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}+\frac {x^4 \sin \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}-\frac {x^6}{12 \pi b} \]

[In]

Int[x^6*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x],x]

[Out]

(15*x^2)/(4*b^5*Pi^3) - x^6/(12*b*Pi) + (7*x^2*Cos[b^2*Pi*x^2])/(4*b^5*Pi^3) + (5*x^3*Cos[(b^2*Pi*x^2)/2]*Fres
nelS[b*x])/(b^4*Pi^2) + (15*FresnelS[b*x]^2)/(2*b^7*Pi^3) - (15*x*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^6*Pi^3
) + (x^5*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi) - (11*Sin[b^2*Pi*x^2])/(2*b^7*Pi^4) + (x^4*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 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 3390

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + ((f_.)*(x_))/2]^2, x_Symbol] :> Dist[1/2, Int[(c + d*x)^m, x], x] -
Dist[1/2, Int[(c + d*x)^m*Cos[2*e + f*x], x], x] /; FreeQ[{c, d, e, f, m}, x]

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 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*} \text {integral}& = \frac {x^5 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {5 \int x^4 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b^2 \pi }-\frac {\int x^5 \sin ^2\left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b \pi } \\ & = \frac {5 x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^4 \pi ^2}+\frac {x^5 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {15 \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x) \, dx}{b^4 \pi ^2}-\frac {5 \int x^3 \sin \left (b^2 \pi x^2\right ) \, dx}{2 b^3 \pi ^2}-\frac {\text {Subst}\left (\int x^2 \sin ^2\left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{2 b \pi } \\ & = \frac {5 x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^4 \pi ^2}-\frac {15 x \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^6 \pi ^3}+\frac {x^5 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }+\frac {15 \int \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b^6 \pi ^3}+\frac {15 \int x \sin ^2\left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b^5 \pi ^3}-\frac {5 \text {Subst}\left (\int x \sin \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b^3 \pi ^2}-\frac {\text {Subst}\left (\int x^2 \, dx,x,x^2\right )}{4 b \pi }+\frac {\text {Subst}\left (\int x^2 \cos \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b \pi } \\ & = -\frac {x^6}{12 b \pi }+\frac {5 x^2 \cos \left (b^2 \pi x^2\right )}{4 b^5 \pi ^3}+\frac {5 x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^4 \pi ^2}-\frac {15 x \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^6 \pi ^3}+\frac {x^5 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }+\frac {x^4 \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {15 \text {Subst}(\int x \, dx,x,\operatorname {FresnelS}(b x))}{b^7 \pi ^3}-\frac {5 \text {Subst}\left (\int \cos \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b^5 \pi ^3}+\frac {15 \text {Subst}\left (\int \sin ^2\left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{2 b^5 \pi ^3}-\frac {\text {Subst}\left (\int x \sin \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{2 b^3 \pi ^2} \\ & = \frac {15 x^2}{4 b^5 \pi ^3}-\frac {x^6}{12 b \pi }+\frac {7 x^2 \cos \left (b^2 \pi x^2\right )}{4 b^5 \pi ^3}+\frac {5 x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^4 \pi ^2}+\frac {15 \operatorname {FresnelS}(b x)^2}{2 b^7 \pi ^3}-\frac {15 x \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^6 \pi ^3}+\frac {x^5 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {5 \sin \left (b^2 \pi x^2\right )}{b^7 \pi ^4}+\frac {x^4 \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac {\text {Subst}\left (\int \cos \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{2 b^5 \pi ^3} \\ & = \frac {15 x^2}{4 b^5 \pi ^3}-\frac {x^6}{12 b \pi }+\frac {7 x^2 \cos \left (b^2 \pi x^2\right )}{4 b^5 \pi ^3}+\frac {5 x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^4 \pi ^2}+\frac {15 \operatorname {FresnelS}(b x)^2}{2 b^7 \pi ^3}-\frac {15 x \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^6 \pi ^3}+\frac {x^5 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {11 \sin \left (b^2 \pi x^2\right )}{2 b^7 \pi ^4}+\frac {x^4 \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00 \[ \int x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x) \, dx=\frac {15 x^2}{4 b^5 \pi ^3}-\frac {x^6}{12 b \pi }+\frac {7 x^2 \cos \left (b^2 \pi x^2\right )}{4 b^5 \pi ^3}+\frac {5 x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^4 \pi ^2}+\frac {15 \operatorname {FresnelS}(b x)^2}{2 b^7 \pi ^3}-\frac {15 x \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^6 \pi ^3}+\frac {x^5 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {11 \sin \left (b^2 \pi x^2\right )}{2 b^7 \pi ^4}+\frac {x^4 \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2} \]

[In]

Integrate[x^6*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x],x]

[Out]

(15*x^2)/(4*b^5*Pi^3) - x^6/(12*b*Pi) + (7*x^2*Cos[b^2*Pi*x^2])/(4*b^5*Pi^3) + (5*x^3*Cos[(b^2*Pi*x^2)/2]*Fres
nelS[b*x])/(b^4*Pi^2) + (15*FresnelS[b*x]^2)/(2*b^7*Pi^3) - (15*x*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^6*Pi^3
) + (x^5*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi) - (11*Sin[b^2*Pi*x^2])/(2*b^7*Pi^4) + (x^4*Sin[b^2*Pi*x^2
])/(4*b^3*Pi^2)

Maple [F]

\[\int x^{6} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right ) \operatorname {FresnelS}\left (b x \right )d x\]

[In]

int(x^6*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x),x)

[Out]

int(x^6*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x),x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.77 \[ \int x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x) \, dx=-\frac {\pi ^{3} b^{6} x^{6} - 60 \, \pi ^{2} b^{3} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {S}\left (b x\right ) - 42 \, \pi b^{2} x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )^{2} - 24 \, \pi b^{2} x^{2} - 90 \, \pi \operatorname {S}\left (b x\right )^{2} - 6 \, {\left ({\left (\pi ^{2} b^{4} x^{4} - 22\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 2 \, {\left (\pi ^{3} b^{5} x^{5} - 15 \, \pi b x\right )} \operatorname {S}\left (b x\right )\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{12 \, \pi ^{4} b^{7}} \]

[In]

integrate(x^6*cos(1/2*b^2*pi*x^2)*fresnel_sin(b*x),x, algorithm="fricas")

[Out]

-1/12*(pi^3*b^6*x^6 - 60*pi^2*b^3*x^3*cos(1/2*pi*b^2*x^2)*fresnel_sin(b*x) - 42*pi*b^2*x^2*cos(1/2*pi*b^2*x^2)
^2 - 24*pi*b^2*x^2 - 90*pi*fresnel_sin(b*x)^2 - 6*((pi^2*b^4*x^4 - 22)*cos(1/2*pi*b^2*x^2) + 2*(pi^3*b^5*x^5 -
15*pi*b*x)*fresnel_sin(b*x))*sin(1/2*pi*b^2*x^2))/(pi^4*b^7)

Sympy [A] (verification not implemented)

Time = 4.60 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.43 \[ \int x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x) \, dx=\begin {cases} - \frac {x^{6} \sin ^{2}{\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{12 \pi b} - \frac {x^{6} \cos ^{2}{\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{12 \pi b} + \frac {x^{5} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )}{\pi b^{2}} + \frac {x^{4} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{2 \pi ^{2} b^{3}} + \frac {5 x^{3} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )}{\pi ^{2} b^{4}} + \frac {2 x^{2} \sin ^{2}{\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{\pi ^{3} b^{5}} + \frac {11 x^{2} \cos ^{2}{\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{2 \pi ^{3} b^{5}} - \frac {15 x \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )}{\pi ^{3} b^{6}} - \frac {11 \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{\pi ^{4} b^{7}} + \frac {15 S^{2}\left (b x\right )}{2 \pi ^{3} b^{7}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

integrate(x**6*cos(1/2*b**2*pi*x**2)*fresnels(b*x),x)

[Out]

Piecewise((-x**6*sin(pi*b**2*x**2/2)**2/(12*pi*b) - x**6*cos(pi*b**2*x**2/2)**2/(12*pi*b) + x**5*sin(pi*b**2*x
**2/2)*fresnels(b*x)/(pi*b**2) + x**4*sin(pi*b**2*x**2/2)*cos(pi*b**2*x**2/2)/(2*pi**2*b**3) + 5*x**3*cos(pi*b
**2*x**2/2)*fresnels(b*x)/(pi**2*b**4) + 2*x**2*sin(pi*b**2*x**2/2)**2/(pi**3*b**5) + 11*x**2*cos(pi*b**2*x**2
/2)**2/(2*pi**3*b**5) - 15*x*sin(pi*b**2*x**2/2)*fresnels(b*x)/(pi**3*b**6) - 11*sin(pi*b**2*x**2/2)*cos(pi*b*
*2*x**2/2)/(pi**4*b**7) + 15*fresnels(b*x)**2/(2*pi**3*b**7), Ne(b, 0)), (0, True))

Maxima [F]

\[ \int x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x) \, dx=\int { x^{6} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {S}\left (b x\right ) \,d x } \]

[In]

integrate(x^6*cos(1/2*b^2*pi*x^2)*fresnel_sin(b*x),x, algorithm="maxima")

[Out]

integrate(x^6*cos(1/2*pi*b^2*x^2)*fresnel_sin(b*x), x)

Giac [F]

\[ \int x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x) \, dx=\int { x^{6} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {S}\left (b x\right ) \,d x } \]

[In]

integrate(x^6*cos(1/2*b^2*pi*x^2)*fresnel_sin(b*x),x, algorithm="giac")

[Out]

integrate(x^6*cos(1/2*pi*b^2*x^2)*fresnel_sin(b*x), x)

Mupad [F(-1)]

Timed out. \[ \int x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x) \, dx=\int x^6\,\mathrm {FresnelS}\left (b\,x\right )\,\cos \left (\frac {\Pi \,b^2\,x^2}{2}\right ) \,d x \]

[In]

int(x^6*FresnelS(b*x)*cos((Pi*b^2*x^2)/2),x)

[Out]

int(x^6*FresnelS(b*x)*cos((Pi*b^2*x^2)/2), x)