\(\int x^6 \operatorname {FresnelS}(b x) \, dx\) [2]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 8, antiderivative size = 109 \[ \int x^6 \operatorname {FresnelS}(b x) \, dx=-\frac {24 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b^5 \pi ^3}+\frac {x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b \pi }+\frac {1}{7} x^7 \operatorname {FresnelS}(b x)+\frac {48 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b^7 \pi ^4}-\frac {6 x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b^3 \pi ^2} \]

[Out]

-24/7*x^2*cos(1/2*b^2*Pi*x^2)/b^5/Pi^3+1/7*x^6*cos(1/2*b^2*Pi*x^2)/b/Pi+1/7*x^7*FresnelS(b*x)+48/7*sin(1/2*b^2
*Pi*x^2)/b^7/Pi^4-6/7*x^4*sin(1/2*b^2*Pi*x^2)/b^3/Pi^2

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6561, 3460, 3377, 2717} \[ \int x^6 \operatorname {FresnelS}(b x) \, dx=\frac {x^6 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{7 \pi b}+\frac {48 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{7 \pi ^4 b^7}-\frac {24 x^2 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{7 \pi ^3 b^5}-\frac {6 x^4 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{7 \pi ^2 b^3}+\frac {1}{7} x^7 \operatorname {FresnelS}(b x) \]

[In]

Int[x^6*FresnelS[b*x],x]

[Out]

(-24*x^2*Cos[(b^2*Pi*x^2)/2])/(7*b^5*Pi^3) + (x^6*Cos[(b^2*Pi*x^2)/2])/(7*b*Pi) + (x^7*FresnelS[b*x])/7 + (48*
Sin[(b^2*Pi*x^2)/2])/(7*b^7*Pi^4) - (6*x^4*Sin[(b^2*Pi*x^2)/2])/(7*b^3*Pi^2)

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 6561

Int[FresnelS[(b_.)*(x_)]*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*(FresnelS[b*x]/(d*(m + 1))), x] -
 Dist[b/(d*(m + 1)), Int[(d*x)^(m + 1)*Sin[(Pi/2)*b^2*x^2], x], x] /; FreeQ[{b, d, m}, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{7} x^7 \operatorname {FresnelS}(b x)-\frac {1}{7} b \int x^7 \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx \\ & = \frac {1}{7} x^7 \operatorname {FresnelS}(b x)-\frac {1}{14} b \text {Subst}\left (\int x^3 \sin \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right ) \\ & = \frac {x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b \pi }+\frac {1}{7} x^7 \operatorname {FresnelS}(b x)-\frac {3 \text {Subst}\left (\int x^2 \cos \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{7 b \pi } \\ & = \frac {x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b \pi }+\frac {1}{7} x^7 \operatorname {FresnelS}(b x)-\frac {6 x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b^3 \pi ^2}+\frac {12 \text {Subst}\left (\int x \sin \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{7 b^3 \pi ^2} \\ & = -\frac {24 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b^5 \pi ^3}+\frac {x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b \pi }+\frac {1}{7} x^7 \operatorname {FresnelS}(b x)-\frac {6 x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b^3 \pi ^2}+\frac {24 \text {Subst}\left (\int \cos \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{7 b^5 \pi ^3} \\ & = -\frac {24 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b^5 \pi ^3}+\frac {x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b \pi }+\frac {1}{7} x^7 \operatorname {FresnelS}(b x)+\frac {48 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b^7 \pi ^4}-\frac {6 x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b^3 \pi ^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.76 \[ \int x^6 \operatorname {FresnelS}(b x) \, dx=\frac {x^2 \left (-24+b^4 \pi ^2 x^4\right ) \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b^5 \pi ^3}+\frac {1}{7} x^7 \operatorname {FresnelS}(b x)-\frac {6 \left (-8+b^4 \pi ^2 x^4\right ) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{7 b^7 \pi ^4} \]

[In]

Integrate[x^6*FresnelS[b*x],x]

[Out]

(x^2*(-24 + b^4*Pi^2*x^4)*Cos[(b^2*Pi*x^2)/2])/(7*b^5*Pi^3) + (x^7*FresnelS[b*x])/7 - (6*(-8 + b^4*Pi^2*x^4)*S
in[(b^2*Pi*x^2)/2])/(7*b^7*Pi^4)

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 0.40 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.27

method result size
meijerg \(\frac {\pi \,b^{3} x^{10} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {5}{2}\right ], \left [\frac {3}{2}, \frac {7}{4}, \frac {7}{2}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{60}\) \(29\)
derivativedivides \(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) b^{7} x^{7}}{7}+\frac {b^{6} x^{6} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{7 \pi }-\frac {6 \left (\frac {b^{4} x^{4} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {4 \left (-\frac {b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}\right )}{\pi }\right )}{7 \pi }}{b^{7}}\) \(107\)
default \(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) b^{7} x^{7}}{7}+\frac {b^{6} x^{6} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{7 \pi }-\frac {6 \left (\frac {b^{4} x^{4} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {4 \left (-\frac {b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}\right )}{\pi }\right )}{7 \pi }}{b^{7}}\) \(107\)
parts \(\frac {x^{7} \operatorname {FresnelS}\left (b x \right )}{7}-\frac {b \left (-\frac {x^{6} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }+\frac {\frac {6 x^{4} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }-\frac {24 \left (-\frac {x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }+\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{4} \pi ^{2}}\right )}{b^{2} \pi }}{b^{2} \pi }\right )}{7}\) \(113\)

[In]

int(x^6*FresnelS(b*x),x,method=_RETURNVERBOSE)

[Out]

1/60*Pi*b^3*x^10*hypergeom([3/4,5/2],[3/2,7/4,7/2],-1/16*x^4*Pi^2*b^4)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.72 \[ \int x^6 \operatorname {FresnelS}(b x) \, dx=\frac {\pi ^{4} b^{7} x^{7} \operatorname {S}\left (b x\right ) + {\left (\pi ^{3} b^{6} x^{6} - 24 \, \pi b^{2} x^{2}\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 6 \, {\left (\pi ^{2} b^{4} x^{4} - 8\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{7 \, \pi ^{4} b^{7}} \]

[In]

integrate(x^6*fresnel_sin(b*x),x, algorithm="fricas")

[Out]

1/7*(pi^4*b^7*x^7*fresnel_sin(b*x) + (pi^3*b^6*x^6 - 24*pi*b^2*x^2)*cos(1/2*pi*b^2*x^2) - 6*(pi^2*b^4*x^4 - 8)
*sin(1/2*pi*b^2*x^2))/(pi^4*b^7)

Sympy [A] (verification not implemented)

Time = 1.03 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.43 \[ \int x^6 \operatorname {FresnelS}(b x) \, dx=\frac {3 x^{7} S\left (b x\right ) \Gamma \left (\frac {3}{4}\right )}{28 \Gamma \left (\frac {7}{4}\right )} + \frac {3 x^{6} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{28 \pi b \Gamma \left (\frac {7}{4}\right )} - \frac {9 x^{4} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{14 \pi ^{2} b^{3} \Gamma \left (\frac {7}{4}\right )} - \frac {18 x^{2} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{7 \pi ^{3} b^{5} \Gamma \left (\frac {7}{4}\right )} + \frac {36 \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{7 \pi ^{4} b^{7} \Gamma \left (\frac {7}{4}\right )} \]

[In]

integrate(x**6*fresnels(b*x),x)

[Out]

3*x**7*fresnels(b*x)*gamma(3/4)/(28*gamma(7/4)) + 3*x**6*cos(pi*b**2*x**2/2)*gamma(3/4)/(28*pi*b*gamma(7/4)) -
 9*x**4*sin(pi*b**2*x**2/2)*gamma(3/4)/(14*pi**2*b**3*gamma(7/4)) - 18*x**2*cos(pi*b**2*x**2/2)*gamma(3/4)/(7*
pi**3*b**5*gamma(7/4)) + 36*sin(pi*b**2*x**2/2)*gamma(3/4)/(7*pi**4*b**7*gamma(7/4))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.68 \[ \int x^6 \operatorname {FresnelS}(b x) \, dx=\frac {1}{7} \, x^{7} \operatorname {S}\left (b x\right ) + \frac {{\left (\pi ^{3} b^{6} x^{6} - 24 \, \pi b^{2} x^{2}\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 6 \, {\left (\pi ^{2} b^{4} x^{4} - 8\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{7 \, \pi ^{4} b^{7}} \]

[In]

integrate(x^6*fresnel_sin(b*x),x, algorithm="maxima")

[Out]

1/7*x^7*fresnel_sin(b*x) + 1/7*((pi^3*b^6*x^6 - 24*pi*b^2*x^2)*cos(1/2*pi*b^2*x^2) - 6*(pi^2*b^4*x^4 - 8)*sin(
1/2*pi*b^2*x^2))/(pi^4*b^7)

Giac [F]

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

[In]

integrate(x^6*fresnel_sin(b*x),x, algorithm="giac")

[Out]

integrate(x^6*fresnel_sin(b*x), x)

Mupad [F(-1)]

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

[In]

int(x^6*FresnelS(b*x),x)

[Out]

int(x^6*FresnelS(b*x), x)