\(\int x^5 \operatorname {FresnelS}(b x) \sin (\frac {1}{2} b^2 \pi x^2) \, dx\) [74]

Optimal result

Integrand size = 20, antiderivative size = 158 \[ \int x^5 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=-\frac {2 x^3}{3 b^3 \pi ^2}-\frac {x^3 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {8 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^6 \pi ^3}-\frac {x^4 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^2 \pi }-\frac {43 \operatorname {FresnelS}\left (\sqrt {2} b x\right )}{8 \sqrt {2} b^6 \pi ^3}+\frac {4 x^2 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac {11 x \sin \left (b^2 \pi x^2\right )}{8 b^5 \pi ^3} \]

[Out]

-2/3*x^3/b^3/Pi^2-1/4*x^3*cos(b^2*Pi*x^2)/b^3/Pi^2+8*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x)/b^6/Pi^3-x^4*cos(1/2*b^
2*Pi*x^2)*FresnelS(b*x)/b^2/Pi+4*x^2*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2)/b^4/Pi^2+11/8*x*sin(b^2*Pi*x^2)/b^5/Pi^
3-43/16*FresnelS(b*x*2^(1/2))/b^6/Pi^3*2^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6589, 6597, 3472, 30, 3467, 3432, 6587, 3466} \[ \int x^5 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=-\frac {43 \operatorname {FresnelS}\left (\sqrt {2} b x\right )}{8 \sqrt {2} \pi ^3 b^6}-\frac {2 x^3}{3 \pi ^2 b^3}-\frac {x^4 \operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {8 \operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^3 b^6}+\frac {11 x \sin \left (\pi b^2 x^2\right )}{8 \pi ^3 b^5}+\frac {4 x^2 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac {x^3 \cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3} \]

[In]

Int[x^5*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2],x]

[Out]

(-2*x^3)/(3*b^3*Pi^2) - (x^3*Cos[b^2*Pi*x^2])/(4*b^3*Pi^2) + (8*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(b^6*Pi^3)
- (x^4*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(b^2*Pi) - (43*FresnelS[Sqrt[2]*b*x])/(8*Sqrt[2]*b^6*Pi^3) + (4*x^2*
FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^4*Pi^2) + (11*x*Sin[b^2*Pi*x^2])/(8*b^5*Pi^3)

Rule 30

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

Rule 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

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 3467

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[e^(n - 1)*(e*x)^(m - n + 1)*(Sin[c + d*
x^n]/(d*n)), x] - Dist[e^n*((m - n + 1)/(d*n)), Int[(e*x)^(m - n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x
] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3472

Int[(x_)^(m_.)*Sin[(a_.) + ((b_.)*(x_)^(n_))/2]^2, x_Symbol] :> Dist[1/2, Int[x^m, x], x] - Dist[1/2, Int[x^m*
Cos[2*a + b*x^n], x], x] /; FreeQ[{a, b, m, n}, x]

Rule 6587

Int[FresnelS[(b_.)*(x_)]*(x_)*Sin[(d_.)*(x_)^2], x_Symbol] :> Simp[(-Cos[d*x^2])*(FresnelS[b*x]/(2*d)), x] + D
ist[1/(2*b*Pi), Int[Sin[2*d*x^2], x], x] /; FreeQ[{b, d}, 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^4 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^2 \pi }+\frac {4 \int x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x) \, dx}{b^2 \pi }+\frac {\int x^4 \sin \left (b^2 \pi x^2\right ) \, dx}{2 b \pi } \\ & = -\frac {x^3 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac {x^4 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^2 \pi }+\frac {4 x^2 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}-\frac {8 \int x \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b^4 \pi ^2}+\frac {3 \int x^2 \cos \left (b^2 \pi x^2\right ) \, dx}{4 b^3 \pi ^2}-\frac {4 \int x^2 \sin ^2\left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b^3 \pi ^2} \\ & = -\frac {x^3 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {8 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^6 \pi ^3}-\frac {x^4 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^2 \pi }+\frac {4 x^2 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac {3 x \sin \left (b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac {3 \int \sin \left (b^2 \pi x^2\right ) \, dx}{8 b^5 \pi ^3}-\frac {4 \int \sin \left (b^2 \pi x^2\right ) \, dx}{b^5 \pi ^3}-\frac {2 \int x^2 \, dx}{b^3 \pi ^2}+\frac {2 \int x^2 \cos \left (b^2 \pi x^2\right ) \, dx}{b^3 \pi ^2} \\ & = -\frac {2 x^3}{3 b^3 \pi ^2}-\frac {x^3 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {8 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^6 \pi ^3}-\frac {x^4 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^2 \pi }-\frac {3 \operatorname {FresnelS}\left (\sqrt {2} b x\right )}{8 \sqrt {2} b^6 \pi ^3}-\frac {2 \sqrt {2} \operatorname {FresnelS}\left (\sqrt {2} b x\right )}{b^6 \pi ^3}+\frac {4 x^2 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac {11 x \sin \left (b^2 \pi x^2\right )}{8 b^5 \pi ^3}-\frac {\int \sin \left (b^2 \pi x^2\right ) \, dx}{b^5 \pi ^3} \\ & = -\frac {2 x^3}{3 b^3 \pi ^2}-\frac {x^3 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {8 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^6 \pi ^3}-\frac {x^4 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^2 \pi }-\frac {11 \operatorname {FresnelS}\left (\sqrt {2} b x\right )}{8 \sqrt {2} b^6 \pi ^3}-\frac {2 \sqrt {2} \operatorname {FresnelS}\left (\sqrt {2} b x\right )}{b^6 \pi ^3}+\frac {4 x^2 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac {11 x \sin \left (b^2 \pi x^2\right )}{8 b^5 \pi ^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.76 \[ \int x^5 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=-\frac {32 b^3 \pi x^3+12 b^3 \pi x^3 \cos \left (b^2 \pi x^2\right )+129 \sqrt {2} \operatorname {FresnelS}\left (\sqrt {2} b x\right )+48 \operatorname {FresnelS}(b x) \left (\left (-8+b^4 \pi ^2 x^4\right ) \cos \left (\frac {1}{2} b^2 \pi x^2\right )-4 b^2 \pi x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )\right )-66 b x \sin \left (b^2 \pi x^2\right )}{48 b^6 \pi ^3} \]

[In]

Integrate[x^5*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2],x]

[Out]

-1/48*(32*b^3*Pi*x^3 + 12*b^3*Pi*x^3*Cos[b^2*Pi*x^2] + 129*Sqrt[2]*FresnelS[Sqrt[2]*b*x] + 48*FresnelS[b*x]*((
-8 + b^4*Pi^2*x^4)*Cos[(b^2*Pi*x^2)/2] - 4*b^2*Pi*x^2*Sin[(b^2*Pi*x^2)/2]) - 66*b*x*Sin[b^2*Pi*x^2])/(b^6*Pi^3
)

Maple [A] (verified)

Time = 1.49 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.28

method	result	size
default	\(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) \left (-\frac {b^{4} x^{4} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {\frac {4 b^{2} x^{2} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {8 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}}{\pi }\right )}{b^{5}}-\frac {\frac {2 b^{3} x^{3}}{3 \pi ^{2}}-\frac {2 \left (\frac {b x \sin \left (b^{2} \pi \,x^{2}\right )}{2 \pi }-\frac {\sqrt {2}\, \operatorname {FresnelS}\left (b x \sqrt {2}\right )}{4 \pi }\right )}{\pi ^{2}}-\frac {-\frac {\pi \,b^{3} x^{3} \cos \left (b^{2} \pi \,x^{2}\right )}{2}+\frac {3 \pi \left (\frac {b x \sin \left (b^{2} \pi \,x^{2}\right )}{2 \pi }-\frac {\sqrt {2}\, \operatorname {FresnelS}\left (b x \sqrt {2}\right )}{4 \pi }\right )}{2}-4 \sqrt {2}\, \operatorname {FresnelS}\left (b x \sqrt {2}\right )}{2 \pi ^{3}}}{b^{5}}}{b}\)	\(202\)

[In]

int(x^5*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.84 \[ \int x^5 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=-\frac {24 \, \pi b^{4} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )^{2} + 20 \, \pi b^{4} x^{3} + 48 \, {\left (\pi ^{2} b^{5} x^{4} - 8 \, b\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {S}\left (b x\right ) + 129 \, \sqrt {2} \sqrt {b^{2}} \operatorname {S}\left (\sqrt {2} \sqrt {b^{2}} x\right ) - 12 \, {\left (16 \, \pi b^{3} x^{2} \operatorname {S}\left (b x\right ) + 11 \, b^{2} x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{48 \, \pi ^{3} b^{7}} \]

[In]

integrate(x^5*fresnel_sin(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="fricas")

[Out]

-1/48*(24*pi*b^4*x^3*cos(1/2*pi*b^2*x^2)^2 + 20*pi*b^4*x^3 + 48*(pi^2*b^5*x^4 - 8*b)*cos(1/2*pi*b^2*x^2)*fresn
el_sin(b*x) + 129*sqrt(2)*sqrt(b^2)*fresnel_sin(sqrt(2)*sqrt(b^2)*x) - 12*(16*pi*b^3*x^2*fresnel_sin(b*x) + 11
*b^2*x*cos(1/2*pi*b^2*x^2))*sin(1/2*pi*b^2*x^2))/(pi^3*b^7)

Sympy [F]

\[ \int x^5 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int x^{5} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )\, dx \]

[In]

integrate(x**5*fresnels(b*x)*sin(1/2*b**2*pi*x**2),x)

[Out]

Integral(x**5*sin(pi*b**2*x**2/2)*fresnels(b*x), x)

Maxima [F]

\[ \int x^5 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int { x^{5} \operatorname {S}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \,d x } \]

[In]

integrate(x^5*fresnel_sin(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="maxima")

[Out]

integrate(x^5*fresnel_sin(b*x)*sin(1/2*pi*b^2*x^2), x)

Giac [F]

\[ \int x^5 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int { x^{5} \operatorname {S}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \,d x } \]

[In]

integrate(x^5*fresnel_sin(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="giac")

[Out]

integrate(x^5*fresnel_sin(b*x)*sin(1/2*pi*b^2*x^2), x)

Mupad [F(-1)]

Timed out. \[ \int x^5 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int x^5\,\mathrm {FresnelS}\left (b\,x\right )\,\sin \left (\frac {\Pi \,b^2\,x^2}{2}\right ) \,d x \]

[In]

int(x^5*FresnelS(b*x)*sin((Pi*b^2*x^2)/2),x)

[Out]

int(x^5*FresnelS(b*x)*sin((Pi*b^2*x^2)/2), x)