\(\int x^4 \operatorname {FresnelS}(b x) \sin (\frac {1}{2} b^2 \pi x^2) \, dx\) [75]

Optimal result

Integrand size = 20, antiderivative size = 120 \[ \int x^4 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=-\frac {3 x^2}{4 b^3 \pi ^2}-\frac {x^2 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac {x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^2 \pi }-\frac {3 \operatorname {FresnelS}(b x)^2}{2 b^5 \pi ^2}+\frac {3 x \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac {\sin \left (b^2 \pi x^2\right )}{b^5 \pi ^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00 \[ \int x^4 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=-\frac {3 x^2}{4 b^3 \pi ^2}-\frac {x^2 \cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac {x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^2 \pi }-\frac {3 \operatorname {FresnelS}(b x)^2}{2 b^5 \pi ^2}+\frac {3 x \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac {\sin \left (b^2 \pi x^2\right )}{b^5 \pi ^3} \] Input:

Integrate[x^4*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2],x]
 

Output:

(-3*x^2)/(4*b^3*Pi^2) - (x^2*Cos[b^2*Pi*x^2])/(4*b^3*Pi^2) - (x^3*Cos[(b^2 
*Pi*x^2)/2]*FresnelS[b*x])/(b^2*Pi) - (3*FresnelS[b*x]^2)/(2*b^5*Pi^2) + ( 
3*x*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^4*Pi^2) + Sin[b^2*Pi*x^2]/(b^5*P 
i^3)
 

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.34, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {7008, 3860, 3042, 3777, 3042, 3117, 7016, 3860, 3042, 3114, 6994, 15}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^4 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right ) \, dx\)

\(\Big \downarrow \) 7008

\(\displaystyle \frac {3 \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)dx}{\pi b^2}+\frac {\int x^3 \sin \left (b^2 \pi x^2\right )dx}{2 \pi b}-\frac {x^3 \operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\)

\(\Big \downarrow \) 3860

\(\displaystyle \frac {3 \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)dx}{\pi b^2}+\frac {\int x^2 \sin \left (b^2 \pi x^2\right )dx^2}{4 \pi b}-\frac {x^3 \operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)dx}{\pi b^2}+\frac {\int x^2 \sin \left (b^2 \pi x^2\right )dx^2}{4 \pi b}-\frac {x^3 \operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\)

\(\Big \downarrow \) 3777

\(\displaystyle \frac {3 \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)dx}{\pi b^2}+\frac {\frac {\int \cos \left (b^2 \pi x^2\right )dx^2}{\pi b^2}-\frac {x^2 \cos \left (\pi b^2 x^2\right )}{\pi b^2}}{4 \pi b}-\frac {x^3 \operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)dx}{\pi b^2}+\frac {\frac {\int \sin \left (b^2 \pi x^2+\frac {\pi }{2}\right )dx^2}{\pi b^2}-\frac {x^2 \cos \left (\pi b^2 x^2\right )}{\pi b^2}}{4 \pi b}-\frac {x^3 \operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\)

\(\Big \downarrow \) 3117

\(\displaystyle \frac {3 \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)dx}{\pi b^2}-\frac {x^3 \operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\frac {\sin \left (\pi b^2 x^2\right )}{\pi ^2 b^4}-\frac {x^2 \cos \left (\pi b^2 x^2\right )}{\pi b^2}}{4 \pi b}\)

\(\Big \downarrow \) 7016

\(\displaystyle \frac {3 \left (-\frac {\int \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b^2}-\frac {\int x \sin ^2\left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b}+\frac {x \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )}{\pi b^2}-\frac {x^3 \operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\frac {\sin \left (\pi b^2 x^2\right )}{\pi ^2 b^4}-\frac {x^2 \cos \left (\pi b^2 x^2\right )}{\pi b^2}}{4 \pi b}\)

\(\Big \downarrow \) 3860

\(\displaystyle \frac {3 \left (-\frac {\int \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b^2}-\frac {\int \sin ^2\left (\frac {1}{2} b^2 \pi x^2\right )dx^2}{2 \pi b}+\frac {x \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )}{\pi b^2}-\frac {x^3 \operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\frac {\sin \left (\pi b^2 x^2\right )}{\pi ^2 b^4}-\frac {x^2 \cos \left (\pi b^2 x^2\right )}{\pi b^2}}{4 \pi b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (-\frac {\int \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b^2}-\frac {\int \sin \left (\frac {1}{2} b^2 \pi x^2\right )^2dx^2}{2 \pi b}+\frac {x \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )}{\pi b^2}-\frac {x^3 \operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\frac {\sin \left (\pi b^2 x^2\right )}{\pi ^2 b^4}-\frac {x^2 \cos \left (\pi b^2 x^2\right )}{\pi b^2}}{4 \pi b}\)

\(\Big \downarrow \) 3114

\(\displaystyle \frac {3 \left (-\frac {\int \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b^2}+\frac {x \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {\frac {x^2}{2}-\frac {\sin \left (\pi b^2 x^2\right )}{2 \pi b^2}}{2 \pi b}\right )}{\pi b^2}-\frac {x^3 \operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\frac {\sin \left (\pi b^2 x^2\right )}{\pi ^2 b^4}-\frac {x^2 \cos \left (\pi b^2 x^2\right )}{\pi b^2}}{4 \pi b}\)

\(\Big \downarrow \) 6994

\(\displaystyle \frac {3 \left (-\frac {\int \operatorname {FresnelS}(b x)d\operatorname {FresnelS}(b x)}{\pi b^3}+\frac {x \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {\frac {x^2}{2}-\frac {\sin \left (\pi b^2 x^2\right )}{2 \pi b^2}}{2 \pi b}\right )}{\pi b^2}-\frac {x^3 \operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\frac {\sin \left (\pi b^2 x^2\right )}{\pi ^2 b^4}-\frac {x^2 \cos \left (\pi b^2 x^2\right )}{\pi b^2}}{4 \pi b}\)

\(\Big \downarrow \) 15

\(\displaystyle -\frac {x^3 \operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\frac {\sin \left (\pi b^2 x^2\right )}{\pi ^2 b^4}-\frac {x^2 \cos \left (\pi b^2 x^2\right )}{\pi b^2}}{4 \pi b}+\frac {3 \left (-\frac {\operatorname {FresnelS}(b x)^2}{2 \pi b^3}+\frac {x \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {\frac {x^2}{2}-\frac {\sin \left (\pi b^2 x^2\right )}{2 \pi b^2}}{2 \pi b}\right )}{\pi b^2}\)

Input:

Int[x^4*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2],x]
 

Output:

-((x^3*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(b^2*Pi)) + (-((x^2*Cos[b^2*Pi*x 
^2])/(b^2*Pi)) + Sin[b^2*Pi*x^2]/(b^4*Pi^2))/(4*b*Pi) + (3*(-1/2*FresnelS[ 
b*x]^2/(b^3*Pi) + (x*FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi) - (x^2/2 
- Sin[b^2*Pi*x^2]/(2*b^2*Pi))/(2*b*Pi)))/(b^2*Pi)
 

Defintions of rubi rules used

rule 15

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3114

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 3117

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

rule 3777

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

rule 3860

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(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[Simplify[ 
(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[ 
(m + 1)/n], 0]))
 

rule 6994

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

rule 7008

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] + (Simp[(m - 1)/(2*d)   Int[ 
x^(m - 2)*Cos[d*x^2]*FresnelS[b*x], x], x] + Simp[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] && IG 
tQ[m, 1]
 

rule 7016

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] + (-Simp[1/(Pi*b)   Int[x^(m - 
1)*Sin[d*x^2]^2, x], x] - Simp[(m - 1)/(2*d)   Int[x^(m - 2)*Sin[d*x^2]*Fre 
snelS[b*x], x], x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2/4)*b^4] && IGtQ[m 
, 1]
 

Maple [F]

\[\int x^{4} \operatorname {FresnelS}\left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )d x\]

Input:

int(x^4*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2),x)
 

Output:

int(x^4*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2),x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.88 \[ \int x^4 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=-\frac {2 \, \pi ^{2} b^{3} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {S}\left (b x\right ) + \pi b^{2} x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )^{2} + \pi b^{2} x^{2} + 3 \, \pi \operatorname {S}\left (b x\right )^{2} - 2 \, {\left (3 \, \pi b x \operatorname {S}\left (b x\right ) + 2 \, \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{2 \, \pi ^{3} b^{5}} \] Input:

integrate(x^4*fresnel_sin(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="fricas")
 

Output:

-1/2*(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*fresnel_sin(b*x)^2 - 2*(3*pi*b*x*fr 
esnel_sin(b*x) + 2*cos(1/2*pi*b^2*x^2))*sin(1/2*pi*b^2*x^2))/(pi^3*b^5)
 

Sympy [A] (verification not implemented)

Time = 1.33 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.26 \[ \int x^4 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\begin {cases} - \frac {x^{3} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )}{\pi b^{2}} - \frac {x^{2} \sin ^{2}{\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{2 \pi ^{2} b^{3}} - \frac {x^{2} \cos ^{2}{\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{\pi ^{2} b^{3}} + \frac {3 x \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )}{\pi ^{2} b^{4}} + \frac {2 \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{\pi ^{3} b^{5}} - \frac {3 S^{2}\left (b x\right )}{2 \pi ^{2} b^{5}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:

integrate(x**4*fresnels(b*x)*sin(1/2*b**2*pi*x**2),x)
 

Output:

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

Maxima [F]

\[ \int x^4 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int { x^{4} \operatorname {S}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \,d x } \] Input:

integrate(x^4*fresnel_sin(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="maxima")
 

Output:

integrate(x^4*fresnel_sin(b*x)*sin(1/2*pi*b^2*x^2), x)
 

Giac [F]

\[ \int x^4 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int { x^{4} \operatorname {S}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \,d x } \] Input:

integrate(x^4*fresnel_sin(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="giac")
 

Output:

integrate(x^4*fresnel_sin(b*x)*sin(1/2*pi*b^2*x^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int x^4 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int x^4\,\mathrm {FresnelS}\left (b\,x\right )\,\sin \left (\frac {\Pi \,b^2\,x^2}{2}\right ) \,d x \] Input:

int(x^4*FresnelS(b*x)*sin((Pi*b^2*x^2)/2),x)
 

Output:

int(x^4*FresnelS(b*x)*sin((Pi*b^2*x^2)/2), x)
 

Reduce [F]

\[ \int x^4 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int x^{4} \mathrm {FresnelS}\left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )d x \] Input:

int(x^4*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2),x)
 

Output:

int(x^4*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2),x)