\(\int x^2 \operatorname {FresnelS}(b x) \sin (\frac {1}{2} b^2 \pi x^2) \, dx\) [77]

Optimal result

Integrand size = 20, antiderivative size = 137 \[ \int x^2 \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=-\frac {\cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac {x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b^2 \pi }+\frac {\operatorname {FresnelC}(b x) \operatorname {FresnelS}(b x)}{2 b^3 \pi }-\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right )}{8 b \pi }+\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i b^2 \pi x^2\right )}{8 b \pi } \] Output:

-1/4*cos(b^2*Pi*x^2)/b^3/Pi^2-x*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x)/b^2/Pi+1 
/2*FresnelC(b*x)*FresnelS(b*x)/b^3/Pi-1/8*I*x^2*hypergeom([1, 1],[3/2, 2], 
-1/2*I*b^2*Pi*x^2)/b/Pi+1/8*I*x^2*hypergeom([1, 1],[3/2, 2],1/2*I*b^2*Pi*x 
^2)/b/Pi
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {7008, 3860, 3042, 3118, 7000}

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.

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

\(\Big \downarrow \) 7008

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

\(\Big \downarrow \) 3860

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3118

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

\(\Big \downarrow \) 7000

\(\displaystyle \frac {-\frac {1}{8} i b x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right )+\frac {1}{8} i b x^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i b^2 \pi x^2\right )+\frac {\operatorname {FresnelC}(b x) \operatorname {FresnelS}(b x)}{2 b}}{\pi b^2}-\frac {x \operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {\cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}\)

Input:

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

Output:

-1/4*Cos[b^2*Pi*x^2]/(b^3*Pi^2) - (x*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(b 
^2*Pi) + ((FresnelC[b*x]*FresnelS[b*x])/(2*b) - (I/8)*b*x^2*Hypergeometric 
PFQ[{1, 1}, {3/2, 2}, (-1/2*I)*b^2*Pi*x^2] + (I/8)*b*x^2*HypergeometricPFQ 
[{1, 1}, {3/2, 2}, (I/2)*b^2*Pi*x^2])/(b^2*Pi)
 

Defintions of rubi rules used

rule 3042

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

rule 3118

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

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 7000

Int[Cos[(d_.)*(x_)^2]*FresnelS[(b_.)*(x_)], x_Symbol] :> Simp[FresnelC[b*x] 
*(FresnelS[b*x]/(2*b)), x] + (-Simp[(1/8)*I*b*x^2*HypergeometricPFQ[{1, 1}, 
 {3/2, 2}, (-2^(-1))*I*b^2*Pi*x^2], x] + Simp[(1/8)*I*b*x^2*HypergeometricP 
FQ[{1, 1}, {3/2, 2}, (1/2)*I*b^2*Pi*x^2], x]) /; FreeQ[{b, d}, 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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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