\(\int x^2 \operatorname {FresnelC}(b x) \sin (\frac {1}{2} b^2 \pi x^2) \, dx\) [206]

Optimal result

Integrand size = 20, antiderivative size = 74 \[ \int x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\frac {x^2}{4 b \pi }-\frac {x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)}{b^2 \pi }+\frac {\operatorname {FresnelC}(b x)^2}{2 b^3 \pi }+\frac {\sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\frac {x^2}{4 b \pi }-\frac {x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)}{b^2 \pi }+\frac {\operatorname {FresnelC}(b x)^2}{2 b^3 \pi }+\frac {\sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {7017, 3861, 3042, 3114, 6995, 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.

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

\(\Big \downarrow \) 7017

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

\(\Big \downarrow \) 3861

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3114

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

\(\Big \downarrow \) 6995

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

\(\Big \downarrow \) 15

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

Input:

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

Output:

-((x*Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x])/(b^2*Pi)) + FresnelC[b*x]^2/(2*b^3 
*Pi) + (x^2/2 + Sin[b^2*Pi*x^2]/(2*b^2*Pi))/(2*b*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 3861

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Cos[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 6995

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

rule 7017

Int[FresnelC[(b_.)*(x_)]*(x_)^(m_)*Sin[(d_.)*(x_)^2], x_Symbol] :> Simp[(-x 
^(m - 1))*Cos[d*x^2]*(FresnelC[b*x]/(2*d)), x] + (Simp[(m - 1)/(2*d)   Int[ 
x^(m - 2)*Cos[d*x^2]*FresnelC[b*x], x], x] + Simp[b/(2*d)   Int[x^(m - 1)*C 
os[d*x^2]^2, x], x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2/4)*b^4] && IGtQ[ 
m, 1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.91 \[ \int x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\frac {\pi b^{2} x^{2} - 4 \, \pi b x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {C}\left (b x\right ) + 2 \, \pi \operatorname {C}\left (b x\right )^{2} + 2 \, \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{4 \, \pi ^{2} b^{3}} \] Input:

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

Output:

1/4*(pi*b^2*x^2 - 4*pi*b*x*cos(1/2*pi*b^2*x^2)*fresnel_cos(b*x) + 2*pi*fre 
snel_cos(b*x)^2 + 2*cos(1/2*pi*b^2*x^2)*sin(1/2*pi*b^2*x^2))/(pi^2*b^3)
 

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.54 \[ \int x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\begin {cases} \frac {x^{2} \sin ^{2}{\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{4 \pi b} + \frac {x^{2} \cos ^{2}{\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{4 \pi b} - \frac {x \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} C\left (b x\right )}{\pi b^{2}} + \frac {\sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{2 \pi ^{2} b^{3}} + \frac {C^{2}\left (b x\right )}{2 \pi b^{3}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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