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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6598, 6576, 30, 3461, 2714} \[ \int x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\frac {\operatorname {FresnelC}(b x)^2}{2 \pi b^3}-\frac {x \operatorname {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\sin \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}+\frac {x^2}{4 \pi b} \]

[In]

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

[Out]

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)

Rule 30

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

Rule 2714

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 3461

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

Rule 6576

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

Rule 6598

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] + (Dist[(m - 1)/(2*d), Int[x^(m - 2)*Cos[d*x^2]*FresnelC[b*x], x], x] + Dist[b/(2*d), Int[x^(m - 1
)*Cos[d*x^2]^2, 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 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)}{b^2 \pi }+\frac {\int \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx}{b^2 \pi }+\frac {\int x \cos ^2\left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{b \pi } \\ & = -\frac {x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)}{b^2 \pi }+\frac {\text {Subst}(\int x \, dx,x,\operatorname {FresnelC}(b x))}{b^3 \pi }+\frac {\text {Subst}\left (\int \cos ^2\left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{2 b \pi } \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\frac {x^2}{4 b \pi }-\frac {x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)}{b^2 \pi }+\frac {\operatorname {FresnelC}(b x)^2}{2 b^3 \pi }+\frac {\sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2} \]

[In]

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

[Out]

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)

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.91 \[ \int x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\frac {\pi b^{2} x^{2} - 4 \, \pi b x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {C}\left (b x\right ) + 2 \, \pi \operatorname {C}\left (b x\right )^{2} + 2 \, \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{4 \, \pi ^{2} b^{3}} \]

[In]

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

[Out]

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*fresnel_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 = 0.46 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.54 \[ \int x^2 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\begin {cases} \frac {x^{2} \sin ^{2}{\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{4 \pi b} + \frac {x^{2} \cos ^{2}{\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{4 \pi b} - \frac {x \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} C\left (b x\right )}{\pi b^{2}} + \frac {\sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )}}{2 \pi ^{2} b^{3}} + \frac {C^{2}\left (b x\right )}{2 \pi b^{3}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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