\(\int x^2 \operatorname {FresnelS}(b x) \, dx\) [6]

Optimal result

Integrand size = 8, antiderivative size = 59 \[ \int x^2 \operatorname {FresnelS}(b x) \, dx=\frac {x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b \pi }+\frac {1}{3} x^3 \operatorname {FresnelS}(b x)-\frac {2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6561, 3460, 3377, 2717} \[ \int x^2 \operatorname {FresnelS}(b x) \, dx=\frac {x^2 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{3 \pi b}-\frac {2 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{3 \pi ^2 b^3}+\frac {1}{3} x^3 \operatorname {FresnelS}(b x) \]

[In]

Int[x^2*FresnelS[b*x],x]

[Out]

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

Rule 2717

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

Rule 3377

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

Rule 3460

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

Rule 6561

Int[FresnelS[(b_.)*(x_)]*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*(FresnelS[b*x]/(d*(m + 1))), x] -
Dist[b/(d*(m + 1)), Int[(d*x)^(m + 1)*Sin[(Pi/2)*b^2*x^2], x], x] /; FreeQ[{b, d, m}, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} x^3 \operatorname {FresnelS}(b x)-\frac {1}{3} b \int x^3 \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx \\ & = \frac {1}{3} x^3 \operatorname {FresnelS}(b x)-\frac {1}{6} b \text {Subst}\left (\int x \sin \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right ) \\ & = \frac {x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b \pi }+\frac {1}{3} x^3 \operatorname {FresnelS}(b x)-\frac {\text {Subst}\left (\int \cos \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{3 b \pi } \\ & = \frac {x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b \pi }+\frac {1}{3} x^3 \operatorname {FresnelS}(b x)-\frac {2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00 \[ \int x^2 \operatorname {FresnelS}(b x) \, dx=\frac {x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b \pi }+\frac {1}{3} x^3 \operatorname {FresnelS}(b x)-\frac {2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2} \]

[In]

Integrate[x^2*FresnelS[b*x],x]

[Out]

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

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.92

method	result	size
derivativedivides	\(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) b^{3} x^{3}}{3}+\frac {b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi }-\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi ^{2}}}{b^{3}}\)	\(54\)
default	\(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) b^{3} x^{3}}{3}+\frac {b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi }-\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi ^{2}}}{b^{3}}\)	\(54\)
parts	\(\frac {x^{3} \operatorname {FresnelS}\left (b x \right )}{3}-\frac {b \left (-\frac {x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }+\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{4} \pi ^{2}}\right )}{3}\)	\(54\)
meijerg	\(\frac {\frac {\sqrt {\pi }\, x^{2} b^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3}-\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \sqrt {\pi }}+\frac {\pi ^{\frac {3}{2}} x^{3} b^{3} \operatorname {FresnelS}\left (b x \right )}{3}}{b^{3} \pi ^{\frac {3}{2}}}\)	\(60\)

[In]

int(x^2*FresnelS(b*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90 \[ \int x^2 \operatorname {FresnelS}(b x) \, dx=\frac {\pi ^{2} b^{3} x^{3} \operatorname {S}\left (b x\right ) + \pi b^{2} x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 2 \, \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{3 \, \pi ^{2} b^{3}} \]

[In]

integrate(x^2*fresnel_sin(b*x),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.36 \[ \int x^2 \operatorname {FresnelS}(b x) \, dx=\frac {x^{3} S\left (b x\right ) \Gamma \left (\frac {3}{4}\right )}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {x^{2} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{4 \pi b \Gamma \left (\frac {7}{4}\right )} - \frac {\sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{2 \pi ^{2} b^{3} \Gamma \left (\frac {7}{4}\right )} \]

[In]

integrate(x**2*fresnels(b*x),x)

[Out]

x**3*fresnels(b*x)*gamma(3/4)/(4*gamma(7/4)) + x**2*cos(pi*b**2*x**2/2)*gamma(3/4)/(4*pi*b*gamma(7/4)) - sin(p
i*b**2*x**2/2)*gamma(3/4)/(2*pi**2*b**3*gamma(7/4))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.83 \[ \int x^2 \operatorname {FresnelS}(b x) \, dx=\frac {1}{3} \, x^{3} \operatorname {S}\left (b x\right ) + \frac {\pi b^{2} x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 2 \, \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{3 \, \pi ^{2} b^{3}} \]

[In]

integrate(x^2*fresnel_sin(b*x),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(x^2*fresnel_sin(b*x),x, algorithm="giac")

[Out]

integrate(x^2*fresnel_sin(b*x), x)

Mupad [F(-1)]

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

[In]

int(x^2*FresnelS(b*x),x)

[Out]

int(x^2*FresnelS(b*x), x)