\(\int \operatorname {FresnelS}(b x)^2 \, dx\) [38]

Optimal result

Integrand size = 6, antiderivative size = 55 \[ \int \operatorname {FresnelS}(b x)^2 \, dx=\frac {2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b \pi }+x \operatorname {FresnelS}(b x)^2-\frac {\operatorname {FresnelS}\left (\sqrt {2} b x\right )}{\sqrt {2} b \pi } \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {6555, 12, 6587, 3432} \[ \int \operatorname {FresnelS}(b x)^2 \, dx=\frac {2 \operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b}+x \operatorname {FresnelS}(b x)^2-\frac {\operatorname {FresnelS}\left (\sqrt {2} b x\right )}{\sqrt {2} \pi b} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 6555

Int[FresnelS[(a_.) + (b_.)*(x_)]^2, x_Symbol] :> Simp[(a + b*x)*(FresnelS[a + b*x]^2/b), x] - Dist[2, Int[(a +
b*x)*Sin[(Pi/2)*(a + b*x)^2]*FresnelS[a + b*x], x], x] /; FreeQ[{a, b}, x]

Rule 6587

Int[FresnelS[(b_.)*(x_)]*(x_)*Sin[(d_.)*(x_)^2], x_Symbol] :> Simp[(-Cos[d*x^2])*(FresnelS[b*x]/(2*d)), x] + D
ist[1/(2*b*Pi), Int[Sin[2*d*x^2], x], x] /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2/4)*b^4]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \operatorname {FresnelS}(b x)^2 \, dx=\frac {2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{b \pi }+x \operatorname {FresnelS}(b x)^2-\frac {\operatorname {FresnelS}\left (\sqrt {2} b x\right )}{\sqrt {2} b \pi } \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {\operatorname {FresnelS}\left (b x \right )^{2} b x +\frac {2 \,\operatorname {FresnelS}\left (b x \right ) \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {\sqrt {2}\, \operatorname {FresnelS}\left (b x \sqrt {2}\right )}{2 \pi }}{b}\)	\(49\)
default	\(\frac {\operatorname {FresnelS}\left (b x \right )^{2} b x +\frac {2 \,\operatorname {FresnelS}\left (b x \right ) \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {\sqrt {2}\, \operatorname {FresnelS}\left (b x \sqrt {2}\right )}{2 \pi }}{b}\)	\(49\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.09 \[ \int \operatorname {FresnelS}(b x)^2 \, dx=\frac {2 \, \pi b^{2} x \operatorname {S}\left (b x\right )^{2} + 4 \, b \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {S}\left (b x\right ) - \sqrt {2} \sqrt {b^{2}} \operatorname {S}\left (\sqrt {2} \sqrt {b^{2}} x\right )}{2 \, \pi b^{2}} \]

[In]

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

[Out]

1/2*(2*pi*b^2*x*fresnel_sin(b*x)^2 + 4*b*cos(1/2*pi*b^2*x^2)*fresnel_sin(b*x) - sqrt(2)*sqrt(b^2)*fresnel_sin(
sqrt(2)*sqrt(b^2)*x))/(pi*b^2)

Sympy [F]

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

[In]

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

[Out]

Integral(fresnels(b*x)**2, x)

Maxima [F]

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

[In]

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

[Out]

integrate(fresnel_sin(b*x)^2, x)

Giac [F]

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

[In]

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

[Out]

integrate(fresnel_sin(b*x)^2, x)

Mupad [F(-1)]

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

[In]

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

[Out]

int(FresnelS(b*x)^2, x)