\(\int \operatorname {FresnelS}(b x) \sin (\frac {1}{2} b^2 \pi x^2) \, dx\) [79]

Optimal result

Integrand size = 17, antiderivative size = 13 \[ \int \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\frac {\operatorname {FresnelS}(b x)^2}{2 b} \]

[Out]

1/2*FresnelS(b*x)^2/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6575, 30} \[ \int \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\frac {\operatorname {FresnelS}(b x)^2}{2 b} \]

[In]

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

[Out]

FresnelS[b*x]^2/(2*b)

Rule 30

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

Rule 6575

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}(\int x \, dx,x,\operatorname {FresnelS}(b x))}{b} \\ & = \frac {\operatorname {FresnelS}(b x)^2}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\frac {\operatorname {FresnelS}(b x)^2}{2 b} \]

[In]

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

[Out]

FresnelS[b*x]^2/(2*b)

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92

method	result	size
derivativedivides	\(\frac {\operatorname {FresnelS}\left (b x \right )^{2}}{2 b}\)	\(12\)
default	\(\frac {\operatorname {FresnelS}\left (b x \right )^{2}}{2 b}\)	\(12\)

[In]

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

[Out]

1/2*FresnelS(b*x)^2/b

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\frac {\operatorname {S}\left (b x\right )^{2}}{2 \, b} \]

[In]

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

[Out]

1/2*fresnel_sin(b*x)^2/b

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\begin {cases} \frac {S^{2}\left (b x\right )}{2 b} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((fresnels(b*x)**2/(2*b), Ne(b, 0)), (0, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\frac {\operatorname {S}\left (b x\right )^{2}}{2 \, b} \]

[In]

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

[Out]

1/2*fresnel_sin(b*x)^2/b

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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