\(\int \frac {\sin (\frac {1}{2} b^2 \pi x^2)}{\operatorname {FresnelS}(b x)} \, dx\) [67]

Optimal result

Integrand size = 19, antiderivative size = 9 \[ \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{\operatorname {FresnelS}(b x)} \, dx=\frac {\log (\operatorname {FresnelS}(b x))}{b} \]

[Out]

ln(FresnelS(b*x))/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {6575, 29} \[ \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{\operatorname {FresnelS}(b x)} \, dx=\frac {\log (\operatorname {FresnelS}(b x))}{b} \]

[In]

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

[Out]

Log[FresnelS[b*x]]/b

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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}\left (\int \frac {1}{x} \, dx,x,\operatorname {FresnelS}(b x)\right )}{b} \\ & = \frac {\log (\operatorname {FresnelS}(b x))}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{\operatorname {FresnelS}(b x)} \, dx=\frac {\log (\operatorname {FresnelS}(b x))}{b} \]

[In]

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

[Out]

Log[FresnelS[b*x]]/b

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.11

method	result	size
derivativedivides	\(\frac {\ln \left (\operatorname {FresnelS}\left (b x \right )\right )}{b}\)	\(10\)
default	\(\frac {\ln \left (\operatorname {FresnelS}\left (b x \right )\right )}{b}\)	\(10\)

[In]

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

[Out]

ln(FresnelS(b*x))/b

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{\operatorname {FresnelS}(b x)} \, dx=\frac {\log \left (\operatorname {S}\left (b x\right )\right )}{b} \]

[In]

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

[Out]

log(fresnel_sin(b*x))/b

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.89 \[ \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{\operatorname {FresnelS}(b x)} \, dx=\begin {cases} \frac {\log {\left (S\left (b x\right ) \right )}}{b} & \text {for}\: b \neq 0 \\\text {NaN} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{\operatorname {FresnelS}(b x)} \, dx=\frac {\log \left (\operatorname {S}\left (b x\right )\right )}{b} \]

[In]

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

[Out]

log(fresnel_sin(b*x))/b

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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