\(\int \frac {\cos (\frac {1}{2} b^2 \pi x^2) \operatorname {FresnelS}(b x)}{x^2} \, dx\) [101]

Optimal result

Integrand size = 20, antiderivative size = 48 \[ \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x^2} \, dx=-\frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x}-\frac {1}{2} b \pi \operatorname {FresnelS}(b x)^2+\frac {1}{4} b \text {Si}\left (b^2 \pi x^2\right ) \]

[Out]

-cos(1/2*b^2*Pi*x^2)*FresnelS(b*x)/x-1/2*b*Pi*FresnelS(b*x)^2+1/4*b*Si(b^2*Pi*x^2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6599, 6575, 30, 3456} \[ \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x^2} \, dx=-\frac {\operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{x}+\frac {1}{4} b \text {Si}\left (b^2 \pi x^2\right )-\frac {1}{2} \pi b \operatorname {FresnelS}(b x)^2 \]

[In]

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

[Out]

-((Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/x) - (b*Pi*FresnelS[b*x]^2)/2 + (b*SinIntegral[b^2*Pi*x^2])/4

Rule 30

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

Rule 3456

Int[Sin[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[SinIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, 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]

Rule 6599

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x}+\frac {1}{2} b \int \frac {\sin \left (b^2 \pi x^2\right )}{x} \, dx-\left (b^2 \pi \right ) \int \operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx \\ & = -\frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x}+\frac {1}{4} b \text {Si}\left (b^2 \pi x^2\right )-(b \pi ) \text {Subst}(\int x \, dx,x,\operatorname {FresnelS}(b x)) \\ & = -\frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x}-\frac {1}{2} b \pi \operatorname {FresnelS}(b x)^2+\frac {1}{4} b \text {Si}\left (b^2 \pi x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x^2} \, dx=-\frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x}-\frac {1}{2} b \pi \operatorname {FresnelS}(b x)^2+\frac {1}{4} b \text {Si}\left (b^2 \pi x^2\right ) \]

[In]

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

[Out]

-((Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/x) - (b*Pi*FresnelS[b*x]^2)/2 + (b*SinIntegral[b^2*Pi*x^2])/4

Maple [F]

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

[In]

int(cos(1/2*b^2*Pi*x^2)*FresnelS(b*x)/x^2,x)

[Out]

int(cos(1/2*b^2*Pi*x^2)*FresnelS(b*x)/x^2,x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.96 \[ \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x^2} \, dx=-\frac {2 \, \pi b x \operatorname {S}\left (b x\right )^{2} - b x \operatorname {Si}\left (\pi b^{2} x^{2}\right ) + 4 \, \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {S}\left (b x\right )}{4 \, x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(cos(pi*b**2*x**2/2)*fresnels(b*x)/x**2, x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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