\(\int x \cos (\frac {1}{2} b^2 \pi x^2) \operatorname {FresnelC}(b x) \, dx\) [187]

Optimal result

Integrand size = 18, antiderivative size = 48 \[ \int x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=-\frac {\operatorname {FresnelS}\left (\sqrt {2} b x\right )}{2 \sqrt {2} b^2 \pi }+\frac {\operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi } \]

[Out]

FresnelC(b*x)*sin(1/2*b^2*Pi*x^2)/b^2/Pi-1/4*FresnelS(b*x*2^(1/2))/b^2/Pi*2^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6588, 3432} \[ \int x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=\frac {\operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {\operatorname {FresnelS}\left (\sqrt {2} b x\right )}{2 \sqrt {2} \pi b^2} \]

[In]

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

[Out]

-1/2*FresnelS[Sqrt[2]*b*x]/(Sqrt[2]*b^2*Pi) + (FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi)

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 6588

Int[Cos[(d_.)*(x_)^2]*FresnelC[(b_.)*(x_)]*(x_), x_Symbol] :> Simp[Sin[d*x^2]*(FresnelC[b*x]/(2*d)), x] - Dist
[b/(4*d), 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}& = \frac {\operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi }-\frac {\int \sin \left (b^2 \pi x^2\right ) \, dx}{2 b \pi } \\ & = -\frac {\operatorname {FresnelS}\left (\sqrt {2} b x\right )}{2 \sqrt {2} b^2 \pi }+\frac {\operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi } \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.92 \[ \int x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=-\frac {\sqrt {2} \operatorname {FresnelS}\left (\sqrt {2} b x\right )-4 \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{4 b^2 \pi } \]

[In]

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

[Out]

-1/4*(Sqrt[2]*FresnelS[Sqrt[2]*b*x] - 4*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi)

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {\frac {\operatorname {FresnelC}\left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b \pi }-\frac {\operatorname {FresnelS}\left (b x \sqrt {2}\right ) \sqrt {2}}{4 b \pi }}{b}\)	\(45\)

[In]

int(x*cos(1/2*b^2*Pi*x^2)*FresnelC(b*x),x,method=_RETURNVERBOSE)

[Out]

(FresnelC(b*x)*sin(1/2*b^2*Pi*x^2)/b/Pi-1/4*FresnelS(b*x*2^(1/2))/b/Pi*2^(1/2))/b

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.98 \[ \int x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=\frac {4 \, b \operatorname {C}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - \sqrt {2} \sqrt {b^{2}} \operatorname {S}\left (\sqrt {2} \sqrt {b^{2}} x\right )}{4 \, \pi b^{3}} \]

[In]

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

[Out]

1/4*(4*b*fresnel_cos(b*x)*sin(1/2*pi*b^2*x^2) - sqrt(2)*sqrt(b^2)*fresnel_sin(sqrt(2)*sqrt(b^2)*x))/(pi*b^3)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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