\(\int x \operatorname {FresnelC}(b x) \sin (\frac {1}{2} b^2 \pi x^2) \, dx\) [207]

Optimal result

Integrand size = 18, antiderivative size = 60 \[ \int x \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\frac {x}{2 b \pi }-\frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)}{b^2 \pi }+\frac {\operatorname {FresnelC}\left (\sqrt {2} b x\right )}{2 \sqrt {2} b^2 \pi } \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6596, 3439, 3433} \[ \int x \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=-\frac {\operatorname {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\operatorname {FresnelC}\left (\sqrt {2} b x\right )}{2 \sqrt {2} \pi b^2}+\frac {x}{2 \pi b} \]

[In]

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

[Out]

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

Rule 3433

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

Rule 3439

Int[((a_.) + Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)]*(b_.))^(p_), x_Symbol] :> Int[ExpandTrigReduce[(a +
b*Cos[c + d*(e + f*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 1] && IGtQ[n, 1]

Rule 6596

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.80 \[ \int x \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\frac {2 b x-4 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x)+\sqrt {2} \operatorname {FresnelC}\left (\sqrt {2} b x\right )}{4 b^2 \pi } \]

[In]

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

[Out]

(2*b*x - 4*Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x] + Sqrt[2]*FresnelC[Sqrt[2]*b*x])/(4*b^2*Pi)

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.87

method	result	size
default	\(\frac {-\frac {\operatorname {FresnelC}\left (b x \right ) \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b \pi }+\frac {\frac {b x}{2}+\frac {\sqrt {2}\, \operatorname {FresnelC}\left (b x \sqrt {2}\right )}{4}}{b \pi }}{b}\)	\(52\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.87 \[ \int x \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\frac {2 \, b^{2} x - 4 \, b \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {C}\left (b x\right ) + \sqrt {2} \sqrt {b^{2}} \operatorname {C}\left (\sqrt {2} \sqrt {b^{2}} x\right )}{4 \, \pi b^{3}} \]

[In]

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

[Out]

1/4*(2*b^2*x - 4*b*cos(1/2*pi*b^2*x^2)*fresnel_cos(b*x) + sqrt(2)*sqrt(b^2)*fresnel_cos(sqrt(2)*sqrt(b^2)*x))/
(pi*b^3)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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