\(\int \cos (\frac {1}{2} b^2 \pi x^2) \operatorname {FresnelC}(b x) \, dx\) [188]

Optimal result

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

[Out]

1/2*FresnelC(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 = {6576, 30} \[ \int \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=\frac {\operatorname {FresnelC}(b x)^2}{2 b} \]

[In]

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

[Out]

FresnelC[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 6576

Int[Cos[(d_.)*(x_)^2]*FresnelC[(b_.)*(x_)]^(n_.), x_Symbol] :> Dist[Pi*(b/(2*d)), Subst[Int[x^n, x], x, Fresne
lC[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 {FresnelC}(b x))}{b} \\ & = \frac {\operatorname {FresnelC}(b x)^2}{2 b} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=\frac {\operatorname {C}\left (b x\right )^{2}}{2 \, b} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=\begin {cases} \frac {C^{2}\left (b x\right )}{2 b} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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