3.2.47 \(\int \text {FresnelC}(b x)^2 \, dx\) [147]

Optimal. Leaf size=54 \[ x \text {FresnelC}(b x)^2+\frac {S\left (\sqrt {2} b x\right )}{\sqrt {2} b \pi }-\frac {2 \text {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b \pi } \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {6556, 12, 6588, 3432} \begin {gather*} -\frac {2 \text {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b}+x \text {FresnelC}(b x)^2+\frac {S\left (\sqrt {2} b x\right )}{\sqrt {2} \pi b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[FresnelC[b*x]^2,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 6556

Int[FresnelC[(a_.) + (b_.)*(x_)]^2, x_Symbol] :> Simp[(a + b*x)*(FresnelC[a + b*x]^2/b), x] - Dist[2, Int[(a +
 b*x)*Cos[(Pi/2)*(a + b*x)^2]*FresnelC[a + b*x], x], x] /; FreeQ[{a, b}, 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*} \int C(b x)^2 \, dx &=x C(b x)^2-2 \int b x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x) \, dx\\ &=x C(b x)^2-(2 b) \int x \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x) \, dx\\ &=x C(b x)^2-\frac {2 C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b \pi }+\frac {\int \sin \left (b^2 \pi x^2\right ) \, dx}{\pi }\\ &=x C(b x)^2+\frac {S\left (\sqrt {2} b x\right )}{\sqrt {2} b \pi }-\frac {2 C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b \pi }\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.01, size = 54, normalized size = 1.00 \begin {gather*} x \text {FresnelC}(b x)^2+\frac {S\left (\sqrt {2} b x\right )}{\sqrt {2} b \pi }-\frac {2 \text {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b \pi } \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[FresnelC[b*x]^2,x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.40, size = 49, normalized size = 0.91

method result size
derivativedivides \(\frac {\FresnelC \left (b x \right )^{2} b x -\frac {2 \FresnelC \left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {\sqrt {2}\, \mathrm {S}\left (b x \sqrt {2}\right )}{2 \pi }}{b}\) \(49\)
default \(\frac {\FresnelC \left (b x \right )^{2} b x -\frac {2 \FresnelC \left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {\sqrt {2}\, \mathrm {S}\left (b x \sqrt {2}\right )}{2 \pi }}{b}\) \(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelC(b*x)^2,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_cos(b*x)^2,x, algorithm="maxima")

[Out]

integrate(fresnel_cos(b*x)^2, x)

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 59, normalized size = 1.09 \begin {gather*} \frac {2 \, \pi b^{2} x \operatorname {C}\left (b x\right )^{2} - 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 )}{2 \, \pi b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_cos(b*x)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int C^{2}\left (b x\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnelc(b*x)**2,x)

[Out]

Integral(fresnelc(b*x)**2, x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_cos(b*x)^2,x, algorithm="giac")

[Out]

integrate(fresnel_cos(b*x)^2, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\mathrm {FresnelC}\left (b\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelC(b*x)^2,x)

[Out]

int(FresnelC(b*x)^2, x)

________________________________________________________________________________________