3.3.10 \(\int \frac {\text {FresnelC}(b x) \sin (\frac {1}{2} b^2 \pi x^2)}{x^2} \, dx\) [210]

Optimal. Leaf size=48 \[ \frac {1}{2} b \pi \text {FresnelC}(b x)^2-\frac {\text {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x}+\frac {1}{4} b \text {Si}\left (b^2 \pi x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6600, 6576, 30, 3456} \begin {gather*} -\frac {\text {FresnelC}(b x) \sin \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 \text {FresnelC}(b x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6600

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

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Mathematica [A]
time = 0.01, size = 48, normalized size = 1.00 \begin {gather*} \frac {1}{2} b \pi \text {FresnelC}(b x)^2-\frac {\text {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x}+\frac {1}{4} b \text {Si}\left (b^2 \pi x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Pi*FresnelC[b*x]^2)/2 - (FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/x + (b*SinIntegral[b^2*Pi*x^2])/4

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\FresnelC \left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [A]
time = 0.40, size = 45, normalized size = 0.94 \begin {gather*} \frac {2 \, \pi b x \operatorname {C}\left (b x\right )^{2} + b x \operatorname {Si}\left (\pi b^{2} x^{2}\right ) - 4 \, \operatorname {C}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*pi*b*x*fresnel_cos(b*x)^2 + b*x*sin_integral(pi*b^2*x^2) - 4*fresnel_cos(b*x)*sin(1/2*pi*b^2*x^2))/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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