3.192 \(\int \frac{\cos (\frac{1}{2} b^2 \pi x^2) \text{FresnelC}(b x)}{x^4} \, dx\)

Optimal. Leaf size=109 \[ \frac{\pi b^2 \text{FresnelC}(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{3 x}-\frac{\text{FresnelC}(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{3 x^3}-\frac{1}{6} \pi ^2 b^3 \text{FresnelC}(b x)^2-\frac{1}{6} \pi b^3 \text{Si}\left (b^2 \pi x^2\right )-\frac{b \cos \left (\pi b^2 x^2\right )}{12 x^2}-\frac{b}{12 x^2} \]

[Out]

-b/(12*x^2) - (b*Cos[b^2*Pi*x^2])/(12*x^2) - (Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x])/(3*x^3) - (b^3*Pi^2*FresnelC[
b*x]^2)/6 + (b^2*Pi*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(3*x) - (b^3*Pi*SinIntegral[b^2*Pi*x^2])/6

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Rubi [A]  time = 0.115158, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6457, 6465, 6441, 30, 3375, 3380, 3297, 3299} \[ \frac{\pi b^2 \text{FresnelC}(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{3 x}-\frac{\text{FresnelC}(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{3 x^3}-\frac{1}{6} \pi ^2 b^3 \text{FresnelC}(b x)^2-\frac{1}{6} \pi b^3 \text{Si}\left (b^2 \pi x^2\right )-\frac{b \cos \left (\pi b^2 x^2\right )}{12 x^2}-\frac{b}{12 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x])/x^4,x]

[Out]

-b/(12*x^2) - (b*Cos[b^2*Pi*x^2])/(12*x^2) - (Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x])/(3*x^3) - (b^3*Pi^2*FresnelC[
b*x]^2)/6 + (b^2*Pi*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(3*x) - (b^3*Pi*SinIntegral[b^2*Pi*x^2])/6

Rule 6457

Int[Cos[(d_.)*(x_)^2]*FresnelC[(b_.)*(x_)]*(x_)^(m_), x_Symbol] :> Simp[(x^(m + 1)*Cos[d*x^2]*FresnelC[b*x])/(
m + 1), x] + (Dist[(2*d)/(m + 1), Int[x^(m + 2)*Sin[d*x^2]*FresnelC[b*x], x], x] - Dist[b/(2*(m + 1)), Int[x^(
m + 1)*Cos[2*d*x^2], x], x] - Simp[(b*x^(m + 2))/(2*(m + 1)*(m + 2)), x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^
2*b^4)/4] && ILtQ[m, -2]

Rule 6465

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*b^4)/4] && ILtQ[m, -1]

Rule 6441

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*b^4)/4]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 3375

Int[Sin[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[SinIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, x]

Rule 3380

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rubi steps

\begin{align*} \int \frac{\cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x)}{x^4} \, dx &=-\frac{b}{12 x^2}-\frac{\cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x)}{3 x^3}+\frac{1}{6} b \int \frac{\cos \left (b^2 \pi x^2\right )}{x^3} \, dx-\frac{1}{3} \left (b^2 \pi \right ) \int \frac{C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{x^2} \, dx\\ &=-\frac{b}{12 x^2}-\frac{\cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x)}{3 x^3}+\frac{b^2 \pi C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{3 x}+\frac{1}{12} b \operatorname{Subst}\left (\int \frac{\cos \left (b^2 \pi x\right )}{x^2} \, dx,x,x^2\right )-\frac{1}{6} \left (b^3 \pi \right ) \int \frac{\sin \left (b^2 \pi x^2\right )}{x} \, dx-\frac{1}{3} \left (b^4 \pi ^2\right ) \int \cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x) \, dx\\ &=-\frac{b}{12 x^2}-\frac{b \cos \left (b^2 \pi x^2\right )}{12 x^2}-\frac{\cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x)}{3 x^3}+\frac{b^2 \pi C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{3 x}-\frac{1}{12} b^3 \pi \text{Si}\left (b^2 \pi x^2\right )-\frac{1}{12} \left (b^3 \pi \right ) \operatorname{Subst}\left (\int \frac{\sin \left (b^2 \pi x\right )}{x} \, dx,x,x^2\right )-\frac{1}{3} \left (b^3 \pi ^2\right ) \operatorname{Subst}(\int x \, dx,x,C(b x))\\ &=-\frac{b}{12 x^2}-\frac{b \cos \left (b^2 \pi x^2\right )}{12 x^2}-\frac{\cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x)}{3 x^3}-\frac{1}{6} b^3 \pi ^2 C(b x)^2+\frac{b^2 \pi C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{3 x}-\frac{1}{6} b^3 \pi \text{Si}\left (b^2 \pi x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0069185, size = 109, normalized size = 1. \[ \frac{\pi b^2 \text{FresnelC}(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{3 x}-\frac{\text{FresnelC}(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{3 x^3}-\frac{1}{6} \pi ^2 b^3 \text{FresnelC}(b x)^2-\frac{1}{6} \pi b^3 \text{Si}\left (b^2 \pi x^2\right )-\frac{b \cos \left (\pi b^2 x^2\right )}{12 x^2}-\frac{b}{12 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x])/x^4,x]

[Out]

-b/(12*x^2) - (b*Cos[b^2*Pi*x^2])/(12*x^2) - (Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x])/(3*x^3) - (b^3*Pi^2*FresnelC[
b*x]^2)/6 + (b^2*Pi*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(3*x) - (b^3*Pi*SinIntegral[b^2*Pi*x^2])/6

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Maple [F]  time = 0.057, size = 0, normalized size = 0. \begin{align*} \int{\frac{{\it FresnelC} \left ( bx \right ) }{{x}^{4}}\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(1/2*b^2*Pi*x^2)*FresnelC(b*x)/x^4,x)

[Out]

int(cos(1/2*b^2*Pi*x^2)*FresnelC(b*x)/x^4,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (\frac{1}{2} \, \pi b^{2} x^{2}\right ){\rm fresnelc}\left (b x\right )}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1/2*b^2*pi*x^2)*fresnelc(b*x)/x^4,x, algorithm="maxima")

[Out]

integrate(cos(1/2*pi*b^2*x^2)*fresnelc(b*x)/x^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\cos \left (\frac{1}{2} \, \pi b^{2} x^{2}\right ){\rm fresnelc}\left (b x\right )}{x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1/2*b^2*pi*x^2)*fresnelc(b*x)/x^4,x, algorithm="fricas")

[Out]

integral(cos(1/2*pi*b^2*x^2)*fresnelc(b*x)/x^4, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos{\left (\frac{\pi b^{2} x^{2}}{2} \right )} C\left (b x\right )}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1/2*b**2*pi*x**2)*fresnelc(b*x)/x**4,x)

[Out]

Integral(cos(pi*b**2*x**2/2)*fresnelc(b*x)/x**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (\frac{1}{2} \, \pi b^{2} x^{2}\right ){\rm fresnelc}\left (b x\right )}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1/2*b^2*pi*x^2)*fresnelc(b*x)/x^4,x, algorithm="giac")

[Out]

integrate(cos(1/2*pi*b^2*x^2)*fresnelc(b*x)/x^4, x)