3.121 \(\int \frac{\text{FresnelC}(b x)}{x^4} \, dx\)

Optimal. Leaf size=52 \[ -\frac{1}{12} \pi b^3 \text{Si}\left (\frac{1}{2} b^2 \pi x^2\right )-\frac{b \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{6 x^2}-\frac{\text{FresnelC}(b x)}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.0636539, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6427, 3380, 3297, 3299} \[ -\frac{1}{12} \pi b^3 \text{Si}\left (\frac{1}{2} b^2 \pi x^2\right )-\frac{b \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{6 x^2}-\frac{\text{FresnelC}(b x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[FresnelC[b*x]/x^4,x]

[Out]

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

Rule 6427

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

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

Mathematica [A]  time = 0.0152894, size = 52, normalized size = 1. \[ -\frac{1}{12} \pi b^3 \text{Si}\left (\frac{1}{2} b^2 \pi x^2\right )-\frac{b \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{6 x^2}-\frac{\text{FresnelC}(b x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[FresnelC[b*x]/x^4,x]

[Out]

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

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Maple [A]  time = 0.047, size = 49, normalized size = 0.9 \begin{align*}{b}^{3} \left ( -{\frac{{\it FresnelC} \left ( bx \right ) }{3\,{x}^{3}{b}^{3}}}-{\frac{1}{6\,{b}^{2}{x}^{2}}\cos \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) }-{\frac{\pi }{12}{\it Si} \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelC(b*x)/x^4,x)

[Out]

b^3*(-1/3*FresnelC(b*x)/b^3/x^3-1/6/b^2/x^2*cos(1/2*b^2*Pi*x^2)-1/12*Pi*Si(1/2*b^2*Pi*x^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnelc(b*x)/x^4,x, algorithm="maxima")

[Out]

integrate(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{{\rm fresnelc}\left (b x\right )}{x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnelc(b*x)/x^4,x, algorithm="fricas")

[Out]

integral(fresnelc(b*x)/x^4, x)

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Sympy [A]  time = 0.734969, size = 42, normalized size = 0.81 \begin{align*} - \frac{b \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{3}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{1}{2}, \frac{1}{2}, \frac{5}{4} \end{matrix}\middle |{- \frac{\pi ^{2} b^{4} x^{4}}{16}} \right )}}{8 x^{2} \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnelc(b*x)/x**4,x)

[Out]

-b*gamma(1/4)*hyper((-1/2, 1/4), (1/2, 1/2, 5/4), -pi**2*b**4*x**4/16)/(8*x**2*gamma(5/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnelc(b*x)/x^4,x, algorithm="giac")

[Out]

integrate(fresnelc(b*x)/x^4, x)