3.2.21 \(\int \frac {\text {FresnelC}(b x)}{x^4} \, dx\) [121]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6562, 3461, 3378, 3380} \begin {gather*} -\frac {b \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{6 x^2}-\frac {1}{12} \pi b^3 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {\text {FresnelC}(b x)}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(b*Cos[(b^2*Pi*x^2)/2])/x^2 - FresnelC[b*x]/(3*x^3) - (b^3*Pi*SinIntegral[(b^2*Pi*x^2)/2])/12

Rule 3378

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 3380

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]

Rule 3461

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 6562

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/2)*b^2*x^2], x], x] /; FreeQ[{b, d, m}, x] && NeQ[m, -1]

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 \text {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 ) \text {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*}

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Mathematica [A]
time = 0.01, size = 52, normalized size = 1.00 \begin {gather*} -\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{6 x^2}-\frac {\text {FresnelC}(b x)}{3 x^3}-\frac {1}{12} b^3 \pi \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(b*Cos[(b^2*Pi*x^2)/2])/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 = 0.32, size = 49, normalized size = 0.94

method	result	size
meijerg	\(-\frac {b \hypergeom \left (\left [-\frac {1}{2}, \frac {1}{4}\right ], \left [\frac {1}{2}, \frac {1}{2}, \frac {5}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{2 x^{2}}\)	\(26\)
derivativedivides	\(b^{3} \left (-\frac {\FresnelC \left (b x \right )}{3 b^{3} x^{3}}-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 b^{2} x^{2}}-\frac {\pi \sinIntegral \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{12}\right )\)	\(49\)
default	\(b^{3} \left (-\frac {\FresnelC \left (b x \right )}{3 b^{3} x^{3}}-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 b^{2} x^{2}}-\frac {\pi \sinIntegral \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{12}\right )\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelC(b*x)/x^4,x,method=_RETURNVERBOSE)

[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 [C] Result contains complex when optimal does not.
time = 0.32, size = 44, normalized size = 0.85 \begin {gather*} -\frac {1}{24} \, {\left (i \, \pi \Gamma \left (-1, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) - i \, \pi \Gamma \left (-1, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{3} - \frac {\operatorname {C}\left (b x\right )}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(I*pi*gamma(-1, 1/2*I*pi*b^2*x^2) - I*pi*gamma(-1, -1/2*I*pi*b^2*x^2))*b^3 - 1/3*fresnel_cos(b*x)/x^3

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Fricas [A]
time = 0.37, size = 44, normalized size = 0.85 \begin {gather*} -\frac {\pi b^{3} x^{3} \operatorname {Si}\left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 2 \, b x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 4 \, \operatorname {C}\left (b x\right )}{12 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(pi*b^3*x^3*sin_integral(1/2*pi*b^2*x^2) + 2*b*x*cos(1/2*pi*b^2*x^2) + 4*fresnel_cos(b*x))/x^3

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Sympy [A]
time = 0.40, size = 42, normalized size = 0.81 \begin {gather*} - \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 {gather*}

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 = 0.00, 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)/x^4,x, algorithm="giac")

[Out]

integrate(fresnel_cos(b*x)/x^4, 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 )}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(FresnelC(b*x)/x^4, x)

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