3.2.27 \(\int \frac {\text {FresnelC}(b x)}{x^{10}} \, dx\) [127]

Optimal. Leaf size=127 \[ -\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{72 x^8}+\frac {b^5 \pi ^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{1728 x^4}+\frac {b^9 \pi ^4 \text {CosIntegral}\left (\frac {1}{2} b^2 \pi x^2\right )}{6912}-\frac {\text {FresnelC}(b x)}{9 x^9}+\frac {b^3 \pi \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{432 x^6}-\frac {b^7 \pi ^3 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3456 x^2} \]

[Out]

1/6912*b^9*Pi^4*Ci(1/2*b^2*Pi*x^2)-1/72*b*cos(1/2*b^2*Pi*x^2)/x^8+1/1728*b^5*Pi^2*cos(1/2*b^2*Pi*x^2)/x^4-1/9*
FresnelC(b*x)/x^9+1/432*b^3*Pi*sin(1/2*b^2*Pi*x^2)/x^6-1/3456*b^7*Pi^3*sin(1/2*b^2*Pi*x^2)/x^2

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Rubi [A]
time = 0.10, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6562, 3461, 3378, 3383} \begin {gather*} -\frac {b \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{72 x^8}+\frac {\pi ^4 b^9 \text {CosIntegral}\left (\frac {1}{2} \pi b^2 x^2\right )}{6912}-\frac {\pi ^3 b^7 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{3456 x^2}+\frac {\pi ^2 b^5 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{1728 x^4}+\frac {\pi b^3 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{432 x^6}-\frac {\text {FresnelC}(b x)}{9 x^9} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[FresnelC[b*x]/x^10,x]

[Out]

-1/72*(b*Cos[(b^2*Pi*x^2)/2])/x^8 + (b^5*Pi^2*Cos[(b^2*Pi*x^2)/2])/(1728*x^4) + (b^9*Pi^4*CosIntegral[(b^2*Pi*
x^2)/2])/6912 - FresnelC[b*x]/(9*x^9) + (b^3*Pi*Sin[(b^2*Pi*x^2)/2])/(432*x^6) - (b^7*Pi^3*Sin[(b^2*Pi*x^2)/2]
)/(3456*x^2)

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 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - 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^{10}} \, dx &=-\frac {C(b x)}{9 x^9}+\frac {1}{9} b \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{x^9} \, dx\\ &=-\frac {C(b x)}{9 x^9}+\frac {1}{18} b \text {Subst}\left (\int \frac {\cos \left (\frac {1}{2} b^2 \pi x\right )}{x^5} \, dx,x,x^2\right )\\ &=-\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{72 x^8}-\frac {C(b x)}{9 x^9}-\frac {1}{144} \left (b^3 \pi \right ) \text {Subst}\left (\int \frac {\sin \left (\frac {1}{2} b^2 \pi x\right )}{x^4} \, dx,x,x^2\right )\\ &=-\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{72 x^8}-\frac {C(b x)}{9 x^9}+\frac {b^3 \pi \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{432 x^6}-\frac {1}{864} \left (b^5 \pi ^2\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {1}{2} b^2 \pi x\right )}{x^3} \, dx,x,x^2\right )\\ &=-\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{72 x^8}+\frac {b^5 \pi ^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{1728 x^4}-\frac {C(b x)}{9 x^9}+\frac {b^3 \pi \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{432 x^6}+\frac {\left (b^7 \pi ^3\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {1}{2} b^2 \pi x\right )}{x^2} \, dx,x,x^2\right )}{3456}\\ &=-\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{72 x^8}+\frac {b^5 \pi ^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{1728 x^4}-\frac {C(b x)}{9 x^9}+\frac {b^3 \pi \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{432 x^6}-\frac {b^7 \pi ^3 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3456 x^2}+\frac {\left (b^9 \pi ^4\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {1}{2} b^2 \pi x\right )}{x} \, dx,x,x^2\right )}{6912}\\ &=-\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{72 x^8}+\frac {b^5 \pi ^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{1728 x^4}+\frac {b^9 \pi ^4 \text {Ci}\left (\frac {1}{2} b^2 \pi x^2\right )}{6912}-\frac {C(b x)}{9 x^9}+\frac {b^3 \pi \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{432 x^6}-\frac {b^7 \pi ^3 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3456 x^2}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 96, normalized size = 0.76 \begin {gather*} \frac {\frac {4 b \left (-24+b^4 \pi ^2 x^4\right ) \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{x^8}+b^9 \pi ^4 \text {CosIntegral}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {768 \text {FresnelC}(b x)}{x^9}-\frac {2 b^3 \pi \left (-8+b^4 \pi ^2 x^4\right ) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6}}{6912} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[FresnelC[b*x]/x^10,x]

[Out]

((4*b*(-24 + b^4*Pi^2*x^4)*Cos[(b^2*Pi*x^2)/2])/x^8 + b^9*Pi^4*CosIntegral[(b^2*Pi*x^2)/2] - (768*FresnelC[b*x
])/x^9 - (2*b^3*Pi*(-8 + b^4*Pi^2*x^4)*Sin[(b^2*Pi*x^2)/2])/x^6)/6912

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Maple [A]
time = 0.42, size = 115, normalized size = 0.91

method	result	size
meijerg	\(\frac {\pi ^{\frac {9}{2}} b^{9} \left (-\frac {\pi ^{\frac {3}{2}} x^{4} b^{4} \hypergeom \left (\left [1, 1, \frac {13}{4}\right ], \left [2, \frac {7}{2}, 4, \frac {17}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{585}+\frac {-\frac {332}{243}+\frac {16 \gamma }{27}-\frac {16 \ln \left (2\right )}{27}+\frac {32 \ln \left (x \right )}{27}+\frac {16 \ln \left (\pi \right )}{27}+\frac {32 \ln \left (b \right )}{27}}{\sqrt {\pi }}-\frac {512}{\pi ^{\frac {9}{2}} x^{8} b^{8}}+\frac {128}{5 \pi ^{\frac {5}{2}} x^{4} b^{4}}\right )}{4096}\)	\(90\)
derivativedivides	\(b^{9} \left (-\frac {\FresnelC \left (b x \right )}{9 b^{9} x^{9}}-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{72 b^{8} x^{8}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 b^{6} x^{6}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 b^{4} x^{4}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b^{2} x^{2}}+\frac {\pi \cosineIntegral \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{4}\right )}{6}\right )}{72}\right )\)	\(115\)
default	\(b^{9} \left (-\frac {\FresnelC \left (b x \right )}{9 b^{9} x^{9}}-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{72 b^{8} x^{8}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 b^{6} x^{6}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 b^{4} x^{4}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b^{2} x^{2}}+\frac {\pi \cosineIntegral \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{4}\right )}{6}\right )}{72}\right )\)	\(115\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelC(b*x)/x^10,x,method=_RETURNVERBOSE)

[Out]

b^9*(-1/9*FresnelC(b*x)/b^9/x^9-1/72/b^8/x^8*cos(1/2*b^2*Pi*x^2)-1/72*Pi*(-1/6/b^6/x^6*sin(1/2*b^2*Pi*x^2)+1/6
*Pi*(-1/4/b^4/x^4*cos(1/2*b^2*Pi*x^2)-1/4*Pi*(-1/2/b^2/x^2*sin(1/2*b^2*Pi*x^2)+1/4*Pi*Ci(1/2*b^2*Pi*x^2)))))

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Maxima [C] Result contains complex when optimal does not.
time = 0.31, size = 46, normalized size = 0.36 \begin {gather*} -\frac {1}{576} \, {\left (\pi ^{4} \Gamma \left (-4, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) + \pi ^{4} \Gamma \left (-4, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{9} - \frac {\operatorname {C}\left (b x\right )}{9 \, x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_cos(b*x)/x^10,x, algorithm="maxima")

[Out]

-1/576*(pi^4*gamma(-4, 1/2*I*pi*b^2*x^2) + pi^4*gamma(-4, -1/2*I*pi*b^2*x^2))*b^9 - 1/9*fresnel_cos(b*x)/x^9

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Fricas [A]
time = 0.36, size = 111, normalized size = 0.87 \begin {gather*} \frac {\pi ^{4} b^{9} x^{9} \operatorname {Ci}\left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + \pi ^{4} b^{9} x^{9} \operatorname {Ci}\left (-\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 8 \, {\left (\pi ^{2} b^{5} x^{5} - 24 \, b x\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 4 \, {\left (\pi ^{3} b^{7} x^{7} - 8 \, \pi b^{3} x^{3}\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 1536 \, \operatorname {C}\left (b x\right )}{13824 \, x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_cos(b*x)/x^10,x, algorithm="fricas")

[Out]

1/13824*(pi^4*b^9*x^9*cos_integral(1/2*pi*b^2*x^2) + pi^4*b^9*x^9*cos_integral(-1/2*pi*b^2*x^2) + 8*(pi^2*b^5*
x^5 - 24*b*x)*cos(1/2*pi*b^2*x^2) - 4*(pi^3*b^7*x^7 - 8*pi*b^3*x^3)*sin(1/2*pi*b^2*x^2) - 1536*fresnel_cos(b*x
))/x^9

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Sympy [A]
time = 7.39, size = 76, normalized size = 0.60 \begin {gather*} - \frac {\pi ^{6} b^{13} x^{4} \Gamma \left (\frac {13}{4}\right ) {{}_{3}F_{4}\left (\begin {matrix} 1, 1, \frac {13}{4} \\ 2, \frac {7}{2}, 4, \frac {17}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{737280 \Gamma \left (\frac {17}{4}\right )} + \frac {\pi ^{4} b^{9} \log {\left (b^{4} x^{4} \right )}}{13824} + \frac {\pi ^{2} b^{5}}{160 x^{4}} - \frac {b}{8 x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnelc(b*x)/x**10,x)

[Out]

-pi**6*b**13*x**4*gamma(13/4)*hyper((1, 1, 13/4), (2, 7/2, 4, 17/4), -pi**2*b**4*x**4/16)/(737280*gamma(17/4))
+ pi**4*b**9*log(b**4*x**4)/13824 + pi**2*b**5/(160*x**4) - b/(8*x**8)

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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^10,x, algorithm="giac")

[Out]

integrate(fresnel_cos(b*x)/x^10, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {FresnelC}\left (b\,x\right )}{x^{10}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelC(b*x)/x^10,x)

[Out]

int(FresnelC(b*x)/x^10, x)

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