\(\int \frac {\operatorname {FresnelS}(b x)}{x^{10}} \, dx\) [18]

Optimal result

Integrand size = 8, antiderivative size = 127 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^{10}} \, dx=-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{432 x^6}+\frac {b^7 \pi ^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{3456 x^2}-\frac {\operatorname {FresnelS}(b x)}{9 x^9}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{72 x^8}+\frac {b^5 \pi ^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{1728 x^4}+\frac {b^9 \pi ^4 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right )}{6912} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6561, 3460, 3378, 3380} \[ \int \frac {\operatorname {FresnelS}(b x)}{x^{10}} \, dx=-\frac {b \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{72 x^8}+\frac {\pi ^4 b^9 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right )}{6912}+\frac {\pi ^3 b^7 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{3456 x^2}+\frac {\pi ^2 b^5 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{1728 x^4}-\frac {\pi b^3 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{432 x^6}-\frac {\operatorname {FresnelS}(b x)}{9 x^9} \]

[In]

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

[Out]

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

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 3460

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[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 6561

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\operatorname {FresnelS}(b x)}{9 x^9}+\frac {1}{9} b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^9} \, dx \\ & = -\frac {\operatorname {FresnelS}(b x)}{9 x^9}+\frac {1}{18} b \text {Subst}\left (\int \frac {\sin \left (\frac {1}{2} b^2 \pi x\right )}{x^5} \, dx,x,x^2\right ) \\ & = -\frac {\operatorname {FresnelS}(b x)}{9 x^9}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{72 x^8}+\frac {1}{144} \left (b^3 \pi \right ) \text {Subst}\left (\int \frac {\cos \left (\frac {1}{2} b^2 \pi x\right )}{x^4} \, dx,x,x^2\right ) \\ & = -\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{432 x^6}-\frac {\operatorname {FresnelS}(b x)}{9 x^9}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{72 x^8}-\frac {1}{864} \left (b^5 \pi ^2\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {1}{2} b^2 \pi x\right )}{x^3} \, dx,x,x^2\right ) \\ & = -\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{432 x^6}-\frac {\operatorname {FresnelS}(b x)}{9 x^9}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{72 x^8}+\frac {b^5 \pi ^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{1728 x^4}-\frac {\left (b^7 \pi ^3\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {1}{2} b^2 \pi x\right )}{x^2} \, dx,x,x^2\right )}{3456} \\ & = -\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{432 x^6}+\frac {b^7 \pi ^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{3456 x^2}-\frac {\operatorname {FresnelS}(b x)}{9 x^9}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{72 x^8}+\frac {b^5 \pi ^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{1728 x^4}+\frac {\left (b^9 \pi ^4\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {1}{2} b^2 \pi x\right )}{x} \, dx,x,x^2\right )}{6912} \\ & = -\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{432 x^6}+\frac {b^7 \pi ^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{3456 x^2}-\frac {\operatorname {FresnelS}(b x)}{9 x^9}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{72 x^8}+\frac {b^5 \pi ^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{1728 x^4}+\frac {b^9 \pi ^4 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right )}{6912} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.76 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^{10}} \, dx=\frac {\frac {2 b^3 \pi \left (-8+b^4 \pi ^2 x^4\right ) \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6}-\frac {768 \operatorname {FresnelS}(b x)}{x^9}+\frac {4 b \left (-24+b^4 \pi ^2 x^4\right ) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^8}+b^9 \pi ^4 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right )}{6912} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 0.45 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.23

method	result	size
meijerg	\(-\frac {\pi \,b^{3} \operatorname {hypergeom}\left (\left [-\frac {3}{2}, \frac {3}{4}\right ], \left [-\frac {1}{2}, \frac {3}{2}, \frac {7}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{36 x^{6}}\)	\(29\)
parts	\(-\frac {\operatorname {FresnelS}\left (b x \right )}{9 x^{9}}+\frac {b \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{8 x^{8}}+\frac {b^{2} \pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 x^{6}}-\frac {b^{2} \pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 x^{4}}+\frac {b^{2} \pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 x^{2}}-\frac {b^{2} \pi \,\operatorname {Si}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{4}\right )}{6}\right )}{8}\right )}{9}\)	\(112\)
derivativedivides	\(b^{9} \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{9 b^{9} x^{9}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{72 b^{8} x^{8}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 b^{6} x^{6}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 b^{4} x^{4}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b^{2} x^{2}}-\frac {\pi \,\operatorname {Si}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{4}\right )}{6}\right )}{72}\right )\)	\(115\)
default	\(b^{9} \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{9 b^{9} x^{9}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{72 b^{8} x^{8}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 b^{6} x^{6}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 b^{4} x^{4}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b^{2} x^{2}}-\frac {\pi \,\operatorname {Si}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{4}\right )}{6}\right )}{72}\right )\)	\(115\)

[In]

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

[Out]

-1/36*Pi*b^3/x^6*hypergeom([-3/2,3/4],[-1/2,3/2,7/4],-1/16*x^4*Pi^2*b^4)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.72 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^{10}} \, dx=\frac {\pi ^{4} b^{9} x^{9} \operatorname {Si}\left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 2 \, {\left (\pi ^{3} b^{7} x^{7} - 8 \, \pi b^{3} x^{3}\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 4 \, {\left (\pi ^{2} b^{5} x^{5} - 24 \, b x\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 768 \, \operatorname {S}\left (b x\right )}{6912 \, x^{9}} \]

[In]

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

[Out]

1/6912*(pi^4*b^9*x^9*sin_integral(1/2*pi*b^2*x^2) + 2*(pi^3*b^7*x^7 - 8*pi*b^3*x^3)*cos(1/2*pi*b^2*x^2) + 4*(p
i^2*b^5*x^5 - 24*b*x)*sin(1/2*pi*b^2*x^2) - 768*fresnel_sin(b*x))/x^9

Sympy [A] (verification not implemented)

Time = 1.58 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.38 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^{10}} \, dx=- \frac {\pi b^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} - \frac {3}{2}, \frac {3}{4} \\ - \frac {1}{2}, \frac {3}{2}, \frac {7}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{48 x^{6} \Gamma \left (\frac {7}{4}\right )} \]

[In]

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

[Out]

-pi*b**3*gamma(3/4)*hyper((-3/2, 3/4), (-1/2, 3/2, 7/4), -pi**2*b**4*x**4/16)/(48*x**6*gamma(7/4))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.38 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^{10}} \, dx=-\frac {1}{576} \, {\left (i \, \pi ^{4} \Gamma \left (-4, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) - i \, \pi ^{4} \Gamma \left (-4, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{9} - \frac {\operatorname {S}\left (b x\right )}{9 \, x^{9}} \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\operatorname {FresnelS}(b x)}{x^{10}} \, dx=\int { \frac {\operatorname {S}\left (b x\right )}{x^{10}} \,d x } \]

[In]

integrate(fresnel_sin(b*x)/x^10,x, algorithm="giac")

[Out]

integrate(fresnel_sin(b*x)/x^10, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\operatorname {FresnelS}(b x)}{x^{10}} \, dx=\int \frac {\mathrm {FresnelS}\left (b\,x\right )}{x^{10}} \,d x \]

[In]

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

[Out]

int(FresnelS(b*x)/x^10, x)