\(\int \frac {\operatorname {FresnelS}(b x)}{x^8} \, dx\) [16]

Optimal result

Integrand size = 8, antiderivative size = 102 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^8} \, dx=-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{168 x^4}-\frac {1}{672} b^7 \pi ^3 \operatorname {CosIntegral}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {\operatorname {FresnelS}(b x)}{7 x^7}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{42 x^6}+\frac {b^5 \pi ^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{336 x^2} \] Output:

-1/168*b^3*Pi*cos(1/2*b^2*Pi*x^2)/x^4-1/672*b^7*Pi^3*Ci(1/2*b^2*Pi*x^2)-1/ 
7*FresnelS(b*x)/x^7-1/42*b*sin(1/2*b^2*Pi*x^2)/x^6+1/336*b^5*Pi^2*sin(1/2* 
b^2*Pi*x^2)/x^2
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.83 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^8} \, dx=\frac {1}{672} \left (-\frac {4 b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{x^4}-b^7 \pi ^3 \operatorname {CosIntegral}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {96 \operatorname {FresnelS}(b x)}{x^7}+\frac {2 b \left (-8+b^4 \pi ^2 x^4\right ) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6}\right ) \] Input:

Integrate[FresnelS[b*x]/x^8,x]
 

Output:

((-4*b^3*Pi*Cos[(b^2*Pi*x^2)/2])/x^4 - b^7*Pi^3*CosIntegral[(b^2*Pi*x^2)/2 
] - (96*FresnelS[b*x])/x^7 + (2*b*(-8 + b^4*Pi^2*x^4)*Sin[(b^2*Pi*x^2)/2]) 
/x^6)/672
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.375, Rules used = {6980, 3860, 3042, 3778, 3042, 3778, 25, 3042, 3778, 3042, 3783}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\operatorname {FresnelS}(b x)}{x^8} \, dx\)

\(\Big \downarrow \) 6980

\(\displaystyle \frac {1}{7} b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^7}dx-\frac {\operatorname {FresnelS}(b x)}{7 x^7}\)

\(\Big \downarrow \) 3860

\(\displaystyle \frac {1}{14} b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^8}dx^2-\frac {\operatorname {FresnelS}(b x)}{7 x^7}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{14} b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^8}dx^2-\frac {\operatorname {FresnelS}(b x)}{7 x^7}\)

\(\Big \downarrow \) 3778

\(\displaystyle \frac {1}{14} b \left (\frac {1}{6} \pi b^2 \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6}dx^2-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{3 x^6}\right )-\frac {\operatorname {FresnelS}(b x)}{7 x^7}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{14} b \left (\frac {1}{6} \pi b^2 \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2+\frac {\pi }{2}\right )}{x^6}dx^2-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{3 x^6}\right )-\frac {\operatorname {FresnelS}(b x)}{7 x^7}\)

\(\Big \downarrow \) 3778

\(\displaystyle \frac {1}{14} b \left (\frac {1}{6} \pi b^2 \left (\frac {1}{4} \pi b^2 \int -\frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^4}dx^2-\frac {\cos \left (\frac {1}{2} \pi b^2 x^2\right )}{2 x^4}\right )-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{3 x^6}\right )-\frac {\operatorname {FresnelS}(b x)}{7 x^7}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{14} b \left (\frac {1}{6} \pi b^2 \left (-\frac {1}{4} \pi b^2 \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^4}dx^2-\frac {\cos \left (\frac {1}{2} \pi b^2 x^2\right )}{2 x^4}\right )-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{3 x^6}\right )-\frac {\operatorname {FresnelS}(b x)}{7 x^7}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{14} b \left (\frac {1}{6} \pi b^2 \left (-\frac {1}{4} \pi b^2 \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^4}dx^2-\frac {\cos \left (\frac {1}{2} \pi b^2 x^2\right )}{2 x^4}\right )-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{3 x^6}\right )-\frac {\operatorname {FresnelS}(b x)}{7 x^7}\)

\(\Big \downarrow \) 3778

\(\displaystyle \frac {1}{14} b \left (\frac {1}{6} \pi b^2 \left (-\frac {1}{4} \pi b^2 \left (\frac {1}{2} \pi b^2 \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2}dx^2-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{x^2}\right )-\frac {\cos \left (\frac {1}{2} \pi b^2 x^2\right )}{2 x^4}\right )-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{3 x^6}\right )-\frac {\operatorname {FresnelS}(b x)}{7 x^7}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{14} b \left (\frac {1}{6} \pi b^2 \left (-\frac {1}{4} \pi b^2 \left (\frac {1}{2} \pi b^2 \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2+\frac {\pi }{2}\right )}{x^2}dx^2-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{x^2}\right )-\frac {\cos \left (\frac {1}{2} \pi b^2 x^2\right )}{2 x^4}\right )-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{3 x^6}\right )-\frac {\operatorname {FresnelS}(b x)}{7 x^7}\)

\(\Big \downarrow \) 3783

\(\displaystyle \frac {1}{14} b \left (\frac {1}{6} \pi b^2 \left (-\frac {1}{4} \pi b^2 \left (\frac {1}{2} \pi b^2 \operatorname {CosIntegral}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{x^2}\right )-\frac {\cos \left (\frac {1}{2} \pi b^2 x^2\right )}{2 x^4}\right )-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{3 x^6}\right )-\frac {\operatorname {FresnelS}(b x)}{7 x^7}\)

Input:

Int[FresnelS[b*x]/x^8,x]
 

Output:

-1/7*FresnelS[b*x]/x^7 + (b*(-1/3*Sin[(b^2*Pi*x^2)/2]/x^6 + (b^2*Pi*(-1/2* 
Cos[(b^2*Pi*x^2)/2]/x^4 - (b^2*Pi*((b^2*Pi*CosIntegral[(b^2*Pi*x^2)/2])/2 
- Sin[(b^2*Pi*x^2)/2]/x^2))/4))/6))/14
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3778

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] - Simp[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 3783

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosInte 
gral[e - Pi/2 + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - 
c*f, 0]
 

rule 3860

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(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[Simplify[ 
(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[ 
(m + 1)/n], 0]))
 

rule 6980

Int[FresnelS[(b_.)*(x_)]*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1 
)*(FresnelS[b*x]/(d*(m + 1))), x] - Simp[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]
 

Maple [C] (verified)

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

Time = 0.58 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.77

method	result	size
meijerg	\(\frac {\pi ^{\frac {7}{2}} b^{7} \left (-\frac {128}{3 \pi ^{\frac {5}{2}} x^{4} b^{4}}-\frac {16 \left (-\frac {89}{21}+2 \gamma -2 \ln \left (2\right )+4 \ln \left (x \right )+2 \ln \left (\pi \right )+4 \ln \left (b \right )\right )}{21 \sqrt {\pi }}+\frac {\pi ^{\frac {3}{2}} x^{4} b^{4} \operatorname {hypergeom}\left (\left [1, 1, \frac {11}{4}\right ], \left [2, 3, \frac {7}{2}, \frac {15}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{165}\right )}{1024}\)	\(79\)
parts	\(-\frac {\operatorname {FresnelS}\left (b x \right )}{7 x^{7}}+\frac {b \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 x^{6}}+\frac {b^{2} \pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 x^{4}}-\frac {b^{2} \pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 x^{2}}+\frac {b^{2} \pi \,\operatorname {Ci}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{4}\right )}{6}\right )}{7}\)	\(90\)
derivativedivides	\(b^{7} \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{7 b^{7} x^{7}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{42 b^{6} x^{6}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 b^{4} x^{4}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b^{2} x^{2}}+\frac {\pi \,\operatorname {Ci}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{4}\right )}{42}\right )\)	\(93\)
default	\(b^{7} \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{7 b^{7} x^{7}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{42 b^{6} x^{6}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 b^{4} x^{4}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b^{2} x^{2}}+\frac {\pi \,\operatorname {Ci}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{4}\right )}{42}\right )\)	\(93\)

Input:

int(FresnelS(b*x)/x^8,x,method=_RETURNVERBOSE)
 

Output:

1/1024*Pi^(7/2)*b^7*(-128/3/Pi^(5/2)/x^4/b^4-16/21*(-89/21+2*gamma-2*ln(2) 
+4*ln(x)+2*ln(Pi)+4*ln(b))/Pi^(1/2)+1/165*Pi^(3/2)*x^4*b^4*hypergeom([1,1, 
11/4],[2,3,7/2,15/4],-1/16*x^4*Pi^2*b^4))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.76 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^8} \, dx=-\frac {\pi ^{3} b^{7} x^{7} \operatorname {Ci}\left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 4 \, \pi b^{3} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 2 \, {\left (\pi ^{2} b^{5} x^{5} - 8 \, b x\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 96 \, \operatorname {S}\left (b x\right )}{672 \, x^{7}} \] Input:

integrate(fresnel_sin(b*x)/x^8,x, algorithm="fricas")
 

Output:

-1/672*(pi^3*b^7*x^7*cos_integral(1/2*pi*b^2*x^2) + 4*pi*b^3*x^3*cos(1/2*p 
i*b^2*x^2) - 2*(pi^2*b^5*x^5 - 8*b*x)*sin(1/2*pi*b^2*x^2) + 96*fresnel_sin 
(b*x))/x^7
 

Sympy [A] (verification not implemented)

Time = 2.75 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.67 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^8} \, dx=\frac {\pi ^{5} b^{11} x^{4} \Gamma \left (\frac {11}{4}\right ) {{}_{3}F_{4}\left (\begin {matrix} 1, 1, \frac {11}{4} \\ 2, 3, \frac {7}{2}, \frac {15}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{61440 \Gamma \left (\frac {15}{4}\right )} - \frac {\pi ^{3} b^{7} \log {\left (b^{4} x^{4} \right )}}{1344} - \frac {\pi b^{3}}{24 x^{4}} \] Input:

integrate(fresnels(b*x)/x**8,x)
 

Output:

pi**5*b**11*x**4*gamma(11/4)*hyper((1, 1, 11/4), (2, 3, 7/2, 15/4), -pi**2 
*b**4*x**4/16)/(61440*gamma(15/4)) - pi**3*b**7*log(b**4*x**4)/1344 - pi*b 
**3/(24*x**4)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.45 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^8} \, dx=-\frac {1}{224} \, {\left (\pi ^{3} \Gamma \left (-3, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) + \pi ^{3} \Gamma \left (-3, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{7} - \frac {\operatorname {S}\left (b x\right )}{7 \, x^{7}} \] Input:

integrate(fresnel_sin(b*x)/x^8,x, algorithm="maxima")
 

Output:

-1/224*(pi^3*gamma(-3, 1/2*I*pi*b^2*x^2) + pi^3*gamma(-3, -1/2*I*pi*b^2*x^ 
2))*b^7 - 1/7*fresnel_sin(b*x)/x^7
 

Giac [F]

\[ \int \frac {\operatorname {FresnelS}(b x)}{x^8} \, dx=\int { \frac {\operatorname {S}\left (b x\right )}{x^{8}} \,d x } \] Input:

integrate(fresnel_sin(b*x)/x^8,x, algorithm="giac")
 

Output:

integrate(fresnel_sin(b*x)/x^8, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\operatorname {FresnelS}(b x)}{x^8} \, dx=\int \frac {\mathrm {FresnelS}\left (b\,x\right )}{x^8} \,d x \] Input:

int(FresnelS(b*x)/x^8,x)
 

Output:

int(FresnelS(b*x)/x^8, x)
 

Reduce [F]

\[ \int \frac {\operatorname {FresnelS}(b x)}{x^8} \, dx=\int \frac {\mathrm {FresnelS}\left (b x \right )}{x^{8}}d x \] Input:

int(FresnelS(b*x)/x^8,x)
 

Output:

int(FresnelS(b*x)/x^8,x)