\(\int \frac {\operatorname {FresnelS}(b x)}{x^5} \, dx\) [13]

Optimal result

Integrand size = 8, antiderivative size = 69 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^5} \, dx=-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x}-\frac {1}{12} b^4 \pi ^2 \operatorname {FresnelS}(b x)-\frac {\operatorname {FresnelS}(b x)}{4 x^4}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x^3} \] Output:

-1/12*b^3*Pi*cos(1/2*b^2*Pi*x^2)/x-1/12*b^4*Pi^2*FresnelS(b*x)-1/4*Fresnel 
S(b*x)/x^4-1/12*b*sin(1/2*b^2*Pi*x^2)/x^3
 

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^5} \, dx=-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x}-\frac {1}{12} b^4 \pi ^2 \operatorname {FresnelS}(b x)-\frac {\operatorname {FresnelS}(b x)}{4 x^4}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x^3} \] Input:

Integrate[FresnelS[b*x]/x^5,x]
 

Output:

-1/12*(b^3*Pi*Cos[(b^2*Pi*x^2)/2])/x - (b^4*Pi^2*FresnelS[b*x])/12 - Fresn 
elS[b*x]/(4*x^4) - (b*Sin[(b^2*Pi*x^2)/2])/(12*x^3)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6980, 3868, 3869, 3832}

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.

\(\displaystyle \int \frac {\operatorname {FresnelS}(b x)}{x^5} \, dx\)

\(\Big \downarrow \) 6980

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

\(\Big \downarrow \) 3868

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

\(\Big \downarrow \) 3869

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

\(\Big \downarrow \) 3832

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

Input:

Int[FresnelS[b*x]/x^5,x]
 

Output:

-1/4*FresnelS[b*x]/x^4 + (b*((b^2*Pi*(-(Cos[(b^2*Pi*x^2)/2]/x) - b*Pi*Fres 
nelS[b*x]))/3 - Sin[(b^2*Pi*x^2)/2]/(3*x^3)))/4
 

Defintions of rubi rules used

rule 3832

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ 
d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
 

rule 3868

Int[((e_.)*(x_))^(m_)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[(e*x) 
^(m + 1)*(Sin[c + d*x^n]/(e*(m + 1))), x] - Simp[d*(n/(e^n*(m + 1)))   Int[ 
(e*x)^(m + n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n, 0] & 
& LtQ[m, -1]
 

rule 3869

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_), x_Symbol] :> Simp[(e*x) 
^(m + 1)*(Cos[c + d*x^n]/(e*(m + 1))), x] + Simp[d*(n/(e^n*(m + 1)))   Int[ 
(e*x)^(m + n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n, 0] & 
& LtQ[m, -1]
 

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 [A] (verified)

Time = 0.45 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(b^{4} \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{4 b^{4} x^{4}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{12 b^{3} x^{3}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b x}-\pi \,\operatorname {FresnelS}\left (b x \right )\right )}{12}\right )\)	\(65\)
default	\(b^{4} \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{4 b^{4} x^{4}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{12 b^{3} x^{3}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b x}-\pi \,\operatorname {FresnelS}\left (b x \right )\right )}{12}\right )\)	\(65\)
meijerg	\(\frac {\pi ^{2} b^{4} \left (-\frac {32 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi x b}-\frac {32 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi ^{2} x^{3} b^{3}}-\frac {32 \left (x^{4} \pi ^{2} b^{4}+3\right ) \operatorname {FresnelS}\left (b x \right )}{3 \pi ^{2} x^{4} b^{4}}\right )}{128}\)	\(79\)
parts	\(-\frac {\operatorname {FresnelS}\left (b x \right )}{4 x^{4}}+\frac {b \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 x^{3}}+\frac {b^{2} \pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{x}-\frac {b^{2} \pi ^{\frac {3}{2}} \operatorname {FresnelS}\left (\frac {\sqrt {\pi }\, b^{2} x}{\sqrt {b^{2} \pi }}\right )}{\sqrt {b^{2} \pi }}\right )}{3}\right )}{4}\)	\(83\)

Input:

int(FresnelS(b*x)/x^5,x,method=_RETURNVERBOSE)
 

Output:

b^4*(-1/4*FresnelS(b*x)/b^4/x^4-1/12/b^3/x^3*sin(1/2*b^2*Pi*x^2)+1/12*Pi*( 
-1/b/x*cos(1/2*b^2*Pi*x^2)-Pi*FresnelS(b*x)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.78 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^5} \, dx=-\frac {\pi b^{3} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + b x \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + {\left (\pi ^{2} b^{4} x^{4} + 3\right )} \operatorname {S}\left (b x\right )}{12 \, x^{4}} \] Input:

integrate(fresnel_sin(b*x)/x^5,x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.67 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.59 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^5} \, dx=\frac {\pi ^{2} b^{4} S\left (b x\right ) \Gamma \left (- \frac {1}{4}\right )}{64 \Gamma \left (\frac {7}{4}\right )} + \frac {\pi b^{3} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {1}{4}\right )}{64 x \Gamma \left (\frac {7}{4}\right )} + \frac {b \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {1}{4}\right )}{64 x^{3} \Gamma \left (\frac {7}{4}\right )} + \frac {3 S\left (b x\right ) \Gamma \left (- \frac {1}{4}\right )}{64 x^{4} \Gamma \left (\frac {7}{4}\right )} \] Input:

integrate(fresnels(b*x)/x**5,x)
 

Output:

pi**2*b**4*fresnels(b*x)*gamma(-1/4)/(64*gamma(7/4)) + pi*b**3*cos(pi*b**2 
*x**2/2)*gamma(-1/4)/(64*x*gamma(7/4)) + b*sin(pi*b**2*x**2/2)*gamma(-1/4) 
/(64*x**3*gamma(7/4)) + 3*fresnels(b*x)*gamma(-1/4)/(64*x**4*gamma(7/4))
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.18 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.88 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^5} \, dx=-\frac {\sqrt {\frac {1}{2}} \left (\pi x^{2}\right )^{\frac {3}{2}} {\left (-\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) + \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{4}}{64 \, x^{3}} - \frac {\operatorname {S}\left (b x\right )}{4 \, x^{4}} \] Input:

integrate(fresnel_sin(b*x)/x^5,x, algorithm="maxima")
 

Output:

-1/64*sqrt(1/2)*(pi*x^2)^(3/2)*(-(I + 1)*sqrt(2)*gamma(-3/2, 1/2*I*pi*b^2* 
x^2) + (I - 1)*sqrt(2)*gamma(-3/2, -1/2*I*pi*b^2*x^2))*b^4/x^3 - 1/4*fresn 
el_sin(b*x)/x^4
 

Giac [F]

\[ \int \frac {\operatorname {FresnelS}(b x)}{x^5} \, dx=\int { \frac {\operatorname {S}\left (b x\right )}{x^{5}} \,d x } \] Input:

integrate(fresnel_sin(b*x)/x^5,x, algorithm="giac")
 

Output:

integrate(fresnel_sin(b*x)/x^5, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\operatorname {FresnelS}(b x)}{x^5} \, dx=\int \frac {\mathrm {FresnelS}\left (b\,x\right )}{x^5} \,d x \] Input:

int(FresnelS(b*x)/x^5,x)
 

Output:

int(FresnelS(b*x)/x^5, x)
 

Reduce [F]

\[ \int \frac {\operatorname {FresnelS}(b x)}{x^5} \, dx=\int \frac {\mathrm {FresnelS}\left (b x \right )}{x^{5}}d x \] Input:

int(FresnelS(b*x)/x^5,x)
 

Output:

int(FresnelS(b*x)/x^5,x)