\(\int x^7 \operatorname {FresnelS}(b x) \, dx\) [1]

Optimal result

Integrand size = 8, antiderivative size = 124 \[ \int x^7 \operatorname {FresnelS}(b x) \, dx=-\frac {35 x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{8 b^5 \pi ^3}+\frac {x^7 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{8 b \pi }-\frac {105 \operatorname {FresnelS}(b x)}{8 b^8 \pi ^4}+\frac {1}{8} x^8 \operatorname {FresnelS}(b x)+\frac {105 x \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{8 b^7 \pi ^4}-\frac {7 x^5 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{8 b^3 \pi ^2} \] Output:

-35/8*x^3*cos(1/2*b^2*Pi*x^2)/b^5/Pi^3+1/8*x^7*cos(1/2*b^2*Pi*x^2)/b/Pi-10 
5/8*FresnelS(b*x)/b^8/Pi^4+1/8*x^8*FresnelS(b*x)+105/8*x*sin(1/2*b^2*Pi*x^ 
2)/b^7/Pi^4-7/8*x^5*sin(1/2*b^2*Pi*x^2)/b^3/Pi^2
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.71 \[ \int x^7 \operatorname {FresnelS}(b x) \, dx=\frac {b^3 \pi x^3 \left (-35+b^4 \pi ^2 x^4\right ) \cos \left (\frac {1}{2} b^2 \pi x^2\right )+\left (-105+b^8 \pi ^4 x^8\right ) \operatorname {FresnelS}(b x)-7 b x \left (-15+b^4 \pi ^2 x^4\right ) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{8 b^8 \pi ^4} \] Input:

Integrate[x^7*FresnelS[b*x],x]
 

Output:

(b^3*Pi*x^3*(-35 + b^4*Pi^2*x^4)*Cos[(b^2*Pi*x^2)/2] + (-105 + b^8*Pi^4*x^ 
8)*FresnelS[b*x] - 7*b*x*(-15 + b^4*Pi^2*x^4)*Sin[(b^2*Pi*x^2)/2])/(8*b^8* 
Pi^4)
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6980, 3866, 3867, 3866, 3867, 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.

\(\displaystyle \int x^7 \operatorname {FresnelS}(b x) \, dx\)

\(\Big \downarrow \) 6980

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

\(\Big \downarrow \) 3866

\(\displaystyle \frac {1}{8} x^8 \operatorname {FresnelS}(b x)-\frac {1}{8} b \left (\frac {7 \int x^6 \cos \left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b^2}-\frac {x^7 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\)

\(\Big \downarrow \) 3867

\(\displaystyle \frac {1}{8} x^8 \operatorname {FresnelS}(b x)-\frac {1}{8} b \left (\frac {7 \left (\frac {x^5 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {5 \int x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b^2}\right )}{\pi b^2}-\frac {x^7 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\)

\(\Big \downarrow \) 3866

\(\displaystyle \frac {1}{8} x^8 \operatorname {FresnelS}(b x)-\frac {1}{8} b \left (\frac {7 \left (\frac {x^5 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {5 \left (\frac {3 \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b^2}-\frac {x^3 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )}{\pi b^2}\right )}{\pi b^2}-\frac {x^7 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\)

\(\Big \downarrow \) 3867

\(\displaystyle \frac {1}{8} x^8 \operatorname {FresnelS}(b x)-\frac {1}{8} b \left (\frac {7 \left (\frac {x^5 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {5 \left (\frac {3 \left (\frac {x \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {\int \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b^2}\right )}{\pi b^2}-\frac {x^3 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )}{\pi b^2}\right )}{\pi b^2}-\frac {x^7 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\)

\(\Big \downarrow \) 3832

\(\displaystyle \frac {1}{8} x^8 \operatorname {FresnelS}(b x)-\frac {1}{8} b \left (\frac {7 \left (\frac {x^5 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {5 \left (\frac {3 \left (\frac {x \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {\operatorname {FresnelS}(b x)}{\pi b^3}\right )}{\pi b^2}-\frac {x^3 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )}{\pi b^2}\right )}{\pi b^2}-\frac {x^7 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\)

Input:

Int[x^7*FresnelS[b*x],x]
 

Output:

(x^8*FresnelS[b*x])/8 - (b*(-((x^7*Cos[(b^2*Pi*x^2)/2])/(b^2*Pi)) + (7*((x 
^5*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi) - (5*(-((x^3*Cos[(b^2*Pi*x^2)/2])/(b^2*Pi 
)) + (3*(-(FresnelS[b*x]/(b^3*Pi)) + (x*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi)))/(b 
^2*Pi)))/(b^2*Pi)))/(b^2*Pi)))/8
 

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 3866

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

rule 3867

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

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

Time = 0.53 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.23

method	result	size
meijerg	\(\frac {\pi \,b^{3} x^{11} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {11}{4}\right ], \left [\frac {3}{2}, \frac {7}{4}, \frac {15}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{66}\)	\(29\)
derivativedivides	\(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) b^{8} x^{8}}{8}+\frac {b^{7} x^{7} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{8 \pi }-\frac {7 \left (\frac {b^{5} x^{5} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {5 \left (-\frac {b^{3} x^{3} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {\frac {3 b x \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {3 \,\operatorname {FresnelS}\left (b x \right )}{\pi }}{\pi }\right )}{\pi }\right )}{8 \pi }}{b^{8}}\)	\(123\)
default	\(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) b^{8} x^{8}}{8}+\frac {b^{7} x^{7} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{8 \pi }-\frac {7 \left (\frac {b^{5} x^{5} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {5 \left (-\frac {b^{3} x^{3} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {\frac {3 b x \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {3 \,\operatorname {FresnelS}\left (b x \right )}{\pi }}{\pi }\right )}{\pi }\right )}{8 \pi }}{b^{8}}\)	\(123\)
parts	\(\frac {x^{8} \operatorname {FresnelS}\left (b x \right )}{8}-\frac {b \left (-\frac {x^{7} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }+\frac {\frac {7 x^{5} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }-\frac {35 \left (-\frac {x^{3} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }+\frac {\frac {3 x \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }-\frac {3 \,\operatorname {FresnelS}\left (\frac {\sqrt {\pi }\, b^{2} x}{\sqrt {b^{2} \pi }}\right )}{b^{2} \sqrt {\pi }\, \sqrt {b^{2} \pi }}}{b^{2} \pi }\right )}{b^{2} \pi }}{b^{2} \pi }\right )}{8}\)	\(153\)

Input:

int(x^7*FresnelS(b*x),x,method=_RETURNVERBOSE)
 

Output:

1/66*Pi*b^3*x^11*hypergeom([3/4,11/4],[3/2,7/4,15/4],-1/16*x^4*Pi^2*b^4)
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.68 \[ \int x^7 \operatorname {FresnelS}(b x) \, dx=\frac {{\left (\pi ^{3} b^{7} x^{7} - 35 \, \pi b^{3} x^{3}\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + {\left (\pi ^{4} b^{8} x^{8} - 105\right )} \operatorname {S}\left (b x\right ) - 7 \, {\left (\pi ^{2} b^{5} x^{5} - 15 \, b x\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{8 \, \pi ^{4} b^{8}} \] Input:

integrate(x^7*fresnel_sin(b*x),x, algorithm="fricas")
 

Output:

1/8*((pi^3*b^7*x^7 - 35*pi*b^3*x^3)*cos(1/2*pi*b^2*x^2) + (pi^4*b^8*x^8 - 
105)*fresnel_sin(b*x) - 7*(pi^2*b^5*x^5 - 15*b*x)*sin(1/2*pi*b^2*x^2))/(pi 
^4*b^8)
 

Sympy [A] (verification not implemented)

Time = 1.15 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.48 \[ \int x^7 \operatorname {FresnelS}(b x) \, dx=\frac {231 x^{8} S\left (b x\right ) \Gamma \left (\frac {3}{4}\right )}{512 \Gamma \left (\frac {15}{4}\right )} + \frac {231 x^{7} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{512 \pi b \Gamma \left (\frac {15}{4}\right )} - \frac {1617 x^{5} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{512 \pi ^{2} b^{3} \Gamma \left (\frac {15}{4}\right )} - \frac {8085 x^{3} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{512 \pi ^{3} b^{5} \Gamma \left (\frac {15}{4}\right )} + \frac {24255 x \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{512 \pi ^{4} b^{7} \Gamma \left (\frac {15}{4}\right )} - \frac {24255 S\left (b x\right ) \Gamma \left (\frac {3}{4}\right )}{512 \pi ^{4} b^{8} \Gamma \left (\frac {15}{4}\right )} \] Input:

integrate(x**7*fresnels(b*x),x)
 

Output:

231*x**8*fresnels(b*x)*gamma(3/4)/(512*gamma(15/4)) + 231*x**7*cos(pi*b**2 
*x**2/2)*gamma(3/4)/(512*pi*b*gamma(15/4)) - 1617*x**5*sin(pi*b**2*x**2/2) 
*gamma(3/4)/(512*pi**2*b**3*gamma(15/4)) - 8085*x**3*cos(pi*b**2*x**2/2)*g 
amma(3/4)/(512*pi**3*b**5*gamma(15/4)) + 24255*x*sin(pi*b**2*x**2/2)*gamma 
(3/4)/(512*pi**4*b**7*gamma(15/4)) - 24255*fresnels(b*x)*gamma(3/4)/(512*p 
i**4*b**8*gamma(15/4))
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.02 \[ \int x^7 \operatorname {FresnelS}(b x) \, dx=\frac {1}{8} \, x^{8} \operatorname {S}\left (b x\right ) - \frac {\sqrt {\frac {1}{2}} {\left (\left (105 i + 105\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {\frac {1}{2} i \, \pi } b x\right ) - \left (105 i - 105\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {-\frac {1}{2} i \, \pi } b x\right ) - 4 \, {\left (\sqrt {\frac {1}{2}} \pi ^{4} b^{7} x^{7} - 35 \, \sqrt {\frac {1}{2}} \pi ^{2} b^{3} x^{3}\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 28 \, {\left (\sqrt {\frac {1}{2}} \pi ^{3} b^{5} x^{5} - 15 \, \sqrt {\frac {1}{2}} \pi b x\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )\right )}}{16 \, \pi ^{5} b^{8}} \] Input:

integrate(x^7*fresnel_sin(b*x),x, algorithm="maxima")
 

Output:

1/8*x^8*fresnel_sin(b*x) - 1/16*sqrt(1/2)*((105*I + 105)*(1/4)^(1/4)*pi*er 
f(sqrt(1/2*I*pi)*b*x) - (105*I - 105)*(1/4)^(1/4)*pi*erf(sqrt(-1/2*I*pi)*b 
*x) - 4*(sqrt(1/2)*pi^4*b^7*x^7 - 35*sqrt(1/2)*pi^2*b^3*x^3)*cos(1/2*pi*b^ 
2*x^2) + 28*(sqrt(1/2)*pi^3*b^5*x^5 - 15*sqrt(1/2)*pi*b*x)*sin(1/2*pi*b^2* 
x^2))/(pi^5*b^8)
 

Giac [F]

\[ \int x^7 \operatorname {FresnelS}(b x) \, dx=\int { x^{7} \operatorname {S}\left (b x\right ) \,d x } \] Input:

integrate(x^7*fresnel_sin(b*x),x, algorithm="giac")
 

Output:

integrate(x^7*fresnel_sin(b*x), x)
 

Mupad [F(-1)]

Timed out. \[ \int x^7 \operatorname {FresnelS}(b x) \, dx=\int x^7\,\mathrm {FresnelS}\left (b\,x\right ) \,d x \] Input:

int(x^7*FresnelS(b*x),x)
 

Output:

int(x^7*FresnelS(b*x), x)
 

Reduce [F]

\[ \int x^7 \operatorname {FresnelS}(b x) \, dx=\int x^{7} \mathrm {FresnelS}\left (b x \right )d x \] Input:

int(x^7*FresnelS(b*x),x)
 

Output:

int(x^7*FresnelS(b*x),x)