\(\int x^4 \operatorname {FresnelS}(b x) \, dx\) [4]

Optimal result

Integrand size = 8, antiderivative size = 84 \[ \int x^4 \operatorname {FresnelS}(b x) \, dx=-\frac {8 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^5 \pi ^3}+\frac {x^4 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac {1}{5} x^5 \operatorname {FresnelS}(b x)-\frac {4 x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.85 \[ \int x^4 \operatorname {FresnelS}(b x) \, dx=\frac {\left (-8+b^4 \pi ^2 x^4\right ) \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^5 \pi ^3}+\frac {1}{5} x^5 \operatorname {FresnelS}(b x)-\frac {4 x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2} \] Input:

Integrate[x^4*FresnelS[b*x],x]
 

Output:

((-8 + b^4*Pi^2*x^4)*Cos[(b^2*Pi*x^2)/2])/(5*b^5*Pi^3) + (x^5*FresnelS[b*x 
])/5 - (4*x^2*Sin[(b^2*Pi*x^2)/2])/(5*b^3*Pi^2)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.125, Rules used = {6980, 3860, 3042, 3777, 3042, 3777, 25, 3042, 3118}

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.

\(\displaystyle \int x^4 \operatorname {FresnelS}(b x) \, dx\)

\(\Big \downarrow \) 6980

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

\(\Big \downarrow \) 3860

\(\displaystyle \frac {1}{5} x^5 \operatorname {FresnelS}(b x)-\frac {1}{10} b \int x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx^2\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} x^5 \operatorname {FresnelS}(b x)-\frac {1}{10} b \int x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx^2\)

\(\Big \downarrow \) 3777

\(\displaystyle \frac {1}{5} x^5 \operatorname {FresnelS}(b x)-\frac {1}{10} b \left (\frac {4 \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )dx^2}{\pi b^2}-\frac {2 x^4 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} x^5 \operatorname {FresnelS}(b x)-\frac {1}{10} b \left (\frac {4 \int x^2 \sin \left (\frac {1}{2} b^2 \pi x^2+\frac {\pi }{2}\right )dx^2}{\pi b^2}-\frac {2 x^4 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\)

\(\Big \downarrow \) 3777

\(\displaystyle \frac {1}{5} x^5 \operatorname {FresnelS}(b x)-\frac {1}{10} b \left (\frac {4 \left (\frac {2 \int -\sin \left (\frac {1}{2} b^2 \pi x^2\right )dx^2}{\pi b^2}+\frac {2 x^2 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )}{\pi b^2}-\frac {2 x^4 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{5} x^5 \operatorname {FresnelS}(b x)-\frac {1}{10} b \left (\frac {4 \left (\frac {2 x^2 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {2 \int \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx^2}{\pi b^2}\right )}{\pi b^2}-\frac {2 x^4 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} x^5 \operatorname {FresnelS}(b x)-\frac {1}{10} b \left (\frac {4 \left (\frac {2 x^2 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {2 \int \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx^2}{\pi b^2}\right )}{\pi b^2}-\frac {2 x^4 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\)

\(\Big \downarrow \) 3118

\(\displaystyle \frac {1}{5} x^5 \operatorname {FresnelS}(b x)-\frac {1}{10} b \left (\frac {4 \left (\frac {2 x^2 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {4 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}\right )}{\pi b^2}-\frac {2 x^4 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\)

Input:

Int[x^4*FresnelS[b*x],x]
 

Output:

(x^5*FresnelS[b*x])/5 - (b*((-2*x^4*Cos[(b^2*Pi*x^2)/2])/(b^2*Pi) + (4*((4 
*Cos[(b^2*Pi*x^2)/2])/(b^4*Pi^2) + (2*x^2*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi)))/ 
(b^2*Pi)))/10
 

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 3118

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ 
[{c, d}, x]
 

rule 3777

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( 
-(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1)*C 
os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 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 = 0.47 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.35

method	result	size
meijerg	\(\frac {\pi \,b^{3} x^{8} \operatorname {hypergeom}\left (\left [\frac {3}{4}, 2\right ], \left [\frac {3}{2}, \frac {7}{4}, 3\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{48}\)	\(29\)
derivativedivides	\(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) b^{5} x^{5}}{5}+\frac {b^{4} x^{4} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi }-\frac {4 \left (\frac {b^{2} x^{2} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {2 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}\right )}{5 \pi }}{b^{5}}\)	\(80\)
default	\(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) b^{5} x^{5}}{5}+\frac {b^{4} x^{4} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi }-\frac {4 \left (\frac {b^{2} x^{2} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {2 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}\right )}{5 \pi }}{b^{5}}\)	\(80\)
parts	\(\frac {x^{5} \operatorname {FresnelS}\left (b x \right )}{5}-\frac {b \left (-\frac {x^{4} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }+\frac {\frac {4 x^{2} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }+\frac {8 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{4} \pi ^{2}}}{b^{2} \pi }\right )}{5}\)	\(83\)

Input:

int(x^4*FresnelS(b*x),x,method=_RETURNVERBOSE)
 

Output:

1/48*Pi*b^3*x^8*hypergeom([3/4,2],[3/2,7/4,3],-1/16*x^4*Pi^2*b^4)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.77 \[ \int x^4 \operatorname {FresnelS}(b x) \, dx=\frac {\pi ^{3} b^{5} x^{5} \operatorname {S}\left (b x\right ) - 4 \, \pi b^{2} x^{2} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + {\left (\pi ^{2} b^{4} x^{4} - 8\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{5 \, \pi ^{3} b^{5}} \] Input:

integrate(x^4*fresnel_sin(b*x),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.95 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.44 \[ \int x^4 \operatorname {FresnelS}(b x) \, dx=\frac {3 x^{5} S\left (b x\right ) \Gamma \left (\frac {3}{4}\right )}{20 \Gamma \left (\frac {7}{4}\right )} + \frac {3 x^{4} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{20 \pi b \Gamma \left (\frac {7}{4}\right )} - \frac {3 x^{2} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{5 \pi ^{2} b^{3} \Gamma \left (\frac {7}{4}\right )} - \frac {6 \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{5 \pi ^{3} b^{5} \Gamma \left (\frac {7}{4}\right )} \] Input:

integrate(x**4*fresnels(b*x),x)
 

Output:

3*x**5*fresnels(b*x)*gamma(3/4)/(20*gamma(7/4)) + 3*x**4*cos(pi*b**2*x**2/ 
2)*gamma(3/4)/(20*pi*b*gamma(7/4)) - 3*x**2*sin(pi*b**2*x**2/2)*gamma(3/4) 
/(5*pi**2*b**3*gamma(7/4)) - 6*cos(pi*b**2*x**2/2)*gamma(3/4)/(5*pi**3*b** 
5*gamma(7/4))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.74 \[ \int x^4 \operatorname {FresnelS}(b x) \, dx=\frac {1}{5} \, x^{5} \operatorname {S}\left (b x\right ) - \frac {4 \, \pi b^{2} x^{2} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - {\left (\pi ^{2} b^{4} x^{4} - 8\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{5 \, \pi ^{3} b^{5}} \] Input:

integrate(x^4*fresnel_sin(b*x),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

1/5*x^5*fresnel_sin(b*x) - 1/5*(4*pi*b^2*x^2*sin(1/2*pi*b^2*x^2) - (pi^2*b 
^4*x^4 - 8)*cos(1/2*pi*b^2*x^2))/(pi^3*b^5)
 

Giac [F]

\[ \int x^4 \operatorname {FresnelS}(b x) \, dx=\int { x^{4} \operatorname {S}\left (b x\right ) \,d x } \] Input:

integrate(x^4*fresnel_sin(b*x),x, algorithm="giac")
 

Output:

integrate(x^4*fresnel_sin(b*x), x)
 

Mupad [F(-1)]

Timed out. \[ \int x^4 \operatorname {FresnelS}(b x) \, dx=\int x^4\,\mathrm {FresnelS}\left (b\,x\right ) \,d x \] Input:

int(x^4*FresnelS(b*x),x)
 

Output:

int(x^4*FresnelS(b*x), x)
 

Reduce [F]

\[ \int x^4 \operatorname {FresnelS}(b x) \, dx=\int x^{4} \mathrm {FresnelS}\left (b x \right )d x \] Input:

int(x^4*FresnelS(b*x),x)
 

Output:

int(x^4*FresnelS(b*x),x)