\(\int x^3 \operatorname {FresnelS}(b x) \, dx\) [5]

Optimal result

Integrand size = 8, antiderivative size = 74 \[ \int x^3 \operatorname {FresnelS}(b x) \, dx=\frac {x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{4 b \pi }+\frac {3 \operatorname {FresnelS}(b x)}{4 b^4 \pi ^2}+\frac {1}{4} x^4 \operatorname {FresnelS}(b x)-\frac {3 x \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{4 b^3 \pi ^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6561, 3466, 3467, 3432} \[ \int x^3 \operatorname {FresnelS}(b x) \, dx=\frac {3 \operatorname {FresnelS}(b x)}{4 \pi ^2 b^4}+\frac {x^3 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{4 \pi b}-\frac {3 x \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{4 \pi ^2 b^3}+\frac {1}{4} x^4 \operatorname {FresnelS}(b x) \]

[In]

Int[x^3*FresnelS[b*x],x]

[Out]

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

Rule 3432

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 3466

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] + Dist[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 3467

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] - Dist[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 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 {1}{4} x^4 \operatorname {FresnelS}(b x)-\frac {1}{4} b \int x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx \\ & = \frac {x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{4 b \pi }+\frac {1}{4} x^4 \operatorname {FresnelS}(b x)-\frac {3 \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{4 b \pi } \\ & = \frac {x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{4 b \pi }+\frac {1}{4} x^4 \operatorname {FresnelS}(b x)-\frac {3 x \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{4 b^3 \pi ^2}+\frac {3 \int \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{4 b^3 \pi ^2} \\ & = \frac {x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{4 b \pi }+\frac {3 \operatorname {FresnelS}(b x)}{4 b^4 \pi ^2}+\frac {1}{4} x^4 \operatorname {FresnelS}(b x)-\frac {3 x \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{4 b^3 \pi ^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int x^3 \operatorname {FresnelS}(b x) \, dx=\frac {x^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{4 b \pi }+\frac {3 \operatorname {FresnelS}(b x)}{4 b^4 \pi ^2}+\frac {1}{4} x^4 \operatorname {FresnelS}(b x)-\frac {3 x \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{4 b^3 \pi ^2} \]

[In]

Integrate[x^3*FresnelS[b*x],x]

[Out]

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

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.84

method	result	size
meijerg	\(\frac {\frac {\pi \,x^{3} b^{3} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}-\frac {3 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right ) b x}{4}+\frac {\left (21 x^{4} \pi ^{2} b^{4}+63\right ) \operatorname {FresnelS}\left (b x \right )}{84}}{\pi ^{2} b^{4}}\)	\(62\)
derivativedivides	\(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) b^{4} x^{4}}{4}+\frac {b^{3} x^{3} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 \pi }-\frac {3 \left (\frac {b x \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {\operatorname {FresnelS}\left (b x \right )}{\pi }\right )}{4 \pi }}{b^{4}}\)	\(70\)
default	\(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) b^{4} x^{4}}{4}+\frac {b^{3} x^{3} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 \pi }-\frac {3 \left (\frac {b x \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {\operatorname {FresnelS}\left (b x \right )}{\pi }\right )}{4 \pi }}{b^{4}}\)	\(70\)
parts	\(\frac {x^{4} \operatorname {FresnelS}\left (b x \right )}{4}-\frac {b \left (-\frac {x^{3} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }+\frac {\frac {3 x \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }-\frac {3 \,\operatorname {FresnelS}\left (\frac {\sqrt {\pi }\, b^{2} x}{\sqrt {b^{2} \pi }}\right )}{b^{2} \sqrt {\pi }\, \sqrt {b^{2} \pi }}}{b^{2} \pi }\right )}{4}\)	\(94\)

[In]

int(x^3*FresnelS(b*x),x,method=_RETURNVERBOSE)

[Out]

2/Pi^2/b^4*(1/8*Pi*x^3*b^3*cos(1/2*b^2*Pi*x^2)-3/8*sin(1/2*b^2*Pi*x^2)*b*x+1/168*(21*Pi^2*b^4*x^4+63)*FresnelS
(b*x))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.78 \[ \int x^3 \operatorname {FresnelS}(b x) \, dx=\frac {\pi b^{3} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 3 \, b x \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + {\left (\pi ^{2} b^{4} x^{4} + 3\right )} \operatorname {S}\left (b x\right )}{4 \, \pi ^{2} b^{4}} \]

[In]

integrate(x^3*fresnel_sin(b*x),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.51 \[ \int x^3 \operatorname {FresnelS}(b x) \, dx=\frac {21 x^{4} S\left (b x\right ) \Gamma \left (\frac {3}{4}\right )}{64 \Gamma \left (\frac {11}{4}\right )} + \frac {21 x^{3} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{64 \pi b \Gamma \left (\frac {11}{4}\right )} - \frac {63 x \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{64 \pi ^{2} b^{3} \Gamma \left (\frac {11}{4}\right )} + \frac {63 S\left (b x\right ) \Gamma \left (\frac {3}{4}\right )}{64 \pi ^{2} b^{4} \Gamma \left (\frac {11}{4}\right )} \]

[In]

integrate(x**3*fresnels(b*x),x)

[Out]

21*x**4*fresnels(b*x)*gamma(3/4)/(64*gamma(11/4)) + 21*x**3*cos(pi*b**2*x**2/2)*gamma(3/4)/(64*pi*b*gamma(11/4
)) - 63*x*sin(pi*b**2*x**2/2)*gamma(3/4)/(64*pi**2*b**3*gamma(11/4)) + 63*fresnels(b*x)*gamma(3/4)/(64*pi**2*b
**4*gamma(11/4))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.27 \[ \int x^3 \operatorname {FresnelS}(b x) \, dx=\frac {1}{4} \, x^{4} \operatorname {S}\left (b x\right ) + \frac {\sqrt {\frac {1}{2}} {\left (4 \, \sqrt {\frac {1}{2}} \pi ^{2} b^{3} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 12 \, \sqrt {\frac {1}{2}} \pi b x \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + \left (3 i + 3\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {\frac {1}{2} i \, \pi } b x\right ) - \left (3 i - 3\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {-\frac {1}{2} i \, \pi } b x\right )\right )}}{8 \, \pi ^{3} b^{4}} \]

[In]

integrate(x^3*fresnel_sin(b*x),x, algorithm="maxima")

[Out]

1/4*x^4*fresnel_sin(b*x) + 1/8*sqrt(1/2)*(4*sqrt(1/2)*pi^2*b^3*x^3*cos(1/2*pi*b^2*x^2) - 12*sqrt(1/2)*pi*b*x*s
in(1/2*pi*b^2*x^2) + (3*I + 3)*(1/4)^(1/4)*pi*erf(sqrt(1/2*I*pi)*b*x) - (3*I - 3)*(1/4)^(1/4)*pi*erf(sqrt(-1/2
*I*pi)*b*x))/(pi^3*b^4)

Giac [F]

\[ \int x^3 \operatorname {FresnelS}(b x) \, dx=\int { x^{3} \operatorname {S}\left (b x\right ) \,d x } \]

[In]

integrate(x^3*fresnel_sin(b*x),x, algorithm="giac")

[Out]

integrate(x^3*fresnel_sin(b*x), x)

Mupad [F(-1)]

Timed out. \[ \int x^3 \operatorname {FresnelS}(b x) \, dx=\int x^3\,\mathrm {FresnelS}\left (b\,x\right ) \,d x \]

[In]

int(x^3*FresnelS(b*x),x)

[Out]

int(x^3*FresnelS(b*x), x)