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

Optimal result

Integrand size = 6, antiderivative size = 49 \[ \int x \operatorname {FresnelS}(b x) \, dx=\frac {x \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{2 b \pi }-\frac {\operatorname {FresnelC}(b x)}{2 b^2 \pi }+\frac {1}{2} x^2 \operatorname {FresnelS}(b x) \]

[Out]

1/2*x*cos(1/2*b^2*Pi*x^2)/b/Pi-1/2*FresnelC(b*x)/b^2/Pi+1/2*x^2*FresnelS(b*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6561, 3466, 3433} \[ \int x \operatorname {FresnelS}(b x) \, dx=-\frac {\operatorname {FresnelC}(b x)}{2 \pi b^2}+\frac {x \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{2 \pi b}+\frac {1}{2} x^2 \operatorname {FresnelS}(b x) \]

[In]

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

[Out]

(x*Cos[(b^2*Pi*x^2)/2])/(2*b*Pi) - FresnelC[b*x]/(2*b^2*Pi) + (x^2*FresnelS[b*x])/2

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[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 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}{2} x^2 \operatorname {FresnelS}(b x)-\frac {1}{2} b \int x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx \\ & = \frac {x \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{2 b \pi }+\frac {1}{2} x^2 \operatorname {FresnelS}(b x)-\frac {\int \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx}{2 b \pi } \\ & = \frac {x \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{2 b \pi }-\frac {\operatorname {FresnelC}(b x)}{2 b^2 \pi }+\frac {1}{2} x^2 \operatorname {FresnelS}(b x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int x \operatorname {FresnelS}(b x) \, dx=\frac {x \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{2 b \pi }-\frac {\operatorname {FresnelC}(b x)}{2 b^2 \pi }+\frac {1}{2} x^2 \operatorname {FresnelS}(b x) \]

[In]

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

[Out]

(x*Cos[(b^2*Pi*x^2)/2])/(2*b*Pi) - FresnelC[b*x]/(2*b^2*Pi) + (x^2*FresnelS[b*x])/2

Maple [C] (verified)

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

Time = 0.43 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.59

method	result	size
meijerg	\(\frac {b^{3} \pi \,x^{5} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {5}{4}\right ], \left [\frac {3}{2}, \frac {7}{4}, \frac {9}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{30}\)	\(29\)
derivativedivides	\(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) b^{2} x^{2}}{2}+\frac {b x \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 \pi }-\frac {\operatorname {FresnelC}\left (b x \right )}{2 \pi }}{b^{2}}\)	\(44\)
default	\(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) b^{2} x^{2}}{2}+\frac {b x \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 \pi }-\frac {\operatorname {FresnelC}\left (b x \right )}{2 \pi }}{b^{2}}\)	\(44\)
parts	\(\frac {x^{2} \operatorname {FresnelS}\left (b x \right )}{2}-\frac {b \left (-\frac {x \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }+\frac {\operatorname {FresnelC}\left (\frac {\sqrt {\pi }\, b^{2} x}{\sqrt {b^{2} \pi }}\right )}{b^{2} \sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{2}\)	\(64\)

[In]

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

[Out]

1/30*b^3*Pi*x^5*hypergeom([3/4,5/4],[3/2,7/4,9/4],-1/16*x^4*Pi^2*b^4)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.04 \[ \int x \operatorname {FresnelS}(b x) \, dx=\frac {\pi b^{3} x^{2} \operatorname {S}\left (b x\right ) + b^{2} x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - \sqrt {b^{2}} \operatorname {C}\left (\sqrt {b^{2}} x\right )}{2 \, \pi b^{3}} \]

[In]

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

[Out]

1/2*(pi*b^3*x^2*fresnel_sin(b*x) + b^2*x*cos(1/2*pi*b^2*x^2) - sqrt(b^2)*fresnel_cos(sqrt(b^2)*x))/(pi*b^3)

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.08 \[ \int x \operatorname {FresnelS}(b x) \, dx=\frac {\pi b^{3} x^{5} \Gamma \left (\frac {3}{4}\right ) \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4} \\ \frac {3}{2}, \frac {7}{4}, \frac {9}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{32 \Gamma \left (\frac {7}{4}\right ) \Gamma \left (\frac {9}{4}\right )} \]

[In]

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

[Out]

pi*b**3*x**5*gamma(3/4)*gamma(5/4)*hyper((3/4, 5/4), (3/2, 7/4, 9/4), -pi**2*b**4*x**4/16)/(32*gamma(7/4)*gamm
a(9/4))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.31 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.43 \[ \int x \operatorname {FresnelS}(b x) \, dx=\frac {1}{2} \, x^{2} \operatorname {S}\left (b x\right ) + \frac {\sqrt {\frac {1}{2}} {\left (4 \, \sqrt {\frac {1}{2}} \pi b x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + \left (i - 1\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {\frac {1}{2} i \, \pi } b x\right ) - \left (i + 1\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {-\frac {1}{2} i \, \pi } b x\right )\right )}}{4 \, \pi ^{2} b^{2}} \]

[In]

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

[Out]

1/2*x^2*fresnel_sin(b*x) + 1/4*sqrt(1/2)*(4*sqrt(1/2)*pi*b*x*cos(1/2*pi*b^2*x^2) + (I - 1)*(1/4)^(1/4)*pi*erf(
sqrt(1/2*I*pi)*b*x) - (I + 1)*(1/4)^(1/4)*pi*erf(sqrt(-1/2*I*pi)*b*x))/(pi^2*b^2)

Giac [F]

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

[In]

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

[Out]

integrate(x*fresnel_sin(b*x), x)

Mupad [F(-1)]

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

[In]

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

[Out]

int(x*FresnelS(b*x), x)