3.1.8 \(\int S(b x) \, dx\) [8]

Optimal. Leaf size=26 \[ \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{b \pi }+x S(b x) \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6553} \begin {gather*} \frac {\cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b}+x S(b x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[FresnelS[b*x],x]

[Out]

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

Rule 6553

Int[FresnelS[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[(a + b*x)*(FresnelS[a + b*x]/b), x] + Simp[Cos[(Pi/2)*(a +
b*x)^2]/(b*Pi), x] /; FreeQ[{a, b}, x]

Rubi steps

\begin {align*} \int S(b x) \, dx &=\frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{b \pi }+x S(b x)\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 26, normalized size = 1.00 \begin {gather*} \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{b \pi }+x S(b x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[FresnelS[b*x],x]

[Out]

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

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Maple [A]
time = 0.42, size = 27, normalized size = 1.04

method	result	size
derivativedivides	\(\frac {\mathrm {S}\left (b x \right ) b x +\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }}{b}\)	\(27\)
default	\(\frac {\mathrm {S}\left (b x \right ) b x +\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }}{b}\)	\(27\)
meijerg	\(\frac {\pi \,b^{3} x^{4} \hypergeom \left (\left [\frac {3}{4}, 1\right ], \left [\frac {3}{2}, \frac {7}{4}, 2\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{24}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 26, normalized size = 1.00 \begin {gather*} \frac {b x \operatorname {S}\left (b x\right ) + \frac {\cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{\pi }}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x*fresnel_sin(b*x) + cos(1/2*pi*b^2*x^2)/pi)/b

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Fricas [A]
time = 0.33, size = 26, normalized size = 1.00 \begin {gather*} \frac {\pi b x \operatorname {S}\left (b x\right ) + \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{\pi b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(pi*b*x*fresnel_sin(b*x) + cos(1/2*pi*b^2*x^2))/(pi*b)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (20) = 40\).
time = 0.47, size = 48, normalized size = 1.85 \begin {gather*} \frac {3 x S\left (b x\right ) \Gamma \left (\frac {3}{4}\right )}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {3 \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{4 \pi b \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x),x)

[Out]

3*x*fresnels(b*x)*gamma(3/4)/(4*gamma(7/4)) + 3*cos(pi*b**2*x**2/2)*gamma(3/4)/(4*pi*b*gamma(7/4))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(fresnel_sin(b*x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \mathrm {FresnelS}\left (b\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x),x)

[Out]

int(FresnelS(b*x), x)

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