3.1.51 \(\int S(a+b x)^2 \, dx\) [51]

Optimal. Leaf size=70 \[ \frac {2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) S(a+b x)}{b \pi }+\frac {(a+b x) S(a+b x)^2}{b}-\frac {S\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} b \pi } \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6555, 6587, 3432} \begin {gather*} \frac {(a+b x) S(a+b x)^2}{b}-\frac {S\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} \pi b}+\frac {2 S(a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[FresnelS[a + b*x]^2,x]

[Out]

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

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 6555

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

Rule 6587

Int[FresnelS[(b_.)*(x_)]*(x_)*Sin[(d_.)*(x_)^2], x_Symbol] :> Simp[(-Cos[d*x^2])*(FresnelS[b*x]/(2*d)), x] + D
ist[1/(2*b*Pi), Int[Sin[2*d*x^2], x], x] /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2/4)*b^4]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 67, normalized size = 0.96 \begin {gather*} \frac {4 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) S(a+b x)+2 \pi (a+b x) S(a+b x)^2-\sqrt {2} S\left (\sqrt {2} (a+b x)\right )}{2 b \pi } \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[FresnelS[a + b*x]^2,x]

[Out]

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

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Maple [A]
time = 0.52, size = 60, normalized size = 0.86

method	result	size
derivativedivides	\(\frac {\mathrm {S}\left (b x +a \right )^{2} \left (b x +a \right )+\frac {2 \,\mathrm {S}\left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {\sqrt {2}\, \mathrm {S}\left (\left (b x +a \right ) \sqrt {2}\right )}{2 \pi }}{b}\)	\(60\)
default	\(\frac {\mathrm {S}\left (b x +a \right )^{2} \left (b x +a \right )+\frac {2 \,\mathrm {S}\left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {\sqrt {2}\, \mathrm {S}\left (\left (b x +a \right ) \sqrt {2}\right )}{2 \pi }}{b}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_sin(b*x+a)^2,x, algorithm="maxima")

[Out]

integrate(fresnel_sin(b*x + a)^2, x)

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Fricas [A]
time = 0.39, size = 89, normalized size = 1.27 \begin {gather*} \frac {4 \, b \cos \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) \operatorname {S}\left (b x + a\right ) + 2 \, {\left (\pi b^{2} x + \pi a b\right )} \operatorname {S}\left (b x + a\right )^{2} - \sqrt {2} \sqrt {b^{2}} \operatorname {S}\left (\frac {\sqrt {2} \sqrt {b^{2}} {\left (b x + a\right )}}{b}\right )}{2 \, \pi b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnel_sin(b*x+a)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int S^{2}\left (a + b x\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x+a)**2,x)

[Out]

Integral(fresnels(a + b*x)**2, x)

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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+a)^2,x, algorithm="giac")

[Out]

integrate(fresnel_sin(b*x + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {FresnelS}\left (a+b\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(a + b*x)^2,x)

[Out]

int(FresnelS(a + b*x)^2, x)

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