3.51 \(\int S(a+b x)^2 \, dx\)

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

[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 = 0.17, 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 = {6420, 6452, 3351} \[ \frac {(a+b x) S(a+b x)^2}{b}-\frac {S\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} \pi b}+\frac {2 S(a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b} \]

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 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 6420

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*(a + b*x)^2)/2]*FresnelS[a + b*x], x], x] /; FreeQ[{a, b}, x]

Rule 6452

Int[FresnelS[(b_.)*(x_)]*(x_)*Sin[(d_.)*(x_)^2], x_Symbol] :> -Simp[(Cos[d*x^2]*FresnelS[b*x])/(2*d), x] + Dis
t[1/(2*b*Pi), Int[Sin[2*d*x^2], x], x] /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2*b^4)/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 \operatorname {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 {\operatorname {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 \[ \frac {2 \pi (a+b x) S(a+b x)^2-\sqrt {2} S\left (\sqrt {2} (a+b x)\right )+4 S(a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 \pi b} \]

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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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\rm fresnels}\left (b x + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(fresnels(b*x + a)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\rm fresnels}\left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(fresnels(b*x + a)^2, x)

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maple [A]  time = 0.01, size = 60, normalized size = 0.86 \[ \frac {\left (b x +a \right ) \mathrm {S}\left (b x +a \right )^{2}+\frac {2 \,\mathrm {S}\left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {\sqrt {2}\, \mathrm {S}\left (\left (b x +a \right ) \sqrt {2}\right )}{2 \pi }}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*((b*x+a)*FresnelS(b*x+a)^2+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 \[ \int {\rm fresnels}\left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(fresnels(b*x + a)^2, x)

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

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

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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