3.160 \(\int \text{FresnelC}(a+b x)^2 \, dx\)

Optimal. Leaf size=69 \[ \frac{(a+b x) \text{FresnelC}(a+b x)^2}{b}-\frac{2 \text{FresnelC}(a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b}+\frac{S\left (\sqrt{2} (a+b x)\right )}{\sqrt{2} \pi b} \]

[Out]

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

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Rubi [A]  time = 0.150686, antiderivative size = 69, normalized size of antiderivative = 1., 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 = {6421, 6453, 3351} \[ \frac{(a+b x) \text{FresnelC}(a+b x)^2}{b}-\frac{2 \text{FresnelC}(a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b}+\frac{S\left (\sqrt{2} (a+b x)\right )}{\sqrt{2} \pi b} \]

Antiderivative was successfully verified.

[In]

Int[FresnelC[a + b*x]^2,x]

[Out]

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

Rule 6421

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

Rule 6453

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

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]

Rubi steps

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

Mathematica [A]  time = 0.0099422, size = 66, normalized size = 0.96 \[ \frac{2 \pi (a+b x) \text{FresnelC}(a+b x)^2-4 \text{FresnelC}(a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )+\sqrt{2} S\left (\sqrt{2} (a+b x)\right )}{2 \pi b} \]

Antiderivative was successfully verified.

[In]

Integrate[FresnelC[a + b*x]^2,x]

[Out]

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

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Maple [A]  time = 0.054, size = 60, normalized size = 0.9 \begin{align*}{\frac{1}{b} \left ( \left ( bx+a \right ) \left ({\it FresnelC} \left ( bx+a \right ) \right ) ^{2}-2\,{\frac{{\it FresnelC} \left ( bx+a \right ) \sin \left ( 1/2\,\pi \, \left ( bx+a \right ) ^{2} \right ) }{\pi }}+{\frac{\sqrt{2}{\it FresnelS} \left ( \left ( bx+a \right ) \sqrt{2} \right ) }{2\,\pi }} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*((b*x+a)*FresnelC(b*x+a)^2-2*FresnelC(b*x+a)/Pi*sin(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., size = 0, normalized size = 0. \begin{align*} \int{\rm fresnelc}\left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(fresnelc(b*x + a)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\rm fresnelc}\left (b x + a\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(fresnelc(b*x + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(fresnelc(a + b*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\rm fresnelc}\left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(fresnelc(b*x + a)^2, x)