3.11 \(\int \cos ^2(a+b x^2) \, dx\)

Optimal. Leaf size=70 \[ \frac{\sqrt{\pi } \cos (2 a) \text{FresnelC}\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )}{4 \sqrt{b}}-\frac{\sqrt{\pi } \sin (2 a) S\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )}{4 \sqrt{b}}+\frac{x}{2} \]

[Out]

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

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Rubi [A]  time = 0.0437522, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3358, 3354, 3352, 3351} \[ \frac{\sqrt{\pi } \cos (2 a) \text{FresnelC}\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )}{4 \sqrt{b}}-\frac{\sqrt{\pi } \sin (2 a) S\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )}{4 \sqrt{b}}+\frac{x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x^2]^2,x]

[Out]

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

Rule 3358

Int[((a_.) + Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)]*(b_.))^(p_), x_Symbol] :> Int[ExpandTrigReduce[(a +
b*Cos[c + d*(e + f*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 1] && IGtQ[n, 1]

Rule 3354

Int[Cos[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Cos[c], Int[Cos[d*(e + f*x)^2], x], x] - Dist[
Sin[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

Rule 3352

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

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 \cos ^2\left (a+b x^2\right ) \, dx &=\int \left (\frac{1}{2}+\frac{1}{2} \cos \left (2 a+2 b x^2\right )\right ) \, dx\\ &=\frac{x}{2}+\frac{1}{2} \int \cos \left (2 a+2 b x^2\right ) \, dx\\ &=\frac{x}{2}+\frac{1}{2} \cos (2 a) \int \cos \left (2 b x^2\right ) \, dx-\frac{1}{2} \sin (2 a) \int \sin \left (2 b x^2\right ) \, dx\\ &=\frac{x}{2}+\frac{\sqrt{\pi } \cos (2 a) C\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )}{4 \sqrt{b}}-\frac{\sqrt{\pi } S\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right ) \sin (2 a)}{4 \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.0608273, size = 67, normalized size = 0.96 \[ \frac{\sqrt{\pi } \cos (2 a) \text{FresnelC}\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )-\sqrt{\pi } \sin (2 a) S\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )+2 \sqrt{b} x}{4 \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x^2]^2,x]

[Out]

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

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Maple [A]  time = 0.035, size = 45, normalized size = 0.6 \begin{align*}{\frac{x}{2}}+{\frac{\sqrt{\pi }}{4} \left ( \cos \left ( 2\,a \right ){\it FresnelC} \left ( 2\,{\frac{x\sqrt{b}}{\sqrt{\pi }}} \right ) -\sin \left ( 2\,a \right ){\it FresnelS} \left ( 2\,{\frac{x\sqrt{b}}{\sqrt{\pi }}} \right ) \right ){\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x+1/4*Pi^(1/2)/b^(1/2)*(cos(2*a)*FresnelC(2*x*b^(1/2)/Pi^(1/2))-sin(2*a)*FresnelS(2*x*b^(1/2)/Pi^(1/2)))

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Maxima [C]  time = 1.91315, size = 338, normalized size = 4.83 \begin{align*} \frac{\sqrt{2} \sqrt{\pi }{\left ({\left ({\left (\cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) - i \, \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + i \, \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right )\right )} \cos \left (2 \, a\right ) -{\left (i \, \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + i \, \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) - \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right )\right )} \sin \left (2 \, a\right )\right )} \operatorname{erf}\left (\sqrt{2 i \, b} x\right ) +{\left ({\left (\cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + i \, \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) - i \, \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right )\right )} \cos \left (2 \, a\right ) -{\left (-i \, \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) - i \, \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) - \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right )\right )} \sin \left (2 \, a\right )\right )} \operatorname{erf}\left (\sqrt{-2 i \, b} x\right )\right )} \sqrt{{\left | b \right |}} + 16 \, x{\left | b \right |}}{32 \,{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(sqrt(2)*sqrt(pi)*(((cos(1/4*pi + 1/2*arctan2(0, b)) + cos(-1/4*pi + 1/2*arctan2(0, b)) - I*sin(1/4*pi +
1/2*arctan2(0, b)) + I*sin(-1/4*pi + 1/2*arctan2(0, b)))*cos(2*a) - (I*cos(1/4*pi + 1/2*arctan2(0, b)) + I*cos
(-1/4*pi + 1/2*arctan2(0, b)) + sin(1/4*pi + 1/2*arctan2(0, b)) - sin(-1/4*pi + 1/2*arctan2(0, b)))*sin(2*a))*
erf(sqrt(2*I*b)*x) + ((cos(1/4*pi + 1/2*arctan2(0, b)) + cos(-1/4*pi + 1/2*arctan2(0, b)) + I*sin(1/4*pi + 1/2
*arctan2(0, b)) - I*sin(-1/4*pi + 1/2*arctan2(0, b)))*cos(2*a) - (-I*cos(1/4*pi + 1/2*arctan2(0, b)) - I*cos(-
1/4*pi + 1/2*arctan2(0, b)) + sin(1/4*pi + 1/2*arctan2(0, b)) - sin(-1/4*pi + 1/2*arctan2(0, b)))*sin(2*a))*er
f(sqrt(-2*I*b)*x))*sqrt(abs(b)) + 16*x*abs(b))/abs(b)

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Fricas [A]  time = 1.63341, size = 163, normalized size = 2.33 \begin{align*} \frac{\pi \sqrt{\frac{b}{\pi }} \cos \left (2 \, a\right ) \operatorname{C}\left (2 \, x \sqrt{\frac{b}{\pi }}\right ) - \pi \sqrt{\frac{b}{\pi }} \operatorname{S}\left (2 \, x \sqrt{\frac{b}{\pi }}\right ) \sin \left (2 \, a\right ) + 2 \, b x}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.04674, size = 56, normalized size = 0.8 \begin{align*} \frac{x}{2} + \frac{\sqrt{\pi } \left (- \sin{\left (2 a \right )} S\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right ) + \cos{\left (2 a \right )} C\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )\right ) \sqrt{\frac{1}{b}}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/2 + sqrt(pi)*(-sin(2*a)*fresnels(2*sqrt(b)*x/sqrt(pi)) + cos(2*a)*fresnelc(2*sqrt(b)*x/sqrt(pi)))*sqrt(1/b)/
4

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Giac [C]  time = 1.16218, size = 111, normalized size = 1.59 \begin{align*} \frac{1}{2} \, x - \frac{\sqrt{\pi } \operatorname{erf}\left (-\sqrt{b} x{\left (-\frac{i \, b}{{\left | b \right |}} + 1\right )}\right ) e^{\left (2 i \, a\right )}}{8 \, \sqrt{b}{\left (-\frac{i \, b}{{\left | b \right |}} + 1\right )}} - \frac{\sqrt{\pi } \operatorname{erf}\left (-\sqrt{b} x{\left (\frac{i \, b}{{\left | b \right |}} + 1\right )}\right ) e^{\left (-2 i \, a\right )}}{8 \, \sqrt{b}{\left (\frac{i \, b}{{\left | b \right |}} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x - 1/8*sqrt(pi)*erf(-sqrt(b)*x*(-I*b/abs(b) + 1))*e^(2*I*a)/(sqrt(b)*(-I*b/abs(b) + 1)) - 1/8*sqrt(pi)*er
f(-sqrt(b)*x*(I*b/abs(b) + 1))*e^(-2*I*a)/(sqrt(b)*(I*b/abs(b) + 1))