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

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.05, size = 67, normalized size = 0.96 \[ \frac {\sqrt {\pi } \cos (2 a) C\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right )-\sqrt {\pi } \sin (2 a) S\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right )+2 \sqrt {b} x}{4 \sqrt {b}} \]

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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fricas [A]  time = 0.95, size = 59, normalized size = 0.84 \[ \frac {\pi \sqrt {\frac {b}{\pi }} \cos \left (2 \, a\right ) \operatorname {C}\left (2 \, x \sqrt {\frac {b}{\pi }}\right ) - \pi \sqrt {\frac {b}{\pi }} \operatorname {S}\left (2 \, x \sqrt {\frac {b}{\pi }}\right ) \sin \left (2 \, a\right ) + 2 \, b x}{4 \, b} \]

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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giac [C]  time = 0.32, size = 82, normalized size = 1.17 \[ \frac {1}{2} \, x - \frac {\sqrt {\pi } \operatorname {erf}\left (-\sqrt {b} x {\left (-\frac {i \, b}{{\left | b \right |}} + 1\right )}\right ) e^{\left (2 i \, a\right )}}{8 \, \sqrt {b} {\left (-\frac {i \, b}{{\left | b \right |}} + 1\right )}} - \frac {\sqrt {\pi } \operatorname {erf}\left (-\sqrt {b} x {\left (\frac {i \, b}{{\left | b \right |}} + 1\right )}\right ) e^{\left (-2 i \, a\right )}}{8 \, \sqrt {b} {\left (\frac {i \, b}{{\left | b \right |}} + 1\right )}} \]

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))

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maple [A]  time = 0.05, size = 45, normalized size = 0.64 \[ \frac {x}{2}+\frac {\sqrt {\pi }\, \left (\cos \left (2 a \right ) \FresnelC \left (\frac {2 x \sqrt {b}}{\sqrt {\pi }}\right )-\sin \left (2 a \right ) \mathrm {S}\left (\frac {2 x \sqrt {b}}{\sqrt {\pi }}\right )\right )}{4 \sqrt {b}} \]

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 = 1.45, size = 70, normalized size = 1.00 \[ -\frac {4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } {\left ({\left (\left (i - 1\right ) \, \cos \left (2 \, a\right ) + \left (i + 1\right ) \, \sin \left (2 \, a\right )\right )} \operatorname {erf}\left (\sqrt {2 i \, b} x\right ) + {\left (-\left (i + 1\right ) \, \cos \left (2 \, a\right ) - \left (i - 1\right ) \, \sin \left (2 \, a\right )\right )} \operatorname {erf}\left (\sqrt {-2 i \, b} x\right )\right )} b^{\frac {3}{2}} - 16 \, b^{2} x}{32 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(4^(1/4)*sqrt(2)*sqrt(pi)*(((I - 1)*cos(2*a) + (I + 1)*sin(2*a))*erf(sqrt(2*I*b)*x) + (-(I + 1)*cos(2*a)
- (I - 1)*sin(2*a))*erf(sqrt(-2*I*b)*x))*b^(3/2) - 16*b^2*x)/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x^2)^2,x)

[Out]

int(cos(a + b*x^2)^2, x)

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sympy [A]  time = 0.80, size = 56, normalized size = 0.80 \[ \frac {x}{2} + \frac {\sqrt {\pi } \left (- \sin {\left (2 a \right )} S\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right ) + \cos {\left (2 a \right )} C\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right )\right ) \sqrt {\frac {1}{b}}}{4} \]

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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