3.18 \(\int \cos ^3(a+b x^2) \, dx\)

Optimal. Leaf size=153 \[ \frac{3 \sqrt{\frac{\pi }{2}} \cos (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right )}{4 \sqrt{b}}+\frac{\sqrt{\frac{\pi }{6}} \cos (3 a) \text{FresnelC}\left (\sqrt{\frac{6}{\pi }} \sqrt{b} x\right )}{4 \sqrt{b}}-\frac{3 \sqrt{\frac{\pi }{2}} \sin (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )}{4 \sqrt{b}}-\frac{\sqrt{\frac{\pi }{6}} \sin (3 a) S\left (\sqrt{b} \sqrt{\frac{6}{\pi }} x\right )}{4 \sqrt{b}} \]

[Out]

(3*Sqrt[Pi/2]*Cos[a]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*x])/(4*Sqrt[b]) + (Sqrt[Pi/6]*Cos[3*a]*FresnelC[Sqrt[b]*Sqrt[
6/Pi]*x])/(4*Sqrt[b]) - (3*Sqrt[Pi/2]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*x]*Sin[a])/(4*Sqrt[b]) - (Sqrt[Pi/6]*Fresnel
S[Sqrt[b]*Sqrt[6/Pi]*x]*Sin[3*a])/(4*Sqrt[b])

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Rubi [A]  time = 0.0890601, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3358, 3354, 3352, 3351} \[ \frac{3 \sqrt{\frac{\pi }{2}} \cos (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right )}{4 \sqrt{b}}+\frac{\sqrt{\frac{\pi }{6}} \cos (3 a) \text{FresnelC}\left (\sqrt{\frac{6}{\pi }} \sqrt{b} x\right )}{4 \sqrt{b}}-\frac{3 \sqrt{\frac{\pi }{2}} \sin (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )}{4 \sqrt{b}}-\frac{\sqrt{\frac{\pi }{6}} \sin (3 a) S\left (\sqrt{b} \sqrt{\frac{6}{\pi }} x\right )}{4 \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x^2]^3,x]

[Out]

(3*Sqrt[Pi/2]*Cos[a]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*x])/(4*Sqrt[b]) + (Sqrt[Pi/6]*Cos[3*a]*FresnelC[Sqrt[b]*Sqrt[
6/Pi]*x])/(4*Sqrt[b]) - (3*Sqrt[Pi/2]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*x]*Sin[a])/(4*Sqrt[b]) - (Sqrt[Pi/6]*Fresnel
S[Sqrt[b]*Sqrt[6/Pi]*x]*Sin[3*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 ^3\left (a+b x^2\right ) \, dx &=\int \left (\frac{3}{4} \cos \left (a+b x^2\right )+\frac{1}{4} \cos \left (3 a+3 b x^2\right )\right ) \, dx\\ &=\frac{1}{4} \int \cos \left (3 a+3 b x^2\right ) \, dx+\frac{3}{4} \int \cos \left (a+b x^2\right ) \, dx\\ &=\frac{1}{4} (3 \cos (a)) \int \cos \left (b x^2\right ) \, dx+\frac{1}{4} \cos (3 a) \int \cos \left (3 b x^2\right ) \, dx-\frac{1}{4} (3 \sin (a)) \int \sin \left (b x^2\right ) \, dx-\frac{1}{4} \sin (3 a) \int \sin \left (3 b x^2\right ) \, dx\\ &=\frac{3 \sqrt{\frac{\pi }{2}} \cos (a) C\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )}{4 \sqrt{b}}+\frac{\sqrt{\frac{\pi }{6}} \cos (3 a) C\left (\sqrt{b} \sqrt{\frac{6}{\pi }} x\right )}{4 \sqrt{b}}-\frac{3 \sqrt{\frac{\pi }{2}} S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right ) \sin (a)}{4 \sqrt{b}}-\frac{\sqrt{\frac{\pi }{6}} S\left (\sqrt{b} \sqrt{\frac{6}{\pi }} x\right ) \sin (3 a)}{4 \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.237082, size = 116, normalized size = 0.76 \[ \frac{\sqrt{\frac{\pi }{6}} \left (3 \sqrt{3} \cos (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right )+\cos (3 a) \text{FresnelC}\left (\sqrt{\frac{6}{\pi }} \sqrt{b} x\right )-3 \sqrt{3} \sin (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )-\sin (3 a) S\left (\sqrt{b} \sqrt{\frac{6}{\pi }} x\right )\right )}{4 \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x^2]^3,x]

[Out]

(Sqrt[Pi/6]*(3*Sqrt[3]*Cos[a]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*x] + Cos[3*a]*FresnelC[Sqrt[b]*Sqrt[6/Pi]*x] - 3*Sqr
t[3]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*x]*Sin[a] - FresnelS[Sqrt[b]*Sqrt[6/Pi]*x]*Sin[3*a]))/(4*Sqrt[b])

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Maple [A]  time = 0.033, size = 101, normalized size = 0.7 \begin{align*}{\frac{3\,\sqrt{2}\sqrt{\pi }}{8} \left ( \cos \left ( a \right ){\it FresnelC} \left ({\frac{x\sqrt{2}}{\sqrt{\pi }}\sqrt{b}} \right ) -\sin \left ( a \right ){\it FresnelS} \left ({\frac{x\sqrt{2}}{\sqrt{\pi }}\sqrt{b}} \right ) \right ){\frac{1}{\sqrt{b}}}}+{\frac{\sqrt{2}\sqrt{\pi }\sqrt{3}}{24} \left ( \cos \left ( 3\,a \right ){\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{3}x}{\sqrt{\pi }}\sqrt{b}} \right ) -\sin \left ( 3\,a \right ){\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{3}x}{\sqrt{\pi }}\sqrt{b}} \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)^3,x)

[Out]

3/8*2^(1/2)*Pi^(1/2)/b^(1/2)*(cos(a)*FresnelC(x*b^(1/2)*2^(1/2)/Pi^(1/2))-sin(a)*FresnelS(x*b^(1/2)*2^(1/2)/Pi
^(1/2)))+1/24*2^(1/2)*Pi^(1/2)*3^(1/2)/b^(1/2)*(cos(3*a)*FresnelC(2^(1/2)/Pi^(1/2)*3^(1/2)*b^(1/2)*x)-sin(3*a)
*FresnelS(2^(1/2)/Pi^(1/2)*3^(1/2)*b^(1/2)*x))

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Maxima [C]  time = 2.21973, size = 653, normalized size = 4.27 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x^2+a)^3,x, algorithm="maxima")

[Out]

1/96*(sqrt(3)*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(3*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(3*a))*
erf(sqrt(3*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(3*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(3*a))*er
f(sqrt(-3*I*b)*x))*sqrt(abs(b)) + sqrt(pi)*(((9*cos(1/4*pi + 1/2*arctan2(0, b)) + 9*cos(-1/4*pi + 1/2*arctan2(
0, b)) - 9*I*sin(1/4*pi + 1/2*arctan2(0, b)) + 9*I*sin(-1/4*pi + 1/2*arctan2(0, b)))*cos(a) - (9*I*cos(1/4*pi
+ 1/2*arctan2(0, b)) + 9*I*cos(-1/4*pi + 1/2*arctan2(0, b)) + 9*sin(1/4*pi + 1/2*arctan2(0, b)) - 9*sin(-1/4*p
i + 1/2*arctan2(0, b)))*sin(a))*erf(sqrt(I*b)*x) + ((9*cos(1/4*pi + 1/2*arctan2(0, b)) + 9*cos(-1/4*pi + 1/2*a
rctan2(0, b)) + 9*I*sin(1/4*pi + 1/2*arctan2(0, b)) - 9*I*sin(-1/4*pi + 1/2*arctan2(0, b)))*cos(a) - (-9*I*cos
(1/4*pi + 1/2*arctan2(0, b)) - 9*I*cos(-1/4*pi + 1/2*arctan2(0, b)) + 9*sin(1/4*pi + 1/2*arctan2(0, b)) - 9*si
n(-1/4*pi + 1/2*arctan2(0, b)))*sin(a))*erf(sqrt(-I*b)*x))*sqrt(abs(b)))/abs(b)

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Fricas [A]  time = 1.62326, size = 373, normalized size = 2.44 \begin{align*} \frac{\sqrt{6} \pi \sqrt{\frac{b}{\pi }} \cos \left (3 \, a\right ) \operatorname{C}\left (\sqrt{6} x \sqrt{\frac{b}{\pi }}\right ) + 9 \, \sqrt{2} \pi \sqrt{\frac{b}{\pi }} \cos \left (a\right ) \operatorname{C}\left (\sqrt{2} x \sqrt{\frac{b}{\pi }}\right ) - \sqrt{6} \pi \sqrt{\frac{b}{\pi }} \operatorname{S}\left (\sqrt{6} x \sqrt{\frac{b}{\pi }}\right ) \sin \left (3 \, a\right ) - 9 \, \sqrt{2} \pi \sqrt{\frac{b}{\pi }} \operatorname{S}\left (\sqrt{2} x \sqrt{\frac{b}{\pi }}\right ) \sin \left (a\right )}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x^2+a)^3,x, algorithm="fricas")

[Out]

1/24*(sqrt(6)*pi*sqrt(b/pi)*cos(3*a)*fresnel_cos(sqrt(6)*x*sqrt(b/pi)) + 9*sqrt(2)*pi*sqrt(b/pi)*cos(a)*fresne
l_cos(sqrt(2)*x*sqrt(b/pi)) - sqrt(6)*pi*sqrt(b/pi)*fresnel_sin(sqrt(6)*x*sqrt(b/pi))*sin(3*a) - 9*sqrt(2)*pi*
sqrt(b/pi)*fresnel_sin(sqrt(2)*x*sqrt(b/pi))*sin(a))/b

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Sympy [A]  time = 1.56403, size = 129, normalized size = 0.84 \begin{align*} \frac{3 \sqrt{2} \sqrt{\pi } \left (- \sin{\left (a \right )} S\left (\frac{\sqrt{2} \sqrt{b} x}{\sqrt{\pi }}\right ) + \cos{\left (a \right )} C\left (\frac{\sqrt{2} \sqrt{b} x}{\sqrt{\pi }}\right )\right ) \sqrt{\frac{1}{b}}}{8} + \frac{\sqrt{6} \sqrt{\pi } \left (- \sin{\left (3 a \right )} S\left (\frac{\sqrt{6} \sqrt{b} x}{\sqrt{\pi }}\right ) + \cos{\left (3 a \right )} C\left (\frac{\sqrt{6} \sqrt{b} x}{\sqrt{\pi }}\right )\right ) \sqrt{\frac{1}{b}}}{24} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x**2+a)**3,x)

[Out]

3*sqrt(2)*sqrt(pi)*(-sin(a)*fresnels(sqrt(2)*sqrt(b)*x/sqrt(pi)) + cos(a)*fresnelc(sqrt(2)*sqrt(b)*x/sqrt(pi))
)*sqrt(1/b)/8 + sqrt(6)*sqrt(pi)*(-sin(3*a)*fresnels(sqrt(6)*sqrt(b)*x/sqrt(pi)) + cos(3*a)*fresnelc(sqrt(6)*s
qrt(b)*x/sqrt(pi)))*sqrt(1/b)/24

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Giac [C]  time = 1.1657, size = 250, normalized size = 1.63 \begin{align*} -\frac{\sqrt{6} \sqrt{\pi } \operatorname{erf}\left (-\frac{1}{2} \, \sqrt{6} \sqrt{b} x{\left (-\frac{i \, b}{{\left | b \right |}} + 1\right )}\right ) e^{\left (3 i \, a\right )}}{48 \, \sqrt{b}{\left (-\frac{i \, b}{{\left | b \right |}} + 1\right )}} - \frac{3 \, \sqrt{2} \sqrt{\pi } \operatorname{erf}\left (-\frac{1}{2} \, \sqrt{2} x{\left (-\frac{i \, b}{{\left | b \right |}} + 1\right )} \sqrt{{\left | b \right |}}\right ) e^{\left (i \, a\right )}}{16 \,{\left (-\frac{i \, b}{{\left | b \right |}} + 1\right )} \sqrt{{\left | b \right |}}} - \frac{3 \, \sqrt{2} \sqrt{\pi } \operatorname{erf}\left (-\frac{1}{2} \, \sqrt{2} x{\left (\frac{i \, b}{{\left | b \right |}} + 1\right )} \sqrt{{\left | b \right |}}\right ) e^{\left (-i \, a\right )}}{16 \,{\left (\frac{i \, b}{{\left | b \right |}} + 1\right )} \sqrt{{\left | b \right |}}} - \frac{\sqrt{6} \sqrt{\pi } \operatorname{erf}\left (-\frac{1}{2} \, \sqrt{6} \sqrt{b} x{\left (\frac{i \, b}{{\left | b \right |}} + 1\right )}\right ) e^{\left (-3 i \, a\right )}}{48 \, \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)^3,x, algorithm="giac")

[Out]

-1/48*sqrt(6)*sqrt(pi)*erf(-1/2*sqrt(6)*sqrt(b)*x*(-I*b/abs(b) + 1))*e^(3*I*a)/(sqrt(b)*(-I*b/abs(b) + 1)) - 3
/16*sqrt(2)*sqrt(pi)*erf(-1/2*sqrt(2)*x*(-I*b/abs(b) + 1)*sqrt(abs(b)))*e^(I*a)/((-I*b/abs(b) + 1)*sqrt(abs(b)
)) - 3/16*sqrt(2)*sqrt(pi)*erf(-1/2*sqrt(2)*x*(I*b/abs(b) + 1)*sqrt(abs(b)))*e^(-I*a)/((I*b/abs(b) + 1)*sqrt(a
bs(b))) - 1/48*sqrt(6)*sqrt(pi)*erf(-1/2*sqrt(6)*sqrt(b)*x*(I*b/abs(b) + 1))*e^(-3*I*a)/(sqrt(b)*(I*b/abs(b) +
1))