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

Optimal. Leaf size=91 \[ -\frac{\sqrt{\pi } \sin (2 a) \text{FresnelC}\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )}{16 b^{3/2}}-\frac{\sqrt{\pi } \cos (2 a) S\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )}{16 b^{3/2}}+\frac{x \sin \left (2 a+2 b x^2\right )}{8 b}+\frac{x^3}{6} \]

[Out]

x^3/6 - (Sqrt[Pi]*Cos[2*a]*FresnelS[(2*Sqrt[b]*x)/Sqrt[Pi]])/(16*b^(3/2)) - (Sqrt[Pi]*FresnelC[(2*Sqrt[b]*x)/S
qrt[Pi]]*Sin[2*a])/(16*b^(3/2)) + (x*Sin[2*a + 2*b*x^2])/(8*b)

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Rubi [A]  time = 0.0973785, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {3404, 3386, 3353, 3352, 3351} \[ -\frac{\sqrt{\pi } \sin (2 a) \text{FresnelC}\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )}{16 b^{3/2}}-\frac{\sqrt{\pi } \cos (2 a) S\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )}{16 b^{3/2}}+\frac{x \sin \left (2 a+2 b x^2\right )}{8 b}+\frac{x^3}{6} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Cos[a + b*x^2]^2,x]

[Out]

x^3/6 - (Sqrt[Pi]*Cos[2*a]*FresnelS[(2*Sqrt[b]*x)/Sqrt[Pi]])/(16*b^(3/2)) - (Sqrt[Pi]*FresnelC[(2*Sqrt[b]*x)/S
qrt[Pi]]*Sin[2*a])/(16*b^(3/2)) + (x*Sin[2*a + 2*b*x^2])/(8*b)

Rule 3404

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandTrigReduce[(e
*x)^m, (a + b*Cos[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]

Rule 3386

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Sin[c + d*
x^n])/(d*n), x] - Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x
] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3353

Int[Sin[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Sin[c], Int[Cos[d*(e + f*x)^2], x], x] + Dist[
Cos[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 x^2 \cos ^2\left (a+b x^2\right ) \, dx &=\int \left (\frac{x^2}{2}+\frac{1}{2} x^2 \cos \left (2 a+2 b x^2\right )\right ) \, dx\\ &=\frac{x^3}{6}+\frac{1}{2} \int x^2 \cos \left (2 a+2 b x^2\right ) \, dx\\ &=\frac{x^3}{6}+\frac{x \sin \left (2 a+2 b x^2\right )}{8 b}-\frac{\int \sin \left (2 a+2 b x^2\right ) \, dx}{8 b}\\ &=\frac{x^3}{6}+\frac{x \sin \left (2 a+2 b x^2\right )}{8 b}-\frac{\cos (2 a) \int \sin \left (2 b x^2\right ) \, dx}{8 b}-\frac{\sin (2 a) \int \cos \left (2 b x^2\right ) \, dx}{8 b}\\ &=\frac{x^3}{6}-\frac{\sqrt{\pi } \cos (2 a) S\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )}{16 b^{3/2}}-\frac{\sqrt{\pi } C\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right ) \sin (2 a)}{16 b^{3/2}}+\frac{x \sin \left (2 a+2 b x^2\right )}{8 b}\\ \end{align*}

Mathematica [A]  time = 0.178047, size = 87, normalized size = 0.96 \[ \frac{-3 \sqrt{\pi } \sin (2 a) \text{FresnelC}\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )-3 \sqrt{\pi } \cos (2 a) S\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )+2 \sqrt{b} x \left (3 \sin \left (2 \left (a+b x^2\right )\right )+4 b x^2\right )}{48 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Cos[a + b*x^2]^2,x]

[Out]

(-3*Sqrt[Pi]*Cos[2*a]*FresnelS[(2*Sqrt[b]*x)/Sqrt[Pi]] - 3*Sqrt[Pi]*FresnelC[(2*Sqrt[b]*x)/Sqrt[Pi]]*Sin[2*a]
+ 2*Sqrt[b]*x*(4*b*x^2 + 3*Sin[2*(a + b*x^2)]))/(48*b^(3/2))

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Maple [A]  time = 0.031, size = 63, normalized size = 0.7 \begin{align*}{\frac{{x}^{3}}{6}}+{\frac{x\sin \left ( 2\,b{x}^{2}+2\,a \right ) }{8\,b}}-{\frac{\sqrt{\pi }}{16} \left ( \cos \left ( 2\,a \right ){\it FresnelS} \left ( 2\,{\frac{x\sqrt{b}}{\sqrt{\pi }}} \right ) +\sin \left ( 2\,a \right ){\it FresnelC} \left ( 2\,{\frac{x\sqrt{b}}{\sqrt{\pi }}} \right ) \right ){b}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3+1/8*x*sin(2*b*x^2+2*a)/b-1/16/b^(3/2)*Pi^(1/2)*(cos(2*a)*FresnelS(2*x*b^(1/2)/Pi^(1/2))+sin(2*a)*Fresn
elC(2*x*b^(1/2)/Pi^(1/2)))

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Maxima [C]  time = 1.89165, size = 383, normalized size = 4.21 \begin{align*} \frac{64 \, b x^{3}{\left | b \right |} + 48 \, x{\left | b \right |} \sin \left (2 \, b x^{2} + 2 \, a\right ) + \sqrt{2} \sqrt{\pi }{\left ({\left ({\left (-3 i \, \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) - 3 i \, \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) - 3 \, \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + 3 \, \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right )\right )} \cos \left (2 \, a\right ) -{\left (3 \, \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + 3 \, \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) - 3 i \, \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + 3 i \, \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 (3 i \, \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + 3 i \, \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) - 3 \, \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + 3 \, \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right )\right )} \cos \left (2 \, a\right ) -{\left (3 \, \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + 3 \, \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + 3 i \, \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) - 3 i \, \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 |}}}{384 \, b{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(64*b*x^3*abs(b) + 48*x*abs(b)*sin(2*b*x^2 + 2*a) + sqrt(2)*sqrt(pi)*(((-3*I*cos(1/4*pi + 1/2*arctan2(0,
b)) - 3*I*cos(-1/4*pi + 1/2*arctan2(0, b)) - 3*sin(1/4*pi + 1/2*arctan2(0, b)) + 3*sin(-1/4*pi + 1/2*arctan2(
0, b)))*cos(2*a) - (3*cos(1/4*pi + 1/2*arctan2(0, b)) + 3*cos(-1/4*pi + 1/2*arctan2(0, b)) - 3*I*sin(1/4*pi +
1/2*arctan2(0, b)) + 3*I*sin(-1/4*pi + 1/2*arctan2(0, b)))*sin(2*a))*erf(sqrt(2*I*b)*x) + ((3*I*cos(1/4*pi + 1
/2*arctan2(0, b)) + 3*I*cos(-1/4*pi + 1/2*arctan2(0, b)) - 3*sin(1/4*pi + 1/2*arctan2(0, b)) + 3*sin(-1/4*pi +
1/2*arctan2(0, b)))*cos(2*a) - (3*cos(1/4*pi + 1/2*arctan2(0, b)) + 3*cos(-1/4*pi + 1/2*arctan2(0, b)) + 3*I*
sin(1/4*pi + 1/2*arctan2(0, b)) - 3*I*sin(-1/4*pi + 1/2*arctan2(0, b)))*sin(2*a))*erf(sqrt(-2*I*b)*x))*sqrt(ab
s(b)))/(b*abs(b))

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Fricas [A]  time = 1.7135, size = 231, normalized size = 2.54 \begin{align*} \frac{8 \, b^{2} x^{3} + 12 \, b x \cos \left (b x^{2} + a\right ) \sin \left (b x^{2} + a\right ) - 3 \, \pi \sqrt{\frac{b}{\pi }} \cos \left (2 \, a\right ) \operatorname{S}\left (2 \, x \sqrt{\frac{b}{\pi }}\right ) - 3 \, \pi \sqrt{\frac{b}{\pi }} \operatorname{C}\left (2 \, x \sqrt{\frac{b}{\pi }}\right ) \sin \left (2 \, a\right )}{48 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(8*b^2*x^3 + 12*b*x*cos(b*x^2 + a)*sin(b*x^2 + a) - 3*pi*sqrt(b/pi)*cos(2*a)*fresnel_sin(2*x*sqrt(b/pi))
- 3*pi*sqrt(b/pi)*fresnel_cos(2*x*sqrt(b/pi))*sin(2*a))/b^2

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Sympy [B]  time = 3.31795, size = 201, normalized size = 2.21 \begin{align*} \frac{b^{\frac{3}{2}} x^{5} \sqrt{\frac{1}{b}} \sin{\left (2 a \right )} \Gamma \left (\frac{3}{4}\right ) \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{3}\left (\begin{matrix} \frac{3}{4}, \frac{5}{4} \\ \frac{3}{2}, \frac{7}{4}, \frac{9}{4} \end{matrix}\middle |{- b^{2} x^{4}} \right )}}{8 \Gamma \left (\frac{7}{4}\right ) \Gamma \left (\frac{9}{4}\right )} - \frac{\sqrt{b} x^{3} \sqrt{\frac{1}{b}} \cos{\left (2 a \right )} \Gamma \left (\frac{1}{4}\right ) \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{3}\left (\begin{matrix} \frac{1}{4}, \frac{3}{4} \\ \frac{1}{2}, \frac{5}{4}, \frac{7}{4} \end{matrix}\middle |{- b^{2} x^{4}} \right )}}{16 \Gamma \left (\frac{5}{4}\right ) \Gamma \left (\frac{7}{4}\right )} + \frac{x^{3}}{6} - \frac{\sqrt{\pi } x^{2} \sqrt{\frac{1}{b}} \sin{\left (2 a \right )} S\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )}{4} + \frac{\sqrt{\pi } x^{2} \sqrt{\frac{1}{b}} \cos{\left (2 a \right )} C\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b**(3/2)*x**5*sqrt(1/b)*sin(2*a)*gamma(3/4)*gamma(5/4)*hyper((3/4, 5/4), (3/2, 7/4, 9/4), -b**2*x**4)/(8*gamma
(7/4)*gamma(9/4)) - sqrt(b)*x**3*sqrt(1/b)*cos(2*a)*gamma(1/4)*gamma(3/4)*hyper((1/4, 3/4), (1/2, 5/4, 7/4), -
b**2*x**4)/(16*gamma(5/4)*gamma(7/4)) + x**3/6 - sqrt(pi)*x**2*sqrt(1/b)*sin(2*a)*fresnels(2*sqrt(b)*x/sqrt(pi
))/4 + sqrt(pi)*x**2*sqrt(1/b)*cos(2*a)*fresnelc(2*sqrt(b)*x/sqrt(pi))/4

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Giac [C]  time = 1.17095, size = 159, normalized size = 1.75 \begin{align*} \frac{1}{6} \, x^{3} - \frac{i \, x e^{\left (2 i \, b x^{2} + 2 i \, a\right )}}{16 \, b} + \frac{i \, x e^{\left (-2 i \, b x^{2} - 2 i \, a\right )}}{16 \, b} - \frac{i \, \sqrt{\pi } \operatorname{erf}\left (-\sqrt{b} x{\left (-\frac{i \, b}{{\left | b \right |}} + 1\right )}\right ) e^{\left (2 i \, a\right )}}{32 \, b^{\frac{3}{2}}{\left (-\frac{i \, b}{{\left | b \right |}} + 1\right )}} + \frac{i \, \sqrt{\pi } \operatorname{erf}\left (-\sqrt{b} x{\left (\frac{i \, b}{{\left | b \right |}} + 1\right )}\right ) e^{\left (-2 i \, a\right )}}{32 \, b^{\frac{3}{2}}{\left (\frac{i \, b}{{\left | b \right |}} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3 - 1/16*I*x*e^(2*I*b*x^2 + 2*I*a)/b + 1/16*I*x*e^(-2*I*b*x^2 - 2*I*a)/b - 1/32*I*sqrt(pi)*erf(-sqrt(b)*
x*(-I*b/abs(b) + 1))*e^(2*I*a)/(b^(3/2)*(-I*b/abs(b) + 1)) + 1/32*I*sqrt(pi)*erf(-sqrt(b)*x*(I*b/abs(b) + 1))*
e^(-2*I*a)/(b^(3/2)*(I*b/abs(b) + 1))