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

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

[Out]

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

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Rubi [A]  time = 0.0694239, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3386, 3353, 3352, 3351} \[ -\frac{\sqrt{\frac{\pi }{2}} \sin (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right )}{2 b^{3/2}}-\frac{\sqrt{\frac{\pi }{2}} \cos (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )}{2 b^{3/2}}+\frac{x \sin \left (a+b x^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Cos[a + b*x^2],x]

[Out]

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

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 \left (a+b x^2\right ) \, dx &=\frac{x \sin \left (a+b x^2\right )}{2 b}-\frac{\int \sin \left (a+b x^2\right ) \, dx}{2 b}\\ &=\frac{x \sin \left (a+b x^2\right )}{2 b}-\frac{\cos (a) \int \sin \left (b x^2\right ) \, dx}{2 b}-\frac{\sin (a) \int \cos \left (b x^2\right ) \, dx}{2 b}\\ &=-\frac{\sqrt{\frac{\pi }{2}} \cos (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )}{2 b^{3/2}}-\frac{\sqrt{\frac{\pi }{2}} C\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right ) \sin (a)}{2 b^{3/2}}+\frac{x \sin \left (a+b x^2\right )}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.144657, size = 82, normalized size = 0.9 \[ \frac{-\sqrt{2 \pi } \sin (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right )-\sqrt{2 \pi } \cos (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )+2 \sqrt{b} x \sin \left (a+b x^2\right )}{4 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Cos[a + b*x^2],x]

[Out]

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

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Maple [A]  time = 0.025, size = 58, normalized size = 0.6 \begin{align*}{\frac{x\sin \left ( b{x}^{2}+a \right ) }{2\,b}}-{\frac{\sqrt{2}\sqrt{\pi }}{4} \left ( \cos \left ( a \right ){\it FresnelS} \left ({\frac{x\sqrt{2}}{\sqrt{\pi }}\sqrt{b}} \right ) +\sin \left ( a \right ){\it FresnelC} \left ({\frac{x\sqrt{2}}{\sqrt{\pi }}\sqrt{b}} \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),x)

[Out]

1/2*x*sin(b*x^2+a)/b-1/4/b^(3/2)*2^(1/2)*Pi^(1/2)*(cos(a)*FresnelS(x*b^(1/2)*2^(1/2)/Pi^(1/2))+sin(a)*FresnelC
(x*b^(1/2)*2^(1/2)/Pi^(1/2)))

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Maxima [C]  time = 2.1017, size = 338, normalized size = 3.71 \begin{align*} \frac{8 \, x{\left | b \right |} \sin \left (b x^{2} + a\right ) + \sqrt{\pi }{\left ({\left ({\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 )} \cos \left (a\right ) -{\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 )} \sin \left (a\right )\right )} \operatorname{erf}\left (\sqrt{i \, b} x\right ) +{\left ({\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 )} \cos \left (a\right ) -{\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 )} \sin \left (a\right )\right )} \operatorname{erf}\left (\sqrt{-i \, b} x\right )\right )} \sqrt{{\left | b \right |}}}{16 \, 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),x, algorithm="maxima")

[Out]

1/16*(8*x*abs(b)*sin(b*x^2 + a) + sqrt(pi)*(((-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)))*cos(a) - (cos(1/4*pi + 1/2*arcta
n2(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)))*sin(a))*erf(sqrt(I*b)*x) + ((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)))*cos(a) - (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)))*
sin(a))*erf(sqrt(-I*b)*x))*sqrt(abs(b)))/(b*abs(b))

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Fricas [A]  time = 1.64268, size = 220, normalized size = 2.42 \begin{align*} -\frac{\sqrt{2} \pi \sqrt{\frac{b}{\pi }} \cos \left (a\right ) \operatorname{S}\left (\sqrt{2} x \sqrt{\frac{b}{\pi }}\right ) + \sqrt{2} \pi \sqrt{\frac{b}{\pi }} \operatorname{C}\left (\sqrt{2} x \sqrt{\frac{b}{\pi }}\right ) \sin \left (a\right ) - 2 \, b x \sin \left (b x^{2} + a\right )}{4 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 2.13041, size = 209, normalized size = 2.3 \begin{align*} \frac{b^{\frac{3}{2}} x^{5} \sqrt{\frac{1}{b}} \sin{\left (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 |{- \frac{b^{2} x^{4}}{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 (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 |{- \frac{b^{2} x^{4}}{4}} \right )}}{8 \Gamma \left (\frac{5}{4}\right ) \Gamma \left (\frac{7}{4}\right )} - \frac{\sqrt{2} \sqrt{\pi } x^{2} \sqrt{\frac{1}{b}} \sin{\left (a \right )} S\left (\frac{\sqrt{2} \sqrt{b} x}{\sqrt{\pi }}\right )}{2} + \frac{\sqrt{2} \sqrt{\pi } x^{2} \sqrt{\frac{1}{b}} \cos{\left (a \right )} C\left (\frac{\sqrt{2} \sqrt{b} x}{\sqrt{\pi }}\right )}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 1.15215, size = 182, normalized size = 2. \begin{align*} -\frac{i \, x e^{\left (i \, b x^{2} + i \, a\right )}}{4 \, b} + \frac{i \, x e^{\left (-i \, b x^{2} - i \, a\right )}}{4 \, b} - \frac{i \, \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 )}}{8 \, b{\left (-\frac{i \, b}{{\left | b \right |}} + 1\right )} \sqrt{{\left | b \right |}}} + \frac{i \, \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 )}}{8 \, b{\left (\frac{i \, b}{{\left | b \right |}} + 1\right )} \sqrt{{\left | b \right |}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*I*x*e^(I*b*x^2 + I*a)/b + 1/4*I*x*e^(-I*b*x^2 - I*a)/b - 1/8*I*sqrt(2)*sqrt(pi)*erf(-1/2*sqrt(2)*x*(-I*b/
abs(b) + 1)*sqrt(abs(b)))*e^(I*a)/(b*(-I*b/abs(b) + 1)*sqrt(abs(b))) + 1/8*I*sqrt(2)*sqrt(pi)*erf(-1/2*sqrt(2)
*x*(I*b/abs(b) + 1)*sqrt(abs(b)))*e^(-I*a)/(b*(I*b/abs(b) + 1)*sqrt(abs(b)))