3.13 \(\int \frac{\cos ^2(a+b x^2)}{x^2} \, dx\)

Optimal. Leaf size=76 \[ -\sqrt{\pi } \sqrt{b} \sin (2 a) \text{FresnelC}\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )-\sqrt{\pi } \sqrt{b} \cos (2 a) S\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )-\frac{\cos ^2\left (a+b x^2\right )}{x} \]

[Out]

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

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Rubi [A]  time = 0.0673088, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3394, 4573, 3373, 3353, 3352, 3351} \[ -\sqrt{\pi } \sqrt{b} \sin (2 a) \text{FresnelC}\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )-\sqrt{\pi } \sqrt{b} \cos (2 a) S\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )-\frac{\cos ^2\left (a+b x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x^2]^2/x^2,x]

[Out]

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

Rule 3394

Int[Cos[(a_.) + (b_.)*(x_)^(n_)]^(p_)*(x_)^(m_.), x_Symbol] :> Simp[(x^(m + 1)*Cos[a + b*x^n]^p)/(m + 1), x] +
Dist[(b*n*p)/(m + 1), Int[Cos[a + b*x^n]^(p - 1)*Sin[a + b*x^n], x], x] /; FreeQ[{a, b}, x] && IGtQ[p, 1] &&
EqQ[m + n, 0] && NeQ[n, 1] && IntegerQ[n]

Rule 4573

Int[Cos[w_]^(p_.)*(u_.)*Sin[v_]^(p_.), x_Symbol] :> Dist[1/2^p, Int[u*Sin[2*v]^p, x], x] /; EqQ[w, v] && Integ
erQ[p]

Rule 3373

Int[((a_.) + (b_.)*Sin[u_])^(p_.), x_Symbol] :> Int[(a + b*Sin[ExpandToSum[u, x]])^p, x] /; FreeQ[{a, b, p}, x
] && BinomialQ[u, x] &&  !BinomialMatchQ[u, x]

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

Mathematica [A]  time = 0.176776, size = 76, normalized size = 1. \[ -\frac{\sqrt{\pi } \sqrt{b} x \sin (2 a) \text{FresnelC}\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )+\sqrt{\pi } \sqrt{b} x \cos (2 a) S\left (\frac{2 \sqrt{b} x}{\sqrt{\pi }}\right )+\cos ^2\left (a+b x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x^2]^2/x^2,x]

[Out]

-((Cos[a + b*x^2]^2 + Sqrt[b]*Sqrt[Pi]*x*Cos[2*a]*FresnelS[(2*Sqrt[b]*x)/Sqrt[Pi]] + Sqrt[b]*Sqrt[Pi]*x*Fresne
lC[(2*Sqrt[b]*x)/Sqrt[Pi]]*Sin[2*a])/x)

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Maple [A]  time = 0.03, size = 62, normalized size = 0.8 \begin{align*} -{\frac{1}{2\,x}}-{\frac{\cos \left ( 2\,b{x}^{2}+2\,a \right ) }{2\,x}}-\sqrt{b}\sqrt{\pi } \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C]  time = 1.45817, size = 369, normalized size = 4.86 \begin{align*} -\frac{\sqrt{2} \sqrt{x^{2}{\left | b \right |}}{\left ({\left ({\left (\Gamma \left (-\frac{1}{2}, 2 i \, b x^{2}\right ) + \Gamma \left (-\frac{1}{2}, -2 i \, b x^{2}\right )\right )} \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) +{\left (\Gamma \left (-\frac{1}{2}, 2 i \, b x^{2}\right ) + \Gamma \left (-\frac{1}{2}, -2 i \, b x^{2}\right )\right )} \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) +{\left (i \, \Gamma \left (-\frac{1}{2}, 2 i \, b x^{2}\right ) - i \, \Gamma \left (-\frac{1}{2}, -2 i \, b x^{2}\right )\right )} \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) +{\left (-i \, \Gamma \left (-\frac{1}{2}, 2 i \, b x^{2}\right ) + i \, \Gamma \left (-\frac{1}{2}, -2 i \, b x^{2}\right )\right )} \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right )\right )} \cos \left (2 \, a\right ) +{\left ({\left (-i \, \Gamma \left (-\frac{1}{2}, 2 i \, b x^{2}\right ) + i \, \Gamma \left (-\frac{1}{2}, -2 i \, b x^{2}\right )\right )} \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) +{\left (-i \, \Gamma \left (-\frac{1}{2}, 2 i \, b x^{2}\right ) + i \, \Gamma \left (-\frac{1}{2}, -2 i \, b x^{2}\right )\right )} \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) +{\left (\Gamma \left (-\frac{1}{2}, 2 i \, b x^{2}\right ) + \Gamma \left (-\frac{1}{2}, -2 i \, b x^{2}\right )\right )} \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) -{\left (\Gamma \left (-\frac{1}{2}, 2 i \, b x^{2}\right ) + \Gamma \left (-\frac{1}{2}, -2 i \, b x^{2}\right )\right )} \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right )\right )} \sin \left (2 \, a\right )\right )} + 8}{16 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(sqrt(2)*sqrt(x^2*abs(b))*(((gamma(-1/2, 2*I*b*x^2) + gamma(-1/2, -2*I*b*x^2))*cos(1/4*pi + 1/2*arctan2(
0, b)) + (gamma(-1/2, 2*I*b*x^2) + gamma(-1/2, -2*I*b*x^2))*cos(-1/4*pi + 1/2*arctan2(0, b)) + (I*gamma(-1/2,
2*I*b*x^2) - I*gamma(-1/2, -2*I*b*x^2))*sin(1/4*pi + 1/2*arctan2(0, b)) + (-I*gamma(-1/2, 2*I*b*x^2) + I*gamma
(-1/2, -2*I*b*x^2))*sin(-1/4*pi + 1/2*arctan2(0, b)))*cos(2*a) + ((-I*gamma(-1/2, 2*I*b*x^2) + I*gamma(-1/2, -
2*I*b*x^2))*cos(1/4*pi + 1/2*arctan2(0, b)) + (-I*gamma(-1/2, 2*I*b*x^2) + I*gamma(-1/2, -2*I*b*x^2))*cos(-1/4
*pi + 1/2*arctan2(0, b)) + (gamma(-1/2, 2*I*b*x^2) + gamma(-1/2, -2*I*b*x^2))*sin(1/4*pi + 1/2*arctan2(0, b))
- (gamma(-1/2, 2*I*b*x^2) + gamma(-1/2, -2*I*b*x^2))*sin(-1/4*pi + 1/2*arctan2(0, b)))*sin(2*a)) + 8)/x

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Fricas [A]  time = 1.67794, size = 180, normalized size = 2.37 \begin{align*} -\frac{\pi x \sqrt{\frac{b}{\pi }} \cos \left (2 \, a\right ) \operatorname{S}\left (2 \, x \sqrt{\frac{b}{\pi }}\right ) + \pi x \sqrt{\frac{b}{\pi }} \operatorname{C}\left (2 \, x \sqrt{\frac{b}{\pi }}\right ) \sin \left (2 \, a\right ) + \cos \left (b x^{2} + a\right )^{2}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos ^{2}{\left (a + b x^{2} \right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(a + b*x**2)**2/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (b x^{2} + a\right )^{2}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cos(b*x^2 + a)^2/x^2, x)