3.44 \(\int \frac{\cos (a+\frac{b}{x^2})}{x^4} \, dx\)

Optimal. Leaf size=97 \[ \frac{\sqrt{\frac{\pi }{2}} \sin (a) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{x}\right )}{2 b^{3/2}}+\frac{\sqrt{\frac{\pi }{2}} \cos (a) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{x}\right )}{2 b^{3/2}}-\frac{\sin \left (a+\frac{b}{x^2}\right )}{2 b x} \]

[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
]*Sin[a])/(2*b^(3/2)) - Sin[a + b/x^2]/(2*b*x)

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Rubi [A]  time = 0.0532183, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {3410, 3386, 3353, 3352, 3351} \[ \frac{\sqrt{\frac{\pi }{2}} \sin (a) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{x}\right )}{2 b^{3/2}}+\frac{\sqrt{\frac{\pi }{2}} \cos (a) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{x}\right )}{2 b^{3/2}}-\frac{\sin \left (a+\frac{b}{x^2}\right )}{2 b x} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b/x^2]/x^4,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
]*Sin[a])/(2*b^(3/2)) - Sin[a + b/x^2]/(2*b*x)

Rule 3410

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> -Subst[Int[(a + b*Cos[c + d/x^
n])^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m] && EqQ[n, -2
]

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 \frac{\cos \left (a+\frac{b}{x^2}\right )}{x^4} \, dx &=-\operatorname{Subst}\left (\int x^2 \cos \left (a+b x^2\right ) \, dx,x,\frac{1}{x}\right )\\ &=-\frac{\sin \left (a+\frac{b}{x^2}\right )}{2 b x}+\frac{\operatorname{Subst}\left (\int \sin \left (a+b x^2\right ) \, dx,x,\frac{1}{x}\right )}{2 b}\\ &=-\frac{\sin \left (a+\frac{b}{x^2}\right )}{2 b x}+\frac{\cos (a) \operatorname{Subst}\left (\int \sin \left (b x^2\right ) \, dx,x,\frac{1}{x}\right )}{2 b}+\frac{\sin (a) \operatorname{Subst}\left (\int \cos \left (b x^2\right ) \, dx,x,\frac{1}{x}\right )}{2 b}\\ &=\frac{\sqrt{\frac{\pi }{2}} \cos (a) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{x}\right )}{2 b^{3/2}}+\frac{\sqrt{\frac{\pi }{2}} C\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{x}\right ) \sin (a)}{2 b^{3/2}}-\frac{\sin \left (a+\frac{b}{x^2}\right )}{2 b x}\\ \end{align*}

Mathematica [A]  time = 0.155535, size = 88, normalized size = 0.91 \[ \frac{\sqrt{2 \pi } x \sin (a) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{x}\right )+\sqrt{2 \pi } x \cos (a) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{x}\right )-2 \sqrt{b} \sin \left (a+\frac{b}{x^2}\right )}{4 b^{3/2} x} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b/x^2]/x^4,x]

[Out]

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

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Maple [A]  time = 0.036, size = 64, normalized size = 0.7 \begin{align*} -{\frac{1}{2\,bx}\sin \left ( a+{\frac{b}{{x}^{2}}} \right ) }+{\frac{\sqrt{2}\sqrt{\pi }}{4} \left ( \cos \left ( a \right ){\it FresnelS} \left ({\frac{\sqrt{2}}{\sqrt{\pi }x}\sqrt{b}} \right ) +\sin \left ( a \right ){\it FresnelC} \left ({\frac{\sqrt{2}}{\sqrt{\pi }x}\sqrt{b}} \right ) \right ){b}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a+b/x^2)/x^4,x)

[Out]

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

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Maxima [C]  time = 1.33186, size = 359, normalized size = 3.7 \begin{align*} \frac{{\left ({\left (\Gamma \left (\frac{3}{2}, \frac{i \, b}{x^{2}}\right ) + \Gamma \left (\frac{3}{2}, -\frac{i \, b}{x^{2}}\right )\right )} \cos \left (\frac{3}{4} \, \pi + \frac{3}{2} \, \arctan \left (0, b\right )\right ) +{\left (\Gamma \left (\frac{3}{2}, \frac{i \, b}{x^{2}}\right ) + \Gamma \left (\frac{3}{2}, -\frac{i \, b}{x^{2}}\right )\right )} \cos \left (-\frac{3}{4} \, \pi + \frac{3}{2} \, \arctan \left (0, b\right )\right ) -{\left (i \, \Gamma \left (\frac{3}{2}, \frac{i \, b}{x^{2}}\right ) - i \, \Gamma \left (\frac{3}{2}, -\frac{i \, b}{x^{2}}\right )\right )} \sin \left (\frac{3}{4} \, \pi + \frac{3}{2} \, \arctan \left (0, b\right )\right ) -{\left (-i \, \Gamma \left (\frac{3}{2}, \frac{i \, b}{x^{2}}\right ) + i \, \Gamma \left (\frac{3}{2}, -\frac{i \, b}{x^{2}}\right )\right )} \sin \left (-\frac{3}{4} \, \pi + \frac{3}{2} \, \arctan \left (0, b\right )\right )\right )} \cos \left (a\right ) -{\left ({\left (i \, \Gamma \left (\frac{3}{2}, \frac{i \, b}{x^{2}}\right ) - i \, \Gamma \left (\frac{3}{2}, -\frac{i \, b}{x^{2}}\right )\right )} \cos \left (\frac{3}{4} \, \pi + \frac{3}{2} \, \arctan \left (0, b\right )\right ) +{\left (i \, \Gamma \left (\frac{3}{2}, \frac{i \, b}{x^{2}}\right ) - i \, \Gamma \left (\frac{3}{2}, -\frac{i \, b}{x^{2}}\right )\right )} \cos \left (-\frac{3}{4} \, \pi + \frac{3}{2} \, \arctan \left (0, b\right )\right ) +{\left (\Gamma \left (\frac{3}{2}, \frac{i \, b}{x^{2}}\right ) + \Gamma \left (\frac{3}{2}, -\frac{i \, b}{x^{2}}\right )\right )} \sin \left (\frac{3}{4} \, \pi + \frac{3}{2} \, \arctan \left (0, b\right )\right ) -{\left (\Gamma \left (\frac{3}{2}, \frac{i \, b}{x^{2}}\right ) + \Gamma \left (\frac{3}{2}, -\frac{i \, b}{x^{2}}\right )\right )} \sin \left (-\frac{3}{4} \, \pi + \frac{3}{2} \, \arctan \left (0, b\right )\right )\right )} \sin \left (a\right )}{8 \, x^{3} \left (\frac{{\left | b \right |}}{x^{2}}\right )^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b/x^2)/x^4,x, algorithm="maxima")

[Out]

1/8*(((gamma(3/2, I*b/x^2) + gamma(3/2, -I*b/x^2))*cos(3/4*pi + 3/2*arctan2(0, b)) + (gamma(3/2, I*b/x^2) + ga
mma(3/2, -I*b/x^2))*cos(-3/4*pi + 3/2*arctan2(0, b)) - (I*gamma(3/2, I*b/x^2) - I*gamma(3/2, -I*b/x^2))*sin(3/
4*pi + 3/2*arctan2(0, b)) - (-I*gamma(3/2, I*b/x^2) + I*gamma(3/2, -I*b/x^2))*sin(-3/4*pi + 3/2*arctan2(0, b))
)*cos(a) - ((I*gamma(3/2, I*b/x^2) - I*gamma(3/2, -I*b/x^2))*cos(3/4*pi + 3/2*arctan2(0, b)) + (I*gamma(3/2, I
*b/x^2) - I*gamma(3/2, -I*b/x^2))*cos(-3/4*pi + 3/2*arctan2(0, b)) + (gamma(3/2, I*b/x^2) + gamma(3/2, -I*b/x^
2))*sin(3/4*pi + 3/2*arctan2(0, b)) - (gamma(3/2, I*b/x^2) + gamma(3/2, -I*b/x^2))*sin(-3/4*pi + 3/2*arctan2(0
, b)))*sin(a))/(x^3*(abs(b)/x^2)^(3/2))

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Fricas [A]  time = 1.66973, size = 235, normalized size = 2.42 \begin{align*} \frac{\sqrt{2} \pi x \sqrt{\frac{b}{\pi }} \cos \left (a\right ) \operatorname{S}\left (\frac{\sqrt{2} \sqrt{\frac{b}{\pi }}}{x}\right ) + \sqrt{2} \pi x \sqrt{\frac{b}{\pi }} \operatorname{C}\left (\frac{\sqrt{2} \sqrt{\frac{b}{\pi }}}{x}\right ) \sin \left (a\right ) - 2 \, b \sin \left (\frac{a x^{2} + b}{x^{2}}\right )}{4 \, b^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b/x^2)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos{\left (a + \frac{b}{x^{2}} \right )}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b/x**2)/x**4,x)

[Out]

Integral(cos(a + b/x**2)/x**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (a + \frac{b}{x^{2}}\right )}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b/x^2)/x^4,x, algorithm="giac")

[Out]

integrate(cos(a + b/x^2)/x^4, x)