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

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

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 97, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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
]

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*}

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Mathematica [A]  time = 0.16, size = 88, normalized size = 0.91 \[ \frac {\sqrt {2 \pi } x \sin (a) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{x}\right )+\sqrt {2 \pi } x \cos (a) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{x}\right )-2 \sqrt {b} \sin \left (a+\frac {b}{x^2}\right )}{4 b^{3/2} x} \]

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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fricas [A]  time = 0.68, size = 84, normalized size = 0.87 \[ \frac {\sqrt {2} \pi x \sqrt {\frac {b}{\pi }} \cos \relax (a) \operatorname {S}\left (\frac {\sqrt {2} \sqrt {\frac {b}{\pi }}}{x}\right ) + \sqrt {2} \pi x \sqrt {\frac {b}{\pi }} \operatorname {C}\left (\frac {\sqrt {2} \sqrt {\frac {b}{\pi }}}{x}\right ) \sin \relax (a) - 2 \, b \sin \left (\frac {a x^{2} + b}{x^{2}}\right )}{4 \, b^{2} x} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (a + \frac {b}{x^{2}}\right )}{x^{4}}\,{d x} \]

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)

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maple [A]  time = 0.02, size = 64, normalized size = 0.66 \[ -\frac {\sin \left (a +\frac {b}{x^{2}}\right )}{2 b x}+\frac {\sqrt {2}\, \sqrt {\pi }\, \left (\cos \relax (a ) \mathrm {S}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, x}\right )+\sin \relax (a ) \FresnelC \left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, x}\right )\right )}{4 b^{\frac {3}{2}}} \]

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 = 1.29, size = 74, normalized size = 0.76 \[ \frac {\sqrt {2} {\left (x^{4}\right )}^{\frac {3}{2}} {\left ({\left (-\left (i + 1\right ) \, \Gamma \left (\frac {3}{2}, \frac {i \, b}{x^{2}}\right ) + \left (i - 1\right ) \, \Gamma \left (\frac {3}{2}, -\frac {i \, b}{x^{2}}\right )\right )} \cos \relax (a) + {\left (\left (i - 1\right ) \, \Gamma \left (\frac {3}{2}, \frac {i \, b}{x^{2}}\right ) - \left (i + 1\right ) \, \Gamma \left (\frac {3}{2}, -\frac {i \, b}{x^{2}}\right )\right )} \sin \relax (a)\right )} \left (\frac {b^{2}}{x^{4}}\right )^{\frac {3}{4}}}{8 \, b^{3} x^{3}} \]

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*sqrt(2)*((-(I + 1)*gamma(3/2, I*b/x^2) + (I - 1)*gamma(3/2, -I*b/x^2))*cos(a) + ((I - 1)*gamma(3/2, I*b/x^
2) - (I + 1)*gamma(3/2, -I*b/x^2))*sin(a))*(x^4)^(3/2)*(b^2/x^4)^(3/4)/(b^3*x^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (a+\frac {b}{x^2}\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b/x^2)/x^4,x)

[Out]

int(cos(a + b/x^2)/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos {\left (a + \frac {b}{x^{2}} \right )}}{x^{4}}\, dx \]

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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