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

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

[Out]

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

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Rubi [A]  time = 0.0585725, 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 = {3409, 3385, 3354, 3352, 3351} \[ -\frac{\sqrt{\frac{\pi }{2}} \cos (a) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{x}\right )}{2 b^{3/2}}+\frac{\sqrt{\frac{\pi }{2}} \sin (a) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{x}\right )}{2 b^{3/2}}+\frac{\cos \left (a+\frac{b}{x^2}\right )}{2 b x} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/x^2]/x^4,x]

[Out]

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

Rule 3409

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> -Subst[Int[(a + b*Sin[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 3385

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

Rule 3354

Int[Cos[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Cos[c], Int[Cos[d*(e + f*x)^2], x], x] - Dist[
Sin[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{\sin \left (a+\frac{b}{x^2}\right )}{x^4} \, dx &=-\operatorname{Subst}\left (\int x^2 \sin \left (a+b x^2\right ) \, dx,x,\frac{1}{x}\right )\\ &=\frac{\cos \left (a+\frac{b}{x^2}\right )}{2 b x}-\frac{\operatorname{Subst}\left (\int \cos \left (a+b x^2\right ) \, dx,x,\frac{1}{x}\right )}{2 b}\\ &=\frac{\cos \left (a+\frac{b}{x^2}\right )}{2 b x}-\frac{\cos (a) \operatorname{Subst}\left (\int \cos \left (b x^2\right ) \, dx,x,\frac{1}{x}\right )}{2 b}+\frac{\sin (a) \operatorname{Subst}\left (\int \sin \left (b x^2\right ) \, dx,x,\frac{1}{x}\right )}{2 b}\\ &=\frac{\cos \left (a+\frac{b}{x^2}\right )}{2 b x}-\frac{\sqrt{\frac{\pi }{2}} \cos (a) C\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{x}\right )}{2 b^{3/2}}+\frac{\sqrt{\frac{\pi }{2}} S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{x}\right ) \sin (a)}{2 b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.155992, size = 89, normalized size = 0.92 \[ \frac{-\sqrt{2 \pi } x \cos (a) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{x}\right )+\sqrt{2 \pi } x \sin (a) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{x}\right )+2 \sqrt{b} \cos \left (a+\frac{b}{x^2}\right )}{4 b^{3/2} x} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/x^2]/x^4,x]

[Out]

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

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Maple [A]  time = 0.009, size = 65, normalized size = 0.7 \begin{align*}{\frac{1}{2\,bx}\cos \left ( a+{\frac{b}{{x}^{2}}} \right ) }-{\frac{\sqrt{2}\sqrt{\pi }}{4} \left ( \cos \left ( a \right ){\it FresnelC} \left ({\frac{\sqrt{2}}{\sqrt{\pi }x}\sqrt{b}} \right ) -\sin \left ( a \right ){\it FresnelS} \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(sin(a+b/x^2)/x^4,x)

[Out]

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

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Maxima [C]  time = 1.16273, size = 359, normalized size = 3.7 \begin{align*} -\frac{{\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 )} \cos \left (a\right ) -{\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 )} \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(sin(a+b/x^2)/x^4,x, algorithm="maxima")

[Out]

-1/8*(((-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
)))*cos(a) - ((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) + 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)
)*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*arctan
2(0, b)))*sin(a))/(x^3*(abs(b)/x^2)^(3/2))

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Fricas [A]  time = 1.79547, size = 236, normalized size = 2.43 \begin{align*} -\frac{\sqrt{2} \pi x \sqrt{\frac{b}{\pi }} \cos \left (a\right ) \operatorname{C}\left (\frac{\sqrt{2} \sqrt{\frac{b}{\pi }}}{x}\right ) - \sqrt{2} \pi x \sqrt{\frac{b}{\pi }} \operatorname{S}\left (\frac{\sqrt{2} \sqrt{\frac{b}{\pi }}}{x}\right ) \sin \left (a\right ) - 2 \, b \cos \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(sin(a+b/x^2)/x^4,x, algorithm="fricas")

[Out]

-1/4*(sqrt(2)*pi*x*sqrt(b/pi)*cos(a)*fresnel_cos(sqrt(2)*sqrt(b/pi)/x) - sqrt(2)*pi*x*sqrt(b/pi)*fresnel_sin(s
qrt(2)*sqrt(b/pi)/x)*sin(a) - 2*b*cos((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{\sin{\left (a + \frac{b}{x^{2}} \right )}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + b/x**2)/x**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \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(sin(a+b/x^2)/x^4,x, algorithm="giac")

[Out]

integrate(sin(a + b/x^2)/x^4, x)