3.2.21 \(\int \frac {\sin (a+\frac {b}{x^2})}{x^2} \, dx\) [121]

Optimal. Leaf size=75 \[ -\frac {\sqrt {\frac {\pi }{2}} \cos (a) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{x}\right )}{\sqrt {b}}-\frac {\sqrt {\frac {\pi }{2}} C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{x}\right ) \sin (a)}{\sqrt {b}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 75, normalized size of antiderivative = 1.00, 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 = {3464, 3434, 3433, 3432} \begin {gather*} -\frac {\sqrt {\frac {\pi }{2}} \sin (a) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {b}}{x}\right )}{\sqrt {b}}-\frac {\sqrt {\frac {\pi }{2}} \cos (a) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{x}\right )}{\sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/x^2]/x^2,x]

[Out]

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

Rule 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3434

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 3464

Int[(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^(n_)], x_Symbol] :> Dist[2/n, Subst[Int[Sin[a + b*x^2], x], x, x^(n/2)],
x] /; FreeQ[{a, b, m, n}, x] && EqQ[m, n/2 - 1]

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 61, normalized size = 0.81 \begin {gather*} -\frac {\sqrt {\frac {\pi }{2}} \left (\cos (a) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{x}\right )+C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{x}\right ) \sin (a)\right )}{\sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/x^2]/x^2,x]

[Out]

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

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Maple [A]
time = 0.04, size = 47, normalized size = 0.63

method	result	size
derivativedivides	\(-\frac {\sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (a \right ) \mathrm {S}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, x}\right )+\sin \left (a \right ) \FresnelC \left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, x}\right )\right )}{2 \sqrt {b}}\)	\(47\)
default	\(-\frac {\sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (a \right ) \mathrm {S}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, x}\right )+\sin \left (a \right ) \FresnelC \left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, x}\right )\right )}{2 \sqrt {b}}\)	\(47\)
meijerg	\(-\frac {\cos \left (a \right ) \mathrm {S}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, x}\right ) \sqrt {2}\, \sqrt {\pi }}{2 \sqrt {b}}-\frac {\FresnelC \left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, x}\right ) \sin \left (a \right ) \sqrt {2}\, \sqrt {\pi }}{2 \sqrt {b}}\)	\(56\)
risch	\(\frac {i {\mathrm e}^{i a} \sqrt {\pi }\, \erf \left (\frac {\sqrt {-i b}}{x}\right )}{4 \sqrt {-i b}}-\frac {i {\mathrm e}^{-i a} \sqrt {\pi }\, \erf \left (\frac {\sqrt {i b}}{x}\right )}{4 \sqrt {i b}}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a+b/x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.34, size = 98, normalized size = 1.31 \begin {gather*} -\frac {\sqrt {2} \sqrt {x^{4}} {\left ({\left (\left (i + 1\right ) \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {\frac {i \, b}{x^{2}}}\right ) - 1\right )} - \left (i - 1\right ) \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-\frac {i \, b}{x^{2}}}\right ) - 1\right )}\right )} \cos \left (a\right ) + {\left (-\left (i - 1\right ) \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {\frac {i \, b}{x^{2}}}\right ) - 1\right )} + \left (i + 1\right ) \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-\frac {i \, b}{x^{2}}}\right ) - 1\right )}\right )} \sin \left (a\right )\right )} \left (\frac {b^{2}}{x^{4}}\right )^{\frac {1}{4}}}{8 \, b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.38, size = 64, normalized size = 0.85 \begin {gather*} -\frac {\sqrt {2} \pi \sqrt {\frac {b}{\pi }} \cos \left (a\right ) \operatorname {S}\left (\frac {\sqrt {2} \sqrt {\frac {b}{\pi }}}{x}\right ) + \sqrt {2} \pi \sqrt {\frac {b}{\pi }} \operatorname {C}\left (\frac {\sqrt {2} \sqrt {\frac {b}{\pi }}}{x}\right ) \sin \left (a\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin {\left (a + \frac {b}{x^{2}} \right )}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + b/x**2)/x**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(a + b/x^2)/x^2, x)

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Mupad [B]
time = 4.80, size = 55, normalized size = 0.73 \begin {gather*} -\frac {\sqrt {2}\,\sqrt {\pi }\,\mathrm {S}\left (\frac {\sqrt {2}\,\sqrt {b}}{x\,\sqrt {\pi }}\right )\,\cos \left (a\right )}{2\,\sqrt {b}}-\frac {\sqrt {2}\,\sqrt {\pi }\,\mathrm {C}\left (\frac {\sqrt {2}\,\sqrt {b}}{x\,\sqrt {\pi }}\right )\,\sin \left (a\right )}{2\,\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b/x^2)/x^2,x)

[Out]

- (2^(1/2)*pi^(1/2)*fresnels((2^(1/2)*b^(1/2))/(x*pi^(1/2)))*cos(a))/(2*b^(1/2)) - (2^(1/2)*pi^(1/2)*fresnelc(
(2^(1/2)*b^(1/2))/(x*pi^(1/2)))*sin(a))/(2*b^(1/2))

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