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

Optimal. Leaf size=80 \[ \sqrt {2 \pi } \left (-\sqrt {b}\right ) \cos (a) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{x}\right )+\sqrt {2 \pi } \sqrt {b} \sin (a) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{x}\right )+x \sin \left (a+\frac {b}{x^2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {3359, 3387, 3354, 3352, 3351} \[ \sqrt {2 \pi } \left (-\sqrt {b}\right ) \cos (a) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {b}}{x}\right )+\sqrt {2 \pi } \sqrt {b} \sin (a) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{x}\right )+x \sin \left (a+\frac {b}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/x^2],x]

[Out]

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

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 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 3359

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

Rule 3387

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

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 81, normalized size = 1.01 \[ -\sqrt {2 \pi } \sqrt {b} \left (\cos (a) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{x}\right )-\sin (a) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{x}\right )\right )+x \sin (a) \cos \left (\frac {b}{x^2}\right )+x \cos (a) \sin \left (\frac {b}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/x^2],x]

[Out]

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

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fricas [A]  time = 0.71, size = 74, normalized size = 0.92 \[ -\sqrt {2} \pi \sqrt {\frac {b}{\pi }} \cos \relax (a) \operatorname {C}\left (\frac {\sqrt {2} \sqrt {\frac {b}{\pi }}}{x}\right ) + \sqrt {2} \pi \sqrt {\frac {b}{\pi }} \operatorname {S}\left (\frac {\sqrt {2} \sqrt {\frac {b}{\pi }}}{x}\right ) \sin \relax (a) + x \sin \left (\frac {a x^{2} + b}{x^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(a + b/x^2), x)

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maple [A]  time = 0.03, size = 59, normalized size = 0.74 \[ x \sin \left (a +\frac {b}{x^{2}}\right )-\sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \left (\cos \relax (a ) \FresnelC \left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, x}\right )-\sin \relax (a ) \mathrm {S}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, x}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*sin(a+b/x^2)-b^(1/2)*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 = 0.39, size = 127, normalized size = 1.59 \[ \frac {\sqrt {2} {\left (2 \, \sqrt {2} b x^{2} \sqrt {\frac {1}{x^{4}}} \sin \left (\frac {a x^{2} + b}{x^{2}}\right ) + {\left ({\left (\left (i - 1\right ) \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {\frac {i \, b}{x^{2}}}\right ) - 1\right )} - \left (i + 1\right ) \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-\frac {i \, b}{x^{2}}}\right ) - 1\right )}\right )} \cos \relax (a) + {\left (\left (i + 1\right ) \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {\frac {i \, b}{x^{2}}}\right ) - 1\right )} - \left (i - 1\right ) \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-\frac {i \, b}{x^{2}}}\right ) - 1\right )}\right )} \sin \relax (a)\right )} b \left (\frac {b^{2}}{x^{4}}\right )^{\frac {1}{4}}\right )} \sqrt {x^{4}}}{4 \, b x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b/x^2),x)

[Out]

int(sin(a + b/x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + b/x**2), x)

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