3.7 \(\int \frac{\cos (a+b x^2)}{x^3} \, dx\)

Optimal. Leaf size=42 \[ -\frac{1}{2} b \sin (a) \text{CosIntegral}\left (b x^2\right )-\frac{1}{2} b \cos (a) \text{Si}\left (b x^2\right )-\frac{\cos \left (a+b x^2\right )}{2 x^2} \]

[Out]

-Cos[a + b*x^2]/(2*x^2) - (b*CosIntegral[b*x^2]*Sin[a])/2 - (b*Cos[a]*SinIntegral[b*x^2])/2

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Rubi [A]  time = 0.0892469, antiderivative size = 42, 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 = {3380, 3297, 3303, 3299, 3302} \[ -\frac{1}{2} b \sin (a) \text{CosIntegral}\left (b x^2\right )-\frac{1}{2} b \cos (a) \text{Si}\left (b x^2\right )-\frac{\cos \left (a+b x^2\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x^2]/x^3,x]

[Out]

-Cos[a + b*x^2]/(2*x^2) - (b*CosIntegral[b*x^2]*Sin[a])/2 - (b*Cos[a]*SinIntegral[b*x^2])/2

Rule 3380

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3297

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

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

\begin{align*} \int \frac{\cos \left (a+b x^2\right )}{x^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\cos (a+b x)}{x^2} \, dx,x,x^2\right )\\ &=-\frac{\cos \left (a+b x^2\right )}{2 x^2}-\frac{1}{2} b \operatorname{Subst}\left (\int \frac{\sin (a+b x)}{x} \, dx,x,x^2\right )\\ &=-\frac{\cos \left (a+b x^2\right )}{2 x^2}-\frac{1}{2} (b \cos (a)) \operatorname{Subst}\left (\int \frac{\sin (b x)}{x} \, dx,x,x^2\right )-\frac{1}{2} (b \sin (a)) \operatorname{Subst}\left (\int \frac{\cos (b x)}{x} \, dx,x,x^2\right )\\ &=-\frac{\cos \left (a+b x^2\right )}{2 x^2}-\frac{1}{2} b \text{Ci}\left (b x^2\right ) \sin (a)-\frac{1}{2} b \cos (a) \text{Si}\left (b x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0684203, size = 42, normalized size = 1. \[ -\frac{b x^2 \sin (a) \text{CosIntegral}\left (b x^2\right )+b x^2 \cos (a) \text{Si}\left (b x^2\right )+\cos \left (a+b x^2\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x^2]/x^3,x]

[Out]

-(Cos[a + b*x^2] + b*x^2*CosIntegral[b*x^2]*Sin[a] + b*x^2*Cos[a]*SinIntegral[b*x^2])/(2*x^2)

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Maple [A]  time = 0.024, size = 39, normalized size = 0.9 \begin{align*} -{\frac{\cos \left ( b{x}^{2}+a \right ) }{2\,{x}^{2}}}-b \left ({\frac{\cos \left ( a \right ){\it Si} \left ( b{x}^{2} \right ) }{2}}+{\frac{{\it Ci} \left ( b{x}^{2} \right ) \sin \left ( a \right ) }{2}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x^2+a)/x^3,x)

[Out]

-1/2*cos(b*x^2+a)/x^2-b*(1/2*cos(a)*Si(b*x^2)+1/2*Ci(b*x^2)*sin(a))

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Maxima [C]  time = 1.54482, size = 65, normalized size = 1.55 \begin{align*} -\frac{1}{4} \,{\left ({\left (i \, \Gamma \left (-1, i \, b x^{2}\right ) - i \, \Gamma \left (-1, -i \, b x^{2}\right )\right )} \cos \left (a\right ) +{\left (\Gamma \left (-1, i \, b x^{2}\right ) + \Gamma \left (-1, -i \, b x^{2}\right )\right )} \sin \left (a\right )\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x^2+a)/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.59318, size = 178, normalized size = 4.24 \begin{align*} -\frac{2 \, b x^{2} \cos \left (a\right ) \operatorname{Si}\left (b x^{2}\right ) +{\left (b x^{2} \operatorname{Ci}\left (b x^{2}\right ) + b x^{2} \operatorname{Ci}\left (-b x^{2}\right )\right )} \sin \left (a\right ) + 2 \, \cos \left (b x^{2} + a\right )}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x^2+a)/x^3,x, algorithm="fricas")

[Out]

-1/4*(2*b*x^2*cos(a)*sin_integral(b*x^2) + (b*x^2*cos_integral(b*x^2) + b*x^2*cos_integral(-b*x^2))*sin(a) + 2
*cos(b*x^2 + a))/x^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos{\left (a + b x^{2} \right )}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x**2+a)/x**3,x)

[Out]

Integral(cos(a + b*x**2)/x**3, x)

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Giac [B]  time = 1.10915, size = 117, normalized size = 2.79 \begin{align*} -\frac{{\left (b x^{2} + a\right )} b^{2} \operatorname{Ci}\left (b x^{2}\right ) \sin \left (a\right ) - a b^{2} \operatorname{Ci}\left (b x^{2}\right ) \sin \left (a\right ) +{\left (b x^{2} + a\right )} b^{2} \cos \left (a\right ) \operatorname{Si}\left (b x^{2}\right ) - a b^{2} \cos \left (a\right ) \operatorname{Si}\left (b x^{2}\right ) + b^{2} \cos \left (b x^{2} + a\right )}{2 \, b^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x^2+a)/x^3,x, algorithm="giac")

[Out]

-1/2*((b*x^2 + a)*b^2*cos_integral(b*x^2)*sin(a) - a*b^2*cos_integral(b*x^2)*sin(a) + (b*x^2 + a)*b^2*cos(a)*s
in_integral(b*x^2) - a*b^2*cos(a)*sin_integral(b*x^2) + b^2*cos(b*x^2 + a))/(b^2*x^2)