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

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

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 42, 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 = {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 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 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]

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 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]))

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*}

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Mathematica [A]  time = 0.07, size = 42, normalized size = 1.00 \[ -\frac {b x^2 \sin (a) \text {Ci}\left (b x^2\right )+b x^2 \cos (a) \text {Si}\left (b x^2\right )+\cos \left (a+b x^2\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x^2]/x^3,x]

[Out]

-1/2*(Cos[a + b*x^2] + b*x^2*CosIntegral[b*x^2]*Sin[a] + b*x^2*Cos[a]*SinIntegral[b*x^2])/x^2

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fricas [A]  time = 0.93, size = 57, normalized size = 1.36 \[ -\frac {2 \, b x^{2} \cos \relax (a) \operatorname {Si}\left (b x^{2}\right ) + {\left (b x^{2} \operatorname {Ci}\left (b x^{2}\right ) + b x^{2} \operatorname {Ci}\left (-b x^{2}\right )\right )} \sin \relax (a) + 2 \, \cos \left (b x^{2} + a\right )}{4 \, x^{2}} \]

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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giac [B]  time = 0.49, size = 87, normalized size = 2.07 \[ -\frac {{\left (b x^{2} + a\right )} b^{2} \operatorname {Ci}\left (b x^{2}\right ) \sin \relax (a) - a b^{2} \operatorname {Ci}\left (b x^{2}\right ) \sin \relax (a) + {\left (b x^{2} + a\right )} b^{2} \cos \relax (a) \operatorname {Si}\left (b x^{2}\right ) - a b^{2} \cos \relax (a) \operatorname {Si}\left (b x^{2}\right ) + b^{2} \cos \left (b x^{2} + a\right )}{2 \, b^{2} x^{2}} \]

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)

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maple [A]  time = 0.02, size = 39, normalized size = 0.93 \[ -\frac {\cos \left (b \,x^{2}+a \right )}{2 x^{2}}-b \left (\frac {\cos \relax (a ) \Si \left (b \,x^{2}\right )}{2}+\frac {\sin \relax (a ) \Ci \left (b \,x^{2}\right )}{2}\right ) \]

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*sin(a)*Ci(b*x^2))

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maxima [C]  time = 1.07, size = 48, normalized size = 1.14 \[ -\frac {1}{4} \, {\left ({\left (i \, \Gamma \left (-1, i \, b x^{2}\right ) - i \, \Gamma \left (-1, -i \, b x^{2}\right )\right )} \cos \relax (a) + {\left (\Gamma \left (-1, i \, b x^{2}\right ) + \Gamma \left (-1, -i \, b x^{2}\right )\right )} \sin \relax (a)\right )} b \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\cos \left (b\,x^2+a\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x^2)/x^3,x)

[Out]

int(cos(a + b*x^2)/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos {\left (a + b x^{2} \right )}}{x^{3}}\, dx \]

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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