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

Optimal. Leaf size=91 \[ -\frac{3}{8} b \sin (a) \text{CosIntegral}\left (b x^2\right )-\frac{3}{8} b \sin (3 a) \text{CosIntegral}\left (3 b x^2\right )-\frac{3}{8} b \cos (a) \text{Si}\left (b x^2\right )-\frac{3}{8} b \cos (3 a) \text{Si}\left (3 b x^2\right )-\frac{3 \cos \left (a+b x^2\right )}{8 x^2}-\frac{\cos \left (3 \left (a+b x^2\right )\right )}{8 x^2} \]

[Out]

(-3*Cos[a + b*x^2])/(8*x^2) - Cos[3*(a + b*x^2)]/(8*x^2) - (3*b*CosIntegral[b*x^2]*Sin[a])/8 - (3*b*CosIntegra
l[3*b*x^2]*Sin[3*a])/8 - (3*b*Cos[a]*SinIntegral[b*x^2])/8 - (3*b*Cos[3*a]*SinIntegral[3*b*x^2])/8

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Rubi [A]  time = 0.201348, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3404, 3380, 3297, 3303, 3299, 3302} \[ -\frac{3}{8} b \sin (a) \text{CosIntegral}\left (b x^2\right )-\frac{3}{8} b \sin (3 a) \text{CosIntegral}\left (3 b x^2\right )-\frac{3}{8} b \cos (a) \text{Si}\left (b x^2\right )-\frac{3}{8} b \cos (3 a) \text{Si}\left (3 b x^2\right )-\frac{3 \cos \left (a+b x^2\right )}{8 x^2}-\frac{\cos \left (3 \left (a+b x^2\right )\right )}{8 x^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x^2]^3/x^3,x]

[Out]

(-3*Cos[a + b*x^2])/(8*x^2) - Cos[3*(a + b*x^2)]/(8*x^2) - (3*b*CosIntegral[b*x^2]*Sin[a])/8 - (3*b*CosIntegra
l[3*b*x^2]*Sin[3*a])/8 - (3*b*Cos[a]*SinIntegral[b*x^2])/8 - (3*b*Cos[3*a]*SinIntegral[3*b*x^2])/8

Rule 3404

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandTrigReduce[(e
*x)^m, (a + b*Cos[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 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]))

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 ^3\left (a+b x^2\right )}{x^3} \, dx &=\int \left (\frac{3 \cos \left (a+b x^2\right )}{4 x^3}+\frac{\cos \left (3 a+3 b x^2\right )}{4 x^3}\right ) \, dx\\ &=\frac{1}{4} \int \frac{\cos \left (3 a+3 b x^2\right )}{x^3} \, dx+\frac{3}{4} \int \frac{\cos \left (a+b x^2\right )}{x^3} \, dx\\ &=\frac{1}{8} \operatorname{Subst}\left (\int \frac{\cos (3 a+3 b x)}{x^2} \, dx,x,x^2\right )+\frac{3}{8} \operatorname{Subst}\left (\int \frac{\cos (a+b x)}{x^2} \, dx,x,x^2\right )\\ &=-\frac{3 \cos \left (a+b x^2\right )}{8 x^2}-\frac{\cos \left (3 \left (a+b x^2\right )\right )}{8 x^2}-\frac{1}{8} (3 b) \operatorname{Subst}\left (\int \frac{\sin (a+b x)}{x} \, dx,x,x^2\right )-\frac{1}{8} (3 b) \operatorname{Subst}\left (\int \frac{\sin (3 a+3 b x)}{x} \, dx,x,x^2\right )\\ &=-\frac{3 \cos \left (a+b x^2\right )}{8 x^2}-\frac{\cos \left (3 \left (a+b x^2\right )\right )}{8 x^2}-\frac{1}{8} (3 b \cos (a)) \operatorname{Subst}\left (\int \frac{\sin (b x)}{x} \, dx,x,x^2\right )-\frac{1}{8} (3 b \cos (3 a)) \operatorname{Subst}\left (\int \frac{\sin (3 b x)}{x} \, dx,x,x^2\right )-\frac{1}{8} (3 b \sin (a)) \operatorname{Subst}\left (\int \frac{\cos (b x)}{x} \, dx,x,x^2\right )-\frac{1}{8} (3 b \sin (3 a)) \operatorname{Subst}\left (\int \frac{\cos (3 b x)}{x} \, dx,x,x^2\right )\\ &=-\frac{3 \cos \left (a+b x^2\right )}{8 x^2}-\frac{\cos \left (3 \left (a+b x^2\right )\right )}{8 x^2}-\frac{3}{8} b \text{Ci}\left (b x^2\right ) \sin (a)-\frac{3}{8} b \text{Ci}\left (3 b x^2\right ) \sin (3 a)-\frac{3}{8} b \cos (a) \text{Si}\left (b x^2\right )-\frac{3}{8} b \cos (3 a) \text{Si}\left (3 b x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.175254, size = 90, normalized size = 0.99 \[ -\frac{3 b x^2 \sin (a) \text{CosIntegral}\left (b x^2\right )+3 b x^2 \sin (3 a) \text{CosIntegral}\left (3 b x^2\right )+3 b x^2 \cos (a) \text{Si}\left (b x^2\right )+3 b x^2 \cos (3 a) \text{Si}\left (3 b x^2\right )+3 \cos \left (a+b x^2\right )+\cos \left (3 \left (a+b x^2\right )\right )}{8 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x^2]^3/x^3,x]

[Out]

-(3*Cos[a + b*x^2] + Cos[3*(a + b*x^2)] + 3*b*x^2*CosIntegral[b*x^2]*Sin[a] + 3*b*x^2*CosIntegral[3*b*x^2]*Sin
[3*a] + 3*b*x^2*Cos[a]*SinIntegral[b*x^2] + 3*b*x^2*Cos[3*a]*SinIntegral[3*b*x^2])/(8*x^2)

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Maple [C]  time = 0.147, size = 162, normalized size = 1.8 \begin{align*}{\frac{3\,{{\rm e}^{-3\,ia}}\pi \,{\it csgn} \left ( b{x}^{2} \right ) b}{16}}-{\frac{3\,{{\rm e}^{-3\,ia}}{\it Si} \left ( 3\,b{x}^{2} \right ) b}{8}}+{\frac{3\,i}{16}}{{\rm e}^{-3\,ia}}{\it Ei} \left ( 1,-3\,ib{x}^{2} \right ) b+{\frac{3\,\pi \,{\it csgn} \left ( b{x}^{2} \right ){{\rm e}^{-ia}}b}{16}}-{\frac{3\,{{\rm e}^{-ia}}{\it Si} \left ( b{x}^{2} \right ) b}{8}}+{\frac{3\,i}{16}}{{\rm e}^{-ia}}{\it Ei} \left ( 1,-ib{x}^{2} \right ) b-{\frac{3\,i}{16}}{{\rm e}^{ia}}b{\it Ei} \left ( 1,-ib{x}^{2} \right ) -{\frac{3\,i}{16}}{{\rm e}^{3\,ia}}b{\it Ei} \left ( 1,-3\,ib{x}^{2} \right ) -{\frac{3\,\cos \left ( b{x}^{2}+a \right ) }{8\,{x}^{2}}}-{\frac{\cos \left ( 3\,b{x}^{2}+3\,a \right ) }{8\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16*exp(-3*I*a)*Pi*csgn(b*x^2)*b-3/8*exp(-3*I*a)*Si(3*b*x^2)*b+3/16*I*exp(-3*I*a)*Ei(1,-3*I*b*x^2)*b+3/16*Pi*
csgn(b*x^2)*exp(-I*a)*b-3/8*exp(-I*a)*Si(b*x^2)*b+3/16*I*exp(-I*a)*Ei(1,-I*b*x^2)*b-3/16*I*exp(I*a)*b*Ei(1,-I*
b*x^2)-3/16*I*exp(3*I*a)*b*Ei(1,-3*I*b*x^2)-3/8*cos(b*x^2+a)/x^2-1/8*cos(3*b*x^2+3*a)/x^2

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Maxima [C]  time = 1.53945, size = 132, normalized size = 1.45 \begin{align*} -\frac{1}{16} \,{\left ({\left (3 i \, \Gamma \left (-1, 3 i \, b x^{2}\right ) - 3 i \, \Gamma \left (-1, -3 i \, b x^{2}\right )\right )} \cos \left (3 \, a\right ) +{\left (3 i \, \Gamma \left (-1, i \, b x^{2}\right ) - 3 i \, \Gamma \left (-1, -i \, b x^{2}\right )\right )} \cos \left (a\right ) + 3 \,{\left (\Gamma \left (-1, 3 i \, b x^{2}\right ) + \Gamma \left (-1, -3 i \, b x^{2}\right )\right )} \sin \left (3 \, a\right ) + 3 \,{\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)^3/x^3,x, algorithm="maxima")

[Out]

-1/16*((3*I*gamma(-1, 3*I*b*x^2) - 3*I*gamma(-1, -3*I*b*x^2))*cos(3*a) + (3*I*gamma(-1, I*b*x^2) - 3*I*gamma(-
1, -I*b*x^2))*cos(a) + 3*(gamma(-1, 3*I*b*x^2) + gamma(-1, -3*I*b*x^2))*sin(3*a) + 3*(gamma(-1, I*b*x^2) + gam
ma(-1, -I*b*x^2))*sin(a))*b

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Fricas [A]  time = 1.68307, size = 340, normalized size = 3.74 \begin{align*} -\frac{6 \, b x^{2} \cos \left (3 \, a\right ) \operatorname{Si}\left (3 \, b x^{2}\right ) + 6 \, b x^{2} \cos \left (a\right ) \operatorname{Si}\left (b x^{2}\right ) + 8 \, \cos \left (b x^{2} + a\right )^{3} + 3 \,{\left (b x^{2} \operatorname{Ci}\left (3 \, b x^{2}\right ) + b x^{2} \operatorname{Ci}\left (-3 \, b x^{2}\right )\right )} \sin \left (3 \, a\right ) + 3 \,{\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 )}{16 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(6*b*x^2*cos(3*a)*sin_integral(3*b*x^2) + 6*b*x^2*cos(a)*sin_integral(b*x^2) + 8*cos(b*x^2 + a)^3 + 3*(b
*x^2*cos_integral(3*b*x^2) + b*x^2*cos_integral(-3*b*x^2))*sin(3*a) + 3*(b*x^2*cos_integral(b*x^2) + b*x^2*cos
_integral(-b*x^2))*sin(a))/x^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos ^{3}{\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)**3/x**3,x)

[Out]

Integral(cos(a + b*x**2)**3/x**3, x)

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Giac [B]  time = 1.17793, size = 250, normalized size = 2.75 \begin{align*} -\frac{3 \,{\left (b x^{2} + a\right )} b^{2} \operatorname{Ci}\left (3 \, b x^{2}\right ) \sin \left (3 \, a\right ) - 3 \, a b^{2} \operatorname{Ci}\left (3 \, b x^{2}\right ) \sin \left (3 \, a\right ) + 3 \,{\left (b x^{2} + a\right )} b^{2} \operatorname{Ci}\left (b x^{2}\right ) \sin \left (a\right ) - 3 \, a b^{2} \operatorname{Ci}\left (b x^{2}\right ) \sin \left (a\right ) + 3 \,{\left (b x^{2} + a\right )} b^{2} \cos \left (a\right ) \operatorname{Si}\left (b x^{2}\right ) - 3 \, a b^{2} \cos \left (a\right ) \operatorname{Si}\left (b x^{2}\right ) - 3 \,{\left (b x^{2} + a\right )} b^{2} \cos \left (3 \, a\right ) \operatorname{Si}\left (-3 \, b x^{2}\right ) + 3 \, a b^{2} \cos \left (3 \, a\right ) \operatorname{Si}\left (-3 \, b x^{2}\right ) + b^{2} \cos \left (3 \, b x^{2} + 3 \, a\right ) + 3 \, b^{2} \cos \left (b x^{2} + a\right )}{8 \, b^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(3*(b*x^2 + a)*b^2*cos_integral(3*b*x^2)*sin(3*a) - 3*a*b^2*cos_integral(3*b*x^2)*sin(3*a) + 3*(b*x^2 + a
)*b^2*cos_integral(b*x^2)*sin(a) - 3*a*b^2*cos_integral(b*x^2)*sin(a) + 3*(b*x^2 + a)*b^2*cos(a)*sin_integral(
b*x^2) - 3*a*b^2*cos(a)*sin_integral(b*x^2) - 3*(b*x^2 + a)*b^2*cos(3*a)*sin_integral(-3*b*x^2) + 3*a*b^2*cos(
3*a)*sin_integral(-3*b*x^2) + b^2*cos(3*b*x^2 + 3*a) + 3*b^2*cos(b*x^2 + a))/(b^2*x^2)