3.81 \(\int x^{-1-n} \cos ^3(a+b x^n) \, dx\)

Optimal. Leaf size=113 \[ -\frac{3 b \sin (a) \text{CosIntegral}\left (b x^n\right )}{4 n}-\frac{3 b \sin (3 a) \text{CosIntegral}\left (3 b x^n\right )}{4 n}-\frac{3 b \cos (a) \text{Si}\left (b x^n\right )}{4 n}-\frac{3 b \cos (3 a) \text{Si}\left (3 b x^n\right )}{4 n}-\frac{3 x^{-n} \cos \left (a+b x^n\right )}{4 n}-\frac{x^{-n} \cos \left (3 \left (a+b x^n\right )\right )}{4 n} \]

[Out]

(-3*Cos[a + b*x^n])/(4*n*x^n) - Cos[3*(a + b*x^n)]/(4*n*x^n) - (3*b*CosIntegral[b*x^n]*Sin[a])/(4*n) - (3*b*Co
sIntegral[3*b*x^n]*Sin[3*a])/(4*n) - (3*b*Cos[a]*SinIntegral[b*x^n])/(4*n) - (3*b*Cos[3*a]*SinIntegral[3*b*x^n
])/(4*n)

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Rubi [A]  time = 0.199548, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3426, 3380, 3297, 3303, 3299, 3302} \[ -\frac{3 b \sin (a) \text{CosIntegral}\left (b x^n\right )}{4 n}-\frac{3 b \sin (3 a) \text{CosIntegral}\left (3 b x^n\right )}{4 n}-\frac{3 b \cos (a) \text{Si}\left (b x^n\right )}{4 n}-\frac{3 b \cos (3 a) \text{Si}\left (3 b x^n\right )}{4 n}-\frac{3 x^{-n} \cos \left (a+b x^n\right )}{4 n}-\frac{x^{-n} \cos \left (3 \left (a+b x^n\right )\right )}{4 n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 - n)*Cos[a + b*x^n]^3,x]

[Out]

(-3*Cos[a + b*x^n])/(4*n*x^n) - Cos[3*(a + b*x^n)]/(4*n*x^n) - (3*b*CosIntegral[b*x^n]*Sin[a])/(4*n) - (3*b*Co
sIntegral[3*b*x^n]*Sin[3*a])/(4*n) - (3*b*Cos[a]*SinIntegral[b*x^n])/(4*n) - (3*b*Cos[3*a]*SinIntegral[3*b*x^n
])/(4*n)

Rule 3426

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

Mathematica [A]  time = 0.234899, size = 95, normalized size = 0.84 \[ -\frac{x^{-n} \left (3 b \sin (a) x^n \text{CosIntegral}\left (b x^n\right )+3 b \sin (3 a) x^n \text{CosIntegral}\left (3 b x^n\right )+3 b \cos (a) x^n \text{Si}\left (b x^n\right )+3 b \cos (3 a) x^n \text{Si}\left (3 b x^n\right )+3 \cos \left (a+b x^n\right )+\cos \left (3 \left (a+b x^n\right )\right )\right )}{4 n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 - n)*Cos[a + b*x^n]^3,x]

[Out]

-(3*Cos[a + b*x^n] + Cos[3*(a + b*x^n)] + 3*b*x^n*CosIntegral[b*x^n]*Sin[a] + 3*b*x^n*CosIntegral[3*b*x^n]*Sin
[3*a] + 3*b*x^n*Cos[a]*SinIntegral[b*x^n] + 3*b*x^n*Cos[3*a]*SinIntegral[3*b*x^n])/(4*n*x^n)

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Maple [A]  time = 0.05, size = 101, normalized size = 0.9 \begin{align*}{\frac{3\,b}{4\,n} \left ( -{\frac{\cos \left ( a+b{x}^{n} \right ) }{b{x}^{n}}}-{\it Si} \left ( b{x}^{n} \right ) \cos \left ( a \right ) -{\it Ci} \left ( b{x}^{n} \right ) \sin \left ( a \right ) \right ) }+{\frac{3\,b}{4\,n} \left ( -{\frac{\cos \left ( 3\,a+3\,b{x}^{n} \right ) }{3\,b{x}^{n}}}-{\it Si} \left ( 3\,b{x}^{n} \right ) \cos \left ( 3\,a \right ) -{\it Ci} \left ( 3\,b{x}^{n} \right ) \sin \left ( 3\,a \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-n-1)*cos(a+b*x^n)^3,x)

[Out]

3/4/n*b*(-cos(a+b*x^n)/(x^n)/b-Si(b*x^n)*cos(a)-Ci(b*x^n)*sin(a))+3/4/n*b*(-1/3*cos(3*a+3*b*x^n)/(x^n)/b-Si(3*
b*x^n)*cos(3*a)-Ci(3*b*x^n)*sin(3*a))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{-n - 1} \cos \left (b x^{n} + a\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-n)*cos(a+b*x^n)^3,x, algorithm="maxima")

[Out]

integrate(x^(-n - 1)*cos(b*x^n + a)^3, x)

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Fricas [A]  time = 1.85088, size = 366, normalized size = 3.24 \begin{align*} -\frac{3 \, b x^{n} \operatorname{Ci}\left (3 \, b x^{n}\right ) \sin \left (3 \, a\right ) + 3 \, b x^{n} \operatorname{Ci}\left (-3 \, b x^{n}\right ) \sin \left (3 \, a\right ) + 3 \, b x^{n} \operatorname{Ci}\left (b x^{n}\right ) \sin \left (a\right ) + 3 \, b x^{n} \operatorname{Ci}\left (-b x^{n}\right ) \sin \left (a\right ) + 6 \, b x^{n} \cos \left (3 \, a\right ) \operatorname{Si}\left (3 \, b x^{n}\right ) + 6 \, b x^{n} \cos \left (a\right ) \operatorname{Si}\left (b x^{n}\right ) + 8 \, \cos \left (b x^{n} + a\right )^{3}}{8 \, n x^{n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-n)*cos(a+b*x^n)^3,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1-n)*cos(a+b*x**n)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{-n - 1} \cos \left (b x^{n} + a\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-n)*cos(a+b*x^n)^3,x, algorithm="giac")

[Out]

integrate(x^(-n - 1)*cos(b*x^n + a)^3, x)