3.80 \(\int x^{-1-n} \cos ^2(a+b x^n) \, dx\)

Optimal. Leaf size=69 \[ -\frac{b \sin (2 a) \text{CosIntegral}\left (2 b x^n\right )}{n}-\frac{b \cos (2 a) \text{Si}\left (2 b x^n\right )}{n}-\frac{x^{-n} \cos \left (2 \left (a+b x^n\right )\right )}{2 n}-\frac{x^{-n}}{2 n} \]

[Out]

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

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Rubi [A]  time = 0.120562, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{b \sin (2 a) \text{CosIntegral}\left (2 b x^n\right )}{n}-\frac{b \cos (2 a) \text{Si}\left (2 b x^n\right )}{n}-\frac{x^{-n} \cos \left (2 \left (a+b x^n\right )\right )}{2 n}-\frac{x^{-n}}{2 n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 - n)*Cos[a + b*x^n]^2,x]

[Out]

-1/(2*n*x^n) - Cos[2*(a + b*x^n)]/(2*n*x^n) - (b*CosIntegral[2*b*x^n]*Sin[2*a])/n - (b*Cos[2*a]*SinIntegral[2*
b*x^n])/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 ^2\left (a+b x^n\right ) \, dx &=\int \left (\frac{x^{-1-n}}{2}+\frac{1}{2} x^{-1-n} \cos \left (2 a+2 b x^n\right )\right ) \, dx\\ &=-\frac{x^{-n}}{2 n}+\frac{1}{2} \int x^{-1-n} \cos \left (2 a+2 b x^n\right ) \, dx\\ &=-\frac{x^{-n}}{2 n}+\frac{\operatorname{Subst}\left (\int \frac{\cos (2 a+2 b x)}{x^2} \, dx,x,x^n\right )}{2 n}\\ &=-\frac{x^{-n}}{2 n}-\frac{x^{-n} \cos \left (2 \left (a+b x^n\right )\right )}{2 n}-\frac{b \operatorname{Subst}\left (\int \frac{\sin (2 a+2 b x)}{x} \, dx,x,x^n\right )}{n}\\ &=-\frac{x^{-n}}{2 n}-\frac{x^{-n} \cos \left (2 \left (a+b x^n\right )\right )}{2 n}-\frac{(b \cos (2 a)) \operatorname{Subst}\left (\int \frac{\sin (2 b x)}{x} \, dx,x,x^n\right )}{n}-\frac{(b \sin (2 a)) \operatorname{Subst}\left (\int \frac{\cos (2 b x)}{x} \, dx,x,x^n\right )}{n}\\ &=-\frac{x^{-n}}{2 n}-\frac{x^{-n} \cos \left (2 \left (a+b x^n\right )\right )}{2 n}-\frac{b \text{Ci}\left (2 b x^n\right ) \sin (2 a)}{n}-\frac{b \cos (2 a) \text{Si}\left (2 b x^n\right )}{n}\\ \end{align*}

Mathematica [A]  time = 0.186908, size = 53, normalized size = 0.77 \[ -\frac{x^{-n} \left (b \sin (2 a) x^n \text{CosIntegral}\left (2 b x^n\right )+b \cos (2 a) x^n \text{Si}\left (2 b x^n\right )+\cos ^2\left (a+b x^n\right )\right )}{n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 - n)*Cos[a + b*x^n]^2,x]

[Out]

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

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Maple [A]  time = 0.044, size = 65, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,n{x}^{n}}}+{\frac{b}{n} \left ( -{\frac{\cos \left ( 2\,a+2\,b{x}^{n} \right ) }{2\,b{x}^{n}}}-{\it Si} \left ( 2\,b{x}^{n} \right ) \cos \left ( 2\,a \right ) -{\it Ci} \left ( 2\,b{x}^{n} \right ) \sin \left ( 2\,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)^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{n x^{n} \int \frac{\cos \left (2 \, b x^{n} + 2 \, a\right )}{x x^{n}}\,{d x} - 1}{2 \, 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)^2,x, algorithm="maxima")

[Out]

1/2*(n*x^n*integrate(cos(2*b*x^n + 2*a)/(x*x^n), x) - 1)/(n*x^n)

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Fricas [A]  time = 1.73607, size = 209, normalized size = 3.03 \begin{align*} -\frac{b x^{n} \operatorname{Ci}\left (2 \, b x^{n}\right ) \sin \left (2 \, a\right ) + b x^{n} \operatorname{Ci}\left (-2 \, b x^{n}\right ) \sin \left (2 \, a\right ) + 2 \, b x^{n} \cos \left (2 \, a\right ) \operatorname{Si}\left (2 \, b x^{n}\right ) + 2 \, \cos \left (b x^{n} + a\right )^{2}}{2 \, 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)^2,x, algorithm="fricas")

[Out]

-1/2*(b*x^n*cos_integral(2*b*x^n)*sin(2*a) + b*x^n*cos_integral(-2*b*x^n)*sin(2*a) + 2*b*x^n*cos(2*a)*sin_inte
gral(2*b*x^n) + 2*cos(b*x^n + a)^2)/(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)**2,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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-n)*cos(a+b*x^n)^2,x, algorithm="giac")

[Out]

integrate(x^(-n - 1)*cos(b*x^n + a)^2, x)