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

Optimal. Leaf size=95 \[ -\frac{b^2 \cos (2 a) \text{CosIntegral}\left (2 b x^n\right )}{n}+\frac{b^2 \sin (2 a) \text{Si}\left (2 b x^n\right )}{n}+\frac{b x^{-n} \sin \left (2 \left (a+b x^n\right )\right )}{2 n}-\frac{x^{-2 n} \cos \left (2 \left (a+b x^n\right )\right )}{4 n}-\frac{x^{-2 n}}{4 n} \]

[Out]

-1/(4*n*x^(2*n)) - Cos[2*(a + b*x^n)]/(4*n*x^(2*n)) - (b^2*Cos[2*a]*CosIntegral[2*b*x^n])/n + (b*Sin[2*(a + b*
x^n)])/(2*n*x^n) + (b^2*Sin[2*a]*SinIntegral[2*b*x^n])/n

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Rubi [A]  time = 0.14742, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ -\frac{b^2 \cos (2 a) \text{CosIntegral}\left (2 b x^n\right )}{n}+\frac{b^2 \sin (2 a) \text{Si}\left (2 b x^n\right )}{n}+\frac{b x^{-n} \sin \left (2 \left (a+b x^n\right )\right )}{2 n}-\frac{x^{-2 n} \cos \left (2 \left (a+b x^n\right )\right )}{4 n}-\frac{x^{-2 n}}{4 n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 - 2*n)*Cos[a + b*x^n]^2,x]

[Out]

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

Mathematica [A]  time = 0.21181, size = 82, normalized size = 0.86 \[ -\frac{x^{-2 n} \left (4 b^2 \cos (2 a) x^{2 n} \text{CosIntegral}\left (2 b x^n\right )-4 b^2 \sin (2 a) x^{2 n} \text{Si}\left (2 b x^n\right )-2 b x^n \sin \left (2 \left (a+b x^n\right )\right )+\cos \left (2 \left (a+b x^n\right )\right )+1\right )}{4 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1 + Cos[2*(a + b*x^n)] + 4*b^2*x^(2*n)*Cos[2*a]*CosIntegral[2*b*x^n] - 2*b*x^n*Sin[2*(a + b*x^n)] - 4*b^2*x^
(2*n)*Sin[2*a]*SinIntegral[2*b*x^n])/(4*n*x^(2*n))

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Maple [A]  time = 0.043, size = 89, normalized size = 0.9 \begin{align*} -{\frac{1}{4\,n \left ({x}^{n} \right ) ^{2}}}+2\,{\frac{{b}^{2}}{n} \left ( -1/8\,{\frac{\cos \left ( 2\,a+2\,b{x}^{n} \right ) }{ \left ({x}^{n} \right ) ^{2}{b}^{2}}}+1/4\,{\frac{\sin \left ( 2\,a+2\,b{x}^{n} \right ) }{b{x}^{n}}}+1/2\,{\it Si} \left ( 2\,b{x}^{n} \right ) \sin \left ( 2\,a \right ) -1/2\,{\it Ci} \left ( 2\,b{x}^{n} \right ) \cos \left ( 2\,a \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/n/(x^n)^2+2/n*b^2*(-1/8*cos(2*a+2*b*x^n)/(x^n)^2/b^2+1/4*sin(2*a+2*b*x^n)/(x^n)/b+1/2*Si(2*b*x^n)*sin(2*a
)-1/2*Ci(2*b*x^n)*cos(2*a))

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Maxima [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^(-1-2*n)*cos(a+b*x^n)^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.12339, size = 290, normalized size = 3.05 \begin{align*} -\frac{b^{2} x^{2 \, n} \cos \left (2 \, a\right ) \operatorname{Ci}\left (2 \, b x^{n}\right ) + b^{2} x^{2 \, n} \cos \left (2 \, a\right ) \operatorname{Ci}\left (-2 \, b x^{n}\right ) - 2 \, b^{2} x^{2 \, n} \sin \left (2 \, a\right ) \operatorname{Si}\left (2 \, b x^{n}\right ) - 2 \, b x^{n} \cos \left (b x^{n} + a\right ) \sin \left (b x^{n} + a\right ) + \cos \left (b x^{n} + a\right )^{2}}{2 \, n x^{2 \, n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^2*x^(2*n)*cos(2*a)*cos_integral(2*b*x^n) + b^2*x^(2*n)*cos(2*a)*cos_integral(-2*b*x^n) - 2*b^2*x^(2*n)
*sin(2*a)*sin_integral(2*b*x^n) - 2*b*x^n*cos(b*x^n + a)*sin(b*x^n + a) + cos(b*x^n + a)^2)/(n*x^(2*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**(-1-2*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^{-2 \, 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^(-1-2*n)*cos(a+b*x^n)^2,x, algorithm="giac")

[Out]

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