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

Optimal. Leaf size=78 \[ -\frac {b^2 \cos (a) \text {Ci}\left (b x^n\right )}{2 n}+\frac {b^2 \sin (a) \text {Si}\left (b x^n\right )}{2 n}+\frac {b x^{-n} \sin \left (a+b x^n\right )}{2 n}-\frac {x^{-2 n} \cos \left (a+b x^n\right )}{2 n} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3380, 3297, 3303, 3299, 3302} \[ -\frac {b^2 \cos (a) \text {CosIntegral}\left (b x^n\right )}{2 n}+\frac {b^2 \sin (a) \text {Si}\left (b x^n\right )}{2 n}+\frac {b x^{-n} \sin \left (a+b x^n\right )}{2 n}-\frac {x^{-2 n} \cos \left (a+b x^n\right )}{2 n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 - 2*n)*Cos[a + b*x^n],x]

[Out]

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

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

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Mathematica [A]  time = 0.14, size = 70, normalized size = 0.90 \[ -\frac {x^{-2 n} \left (b^2 \cos (a) x^{2 n} \text {Ci}\left (b x^n\right )-b^2 \sin (a) x^{2 n} \text {Si}\left (b x^n\right )-b x^n \sin \left (a+b x^n\right )+\cos \left (a+b x^n\right )\right )}{2 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.92, size = 90, normalized size = 1.15 \[ -\frac {b^{2} x^{2 \, n} \cos \relax (a) \operatorname {Ci}\left (b x^{n}\right ) + b^{2} x^{2 \, n} \cos \relax (a) \operatorname {Ci}\left (-b x^{n}\right ) - 2 \, b^{2} x^{2 \, n} \sin \relax (a) \operatorname {Si}\left (b x^{n}\right ) - 2 \, b x^{n} \sin \left (b x^{n} + a\right ) + 2 \, \cos \left (b x^{n} + a\right )}{4 \, n x^{2 \, n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b^2*x^(2*n)*cos(a)*cos_integral(b*x^n) + b^2*x^(2*n)*cos(a)*cos_integral(-b*x^n) - 2*b^2*x^(2*n)*sin(a)*
sin_integral(b*x^n) - 2*b*x^n*sin(b*x^n + a) + 2*cos(b*x^n + a))/(n*x^(2*n))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{-2 \, n - 1} \cos \left (b x^{n} + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 65, normalized size = 0.83 \[ \frac {b^{2} \left (-\frac {\cos \left (a +b \,x^{n}\right ) x^{-2 n}}{2 b^{2}}+\frac {\sin \left (a +b \,x^{n}\right ) x^{-n}}{2 b}+\frac {\Si \left (b \,x^{n}\right ) \sin \relax (a )}{2}-\frac {\Ci \left (b \,x^{n}\right ) \cos \relax (a )}{2}\right )}{n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{-2 \, n - 1} \cos \left (b x^{n} + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (a+b\,x^n\right )}{x^{2\,n+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x^n)/x^(2*n + 1),x)

[Out]

int(cos(a + b*x^n)/x^(2*n + 1), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{- 2 n - 1} \cos {\left (a + b x^{n} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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