\(\int x^{-1-n} \cos ^2(a+b x^n) \, dx\) [80]

Optimal result

Integrand size = 18, antiderivative size = 69 \[ \int x^{-1-n} \cos ^2\left (a+b x^n\right ) \, dx=-\frac {x^{-n}}{2 n}-\frac {x^{-n} \cos \left (2 \left (a+b x^n\right )\right )}{2 n}-\frac {b \operatorname {CosIntegral}\left (2 b x^n\right ) \sin (2 a)}{n}-\frac {b \cos (2 a) \text {Si}\left (2 b x^n\right )}{n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3507, 3461, 3378, 3384, 3380, 3383} \[ \int x^{-1-n} \cos ^2\left (a+b x^n\right ) \, dx=-\frac {b \sin (2 a) \operatorname {CosIntegral}\left (2 b x^n\right )}{n}-\frac {b \cos (2 a) \text {Si}\left (2 b x^n\right )}{n}-\frac {x^{-n} \cos \left (2 \left (a+b x^n\right )\right )}{2 n}-\frac {x^{-n}}{2 n} \]

[In]

Int[x^(-1 - n)*Cos[a + b*x^n]^2,x]

[Out]

-1/2*1/(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 3378

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 3380

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 3383

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 3384

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 3461

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 3507

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]

Rubi steps \begin{align*} \text {integral}& = \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 {\text {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 \text {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)) \text {Subst}\left (\int \frac {\sin (2 b x)}{x} \, dx,x,x^n\right )}{n}-\frac {(b \sin (2 a)) \text {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 \operatorname {CosIntegral}\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] (verified)

Time = 0.20 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77 \[ \int x^{-1-n} \cos ^2\left (a+b x^n\right ) \, dx=-\frac {x^{-n} \left (\cos ^2\left (a+b x^n\right )+b x^n \operatorname {CosIntegral}\left (2 b x^n\right ) \sin (2 a)+b x^n \cos (2 a) \text {Si}\left (2 b x^n\right )\right )}{n} \]

[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))

Maple [A] (verified)

Time = 2.57 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {x^{-n}}{2 n}+\frac {b \left (-\frac {\cos \left (2 a +2 b \,x^{n}\right ) x^{-n}}{2 b}-\operatorname {Si}\left (2 b \,x^{n}\right ) \cos \left (2 a \right )-\operatorname {Ci}\left (2 b \,x^{n}\right ) \sin \left (2 a \right )\right )}{n}\)	\(65\)
risch	\(\frac {\left (b \,{\mathrm e}^{-2 i a} \pi \,\operatorname {csgn}\left (b \,x^{n}\right ) x^{n}+i b \,{\mathrm e}^{-2 i a} \operatorname {Ei}_{1}\left (-2 i b \,x^{n}\right ) x^{n}-i b \,{\mathrm e}^{2 i a} \operatorname {Ei}_{1}\left (-2 i b \,x^{n}\right ) x^{n}-2 b \,{\mathrm e}^{-2 i a} \operatorname {Si}\left (2 b \,x^{n}\right ) x^{n}-\cos \left (2 a +2 b \,x^{n}\right )-1\right ) x^{-n}}{2 n}\)	\(103\)

[In]

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

[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))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77 \[ \int x^{-1-n} \cos ^2\left (a+b x^n\right ) \, dx=-\frac {b x^{n} \operatorname {Ci}\left (2 \, b x^{n}\right ) \sin \left (2 \, a\right ) + b x^{n} \cos \left (2 \, a\right ) \operatorname {Si}\left (2 \, b x^{n}\right ) + \cos \left (b x^{n} + a\right )^{2}}{n x^{n}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int x^{-1-n} \cos ^2\left (a+b x^n\right ) \, dx=\int x^{- n - 1} \cos ^{2}{\left (a + b x^{n} \right )}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int x^{-1-n} \cos ^2\left (a+b x^n\right ) \, dx=\int { x^{-n - 1} \cos \left (b x^{n} + a\right )^{2} \,d x } \]

[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)

Giac [F]

\[ \int x^{-1-n} \cos ^2\left (a+b x^n\right ) \, dx=\int { x^{-n - 1} \cos \left (b x^{n} + a\right )^{2} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int x^{-1-n} \cos ^2\left (a+b x^n\right ) \, dx=\int \frac {{\cos \left (a+b\,x^n\right )}^2}{x^{n+1}} \,d x \]

[In]

int(cos(a + b*x^n)^2/x^(n + 1),x)

[Out]

int(cos(a + b*x^n)^2/x^(n + 1), x)