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

Optimal. Leaf size=34 \[ \frac{\cos \left (a+b x^n\right )}{b^2 n}+\frac{x^n \sin \left (a+b x^n\right )}{b n} \]

[Out]

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

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Rubi [A]  time = 0.030392, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {3380, 3296, 2638} \[ \frac{\cos \left (a+b x^n\right )}{b^2 n}+\frac{x^n \sin \left (a+b x^n\right )}{b n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 2*n)*Cos[a + b*x^n],x]

[Out]

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

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 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.0628097, size = 29, normalized size = 0.85 \[ \frac{b x^n \sin \left (a+b x^n\right )+\cos \left (a+b x^n\right )}{b^2 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.013, size = 44, normalized size = 1.3 \begin{align*}{\frac{\cos \left ( a+b{x}^{n} \right ) + \left ( a+b{x}^{n} \right ) \sin \left ( a+b{x}^{n} \right ) -a\sin \left ( a+b{x}^{n} \right ) }{n{b}^{2}}} \end{align*}

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*(cos(a+b*x^n)+(a+b*x^n)*sin(a+b*x^n)-a*sin(a+b*x^n))

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Maxima [A]  time = 1.00682, size = 39, normalized size = 1.15 \begin{align*} \frac{b x^{n} \sin \left (b x^{n} + a\right ) + \cos \left (b x^{n} + a\right )}{b^{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),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.84286, size = 66, normalized size = 1.94 \begin{align*} \frac{b x^{n} \sin \left (b x^{n} + a\right ) + \cos \left (b x^{n} + a\right )}{b^{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),x, algorithm="fricas")

[Out]

(b*x^n*sin(b*x^n + a) + cos(b*x^n + a))/(b^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),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 )\,{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),x, algorithm="giac")

[Out]

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