3.264 \(\int \sec ^p(a+\frac{i \log (c x^n)}{n (-2+p)}) \, dx\)

Optimal. Leaf size=95 \[ \frac{e^{-2 i a} (2-p) x \left (c x^n\right )^{-\frac{2}{n (2-p)}} \left (1+e^{2 i a} \left (c x^n\right )^{\frac{2}{n (2-p)}}\right ) \sec ^p\left (a-\frac{i \log \left (c x^n\right )}{n (2-p)}\right )}{2 (1-p)} \]

[Out]

((2 - p)*x*(1 + E^((2*I)*a)*(c*x^n)^(2/(n*(2 - p))))*Sec[a - (I*Log[c*x^n])/(n*(2 - p))]^p)/(2*E^((2*I)*a)*(1
- p)*(c*x^n)^(2/(n*(2 - p))))

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Rubi [A]  time = 0.0910376, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4503, 4507, 261} \[ \frac{e^{-2 i a} (2-p) x \left (c x^n\right )^{-\frac{2}{n (2-p)}} \left (1+e^{2 i a} \left (c x^n\right )^{\frac{2}{n (2-p)}}\right ) \sec ^p\left (a-\frac{i \log \left (c x^n\right )}{n (2-p)}\right )}{2 (1-p)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + (I*Log[c*x^n])/(n*(-2 + p))]^p,x]

[Out]

((2 - p)*x*(1 + E^((2*I)*a)*(c*x^n)^(2/(n*(2 - p))))*Sec[a - (I*Log[c*x^n])/(n*(2 - p))]^p)/(2*E^((2*I)*a)*(1
- p)*(c*x^n)^(2/(n*(2 - p))))

Rule 4503

Int[Sec[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[x
^(1/n - 1)*Sec[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a, b, c, d, n, p}, x] && (NeQ[c, 1] || NeQ[n,
1])

Rule 4507

Int[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[(Sec[d*(a + b*Log[x])]^p*(1
+ E^(2*I*a*d)*x^(2*I*b*d))^p)/x^(I*b*d*p), Int[((e*x)^m*x^(I*b*d*p))/(1 + E^(2*I*a*d)*x^(2*I*b*d))^p, x], x]
/; FreeQ[{a, b, d, e, m, p}, x] &&  !IntegerQ[p]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \sec ^p\left (a+\frac{i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx &=\frac{\left (x \left (c x^n\right )^{-1/n}\right ) \operatorname{Subst}\left (\int x^{-1+\frac{1}{n}} \sec ^p\left (a+\frac{i \log (x)}{n (-2+p)}\right ) \, dx,x,c x^n\right )}{n}\\ &=\frac{\left (x \left (c x^n\right )^{-\frac{1}{n}+\frac{p}{n (-2+p)}} \left (1+e^{2 i a} \left (c x^n\right )^{-\frac{2}{n (-2+p)}}\right )^p \sec ^p\left (a+\frac{i \log \left (c x^n\right )}{n (-2+p)}\right )\right ) \operatorname{Subst}\left (\int x^{-1+\frac{1}{n}-\frac{p}{n (-2+p)}} \left (1+e^{2 i a} x^{-\frac{2}{n (-2+p)}}\right )^{-p} \, dx,x,c x^n\right )}{n}\\ &=\frac{e^{-2 i a} (2-p) x \left (c x^n\right )^{-\frac{2}{n (2-p)}} \left (1+e^{2 i a} \left (c x^n\right )^{\frac{2}{n (2-p)}}\right ) \sec ^p\left (a-\frac{i \log \left (c x^n\right )}{n (2-p)}\right )}{2 (1-p)}\\ \end{align*}

Mathematica [A]  time = 0.858004, size = 67, normalized size = 0.71 \[ \frac{e^{-2 i a} (p-2) x \left (\left (c x^n\right )^{\frac{2}{n (p-2)}}+e^{2 i a}\right ) \sec ^p\left (a+\frac{i \log \left (c x^n\right )}{n (p-2)}\right )}{2 (p-1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + (I*Log[c*x^n])/(n*(-2 + p))]^p,x]

[Out]

((-2 + p)*x*(E^((2*I)*a) + (c*x^n)^(2/(n*(-2 + p))))*Sec[a + (I*Log[c*x^n])/(n*(-2 + p))]^p)/(2*E^((2*I)*a)*(-
1 + p))

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Maple [F]  time = 0.372, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( a+{\frac{i\ln \left ( c{x}^{n} \right ) }{n \left ( p-2 \right ) }} \right ) \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(a+I*ln(c*x^n)/n/(p-2))^p,x)

[Out]

int(sec(a+I*ln(c*x^n)/n/(p-2))^p,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sec \left (a + \frac{i \, \log \left (c x^{n}\right )}{n{\left (p - 2\right )}}\right )^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(a+I*log(c*x^n)/n/(-2+p))^p,x, algorithm="maxima")

[Out]

integrate(sec(a + I*log(c*x^n)/(n*(p - 2)))^p, x)

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Fricas [A]  time = 0.505553, size = 340, normalized size = 3.58 \begin{align*} \frac{{\left ({\left (p - 2\right )} x e^{\left (\frac{2 \,{\left (i \, a n p - 2 i \, a n - \log \left (c x^{n}\right )\right )}}{n p - 2 \, n}\right )} +{\left (p - 2\right )} x\right )} \left (\frac{2 \, e^{\left (\frac{i \, a n p - 2 i \, a n - \log \left (c x^{n}\right )}{n p - 2 \, n}\right )}}{e^{\left (\frac{2 \,{\left (i \, a n p - 2 i \, a n - \log \left (c x^{n}\right )\right )}}{n p - 2 \, n}\right )} + 1}\right )^{p} e^{\left (-\frac{2 \,{\left (i \, a n p - 2 i \, a n - \log \left (c x^{n}\right )\right )}}{n p - 2 \, n}\right )}}{2 \,{\left (p - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(a+I*log(c*x^n)/n/(-2+p))^p,x, algorithm="fricas")

[Out]

1/2*((p - 2)*x*e^(2*(I*a*n*p - 2*I*a*n - log(c*x^n))/(n*p - 2*n)) + (p - 2)*x)*(2*e^((I*a*n*p - 2*I*a*n - log(
c*x^n))/(n*p - 2*n))/(e^(2*(I*a*n*p - 2*I*a*n - log(c*x^n))/(n*p - 2*n)) + 1))^p*e^(-2*(I*a*n*p - 2*I*a*n - lo
g(c*x^n))/(n*p - 2*n))/(p - 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sec ^{p}{\left (a + \frac{i \log{\left (c x^{n} \right )}}{n \left (p - 2\right )} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(a+I*ln(c*x**n)/n/(-2+p))**p,x)

[Out]

Integral(sec(a + I*log(c*x**n)/(n*(p - 2)))**p, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sec \left (a + \frac{i \, \log \left (c x^{n}\right )}{n{\left (p - 2\right )}}\right )^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(a+I*log(c*x^n)/n/(-2+p))^p,x, algorithm="giac")

[Out]

integrate(sec(a + I*log(c*x^n)/(n*(p - 2)))^p, x)