3.3.65 \(\int \sec ^p(a-\frac {i \log (c x^n)}{n (-2+p)}) \, dx\) [265]

Optimal. Leaf size=70 \[ \frac {(2-p) x \left (1+e^{2 i a} \left (c x^n\right )^{-\frac {2}{n (2-p)}}\right ) \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (2-p)}\right )}{2 (1-p)} \]

[Out]

1/2*(2-p)*x*(1+exp(2*I*a)/((c*x^n)^(2/n/(2-p))))*sec(a+I*ln(c*x^n)/n/(2-p))^p/(1-p)

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4599, 4603, 270} \begin {gather*} \frac {(2-p) x \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 {gather*}

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*(1 - p))

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 4599

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 4603

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]

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 ) \text {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 ) \text {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 {(2-p) x \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 = 1.96, size = 117, normalized size = 1.67 \begin {gather*} \frac {2^{-1+p} e^{i a} (-2+p) x \left (c x^n\right )^{\frac {1}{n (-2+p)}} \left (\frac {e^{\frac {i a (2+p)}{-2+p}} \left (c x^n\right )^{\frac {1}{n (-2+p)}}}{e^{\frac {4 i a}{-2+p}}+e^{\frac {2 i a p}{-2+p}} \left (c x^n\right )^{\frac {2}{n (-2+p)}}}\right )^{-1+p}}{-1+p} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2^(-1 + p)*E^(I*a)*(-2 + p)*x*(c*x^n)^(1/(n*(-2 + p)))*((E^((I*a*(2 + p))/(-2 + p))*(c*x^n)^(1/(n*(-2 + p))))
/(E^(((4*I)*a)/(-2 + p)) + E^(((2*I)*a*p)/(-2 + p))*(c*x^n)^(2/(n*(-2 + p)))))^(-1 + p))/(-1 + p)

________________________________________________________________________________________

Maple [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int \sec ^{p}\left (a -\frac {i \ln \left (c \,x^{n}\right )}{n \left (-2+p \right )}\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (55) = 110\).
time = 2.16, size = 149, normalized size = 2.13 \begin {gather*} \frac {{\left ({\left (p - 2\right )} x e^{\left (\frac {2 \, {\left (-i \, a n p + 2 i \, a n - n \log \left (x\right ) - \log \left (c\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 - n \log \left (x\right ) - \log \left (c\right )}{n p - 2 \, n}\right )}}{e^{\left (\frac {2 \, {\left (-i \, a n p + 2 i \, a n - n \log \left (x\right ) - \log \left (c\right )\right )}}{n p - 2 \, n}\right )} + 1}\right )^{p} e^{\left (-\frac {2 \, {\left (-i \, a n p + 2 i \, a n - n \log \left (x\right ) - \log \left (c\right )\right )}}{n p - 2 \, n}\right )}}{2 \, {\left (p - 1\right )}} \end {gather*}

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sec ^{p}{\left (a - \frac {i \log {\left (c x^{n} \right )}}{n \left (p - 2\right )} \right )}\, dx \end {gather*}

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)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (\frac {1}{\cos \left (a-\frac {\ln \left (c\,x^n\right )\,1{}\mathrm {i}}{n\,\left (p-2\right )}\right )}\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(a - (log(c*x^n)*1i)/(n*(p - 2))))^p,x)

[Out]

int((1/cos(a - (log(c*x^n)*1i)/(n*(p - 2))))^p, x)

________________________________________________________________________________________