\(\int \sec ^2(a+b \log (c x^n)) \, dx\) [245]

Optimal result

Integrand size = 13, antiderivative size = 85 \[ \int \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {4 e^{2 i a} x \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (2-\frac {i}{b n}\right ),\frac {1}{2} \left (4-\frac {i}{b n}\right ),-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{1+2 i b n} \]

[Out]

4*exp(2*I*a)*x*(c*x^n)^(2*I*b)*hypergeom([2, 1-1/2*I/b/n],[2-1/2*I/b/n],-exp(2*I*a)*(c*x^n)^(2*I*b))/(1+2*I*b*
n)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {4599, 4601, 371} \[ \int \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {4 e^{2 i a} x \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (2-\frac {i}{b n}\right ),\frac {1}{2} \left (4-\frac {i}{b n}\right ),-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{1+2 i b n} \]

[In]

Int[Sec[a + b*Log[c*x^n]]^2,x]

[Out]

(4*E^((2*I)*a)*x*(c*x^n)^((2*I)*b)*Hypergeometric2F1[2, (2 - I/(b*n))/2, (4 - I/(b*n))/2, -(E^((2*I)*a)*(c*x^n
)^((2*I)*b))])/(1 + (2*I)*b*n)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

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 4601

Int[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[2^p*E^(I*a*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}, x] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (x \left (c x^n\right )^{-1/n}\right ) \text {Subst}\left (\int x^{-1+\frac {1}{n}} \sec ^2(a+b \log (x)) \, dx,x,c x^n\right )}{n} \\ & = \frac {\left (4 e^{2 i a} x \left (c x^n\right )^{-1/n}\right ) \text {Subst}\left (\int \frac {x^{-1+2 i b+\frac {1}{n}}}{\left (1+e^{2 i a} x^{2 i b}\right )^2} \, dx,x,c x^n\right )}{n} \\ & = \frac {4 e^{2 i a} x \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (2-\frac {i}{b n}\right ),\frac {1}{2} \left (4-\frac {i}{b n}\right ),-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{1+2 i b n} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.81 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.73 \[ \int \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x \left (\frac {e^{2 i a} \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (1,1-\frac {i}{2 b n},2-\frac {i}{2 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )}{-i+2 b n}-i \operatorname {Hypergeometric2F1}\left (1,-\frac {i}{2 b n},1-\frac {i}{2 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )+\tan \left (a+b \log \left (c x^n\right )\right )\right )}{b n} \]

[In]

Integrate[Sec[a + b*Log[c*x^n]]^2,x]

[Out]

(x*((E^((2*I)*a)*(c*x^n)^((2*I)*b)*Hypergeometric2F1[1, 1 - (I/2)/(b*n), 2 - (I/2)/(b*n), -E^((2*I)*(a + b*Log
[c*x^n]))])/(-I + 2*b*n) - I*Hypergeometric2F1[1, (-1/2*I)/(b*n), 1 - (I/2)/(b*n), -E^((2*I)*(a + b*Log[c*x^n]
))] + Tan[a + b*Log[c*x^n]]))/(b*n)

Maple [F]

\[\int {\sec \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}d x\]

[In]

int(sec(a+b*ln(c*x^n))^2,x)

[Out]

int(sec(a+b*ln(c*x^n))^2,x)

Fricas [F]

\[ \int \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \sec \left (b \log \left (c x^{n}\right ) + a\right )^{2} \,d x } \]

[In]

integrate(sec(a+b*log(c*x^n))^2,x, algorithm="fricas")

[Out]

integral(sec(b*log(c*x^n) + a)^2, x)

Sympy [F]

\[ \int \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int \sec ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}\, dx \]

[In]

integrate(sec(a+b*ln(c*x**n))**2,x)

[Out]

Integral(sec(a + b*log(c*x**n))**2, x)

Maxima [F]

\[ \int \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \sec \left (b \log \left (c x^{n}\right ) + a\right )^{2} \,d x } \]

[In]

integrate(sec(a+b*log(c*x^n))^2,x, algorithm="maxima")

[Out]

2*(x*cos(2*b*log(x^n) + 2*a)*sin(2*b*log(c)) + x*cos(2*b*log(c))*sin(2*b*log(x^n) + 2*a) - (2*b^2*n^2*cos(2*b*
log(c))*cos(2*b*log(x^n) + 2*a) - 2*b^2*n^2*sin(2*b*log(c))*sin(2*b*log(x^n) + 2*a) + (b^2*cos(2*b*log(c))^2 +
b^2*sin(2*b*log(c))^2)*n^2*cos(2*b*log(x^n) + 2*a)^2 + (b^2*cos(2*b*log(c))^2 + b^2*sin(2*b*log(c))^2)*n^2*si
n(2*b*log(x^n) + 2*a)^2 + b^2*n^2)*integrate((cos(2*b*log(x^n) + 2*a)*sin(2*b*log(c)) + cos(2*b*log(c))*sin(2*
b*log(x^n) + 2*a))/(2*b^2*n^2*cos(2*b*log(c))*cos(2*b*log(x^n) + 2*a) - 2*b^2*n^2*sin(2*b*log(c))*sin(2*b*log(
x^n) + 2*a) + (b^2*cos(2*b*log(c))^2 + b^2*sin(2*b*log(c))^2)*n^2*cos(2*b*log(x^n) + 2*a)^2 + (b^2*cos(2*b*log
(c))^2 + b^2*sin(2*b*log(c))^2)*n^2*sin(2*b*log(x^n) + 2*a)^2 + b^2*n^2), x))/(2*b*n*cos(2*b*log(c))*cos(2*b*l
og(x^n) + 2*a) + (b*cos(2*b*log(c))^2 + b*sin(2*b*log(c))^2)*n*cos(2*b*log(x^n) + 2*a)^2 - 2*b*n*sin(2*b*log(c
))*sin(2*b*log(x^n) + 2*a) + (b*cos(2*b*log(c))^2 + b*sin(2*b*log(c))^2)*n*sin(2*b*log(x^n) + 2*a)^2 + b*n)

Giac [F]

\[ \int \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \sec \left (b \log \left (c x^{n}\right ) + a\right )^{2} \,d x } \]

[In]

integrate(sec(a+b*log(c*x^n))^2,x, algorithm="giac")

[Out]

integrate(sec(b*log(c*x^n) + a)^2, x)

Mupad [F(-1)]

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

[In]

int(1/cos(a + b*log(c*x^n))^2,x)

[Out]

int(1/cos(a + b*log(c*x^n))^2, x)