Integrand size = 15, antiderivative size = 110 \[ \int \frac {1}{\sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}} \, dx=\frac {2 x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {2 i+b n}{4 b n},\frac {1}{4} \left (3-\frac {2 i}{b n}\right ),-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(2-i b n) \sqrt {1+e^{2 i a} \left (c x^n\right )^{2 i b}} \sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}} \]
2*x*hypergeom([-1/2, 1/4*(-2*I-b*n)/b/n],[3/4-1/2*I/b/n],-exp(2*I*a)*(c*x^ n)^(2*I*b))/(2-I*b*n)/(1+exp(2*I*a)*(c*x^n)^(2*I*b))^(1/2)/sec(a+b*ln(c*x^ n))^(1/2)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(380\) vs. \(2(110)=220\).
Time = 3.48 (sec) , antiderivative size = 380, normalized size of antiderivative = 3.45 \[ \int \frac {1}{\sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}} \, dx=\frac {2 b e^{2 i a} n x \left (c x^n\right )^{2 i b} \left ((2 i+b n) x^{2 i b n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4}-\frac {i}{2 b n},\frac {7}{4}-\frac {i}{2 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )+(-2 i+3 b n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {2 i+b n}{4 b n},\frac {3}{4}-\frac {i}{2 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )\right )}{(2 i+b n) (-2 i+3 b n) \sqrt {1+e^{2 i a} \left (c x^n\right )^{2 i b}} \sqrt {\frac {e^{i a} \left (c x^n\right )^{i b}}{2+2 e^{2 i a} \left (c x^n\right )^{2 i b}}} \left ((-2+i b n) x^{2 i b n}-i e^{2 i a} (-2 i+b n) \left (c x^n\right )^{2 i b}\right )}-\frac {2 x \cos \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{\sqrt {\sec \left (a+b \log \left (c x^n\right )\right )} \left (-2 \cos \left (a-b n \log (x)+b \log \left (c x^n\right )\right )+b n \sin \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )} \]
(2*b*E^((2*I)*a)*n*x*(c*x^n)^((2*I)*b)*((2*I + b*n)*x^((2*I)*b*n)*Hypergeo metric2F1[1/2, 3/4 - (I/2)/(b*n), 7/4 - (I/2)/(b*n), -(E^((2*I)*a)*(c*x^n) ^((2*I)*b))] + (-2*I + 3*b*n)*Hypergeometric2F1[1/2, -1/4*(2*I + b*n)/(b*n ), 3/4 - (I/2)/(b*n), -(E^((2*I)*a)*(c*x^n)^((2*I)*b))]))/((2*I + b*n)*(-2 *I + 3*b*n)*Sqrt[1 + E^((2*I)*a)*(c*x^n)^((2*I)*b)]*Sqrt[(E^(I*a)*(c*x^n)^ (I*b))/(2 + 2*E^((2*I)*a)*(c*x^n)^((2*I)*b))]*((-2 + I*b*n)*x^((2*I)*b*n) - I*E^((2*I)*a)*(-2*I + b*n)*(c*x^n)^((2*I)*b))) - (2*x*Cos[a - b*n*Log[x] + b*Log[c*x^n]])/(Sqrt[Sec[a + b*Log[c*x^n]]]*(-2*Cos[a - b*n*Log[x] + b* Log[c*x^n]] + b*n*Sin[a - b*n*Log[x] + b*Log[c*x^n]]))
Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5014, 5018, 888}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}} \, dx\) |
\(\Big \downarrow \) 5014 |
\(\displaystyle \frac {x \left (c x^n\right )^{-1/n} \int \frac {\left (c x^n\right )^{\frac {1}{n}-1}}{\sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}}d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 5018 |
\(\displaystyle \frac {x \left (c x^n\right )^{-\frac {1}{n}+\frac {i b}{2}} \int \left (c x^n\right )^{-\frac {i b}{2}+\frac {1}{n}-1} \sqrt {e^{2 i a} \left (c x^n\right )^{2 i b}+1}d\left (c x^n\right )}{n \sqrt {1+e^{2 i a} \left (c x^n\right )^{2 i b}} \sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {2 x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {b n+2 i}{4 b n},\frac {1}{4} \left (3-\frac {2 i}{b n}\right ),-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(2-i b n) \sqrt {1+e^{2 i a} \left (c x^n\right )^{2 i b}} \sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}}\) |
(2*x*Hypergeometric2F1[-1/2, -1/4*(2*I + b*n)/(b*n), (3 - (2*I)/(b*n))/4, -(E^((2*I)*a)*(c*x^n)^((2*I)*b))])/((2 - I*b*n)*Sqrt[1 + E^((2*I)*a)*(c*x^ n)^((2*I)*b)]*Sqrt[Sec[a + b*Log[c*x^n]]])
3.3.72.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[Sec[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Si mp[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])
Int[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Simp[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] /; F reeQ[{a, b, d, e, m, p}, x] && !IntegerQ[p]
\[\int \frac {1}{\sqrt {\sec \left (a +b \ln \left (c \,x^{n}\right )\right )}}d x\]
Exception generated. \[ \int \frac {1}{\sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (has polynomial part)
\[ \int \frac {1}{\sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}} \, dx=\int \frac {1}{\sqrt {\sec {\left (a + b \log {\left (c x^{n} \right )} \right )}}}\, dx \]
\[ \int \frac {1}{\sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}} \, dx=\int { \frac {1}{\sqrt {\sec \left (b \log \left (c x^{n}\right ) + a\right )}} \,d x } \]
\[ \int \frac {1}{\sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}} \, dx=\int { \frac {1}{\sqrt {\sec \left (b \log \left (c x^{n}\right ) + a\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {\sec \left (a+b \log \left (c x^n\right )\right )}} \, dx=\int \frac {1}{\sqrt {\frac {1}{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}}} \,d x \]