Integrand size = 15, antiderivative size = 110 \[ \int \frac {1}{\sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {2 x \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1}{4} \left (-5-\frac {2 i}{b n}\right ),-\frac {2 i+b n}{4 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(2-5 i b n) \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \]
2*x*hypergeom([-5/2, -5/4-1/2*I/b/n],[1/4*(-2*I-b*n)/b/n],-exp(2*I*a)*(c*x ^n)^(2*I*b))/(2-5*I*b*n)/(1+exp(2*I*a)*(c*x^n)^(2*I*b))^(5/2)/sec(a+b*ln(c *x^n))^(5/2)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(867\) vs. \(2(110)=220\).
Time = 7.96 (sec) , antiderivative size = 867, normalized size of antiderivative = 7.88 \[ \int \frac {1}{\sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {30 b^3 e^{2 i \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )} n^3 x \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 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )} x^{2 i b n}\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 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )} x^{2 i b n}\right )\right )}{(2-5 i b n) (2 i+b n) (-2 i+3 b n) (-2 i+5 b n) \left (-2 i-b n+e^{2 i \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )} (-2 i+b n)\right ) \sqrt {1+e^{2 i \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )} x^{2 i b n}} \sqrt {\frac {e^{i \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )} x^{i b n}}{2+2 e^{2 i \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )} x^{2 i b n}}}}+\sqrt {\sec \left (a+b n \log (x)+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )} \left (-\frac {x \cos (b n \log (x)) \left (12+55 b^2 n^2+12 \cos \left (2 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )+65 b^2 n^2 \cos \left (2 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )+4 b n \sin \left (2 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )\right )}{4 (-2 i+5 b n) (2 i+5 b n) \left (-2 \cos \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )+b n \sin \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )}+\frac {x \sin (b n \log (x)) \left (-16 b n-4 b n \cos \left (2 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )+12 \sin \left (2 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )+65 b^2 n^2 \sin \left (2 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )\right )}{4 (-2 i+5 b n) (2 i+5 b n) \left (-2 \cos \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )+b n \sin \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )}+\frac {x \sin (3 b n \log (x)) \left (5 b n \cos \left (3 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )-2 \sin \left (3 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )\right )}{2 (-2 i+5 b n) (2 i+5 b n)}+\frac {x \cos (3 b n \log (x)) \left (2 \cos \left (3 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )+5 b n \sin \left (3 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )\right )}{2 (-2 i+5 b n) (2 i+5 b n)}\right ) \]
(30*b^3*E^((2*I)*(a + b*(-(n*Log[x]) + Log[c*x^n])))*n^3*x*((2*I + b*n)*x^ ((2*I)*b*n)*Hypergeometric2F1[1/2, 3/4 - (I/2)/(b*n), 7/4 - (I/2)/(b*n), - (E^((2*I)*(a + b*(-(n*Log[x]) + Log[c*x^n])))*x^((2*I)*b*n))] + (-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 + b*(-(n*Log[x]) + Log[c*x^n])))*x^((2*I)*b*n))]))/((2 - (5*I)* b*n)*(2*I + b*n)*(-2*I + 3*b*n)*(-2*I + 5*b*n)*(-2*I - b*n + E^((2*I)*(a + b*(-(n*Log[x]) + Log[c*x^n])))*(-2*I + b*n))*Sqrt[1 + E^((2*I)*(a + b*(-( n*Log[x]) + Log[c*x^n])))*x^((2*I)*b*n)]*Sqrt[(E^(I*(a + b*(-(n*Log[x]) + Log[c*x^n])))*x^(I*b*n))/(2 + 2*E^((2*I)*(a + b*(-(n*Log[x]) + Log[c*x^n]) ))*x^((2*I)*b*n))]) + Sqrt[Sec[a + b*n*Log[x] + b*(-(n*Log[x]) + Log[c*x^n ])]]*(-1/4*(x*Cos[b*n*Log[x]]*(12 + 55*b^2*n^2 + 12*Cos[2*(a + b*(-(n*Log[ x]) + Log[c*x^n]))] + 65*b^2*n^2*Cos[2*(a + b*(-(n*Log[x]) + Log[c*x^n]))] + 4*b*n*Sin[2*(a + b*(-(n*Log[x]) + Log[c*x^n]))]))/((-2*I + 5*b*n)*(2*I + 5*b*n)*(-2*Cos[a + b*(-(n*Log[x]) + Log[c*x^n])] + b*n*Sin[a + b*(-(n*Lo g[x]) + Log[c*x^n])])) + (x*Sin[b*n*Log[x]]*(-16*b*n - 4*b*n*Cos[2*(a + b* (-(n*Log[x]) + Log[c*x^n]))] + 12*Sin[2*(a + b*(-(n*Log[x]) + Log[c*x^n])) ] + 65*b^2*n^2*Sin[2*(a + b*(-(n*Log[x]) + Log[c*x^n]))]))/(4*(-2*I + 5*b* n)*(2*I + 5*b*n)*(-2*Cos[a + b*(-(n*Log[x]) + Log[c*x^n])] + b*n*Sin[a + b *(-(n*Log[x]) + Log[c*x^n])])) + (x*Sin[3*b*n*Log[x]]*(5*b*n*Cos[3*(a + b* (-(n*Log[x]) + Log[c*x^n]))] - 2*Sin[3*(a + b*(-(n*Log[x]) + 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.
| \(\displaystyle \int \frac {1}{\sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx\) |
| \(\Big \downarrow \) 5014 |
| \(\displaystyle \frac {x \left (c x^n\right )^{-1/n} \int \frac {\left (c x^n\right )^{\frac {1}{n}-1}}{\sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}d\left (c x^n\right )}{n}\) |
| \(\Big \downarrow \) 5018 |
| \(\displaystyle \frac {x \left (c x^n\right )^{-\frac {1}{n}+\frac {5 i b}{2}} \int \left (c x^n\right )^{-\frac {5 i b}{2}+\frac {1}{n}-1} \left (e^{2 i a} \left (c x^n\right )^{2 i b}+1\right )^{5/2}d\left (c x^n\right )}{n \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}\) |
| \(\Big \downarrow \) 888 |
| \(\displaystyle \frac {2 x \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1}{4} \left (-5-\frac {2 i}{b n}\right ),-\frac {b n+2 i}{4 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(2-5 i b n) \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}\) |
(2*x*Hypergeometric2F1[-5/2, (-5 - (2*I)/(b*n))/4, -1/4*(2*I + b*n)/(b*n), -(E^((2*I)*a)*(c*x^n)^((2*I)*b))])/((2 - (5*I)*b*n)*(1 + E^((2*I)*a)*(c*x ^n)^((2*I)*b))^(5/2)*Sec[a + b*Log[c*x^n]]^(5/2))
3.3.76.3.1 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]
\[\int \frac {1}{{\sec \left (a +b \ln \left (c \,x^{n}\right )\right )}^{\frac {5}{2}}}d x\]
Exception generated. \[ \int \frac {1}{\sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (has polynomial part)
Timed out. \[ \int \frac {1}{\sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\text {Timed out} \]
\[ \int \frac {1}{\sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\int { \frac {1}{\sec \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {1}{\sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\int { \frac {1}{\sec \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\sec ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\int \frac {1}{{\left (\frac {1}{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}\right )}^{5/2}} \,d x \]