Integrand size = 17, antiderivative size = 79 \[ \int \frac {\sec ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {8 e^{4 i a} \left (c x^n\right )^{4 i b} \operatorname {Hypergeometric2F1}\left (4,2+\frac {i}{b n},3+\frac {i}{b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(1-2 i b n) x^2} \]
-8*exp(4*I*a)*(c*x^n)^(4*I*b)*hypergeom([4, 2+I/b/n],[3+I/b/n],-exp(2*I*a) *(c*x^n)^(2*I*b))/(1-2*I*b*n)/x^2
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(203\) vs. \(2(79)=158\).
Time = 7.06 (sec) , antiderivative size = 203, normalized size of antiderivative = 2.57 \[ \int \frac {\sec ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\frac {-2 e^{2 i a} (-i+b n) \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (1,1+\frac {i}{b n},2+\frac {i}{b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )-2 i \left (1+b^2 n^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {i}{b n},1+\frac {i}{b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )+\sec ^2\left (a+b \log \left (c x^n\right )\right ) \left (b n+\left (1+2 b^2 n^2+\left (1+b^2 n^2\right ) \cos \left (2 \left (a+b \log \left (c x^n\right )\right )\right )\right ) \tan \left (a+b \log \left (c x^n\right )\right )\right )}{3 b^3 n^3 x^2} \]
(-2*E^((2*I)*a)*(-I + b*n)*(c*x^n)^((2*I)*b)*Hypergeometric2F1[1, 1 + I/(b *n), 2 + I/(b*n), -E^((2*I)*(a + b*Log[c*x^n]))] - (2*I)*(1 + b^2*n^2)*Hyp ergeometric2F1[1, I/(b*n), 1 + I/(b*n), -E^((2*I)*(a + b*Log[c*x^n]))] + S ec[a + b*Log[c*x^n]]^2*(b*n + (1 + 2*b^2*n^2 + (1 + b^2*n^2)*Cos[2*(a + b* Log[c*x^n])])*Tan[a + b*Log[c*x^n]]))/(3*b^3*n^3*x^2)
Time = 0.29 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {5020, 5016, 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.
\(\displaystyle \int \frac {\sec ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 5020 |
\(\displaystyle \frac {\left (c x^n\right )^{2/n} \int \left (c x^n\right )^{-1-\frac {2}{n}} \sec ^4\left (a+b \log \left (c x^n\right )\right )d\left (c x^n\right )}{n x^2}\) |
\(\Big \downarrow \) 5016 |
\(\displaystyle \frac {16 e^{4 i a} \left (c x^n\right )^{2/n} \int \frac {\left (c x^n\right )^{4 i b-\frac {2}{n}-1}}{\left (e^{2 i a} \left (c x^n\right )^{2 i b}+1\right )^4}d\left (c x^n\right )}{n x^2}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle -\frac {8 e^{4 i a} \left (c x^n\right )^{4 i b} \operatorname {Hypergeometric2F1}\left (4,2+\frac {i}{b n},3+\frac {i}{b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{x^2 (1-2 i b n)}\) |
(-8*E^((4*I)*a)*(c*x^n)^((4*I)*b)*Hypergeometric2F1[4, 2 + I/(b*n), 3 + I/ (b*n), -(E^((2*I)*a)*(c*x^n)^((2*I)*b))])/((1 - (2*I)*b*n)*x^2)
3.3.58.3.1 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[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Simp[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]
Int[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_ .), x_Symbol] :> Simp[(e*x)^(m + 1)/(e*n*(c*x^n)^((m + 1)/n)) Subst[Int[x ^((m + 1)/n - 1)*Sec[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && (NeQ[c, 1] || NeQ[n, 1])
\[\int \frac {{\sec \left (a +b \ln \left (c \,x^{n}\right )\right )}^{4}}{x^{3}}d x\]
\[ \int \frac {\sec ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int { \frac {\sec \left (b \log \left (c x^{n}\right ) + a\right )^{4}}{x^{3}} \,d x } \]
\[ \int \frac {\sec ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int \frac {\sec ^{4}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{x^{3}}\, dx \]
\[ \int \frac {\sec ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int { \frac {\sec \left (b \log \left (c x^{n}\right ) + a\right )^{4}}{x^{3}} \,d x } \]
4/3*(3*(b*cos(4*b*log(c))^2 + b*sin(4*b*log(c))^2)*n*cos(4*b*log(x^n) + 4* a)^2 + 3*(b*cos(2*b*log(c))^2 + b*sin(2*b*log(c))^2)*n*cos(2*b*log(x^n) + 2*a)^2 + 3*(b*cos(4*b*log(c))^2 + b*sin(4*b*log(c))^2)*n*sin(4*b*log(x^n) + 4*a)^2 + 3*(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^2*n^2*sin(6*b*log(c)) + ((b*cos(6*b*log(c))*cos(4*b*log(c) ) + b*sin(6*b*log(c))*sin(4*b*log(c)))*n + cos(4*b*log(c))*sin(6*b*log(c)) - cos(6*b*log(c))*sin(4*b*log(c)))*cos(4*b*log(x^n) + 4*a) + (3*(b^2*cos( 2*b*log(c))*sin(6*b*log(c)) - b^2*cos(6*b*log(c))*sin(2*b*log(c)))*n^2 + ( b*cos(6*b*log(c))*cos(2*b*log(c)) + b*sin(6*b*log(c))*sin(2*b*log(c)))*n + 2*cos(2*b*log(c))*sin(6*b*log(c)) - 2*cos(6*b*log(c))*sin(2*b*log(c)))*co s(2*b*log(x^n) + 2*a) + ((b*cos(4*b*log(c))*sin(6*b*log(c)) - b*cos(6*b*lo g(c))*sin(4*b*log(c)))*n - cos(6*b*log(c))*cos(4*b*log(c)) - sin(6*b*log(c ))*sin(4*b*log(c)))*sin(4*b*log(x^n) + 4*a) - (3*(b^2*cos(6*b*log(c))*cos( 2*b*log(c)) + b^2*sin(6*b*log(c))*sin(2*b*log(c)))*n^2 - (b*cos(2*b*log(c) )*sin(6*b*log(c)) - b*cos(6*b*log(c))*sin(2*b*log(c)))*n + 2*cos(6*b*log(c ))*cos(2*b*log(c)) + 2*sin(6*b*log(c))*sin(2*b*log(c)))*sin(2*b*log(x^n) + 2*a) + sin(6*b*log(c)))*cos(6*b*log(x^n) + 6*a) + (3*b^2*n^2*sin(4*b*log( c)) + b*n*cos(4*b*log(c)) + 3*(3*(b^2*cos(2*b*log(c))*sin(4*b*log(c)) - b^ 2*cos(4*b*log(c))*sin(2*b*log(c)))*n^2 + 2*(b*cos(4*b*log(c))*cos(2*b*log( c)) + b*sin(4*b*log(c))*sin(2*b*log(c)))*n + cos(2*b*log(c))*sin(4*b*lo...
\[ \int \frac {\sec ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int { \frac {\sec \left (b \log \left (c x^{n}\right ) + a\right )^{4}}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {\sec ^4\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int \frac {1}{x^3\,{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}^4} \,d x \]