Integrand size = 17, antiderivative size = 87 \[ \int \frac {\sec ^2\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {4 e^{2 i a} \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (2+\frac {i}{b n}\right ),\frac {1}{2} \left (4+\frac {i}{b n}\right ),-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{(1-2 i b n) x} \] Output:
-4*exp(2*I*a)*(c*x^n)^(2*I*b)*hypergeom([2, 1+1/2*I/b/n],[2+1/2*I/b/n],-ex p(2*I*a)*(c*x^n)^(2*I*b))/(1-2*I*b*n)/x
Time = 2.77 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.84 \[ \int \frac {\sec ^2\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\frac {-e^{2 i a} \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (1,1+\frac {i}{2 b n},2+\frac {i}{2 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )+(1-2 i b n) \left (\operatorname {Hypergeometric2F1}\left (1,\frac {i}{2 b n},1+\frac {i}{2 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )+i \tan \left (a+b \log \left (c x^n\right )\right )\right )}{b n (i+2 b n) x} \] Input:
Integrate[Sec[a + b*Log[c*x^n]]^2/x^2,x]
Output:
(-(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]))]) + (1 - (2*I)*b*n)*(Hypergeom etric2F1[1, (I/2)/(b*n), 1 + (I/2)/(b*n), -E^((2*I)*(a + b*Log[c*x^n]))] + I*Tan[a + b*Log[c*x^n]]))/(b*n*(I + 2*b*n)*x)
Time = 0.30 (sec) , antiderivative size = 87, 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.
\(\displaystyle \int \frac {\sec ^2\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 5020 |
\(\displaystyle \frac {\left (c x^n\right )^{\frac {1}{n}} \int \left (c x^n\right )^{-1-\frac {1}{n}} \sec ^2\left (a+b \log \left (c x^n\right )\right )d\left (c x^n\right )}{n x}\) |
\(\Big \downarrow \) 5016 |
\(\displaystyle \frac {4 e^{2 i a} \left (c x^n\right )^{\frac {1}{n}} \int \frac {\left (c x^n\right )^{2 i b-\frac {1}{n}-1}}{\left (e^{2 i a} \left (c x^n\right )^{2 i b}+1\right )^2}d\left (c x^n\right )}{n x}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle -\frac {4 e^{2 i a} \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (2+\frac {i}{b n}\right ),\frac {1}{2} \left (4+\frac {i}{b n}\right ),-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{x (1-2 i b n)}\) |
Input:
Int[Sec[a + b*Log[c*x^n]]^2/x^2,x]
Output:
(-4*E^((2*I)*a)*(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)*x)
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])
\[\int \frac {{\sec \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}}{x^{2}}d x\]
Input:
int(sec(a+b*ln(c*x^n))^2/x^2,x)
Output:
int(sec(a+b*ln(c*x^n))^2/x^2,x)
\[ \int \frac {\sec ^2\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int { \frac {\sec \left (b \log \left (c x^{n}\right ) + a\right )^{2}}{x^{2}} \,d x } \] Input:
integrate(sec(a+b*log(c*x^n))^2/x^2,x, algorithm="fricas")
Output:
integral(sec(b*log(c*x^n) + a)^2/x^2, x)
\[ \int \frac {\sec ^2\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int \frac {\sec ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{x^{2}}\, dx \] Input:
integrate(sec(a+b*ln(c*x**n))**2/x**2,x)
Output:
Integral(sec(a + b*log(c*x**n))**2/x**2, x)
\[ \int \frac {\sec ^2\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int { \frac {\sec \left (b \log \left (c x^{n}\right ) + a\right )^{2}}{x^{2}} \,d x } \] Input:
integrate(sec(a+b*log(c*x^n))^2/x^2,x, algorithm="maxima")
Output:
2*((2*b^2*n^2*x*cos(2*b*log(c))*cos(2*b*log(x^n) + 2*a) - 2*b^2*n^2*x*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*x*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*x*sin(2*b*log(x^n) + 2*a)^2 + b^2*n^2*x)*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*x^2*cos(2*b*log(c))*cos(2*b*log(x^n) + 2*a) - 2*b^2* n^2*x^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*x^2*cos(2*b*log(x^n) + 2*a)^2 + (b^2*cos(2*b*l og(c))^2 + b^2*sin(2*b*log(c))^2)*n^2*x^2*sin(2*b*log(x^n) + 2*a)^2 + b^2* n^2*x^2), x) + cos(2*b*log(x^n) + 2*a)*sin(2*b*log(c)) + cos(2*b*log(c))*s in(2*b*log(x^n) + 2*a))/(2*b*n*x*cos(2*b*log(c))*cos(2*b*log(x^n) + 2*a) + (b*cos(2*b*log(c))^2 + b*sin(2*b*log(c))^2)*n*x*cos(2*b*log(x^n) + 2*a)^2 - 2*b*n*x*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*x*sin(2*b*log(x^n) + 2*a)^2 + b*n*x)
\[ \int \frac {\sec ^2\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int { \frac {\sec \left (b \log \left (c x^{n}\right ) + a\right )^{2}}{x^{2}} \,d x } \] Input:
integrate(sec(a+b*log(c*x^n))^2/x^2,x, algorithm="giac")
Output:
integrate(sec(b*log(c*x^n) + a)^2/x^2, x)
Timed out. \[ \int \frac {\sec ^2\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int \frac {1}{x^2\,{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}^2} \,d x \] Input:
int(1/(x^2*cos(a + b*log(c*x^n))^2),x)
Output:
int(1/(x^2*cos(a + b*log(c*x^n))^2), x)
\[ \int \frac {\sec ^2\left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\text {too large to display} \] Input:
int(sec(a+b*log(c*x^n))^2/x^2,x)
Output:
( - 14*cos(log(x**n*c)*b + a)*sin(log(x**n*c)*b + a)*tan((log(x**n*c)*b + a)/2)**4*b**3*n**3 - 2*cos(log(x**n*c)*b + a)*sin(log(x**n*c)*b + a)*tan(( log(x**n*c)*b + a)/2)**4*b*n + 28*cos(log(x**n*c)*b + a)*sin(log(x**n*c)*b + a)*tan((log(x**n*c)*b + a)/2)**2*b**3*n**3 + 4*cos(log(x**n*c)*b + a)*s in(log(x**n*c)*b + a)*tan((log(x**n*c)*b + a)/2)**2*b*n - 14*cos(log(x**n* c)*b + a)*sin(log(x**n*c)*b + a)*b**3*n**3 - 2*cos(log(x**n*c)*b + a)*sin( log(x**n*c)*b + a)*b*n + 28*cos(log(x**n*c)*b + a)*tan((log(x**n*c)*b + a) /2)**4*b**2*n**2 - 56*cos(log(x**n*c)*b + a)*tan((log(x**n*c)*b + a)/2)**2 *b**2*n**2 + 28*cos(log(x**n*c)*b + a)*b**2*n**2 + 1792*int(tan((log(x**n* c)*b + a)/2)**3/(14*tan((log(x**n*c)*b + a)/2)**6*b**4*n**4*x**2 - 15*tan( (log(x**n*c)*b + a)/2)**6*b**2*n**2*x**2 + tan((log(x**n*c)*b + a)/2)**6*x **2 - 42*tan((log(x**n*c)*b + a)/2)**4*b**4*n**4*x**2 + 45*tan((log(x**n*c )*b + a)/2)**4*b**2*n**2*x**2 - 3*tan((log(x**n*c)*b + a)/2)**4*x**2 + 42* tan((log(x**n*c)*b + a)/2)**2*b**4*n**4*x**2 - 45*tan((log(x**n*c)*b + a)/ 2)**2*b**2*n**2*x**2 + 3*tan((log(x**n*c)*b + a)/2)**2*x**2 - 14*b**4*n**4 *x**2 + 15*b**2*n**2*x**2 - x**2),x)*sin(log(x**n*c)*b + a)**2*tan((log(x* *n*c)*b + a)/2)**4*b**7*n**7*x - 1920*int(tan((log(x**n*c)*b + a)/2)**3/(1 4*tan((log(x**n*c)*b + a)/2)**6*b**4*n**4*x**2 - 15*tan((log(x**n*c)*b + a )/2)**6*b**2*n**2*x**2 + tan((log(x**n*c)*b + a)/2)**6*x**2 - 42*tan((log( x**n*c)*b + a)/2)**4*b**4*n**4*x**2 + 45*tan((log(x**n*c)*b + a)/2)**4*...