Integrand size = 17, antiderivative size = 102 \[ \int x^m \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {4 e^{2 i a} x^{1+m} \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (2,-\frac {i (1+m)-2 b n}{2 b n},-\frac {i (1+m)-4 b n}{2 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{1+m+2 i b n} \]
4*exp(2*I*a)*x^(1+m)*(c*x^n)^(2*I*b)*hypergeom([2, 1/2*(-I*(1+m)+2*b*n)/b/ n],[1/2*(-I*(1+m)+4*b*n)/b/n],-exp(2*I*a)*(c*x^n)^(2*I*b))/(1+m+2*I*b*n)
Time = 15.27 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.94 \[ \int x^m \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {i x^{1+m} \left ((1+m+2 i b n) \operatorname {Hypergeometric2F1}\left (1,-\frac {i (1+m)}{2 b n},1-\frac {i (1+m)}{2 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )-e^{2 i a} (1+m) \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (1,-\frac {i (1+m+2 i b n)}{2 b n},-\frac {i (1+m+4 i b n)}{2 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )+i (1+m+2 i b n) \tan \left (a+b \log \left (c x^n\right )\right )\right )}{b n (1+m+2 i b n)} \]
((-I)*x^(1 + m)*((1 + m + (2*I)*b*n)*Hypergeometric2F1[1, ((-1/2*I)*(1 + m ))/(b*n), 1 - ((I/2)*(1 + m))/(b*n), -E^((2*I)*(a + b*Log[c*x^n]))] - E^(( 2*I)*a)*(1 + m)*(c*x^n)^((2*I)*b)*Hypergeometric2F1[1, ((-1/2*I)*(1 + m + (2*I)*b*n))/(b*n), ((-1/2*I)*(1 + m + (4*I)*b*n))/(b*n), -E^((2*I)*(a + b* Log[c*x^n]))] + I*(1 + m + (2*I)*b*n)*Tan[a + b*Log[c*x^n]]))/(b*n*(1 + m + (2*I)*b*n))
Time = 0.31 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.17, 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.
| \(\displaystyle \int x^m \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 5020 |
| \(\displaystyle \frac {x^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \left (c x^n\right )^{\frac {m+1}{n}-1} \sec ^2\left (a+b \log \left (c x^n\right )\right )d\left (c x^n\right )}{n}\) |
| \(\Big \downarrow \) 5016 |
| \(\displaystyle \frac {4 e^{2 i a} x^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \frac {\left (c x^n\right )^{2 i b+\frac {m+1}{n}-1}}{\left (e^{2 i a} \left (c x^n\right )^{2 i b}+1\right )^2}d\left (c x^n\right )}{n}\) |
| \(\Big \downarrow \) 888 |
| \(\displaystyle \frac {4 e^{2 i a} x^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}+\frac {2 i b n+m+1}{n}} \operatorname {Hypergeometric2F1}\left (2,-\frac {i (m+1)-2 b n}{2 b n},-\frac {i (m+1)-4 b n}{2 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{2 i b n+m+1}\) |
(4*E^((2*I)*a)*x^(1 + m)*(c*x^n)^(-((1 + m)/n) + (1 + m + (2*I)*b*n)/n)*Hy pergeometric2F1[2, -1/2*(I*(1 + m) - 2*b*n)/(b*n), -1/2*(I*(1 + m) - 4*b*n )/(b*n), -(E^((2*I)*a)*(c*x^n)^((2*I)*b))])/(1 + m + (2*I)*b*n)
3.3.79.3.1 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])
\[\int x^{m} {\sec \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}d x\]
\[ \int x^m \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right )^{2} \,d x } \]
\[ \int x^m \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^{m} \sec ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}\, dx \]
\[ \int x^m \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right )^{2} \,d x } \]
2*(x*x^m*cos(2*b*log(x^n) + 2*a)*sin(2*b*log(c)) + x*x^m*cos(2*b*log(c))*s in(2*b*log(x^n) + 2*a) - ((b^2*cos(2*b*log(c))^2 + b^2*sin(2*b*log(c))^2 + (b^2*cos(2*b*log(c))^2 + b^2*sin(2*b*log(c))^2)*m)*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 + (b^2*cos(2*b*lo g(c))^2 + b^2*sin(2*b*log(c))^2)*m)*n^2*sin(2*b*log(x^n) + 2*a)^2 + 2*(b^2 *m*cos(2*b*log(c)) + b^2*cos(2*b*log(c)))*n^2*cos(2*b*log(x^n) + 2*a) - 2* (b^2*m*sin(2*b*log(c)) + b^2*sin(2*b*log(c)))*n^2*sin(2*b*log(x^n) + 2*a) + (b^2*m + b^2)*n^2)*integrate((x^m*cos(2*b*log(x^n) + 2*a)*sin(2*b*log(c) ) + x^m*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*lo g(x^n) + 2*a)^2 + b^2*n^2), x))/(2*b*n*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*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)
\[ \int x^m \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right )^{2} \,d x } \]
Timed out. \[ \int x^m \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int \frac {x^m}{{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}^2} \,d x \]