Integrand size = 15, antiderivative size = 79 \[ \int x \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 e^{2 i a} x^2 \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i}{b n},2-\frac {i}{b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{1+i b n} \]
2*exp(2*I*a)*x^2*(c*x^n)^(2*I*b)*hypergeom([2, 1-I/b/n],[2-I/b/n],-exp(2*I *a)*(c*x^n)^(2*I*b))/(1+I*b*n)
Time = 4.05 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.89 \[ \int x \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^2 \left (e^{2 i a} \left (c x^n\right )^{2 i b} \operatorname {Hypergeometric2F1}\left (1,1-\frac {i}{b n},2-\frac {i}{b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )+(-i+b n) \left (-i \operatorname {Hypergeometric2F1}\left (1,-\frac {i}{b n},1-\frac {i}{b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )+\tan \left (a+b \log \left (c x^n\right )\right )\right )\right )}{b n (-i+b n)} \]
(x^2*(E^((2*I)*a)*(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]))] + (-I + b*n)*((-I)*Hypergeometric2 F1[1, (-I)/(b*n), 1 - I/(b*n), -E^((2*I)*(a + b*Log[c*x^n]))] + Tan[a + b* Log[c*x^n]])))/(b*n*(-I + b*n))
Time = 0.28 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.15, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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.
\(\displaystyle \int x \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 5020 |
\(\displaystyle \frac {x^2 \left (c x^n\right )^{-2/n} \int \left (c x^n\right )^{\frac {2}{n}-1} \sec ^2\left (a+b \log \left (c x^n\right )\right )d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 5016 |
\(\displaystyle \frac {4 e^{2 i a} x^2 \left (c x^n\right )^{-2/n} \int \frac {\left (c x^n\right )^{2 i b+\frac {2}{n}-1}}{\left (e^{2 i a} \left (c x^n\right )^{2 i b}+1\right )^2}d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {2 e^{2 i a} x^2 \left (c x^n\right )^{-\frac {2}{n}+2 \left (\frac {1}{n}+i b\right )} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i}{b n},2-\frac {i}{b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{1+i b n}\) |
(2*E^((2*I)*a)*x^2*(c*x^n)^(2*(I*b + n^(-1)) - 2/n)*Hypergeometric2F1[2, 1 - I/(b*n), 2 - I/(b*n), -(E^((2*I)*a)*(c*x^n)^((2*I)*b))])/(1 + I*b*n)
3.3.44.3.1 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])
\[\int x {\sec \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}d x\]
\[ \int x \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x \sec \left (b \log \left (c x^{n}\right ) + a\right )^{2} \,d x } \]
\[ \int x \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int x \sec ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}\, dx \]
\[ \int x \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x \sec \left (b \log \left (c x^{n}\right ) + a\right )^{2} \,d x } \]
2*(x^2*cos(2*b*log(x^n) + 2*a)*sin(2*b*log(c)) + x^2*cos(2*b*log(c))*sin(2 *b*log(x^n) + 2*a) - 2*(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*l og(c))^2 + b^2*sin(2*b*log(c))^2)*n^2*sin(2*b*log(x^n) + 2*a)^2 + b^2*n^2) *integrate((x*cos(2*b*log(x^n) + 2*a)*sin(2*b*log(c)) + x*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*log(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)
\[ \int x \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x \sec \left (b \log \left (c x^{n}\right ) + a\right )^{2} \,d x } \]
Timed out. \[ \int x \sec ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int \frac {x}{{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}^2} \,d x \]