Integrand size = 13, antiderivative size = 69 \[ \int \text {sech}^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {4 e^{2 a} x \left (c x^n\right )^{2 b} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (2+\frac {1}{b n}\right ),\frac {1}{2} \left (4+\frac {1}{b n}\right ),-e^{2 a} \left (c x^n\right )^{2 b}\right )}{1+2 b n} \]
4*exp(2*a)*x*(c*x^n)^(2*b)*hypergeom([2, 1+1/2/b/n],[2+1/2/b/n],-exp(2*a)* (c*x^n)^(2*b))/(2*b*n+1)
Time = 5.61 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.83 \[ \int \text {sech}^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x \left (-\frac {e^{2 a} \left (c x^n\right )^{2 b} \operatorname {Hypergeometric2F1}\left (1,1+\frac {1}{2 b n},2+\frac {1}{2 b n},-e^{2 a} \left (c x^n\right )^{2 b}\right )}{1+2 b n}+\operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 b n},1+\frac {1}{2 b n},-e^{2 a} \left (c x^n\right )^{2 b}\right )+\tanh \left (a+b \log \left (c x^n\right )\right )\right )}{b n} \]
(x*(-((E^(2*a)*(c*x^n)^(2*b)*Hypergeometric2F1[1, 1 + 1/(2*b*n), 2 + 1/(2* b*n), -(E^(2*a)*(c*x^n)^(2*b))])/(1 + 2*b*n)) + Hypergeometric2F1[1, 1/(2* b*n), 1 + 1/(2*b*n), -(E^(2*a)*(c*x^n)^(2*b))] + Tanh[a + b*Log[c*x^n]]))/ (b*n)
Time = 0.32 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {6079, 6081, 795, 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.
\(\displaystyle \int \text {sech}^2\left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 6079 |
\(\displaystyle \frac {x \left (c x^n\right )^{-1/n} \int \left (c x^n\right )^{\frac {1}{n}-1} \text {sech}^2\left (a+b \log \left (c x^n\right )\right )d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 6081 |
\(\displaystyle \frac {4 e^{-2 a} x \left (c x^n\right )^{-1/n} \int \frac {\left (c x^n\right )^{-2 b+\frac {1}{n}-1}}{\left (e^{-2 a} \left (c x^n\right )^{-2 b}+1\right )^2}d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \frac {4 e^{-2 a} x \left (c x^n\right )^{-1/n} \int \frac {\left (c x^n\right )^{2 b+\frac {1}{n}-1}}{\left (\left (c x^n\right )^{2 b}+e^{-2 a}\right )^2}d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {4 e^{2 a} x \left (c x^n\right )^{2 b} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (2+\frac {1}{b n}\right ),\frac {1}{2} \left (4+\frac {1}{b n}\right ),-e^{2 a} \left (c x^n\right )^{2 b}\right )}{2 b n+1}\) |
(4*E^(2*a)*x*(c*x^n)^(2*b)*Hypergeometric2F1[2, (2 + 1/(b*n))/2, (4 + 1/(b *n))/2, -(E^(2*a)*(c*x^n)^(2*b))])/(1 + 2*b*n)
3.2.83.3.1 Defintions of rubi rules used
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)* (b + a/x^n)^p, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && NegQ[n]
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[Sech[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> S imp[x/(n*(c*x^n)^(1/n)) Subst[Int[x^(1/n - 1)*Sech[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_.)*Sech[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Simp[2^p/E^(a*d*p) Int[(e*x)^m*(1/(x^(b*d*p)*(1 + 1/(E^(2*a*d)*x^(2*b *d)))^p)), x], x] /; FreeQ[{a, b, d, e, m}, x] && IntegerQ[p]
\[\int {\operatorname {sech}\left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}d x\]
\[ \int \text {sech}^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \operatorname {sech}\left (b \log \left (c x^{n}\right ) + a\right )^{2} \,d x } \]
\[ \int \text {sech}^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int \operatorname {sech}^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}\, dx \]
\[ \int \text {sech}^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \operatorname {sech}\left (b \log \left (c x^{n}\right ) + a\right )^{2} \,d x } \]
-2*x/(b*c^(2*b)*n*e^(2*b*log(x^n) + 2*a) + b*n) + 4*integrate(1/2/(b*c^(2* b)*n*e^(2*b*log(x^n) + 2*a) + b*n), x)
\[ \int \text {sech}^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \operatorname {sech}\left (b \log \left (c x^{n}\right ) + a\right )^{2} \,d x } \]
Timed out. \[ \int \text {sech}^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\int \frac {1}{{\mathrm {cosh}\left (a+b\,\ln \left (c\,x^n\right )\right )}^2} \,d x \]