Integrand size = 13, antiderivative size = 70 \[ \int \text {sech}^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {8 e^{3 a} x \left (c x^n\right )^{3 b} \operatorname {Hypergeometric2F1}\left (3,\frac {3 b+\frac {1}{n}}{2 b},\frac {1}{2} \left (5+\frac {1}{b n}\right ),-e^{2 a} \left (c x^n\right )^{2 b}\right )}{1+3 b n} \]
8*exp(3*a)*x*(c*x^n)^(3*b)*hypergeom([3, 1/2*(3*b+1/n)/b],[5/2+1/2/b/n],-e xp(2*a)*(c*x^n)^(2*b))/(3*b*n+1)
Time = 1.59 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.44 \[ \int \text {sech}^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x \left (2 e^a (-1+b n) \left (c x^n\right )^b \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (1+\frac {1}{b n}\right ),\frac {1}{2} \left (3+\frac {1}{b n}\right ),-e^{2 a} \left (c x^n\right )^{2 b}\right )+\text {sech}\left (a+b \log \left (c x^n\right )\right ) \left (1+b n \tanh \left (a+b \log \left (c x^n\right )\right )\right )\right )}{2 b^2 n^2} \]
(x*(2*E^a*(-1 + b*n)*(c*x^n)^b*Hypergeometric2F1[1, (1 + 1/(b*n))/2, (3 + 1/(b*n))/2, -(E^(2*a)*(c*x^n)^(2*b))] + Sech[a + b*Log[c*x^n]]*(1 + b*n*Ta nh[a + b*Log[c*x^n]])))/(2*b^2*n^2)
Time = 0.32 (sec) , antiderivative size = 70, 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.
\(\displaystyle \int \text {sech}^3\left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 6079 |
\(\displaystyle \frac {x \left (c x^n\right )^{-1/n} \int \left (c x^n\right )^{\frac {1}{n}-1} \text {sech}^3\left (a+b \log \left (c x^n\right )\right )d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 6081 |
\(\displaystyle \frac {8 e^{-3 a} x \left (c x^n\right )^{-1/n} \int \frac {\left (c x^n\right )^{-3 b+\frac {1}{n}-1}}{\left (e^{-2 a} \left (c x^n\right )^{-2 b}+1\right )^3}d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \frac {8 e^{-3 a} x \left (c x^n\right )^{-1/n} \int \frac {\left (c x^n\right )^{3 b+\frac {1}{n}-1}}{\left (\left (c x^n\right )^{2 b}+e^{-2 a}\right )^3}d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {8 e^{3 a} x \left (c x^n\right )^{3 b} \operatorname {Hypergeometric2F1}\left (3,\frac {3 b+\frac {1}{n}}{2 b},\frac {1}{2} \left (5+\frac {1}{b n}\right ),-e^{2 a} \left (c x^n\right )^{2 b}\right )}{3 b n+1}\) |
(8*E^(3*a)*x*(c*x^n)^(3*b)*Hypergeometric2F1[3, (3*b + n^(-1))/(2*b), (5 + 1/(b*n))/2, -(E^(2*a)*(c*x^n)^(2*b))])/(1 + 3*b*n)
3.2.84.3.1 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]
\[\int {\operatorname {sech}\left (a +b \ln \left (c \,x^{n}\right )\right )}^{3}d x\]
\[ \int \text {sech}^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \operatorname {sech}\left (b \log \left (c x^{n}\right ) + a\right )^{3} \,d x } \]
\[ \int \text {sech}^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\int \operatorname {sech}^{3}{\left (a + b \log {\left (c x^{n} \right )} \right )}\, dx \]
\[ \int \text {sech}^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \operatorname {sech}\left (b \log \left (c x^{n}\right ) + a\right )^{3} \,d x } \]
8*(b^2*c^b*n^2 - c^b)*integrate(1/8*e^(b*log(x^n) + a)/(b^2*c^(2*b)*n^2*e^ (2*b*log(x^n) + 2*a) + b^2*n^2), x) + ((b*c^(3*b)*n + c^(3*b))*x*e^(3*b*lo g(x^n) + 3*a) - (b*c^b*n - c^b)*x*e^(b*log(x^n) + a))/(b^2*c^(4*b)*n^2*e^( 4*b*log(x^n) + 4*a) + 2*b^2*c^(2*b)*n^2*e^(2*b*log(x^n) + 2*a) + b^2*n^2)
\[ \int \text {sech}^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \operatorname {sech}\left (b \log \left (c x^{n}\right ) + a\right )^{3} \,d x } \]
Timed out. \[ \int \text {sech}^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\int \frac {1}{{\mathrm {cosh}\left (a+b\,\ln \left (c\,x^n\right )\right )}^3} \,d x \]