Integrand size = 24, antiderivative size = 363 \[ \int (e x)^{-1+3 n} \left (a+b \text {sech}\left (c+d x^n\right )\right )^2 \, dx=\frac {a^2 (e x)^{3 n}}{3 e n}+\frac {b^2 x^{-n} (e x)^{3 n}}{d e n}+\frac {4 a b x^{-n} (e x)^{3 n} \arctan \left (e^{c+d x^n}\right )}{d e n}-\frac {2 b^2 x^{-2 n} (e x)^{3 n} \log \left (1+e^{2 \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {4 i a b x^{-2 n} (e x)^{3 n} \operatorname {PolyLog}\left (2,-i e^{c+d x^n}\right )}{d^2 e n}+\frac {4 i a b x^{-2 n} (e x)^{3 n} \operatorname {PolyLog}\left (2,i e^{c+d x^n}\right )}{d^2 e n}-\frac {b^2 x^{-3 n} (e x)^{3 n} \operatorname {PolyLog}\left (2,-e^{2 \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {4 i a b x^{-3 n} (e x)^{3 n} \operatorname {PolyLog}\left (3,-i e^{c+d x^n}\right )}{d^3 e n}-\frac {4 i a b x^{-3 n} (e x)^{3 n} \operatorname {PolyLog}\left (3,i e^{c+d x^n}\right )}{d^3 e n}+\frac {b^2 x^{-n} (e x)^{3 n} \tanh \left (c+d x^n\right )}{d e n} \]
1/3*a^2*(e*x)^(3*n)/e/n+b^2*(e*x)^(3*n)/d/e/n/(x^n)+4*a*b*(e*x)^(3*n)*arct an(exp(c+d*x^n))/d/e/n/(x^n)-2*b^2*(e*x)^(3*n)*ln(1+exp(2*c+2*d*x^n))/d^2/ e/n/(x^(2*n))-4*I*a*b*(e*x)^(3*n)*polylog(2,-I*exp(c+d*x^n))/d^2/e/n/(x^(2 *n))+4*I*a*b*(e*x)^(3*n)*polylog(2,I*exp(c+d*x^n))/d^2/e/n/(x^(2*n))-b^2*( e*x)^(3*n)*polylog(2,-exp(2*c+2*d*x^n))/d^3/e/n/(x^(3*n))+4*I*a*b*(e*x)^(3 *n)*polylog(3,-I*exp(c+d*x^n))/d^3/e/n/(x^(3*n))-4*I*a*b*(e*x)^(3*n)*polyl og(3,I*exp(c+d*x^n))/d^3/e/n/(x^(3*n))+b^2*(e*x)^(3*n)*tanh(c+d*x^n)/d/e/n /(x^n)
\[ \int (e x)^{-1+3 n} \left (a+b \text {sech}\left (c+d x^n\right )\right )^2 \, dx=\int (e x)^{-1+3 n} \left (a+b \text {sech}\left (c+d x^n\right )\right )^2 \, dx \]
Time = 0.66 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.65, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {5963, 5959, 3042, 4678, 2009}
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.
| \(\displaystyle \int (e x)^{3 n-1} \left (a+b \text {sech}\left (c+d x^n\right )\right )^2 \, dx\) |
| \(\Big \downarrow \) 5963 |
| \(\displaystyle \frac {x^{-3 n} (e x)^{3 n} \int x^{3 n-1} \left (a+b \text {sech}\left (d x^n+c\right )\right )^2dx}{e}\) |
| \(\Big \downarrow \) 5959 |
| \(\displaystyle \frac {x^{-3 n} (e x)^{3 n} \int x^{2 n} \left (a+b \text {sech}\left (d x^n+c\right )\right )^2dx^n}{e n}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {x^{-3 n} (e x)^{3 n} \int x^{2 n} \left (a+b \csc \left (i d x^n+i c+\frac {\pi }{2}\right )\right )^2dx^n}{e n}\) |
| \(\Big \downarrow \) 4678 |
| \(\displaystyle \frac {x^{-3 n} (e x)^{3 n} \int \left (a^2 x^{2 n}+b^2 \text {sech}^2\left (d x^n+c\right ) x^{2 n}+2 a b \text {sech}\left (d x^n+c\right ) x^{2 n}\right )dx^n}{e n}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x^{-3 n} (e x)^{3 n} \left (\frac {1}{3} a^2 x^{3 n}+\frac {4 a b x^{2 n} \arctan \left (e^{c+d x^n}\right )}{d}+\frac {4 i a b \operatorname {PolyLog}\left (3,-i e^{d x^n+c}\right )}{d^3}-\frac {4 i a b \operatorname {PolyLog}\left (3,i e^{d x^n+c}\right )}{d^3}-\frac {4 i a b x^n \operatorname {PolyLog}\left (2,-i e^{d x^n+c}\right )}{d^2}+\frac {4 i a b x^n \operatorname {PolyLog}\left (2,i e^{d x^n+c}\right )}{d^2}-\frac {b^2 \operatorname {PolyLog}\left (2,-e^{2 \left (d x^n+c\right )}\right )}{d^3}-\frac {2 b^2 x^n \log \left (e^{2 \left (c+d x^n\right )}+1\right )}{d^2}+\frac {b^2 x^{2 n} \tanh \left (c+d x^n\right )}{d}+\frac {b^2 x^{2 n}}{d}\right )}{e n}\) |
((e*x)^(3*n)*((b^2*x^(2*n))/d + (a^2*x^(3*n))/3 + (4*a*b*x^(2*n)*ArcTan[E^ (c + d*x^n)])/d - (2*b^2*x^n*Log[1 + E^(2*(c + d*x^n))])/d^2 - ((4*I)*a*b* x^n*PolyLog[2, (-I)*E^(c + d*x^n)])/d^2 + ((4*I)*a*b*x^n*PolyLog[2, I*E^(c + d*x^n)])/d^2 - (b^2*PolyLog[2, -E^(2*(c + d*x^n))])/d^3 + ((4*I)*a*b*Po lyLog[3, (-I)*E^(c + d*x^n)])/d^3 - ((4*I)*a*b*PolyLog[3, I*E^(c + d*x^n)] )/d^3 + (b^2*x^(2*n)*Tanh[c + d*x^n])/d))/(e*n*x^(3*n))
3.1.78.3.1 Defintions of rubi rules used
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) , x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]
Int[(x_)^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbo l] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sech[c + d*x] )^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m + 1)/n], 0] && IntegerQ[p]
Int[((e_)*(x_))^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Simp[e^IntPart[m]*((e*x)^FracPart[m]/x^FracPart[m]) Int[x^m* (a + b*Sech[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x]
\[\int \left (e x \right )^{-1+3 n} {\left (a +b \,\operatorname {sech}\left (c +d \,x^{n}\right )\right )}^{2}d x\]
Timed out. \[ \int (e x)^{-1+3 n} \left (a+b \text {sech}\left (c+d x^n\right )\right )^2 \, dx=\text {Timed out} \]
\[ \int (e x)^{-1+3 n} \left (a+b \text {sech}\left (c+d x^n\right )\right )^2 \, dx=\int \left (e x\right )^{3 n - 1} \left (a + b \operatorname {sech}{\left (c + d x^{n} \right )}\right )^{2}\, dx \]
\[ \int (e x)^{-1+3 n} \left (a+b \text {sech}\left (c+d x^n\right )\right )^2 \, dx=\int { {\left (b \operatorname {sech}\left (d x^{n} + c\right ) + a\right )}^{2} \left (e x\right )^{3 \, n - 1} \,d x } \]
-2*b^2*e^(3*n)*x^(2*n)/(d*e*n*e^(2*d*x^n + 2*c) + d*e*n) + 1/3*(e*x)^(3*n) *a^2/(e*n) + integrate(4*(a*b*d*e^(3*n)*e^(d*x^n + 3*n*log(x) + c) + b^2*e ^(3*n)*x^(2*n))/(d*e*x*e^(2*d*x^n + 2*c) + d*e*x), x)
\[ \int (e x)^{-1+3 n} \left (a+b \text {sech}\left (c+d x^n\right )\right )^2 \, dx=\int { {\left (b \operatorname {sech}\left (d x^{n} + c\right ) + a\right )}^{2} \left (e x\right )^{3 \, n - 1} \,d x } \]
Timed out. \[ \int (e x)^{-1+3 n} \left (a+b \text {sech}\left (c+d x^n\right )\right )^2 \, dx=\int {\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,x^n\right )}\right )}^2\,{\left (e\,x\right )}^{3\,n-1} \,d x \]