Integrand size = 23, antiderivative size = 92 \[ \int \text {sech}^4(e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{2},\frac {5}{2},-p,\frac {3}{2},-\sinh ^2(e+f x),-\frac {b \sinh ^2(e+f x)}{a}\right ) \sqrt {\cosh ^2(e+f x)} \left (a+b \sinh ^2(e+f x)\right )^p \left (1+\frac {b \sinh ^2(e+f x)}{a}\right )^{-p} \tanh (e+f x)}{f} \]
AppellF1(1/2,5/2,-p,3/2,-sinh(f*x+e)^2,-b*sinh(f*x+e)^2/a)*(a+b*sinh(f*x+e )^2)^p*(cosh(f*x+e)^2)^(1/2)*tanh(f*x+e)/f/((1+b*sinh(f*x+e)^2/a)^p)
\[ \int \text {sech}^4(e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \, dx=\int \text {sech}^4(e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \, dx \]
Time = 0.26 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3671, 334, 333}
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.
\(\displaystyle \int \text {sech}^4(e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a-b \sin (i e+i f x)^2\right )^p}{\cos (i e+i f x)^4}dx\) |
\(\Big \downarrow \) 3671 |
\(\displaystyle \frac {\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x) \int \frac {\left (b \sinh ^2(e+f x)+a\right )^p}{\left (\sinh ^2(e+f x)+1\right )^{5/2}}d\sinh (e+f x)}{f}\) |
\(\Big \downarrow \) 334 |
\(\displaystyle \frac {\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \left (\frac {b \sinh ^2(e+f x)}{a}+1\right )^{-p} \int \frac {\left (\frac {b \sinh ^2(e+f x)}{a}+1\right )^p}{\left (\sinh ^2(e+f x)+1\right )^{5/2}}d\sinh (e+f x)}{f}\) |
\(\Big \downarrow \) 333 |
\(\displaystyle \frac {\sqrt {\cosh ^2(e+f x)} \tanh (e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \left (\frac {b \sinh ^2(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},\frac {5}{2},-p,\frac {3}{2},-\sinh ^2(e+f x),-\frac {b \sinh ^2(e+f x)}{a}\right )}{f}\) |
(AppellF1[1/2, 5/2, -p, 3/2, -Sinh[e + f*x]^2, -((b*Sinh[e + f*x]^2)/a)]*S qrt[Cosh[e + f*x]^2]*(a + b*Sinh[e + f*x]^2)^p*Tanh[e + f*x])/(f*(1 + (b*S inh[e + f*x]^2)/a)^p)
3.5.10.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[ (1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && !(IntegerQ[p] || GtQ[a, 0])
Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff*(Sqrt[ Cos[e + f*x]^2]/(f*Cos[e + f*x])) Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && !IntegerQ[p]
\[\int \operatorname {sech}\left (f x +e \right )^{4} \left (a +b \sinh \left (f x +e \right )^{2}\right )^{p}d x\]
\[ \int \text {sech}^4(e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \, dx=\int { {\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{p} \operatorname {sech}\left (f x + e\right )^{4} \,d x } \]
Timed out. \[ \int \text {sech}^4(e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \, dx=\text {Timed out} \]
\[ \int \text {sech}^4(e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \, dx=\int { {\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{p} \operatorname {sech}\left (f x + e\right )^{4} \,d x } \]
\[ \int \text {sech}^4(e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \, dx=\int { {\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{p} \operatorname {sech}\left (f x + e\right )^{4} \,d x } \]
Timed out. \[ \int \text {sech}^4(e+f x) \left (a+b \sinh ^2(e+f x)\right )^p \, dx=\int \frac {{\left (b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^p}{{\mathrm {cosh}\left (e+f\,x\right )}^4} \,d x \]