Integrand size = 18, antiderivative size = 133 \[ \int F^{c (a+b x)} \text {sech}^4(d+e x) \, dx=\frac {2 e^{2 (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1+\frac {b c \log (F)}{2 e},2+\frac {b c \log (F)}{2 e},-e^{2 (d+e x)}\right ) (2 e-b c \log (F))}{3 e^2}+\frac {b c F^{c (a+b x)} \log (F) \text {sech}^2(d+e x)}{6 e^2}+\frac {F^{c (a+b x)} \text {sech}^2(d+e x) \tanh (d+e x)}{3 e} \]
2/3*exp(2*e*x+2*d)*F^(c*(b*x+a))*hypergeom([2, 1+1/2*b*c*ln(F)/e],[2+1/2*b *c*ln(F)/e],-exp(2*e*x+2*d))*(2*e-b*c*ln(F))/e^2+1/6*b*c*F^(c*(b*x+a))*ln( F)*sech(e*x+d)^2/e^2+1/3*F^(c*(b*x+a))*sech(e*x+d)^2*tanh(e*x+d)/e
Time = 0.14 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.76 \[ \int F^{c (a+b x)} \text {sech}^4(d+e x) \, dx=\frac {F^{c (a+b x)} \left (4 e^{2 (d+e x)} \operatorname {Hypergeometric2F1}\left (2,1+\frac {b c \log (F)}{2 e},2+\frac {b c \log (F)}{2 e},-e^{2 (d+e x)}\right ) (2 e-b c \log (F))+\text {sech}^2(d+e x) (b c \log (F)+2 e \tanh (d+e x))\right )}{6 e^2} \]
(F^(c*(a + b*x))*(4*E^(2*(d + e*x))*Hypergeometric2F1[2, 1 + (b*c*Log[F])/ (2*e), 2 + (b*c*Log[F])/(2*e), -E^(2*(d + e*x))]*(2*e - b*c*Log[F]) + Sech [d + e*x]^2*(b*c*Log[F] + 2*e*Tanh[d + e*x])))/(6*e^2)
Time = 0.36 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.11, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6013, 6015}
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.
\(\displaystyle \int \text {sech}^4(d+e x) F^{c (a+b x)} \, dx\) |
\(\Big \downarrow \) 6013 |
\(\displaystyle \frac {1}{6} \left (4-\frac {b^2 c^2 \log ^2(F)}{e^2}\right ) \int F^{c (a+b x)} \text {sech}^2(d+e x)dx+\frac {b c \log (F) \text {sech}^2(d+e x) F^{c (a+b x)}}{6 e^2}+\frac {\tanh (d+e x) \text {sech}^2(d+e x) F^{c (a+b x)}}{3 e}\) |
\(\Big \downarrow \) 6015 |
\(\displaystyle \frac {2 e^{2 (d+e x)} F^{c (a+b x)} \left (4-\frac {b^2 c^2 \log ^2(F)}{e^2}\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {b c \log (F)}{2 e}+1,\frac {b c \log (F)}{2 e}+2,-e^{2 (d+e x)}\right )}{3 (b c \log (F)+2 e)}+\frac {b c \log (F) \text {sech}^2(d+e x) F^{c (a+b x)}}{6 e^2}+\frac {\tanh (d+e x) \text {sech}^2(d+e x) F^{c (a+b x)}}{3 e}\) |
(2*E^(2*(d + e*x))*F^(c*(a + b*x))*Hypergeometric2F1[2, 1 + (b*c*Log[F])/( 2*e), 2 + (b*c*Log[F])/(2*e), -E^(2*(d + e*x))]*(4 - (b^2*c^2*Log[F]^2)/e^ 2))/(3*(2*e + b*c*Log[F])) + (b*c*F^(c*(a + b*x))*Log[F]*Sech[d + e*x]^2)/ (6*e^2) + (F^(c*(a + b*x))*Sech[d + e*x]^2*Tanh[d + e*x])/(3*e)
3.3.91.3.1 Defintions of rubi rules used
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sech[(d_.) + (e_.)*(x_)]^(n_), x_Symb ol] :> Simp[b*c*Log[F]*F^(c*(a + b*x))*(Sech[d + e*x]^(n - 2)/(e^2*(n - 1)* (n - 2))), x] + (Simp[F^(c*(a + b*x))*Sech[d + e*x]^(n - 1)*(Sinh[d + e*x]/ (e*(n - 1))), x] + Simp[(e^2*(n - 2)^2 - b^2*c^2*Log[F]^2)/(e^2*(n - 1)*(n - 2)) Int[F^(c*(a + b*x))*Sech[d + e*x]^(n - 2), x], x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*(n - 2)^2 - b^2*c^2*Log[F]^2, 0] && GtQ[n, 1] && NeQ[n, 2]
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sech[(d_.) + (e_.)*(x_)]^(n_.), x_Sym bol] :> Simp[2^n*E^(n*(d + e*x))*(F^(c*(a + b*x))/(e*n + b*c*Log[F]))*Hyper geometric2F1[n, n/2 + b*c*(Log[F]/(2*e)), 1 + n/2 + b*c*(Log[F]/(2*e)), -E^ (2*(d + e*x))], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]
\[\int F^{c \left (b x +a \right )} \operatorname {sech}\left (e x +d \right )^{4}d x\]
\[ \int F^{c (a+b x)} \text {sech}^4(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {sech}\left (e x + d\right )^{4} \,d x } \]
\[ \int F^{c (a+b x)} \text {sech}^4(d+e x) \, dx=\int F^{c \left (a + b x\right )} \operatorname {sech}^{4}{\left (d + e x \right )}\, dx \]
\[ \int F^{c (a+b x)} \text {sech}^4(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {sech}\left (e x + d\right )^{4} \,d x } \]
-128*(F^(a*c)*b^2*c^2*e*log(F)^2 + 2*F^(a*c)*b*c*e^2*log(F))*integrate(F^( b*c*x)/(b^3*c^3*log(F)^3 - 18*b^2*c^2*e*log(F)^2 + 104*b*c*e^2*log(F) - 19 2*e^3 + (b^3*c^3*e^(10*d)*log(F)^3 - 18*b^2*c^2*e*e^(10*d)*log(F)^2 + 104* b*c*e^2*e^(10*d)*log(F) - 192*e^3*e^(10*d))*e^(10*e*x) + 5*(b^3*c^3*e^(8*d )*log(F)^3 - 18*b^2*c^2*e*e^(8*d)*log(F)^2 + 104*b*c*e^2*e^(8*d)*log(F) - 192*e^3*e^(8*d))*e^(8*e*x) + 10*(b^3*c^3*e^(6*d)*log(F)^3 - 18*b^2*c^2*e*e ^(6*d)*log(F)^2 + 104*b*c*e^2*e^(6*d)*log(F) - 192*e^3*e^(6*d))*e^(6*e*x) + 10*(b^3*c^3*e^(4*d)*log(F)^3 - 18*b^2*c^2*e*e^(4*d)*log(F)^2 + 104*b*c*e ^2*e^(4*d)*log(F) - 192*e^3*e^(4*d))*e^(4*e*x) + 5*(b^3*c^3*e^(2*d)*log(F) ^3 - 18*b^2*c^2*e*e^(2*d)*log(F)^2 + 104*b*c*e^2*e^(2*d)*log(F) - 192*e^3* e^(2*d))*e^(2*e*x)), x) + 16*(8*F^(a*c)*b*c*e*log(F) + 16*F^(a*c)*e^2 + (F ^(a*c)*b^2*c^2*e^(4*d)*log(F)^2 - 14*F^(a*c)*b*c*e*e^(4*d)*log(F) + 48*F^( a*c)*e^2*e^(4*d))*e^(4*e*x) - 8*(F^(a*c)*b*c*e*e^(2*d)*log(F) - 8*F^(a*c)* e^2*e^(2*d))*e^(2*e*x))*F^(b*c*x)/(b^3*c^3*log(F)^3 - 18*b^2*c^2*e*log(F)^ 2 + 104*b*c*e^2*log(F) - 192*e^3 + (b^3*c^3*e^(8*d)*log(F)^3 - 18*b^2*c^2* e*e^(8*d)*log(F)^2 + 104*b*c*e^2*e^(8*d)*log(F) - 192*e^3*e^(8*d))*e^(8*e* x) + 4*(b^3*c^3*e^(6*d)*log(F)^3 - 18*b^2*c^2*e*e^(6*d)*log(F)^2 + 104*b*c *e^2*e^(6*d)*log(F) - 192*e^3*e^(6*d))*e^(6*e*x) + 6*(b^3*c^3*e^(4*d)*log( F)^3 - 18*b^2*c^2*e*e^(4*d)*log(F)^2 + 104*b*c*e^2*e^(4*d)*log(F) - 192*e^ 3*e^(4*d))*e^(4*e*x) + 4*(b^3*c^3*e^(2*d)*log(F)^3 - 18*b^2*c^2*e*e^(2*...
\[ \int F^{c (a+b x)} \text {sech}^4(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {sech}\left (e x + d\right )^{4} \,d x } \]
Timed out. \[ \int F^{c (a+b x)} \text {sech}^4(d+e x) \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\mathrm {cosh}\left (d+e\,x\right )}^4} \,d x \]