\(\int F^{c (a+b x)} \text {sech}^n(d+e x) \, dx\) [891]

Optimal result

Integrand size = 18, antiderivative size = 90 \[ \int F^{c (a+b x)} \text {sech}^n(d+e x) \, dx=\frac {\left (1+e^{2 (d+e x)}\right )^n F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (n,\frac {e n+b c \log (F)}{2 e},1+\frac {e n+b c \log (F)}{2 e},-e^{2 (d+e x)}\right ) \text {sech}^n(d+e x)}{e n+b c \log (F)} \]

[Out]

(1+exp(2*e*x+2*d))^n*F^(b*c*x+a*c)*hypergeom([n, 1/2*(e*n+b*c*ln(F))/e],[1+1/2*(e*n+b*c*ln(F))/e],-exp(2*e*x+2
*d))*sech(e*x+d)^n/(e*n+b*c*ln(F))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {5602, 2291} \[ \int F^{c (a+b x)} \text {sech}^n(d+e x) \, dx=\frac {\left (e^{2 (d+e x)}+1\right )^n F^{a c+b c x} \text {sech}^n(d+e x) \operatorname {Hypergeometric2F1}\left (n,\frac {e n+b c \log (F)}{2 e},\frac {e n+b c \log (F)}{2 e}+1,-e^{2 (d+e x)}\right )}{b c \log (F)+e n} \]

[In]

Int[F^(c*(a + b*x))*Sech[d + e*x]^n,x]

[Out]

((1 + E^(2*(d + e*x)))^n*F^(a*c + b*c*x)*Hypergeometric2F1[n, (e*n + b*c*Log[F])/(2*e), 1 + (e*n + b*c*Log[F])
/(2*e), -E^(2*(d + e*x))]*Sech[d + e*x]^n)/(e*n + b*c*Log[F])

Rule 2291

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_)))*(H_)^((t_.)*((r_.)
+ (s_.)*(x_))), x_Symbol] :> Simp[G^(h*(f + g*x))*H^(t*(r + s*x))*((a + b*F^(e*(c + d*x)))^p/((g*h*Log[G] + s
*t*Log[H])*((a + b*F^(e*(c + d*x)))/a)^p))*Hypergeometric2F1[-p, (g*h*Log[G] + s*t*Log[H])/(d*e*Log[F]), (g*h*
Log[G] + s*t*Log[H])/(d*e*Log[F]) + 1, Simplify[(-b/a)*F^(e*(c + d*x))]], x] /; FreeQ[{F, G, H, a, b, c, d, e,
f, g, h, r, s, t, p}, x] &&  !IntegerQ[p]

Rule 5602

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sech[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Dist[(1 + E^(2*(d + e*x)))^
n*(Sech[d + e*x]^n/E^(n*(d + e*x))), Int[SimplifyIntegrand[F^(c*(a + b*x))*(E^(n*(d + e*x))/(1 + E^(2*(d + e*x
)))^n), x], x], x] /; FreeQ[{F, a, b, c, d, e}, x] &&  !IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \left (e^{-n (d+e x)} \left (1+e^{2 (d+e x)}\right )^n \text {sech}^n(d+e x)\right ) \int e^{d n+e n x} \left (1+e^{2 (d+e x)}\right )^{-n} F^{a c+b c x} \, dx \\ & = \frac {\left (1+e^{2 (d+e x)}\right )^n F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (n,\frac {e n+b c \log (F)}{2 e},1+\frac {e n+b c \log (F)}{2 e},-e^{2 (d+e x)}\right ) \text {sech}^n(d+e x)}{e n+b c \log (F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.99 \[ \int F^{c (a+b x)} \text {sech}^n(d+e x) \, dx=\frac {\left (1+e^{2 (d+e x)}\right )^n F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (n,\frac {e n+b c \log (F)}{2 e},1+\frac {e n+b c \log (F)}{2 e},-e^{2 (d+e x)}\right ) \text {sech}^n(d+e x)}{e n+b c \log (F)} \]

[In]

Integrate[F^(c*(a + b*x))*Sech[d + e*x]^n,x]

[Out]

((1 + E^(2*(d + e*x)))^n*F^(c*(a + b*x))*Hypergeometric2F1[n, (e*n + b*c*Log[F])/(2*e), 1 + (e*n + b*c*Log[F])
/(2*e), -E^(2*(d + e*x))]*Sech[d + e*x]^n)/(e*n + b*c*Log[F])

Maple [F]

\[\int F^{c \left (b x +a \right )} \operatorname {sech}\left (e x +d \right )^{n}d x\]

[In]

int(F^(c*(b*x+a))*sech(e*x+d)^n,x)

[Out]

int(F^(c*(b*x+a))*sech(e*x+d)^n,x)

Fricas [F]

\[ \int F^{c (a+b x)} \text {sech}^n(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {sech}\left (e x + d\right )^{n} \,d x } \]

[In]

integrate(F^(c*(b*x+a))*sech(e*x+d)^n,x, algorithm="fricas")

[Out]

integral(F^(b*c*x + a*c)*sech(e*x + d)^n, x)

Sympy [F]

\[ \int F^{c (a+b x)} \text {sech}^n(d+e x) \, dx=\int F^{c \left (a + b x\right )} \operatorname {sech}^{n}{\left (d + e x \right )}\, dx \]

[In]

integrate(F**(c*(b*x+a))*sech(e*x+d)**n,x)

[Out]

Integral(F**(c*(a + b*x))*sech(d + e*x)**n, x)

Maxima [F]

\[ \int F^{c (a+b x)} \text {sech}^n(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {sech}\left (e x + d\right )^{n} \,d x } \]

[In]

integrate(F^(c*(b*x+a))*sech(e*x+d)^n,x, algorithm="maxima")

[Out]

integrate(F^((b*x + a)*c)*sech(e*x + d)^n, x)

Giac [F]

\[ \int F^{c (a+b x)} \text {sech}^n(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {sech}\left (e x + d\right )^{n} \,d x } \]

[In]

integrate(F^(c*(b*x+a))*sech(e*x+d)^n,x, algorithm="giac")

[Out]

integrate(F^((b*x + a)*c)*sech(e*x + d)^n, x)

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \text {sech}^n(d+e x) \, dx=\int F^{c\,\left (a+b\,x\right )}\,{\left (\frac {1}{\mathrm {cosh}\left (d+e\,x\right )}\right )}^n \,d x \]

[In]

int(F^(c*(a + b*x))*(1/cosh(d + e*x))^n,x)

[Out]

int(F^(c*(a + b*x))*(1/cosh(d + e*x))^n, x)