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

Optimal. Leaf size=90 \[ \frac{\left (e^{2 (d+e x)}+1\right )^n F^{a c+b c x} \text{sech}^n(d+e x) \, _2F_1\left (n,\frac{e n+b c \log (F)}{2 e};\frac{e n+b c \log (F)}{2 e}+1;-e^{2 (d+e x)}\right )}{b c \log (F)+e n} \]

[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])

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Rubi [A]  time = 0.137385, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {5494, 2259} \[ \frac{\left (e^{2 (d+e x)}+1\right )^n F^{a c+b c x} \text{sech}^n(d+e x) \, _2F_1\left (n,\frac{e n+b c \log (F)}{2 e};\frac{e n+b c \log (F)}{2 e}+1;-e^{2 (d+e x)}\right )}{b c \log (F)+e n} \]

Antiderivative was successfully verified.

[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 5494

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]

Rule 2259

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*Hypergeometric2F
1[-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*F^(e*(
c + d*x)))/a)]])/((g*h*Log[G] + s*t*Log[H])*((a + b*F^(e*(c + d*x)))/a)^p), x] /; FreeQ[{F, G, H, a, b, c, d,
e, f, g, h, r, s, t, p}, x] &&  !IntegerQ[p]

Rubi steps

\begin{align*} \int F^{c (a+b x)} \text{sech}^n(d+e x) \, dx &=\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} \, _2F_1\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]  time = 0.0737386, size = 89, normalized size = 0.99 \[ \frac{\left (e^{2 (d+e x)}+1\right )^n F^{c (a+b x)} \text{sech}^n(d+e x) \, _2F_1\left (n,\frac{e n+b c \log (F)}{2 e};\frac{e n+b c \log (F)}{2 e}+1;-e^{2 (d+e x)}\right )}{b c \log (F)+e n} \]

Antiderivative was successfully verified.

[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])

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Maple [F]  time = 0.053, size = 0, normalized size = 0. \begin{align*} \int{F}^{c \left ( bx+a \right ) } \left ({\rm sech} \left (ex+d\right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int F^{{\left (b x + a\right )} c} \operatorname{sech}\left (e x + d\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (F^{b c x + a c} \operatorname{sech}\left (e x + d\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int F^{c \left (a + b x\right )} \operatorname{sech}^{n}{\left (d + e x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int F^{{\left (b x + a\right )} c} \operatorname{sech}\left (e x + d\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)