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

Optimal. Leaf size=68 \[ \frac{2 e^{d+e x} F^{c (a+b x)} \, _2F_1\left (1,\frac{e+b c \log (F)}{2 e};\frac{1}{2} \left (\frac{b c \log (F)}{e}+3\right );-e^{2 (d+e x)}\right )}{b c \log (F)+e} \]

[Out]

(2*E^(d + e*x)*F^(c*(a + b*x))*Hypergeometric2F1[1, (e + b*c*Log[F])/(2*e), (3 + (b*c*Log[F])/e)/2, -E^(2*(d +
e*x))])/(e + b*c*Log[F])

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Rubi [A]  time = 0.0217432, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {5492} \[ \frac{2 e^{d+e x} F^{c (a+b x)} \, _2F_1\left (1,\frac{e+b c \log (F)}{2 e};\frac{1}{2} \left (\frac{b c \log (F)}{e}+3\right );-e^{2 (d+e x)}\right )}{b c \log (F)+e} \]

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))*Sech[d + e*x],x]

[Out]

(2*E^(d + e*x)*F^(c*(a + b*x))*Hypergeometric2F1[1, (e + b*c*Log[F])/(2*e), (3 + (b*c*Log[F])/e)/2, -E^(2*(d +
e*x))])/(e + b*c*Log[F])

Rule 5492

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sech[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Simp[(2^n*E^(n*(d + e*x))*F
^(c*(a + b*x))*Hypergeometric2F1[n, n/2 + (b*c*Log[F])/(2*e), 1 + n/2 + (b*c*Log[F])/(2*e), -E^(2*(d + e*x))])
/(e*n + b*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]

Rubi steps

\begin{align*} \int F^{c (a+b x)} \text{sech}(d+e x) \, dx &=\frac{2 e^{d+e x} F^{c (a+b x)} \, _2F_1\left (1,\frac{e+b c \log (F)}{2 e};\frac{1}{2} \left (3+\frac{b c \log (F)}{e}\right );-e^{2 (d+e x)}\right )}{e+b c \log (F)}\\ \end{align*}

Mathematica [A]  time = 0.0186107, size = 70, normalized size = 1.03 \[ \frac{2 e^{d+e x} F^{c (a+b x)} \, _2F_1\left (1,\frac{b c \log (F)}{2 e}+\frac{1}{2};\frac{b c \log (F)}{2 e}+\frac{3}{2};-e^{2 (d+e x)}\right )}{b c \log (F)+e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*E^(d + e*x)*F^(c*(a + b*x))*Hypergeometric2F1[1, 1/2 + (b*c*Log[F])/(2*e), 3/2 + (b*c*Log[F])/(2*e), -E^(2*
(d + e*x))])/(e + b*c*Log[F])

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Maple [F]  time = 0.016, size = 0, normalized size = 0. \begin{align*} \int{F}^{c \left ( bx+a \right ) }{\rm sech} \left (ex+d\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -4 \, F^{a c} e \int \frac{e^{\left (b c x \log \left (F\right ) + e x + d\right )}}{b c \log \left (F\right ) +{\left (b c e^{\left (4 \, d\right )} \log \left (F\right ) - e e^{\left (4 \, d\right )}\right )} e^{\left (4 \, e x\right )} + 2 \,{\left (b c e^{\left (2 \, d\right )} \log \left (F\right ) - e e^{\left (2 \, d\right )}\right )} e^{\left (2 \, e x\right )} - e}\,{d x} + \frac{2 \, F^{a c} e^{\left (b c x \log \left (F\right ) + e x + d\right )}}{b c \log \left (F\right ) +{\left (b c e^{\left (2 \, d\right )} \log \left (F\right ) - e e^{\left (2 \, d\right )}\right )} e^{\left (2 \, e x\right )} - e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*F^(a*c)*e*integrate(e^(b*c*x*log(F) + e*x + d)/(b*c*log(F) + (b*c*e^(4*d)*log(F) - e*e^(4*d))*e^(4*e*x) + 2
*(b*c*e^(2*d)*log(F) - e*e^(2*d))*e^(2*e*x) - e), x) + 2*F^(a*c)*e^(b*c*x*log(F) + e*x + d)/(b*c*log(F) + (b*c
*e^(2*d)*log(F) - e*e^(2*d))*e^(2*e*x) - e)

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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 ), 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),x, algorithm="fricas")

[Out]

integral(F^(b*c*x + a*c)*sech(e*x + d), 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}{\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),x)

[Out]

Integral(F**(c*(a + b*x))*sech(d + e*x), 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 )\,{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),x, algorithm="giac")

[Out]

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