3.290 \(\int F^{c (a+b x)} \text{sech}^3(d+e x) \, dx\)

Optimal. Leaf size=124 \[ \frac{e^{d+e x} F^{c (a+b x)} (e-b c \log (F)) \, _2F_1\left (1,\frac{e+b c \log (F)}{2 e};\frac{1}{2} \left (\frac{b c \log (F)}{e}+3\right );-e^{2 (d+e x)}\right )}{e^2}+\frac{b c \log (F) \text{sech}(d+e x) F^{c (a+b x)}}{2 e^2}+\frac{\tanh (d+e x) \text{sech}(d+e x) F^{c (a+b x)}}{2 e} \]

[Out]

(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]))/e^2 + (b*c*F^(c*(a + b*x))*Log[F]*Sech[d + e*x])/(2*e^2) + (F^(c*(a + b*x))*Sech[d + e
*x]*Tanh[d + e*x])/(2*e)

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Rubi [A]  time = 0.0524655, antiderivative size = 124, 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 = {5490, 5492} \[ \frac{e^{d+e x} F^{c (a+b x)} (e-b c \log (F)) \, _2F_1\left (1,\frac{e+b c \log (F)}{2 e};\frac{1}{2} \left (\frac{b c \log (F)}{e}+3\right );-e^{2 (d+e x)}\right )}{e^2}+\frac{b c \log (F) \text{sech}(d+e x) F^{c (a+b x)}}{2 e^2}+\frac{\tanh (d+e x) \text{sech}(d+e x) F^{c (a+b x)}}{2 e} \]

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))*Sech[d + e*x]^3,x]

[Out]

(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]))/e^2 + (b*c*F^(c*(a + b*x))*Log[F]*Sech[d + e*x])/(2*e^2) + (F^(c*(a + b*x))*Sech[d + e
*x]*Tanh[d + e*x])/(2*e)

Rule 5490

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sech[(d_.) + (e_.)*(x_)]^(n_), x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b
*x))*Sech[d + e*x]^(n - 2))/(e^2*(n - 1)*(n - 2)), x] + (Dist[(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] + Simp[(F^(c*(a + b*x))*Sech[d + e*x]^(n - 1)*Sinh
[d + e*x])/(e*(n - 1)), 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]

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

Mathematica [A]  time = 0.216721, size = 96, normalized size = 0.77 \[ \frac{F^{c (a+b x)} \left (2 e^{d+e x} (e-b c \log (F)) \, _2F_1\left (1,\frac{e+b c \log (F)}{2 e};\frac{1}{2} \left (\frac{b c \log (F)}{e}+3\right );-e^{2 (d+e x)}\right )+\text{sech}(d+e x) (b c \log (F)+e \tanh (d+e x))\right )}{2 e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(c*(a + b*x))*Sech[d + e*x]^3,x]

[Out]

(F^(c*(a + b*x))*(2*E^(d + e*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]) + Sech[d + e*x]*(b*c*Log[F] + e*Tanh[d + e*x])))/(2*e^2)

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Maple [F]  time = 0.029, size = 0, normalized size = 0. \begin{align*} \int{F}^{c \left ( bx+a \right ) } \left ({\rm sech} \left (ex+d\right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(b*x+a))*sech(e*x+d)^3,x)

[Out]

int(F^(c*(b*x+a))*sech(e*x+d)^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 48 \,{\left (F^{a c} b c e e^{d} \log \left (F\right ) + F^{a c} e^{2} e^{d}\right )} \int \frac{e^{\left (b c x \log \left (F\right ) + e x\right )}}{b^{2} c^{2} \log \left (F\right )^{2} - 8 \, b c e \log \left (F\right ) + 15 \, e^{2} +{\left (b^{2} c^{2} e^{\left (8 \, d\right )} \log \left (F\right )^{2} - 8 \, b c e e^{\left (8 \, d\right )} \log \left (F\right ) + 15 \, e^{2} e^{\left (8 \, d\right )}\right )} e^{\left (8 \, e x\right )} + 4 \,{\left (b^{2} c^{2} e^{\left (6 \, d\right )} \log \left (F\right )^{2} - 8 \, b c e e^{\left (6 \, d\right )} \log \left (F\right ) + 15 \, e^{2} e^{\left (6 \, d\right )}\right )} e^{\left (6 \, e x\right )} + 6 \,{\left (b^{2} c^{2} e^{\left (4 \, d\right )} \log \left (F\right )^{2} - 8 \, b c e e^{\left (4 \, d\right )} \log \left (F\right ) + 15 \, e^{2} e^{\left (4 \, d\right )}\right )} e^{\left (4 \, e x\right )} + 4 \,{\left (b^{2} c^{2} e^{\left (2 \, d\right )} \log \left (F\right )^{2} - 8 \, b c e e^{\left (2 \, d\right )} \log \left (F\right ) + 15 \, e^{2} e^{\left (2 \, d\right )}\right )} e^{\left (2 \, e x\right )}}\,{d x} - \frac{8 \,{\left (6 \, F^{a c} e e^{\left (e x + d\right )} -{\left (F^{a c} b c e^{\left (3 \, d\right )} \log \left (F\right ) - 5 \, F^{a c} e e^{\left (3 \, d\right )}\right )} e^{\left (3 \, e x\right )}\right )} F^{b c x}}{b^{2} c^{2} \log \left (F\right )^{2} - 8 \, b c e \log \left (F\right ) + 15 \, e^{2} +{\left (b^{2} c^{2} e^{\left (6 \, d\right )} \log \left (F\right )^{2} - 8 \, b c e e^{\left (6 \, d\right )} \log \left (F\right ) + 15 \, e^{2} e^{\left (6 \, d\right )}\right )} e^{\left (6 \, e x\right )} + 3 \,{\left (b^{2} c^{2} e^{\left (4 \, d\right )} \log \left (F\right )^{2} - 8 \, b c e e^{\left (4 \, d\right )} \log \left (F\right ) + 15 \, e^{2} e^{\left (4 \, d\right )}\right )} e^{\left (4 \, e x\right )} + 3 \,{\left (b^{2} c^{2} e^{\left (2 \, d\right )} \log \left (F\right )^{2} - 8 \, b c e e^{\left (2 \, d\right )} \log \left (F\right ) + 15 \, e^{2} e^{\left (2 \, d\right )}\right )} e^{\left (2 \, e 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)^3,x, algorithm="maxima")

[Out]

48*(F^(a*c)*b*c*e*e^d*log(F) + F^(a*c)*e^2*e^d)*integrate(e^(b*c*x*log(F) + e*x)/(b^2*c^2*log(F)^2 - 8*b*c*e*l
og(F) + 15*e^2 + (b^2*c^2*e^(8*d)*log(F)^2 - 8*b*c*e*e^(8*d)*log(F) + 15*e^2*e^(8*d))*e^(8*e*x) + 4*(b^2*c^2*e
^(6*d)*log(F)^2 - 8*b*c*e*e^(6*d)*log(F) + 15*e^2*e^(6*d))*e^(6*e*x) + 6*(b^2*c^2*e^(4*d)*log(F)^2 - 8*b*c*e*e
^(4*d)*log(F) + 15*e^2*e^(4*d))*e^(4*e*x) + 4*(b^2*c^2*e^(2*d)*log(F)^2 - 8*b*c*e*e^(2*d)*log(F) + 15*e^2*e^(2
*d))*e^(2*e*x)), x) - 8*(6*F^(a*c)*e*e^(e*x + d) - (F^(a*c)*b*c*e^(3*d)*log(F) - 5*F^(a*c)*e*e^(3*d))*e^(3*e*x
))*F^(b*c*x)/(b^2*c^2*log(F)^2 - 8*b*c*e*log(F) + 15*e^2 + (b^2*c^2*e^(6*d)*log(F)^2 - 8*b*c*e*e^(6*d)*log(F)
+ 15*e^2*e^(6*d))*e^(6*e*x) + 3*(b^2*c^2*e^(4*d)*log(F)^2 - 8*b*c*e*e^(4*d)*log(F) + 15*e^2*e^(4*d))*e^(4*e*x)
+ 3*(b^2*c^2*e^(2*d)*log(F)^2 - 8*b*c*e*e^(2*d)*log(F) + 15*e^2*e^(2*d))*e^(2*e*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 )^{3}, 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)^3,x, algorithm="fricas")

[Out]

integral(F^(b*c*x + a*c)*sech(e*x + d)^3, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(c*(b*x+a))*sech(e*x+d)**3,x)

[Out]

Timed out

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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 )^{3}\,{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)^3,x, algorithm="giac")

[Out]

integrate(F^((b*x + a)*c)*sech(e*x + d)^3, x)