\(\int F^{c (a+b x)} \text {sech}^3(d+e x) \, dx\) [291]

Optimal result

Integrand size = 18, antiderivative size = 124 \[ \int F^{c (a+b x)} \text {sech}^3(d+e x) \, dx=\frac {e^{d+e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),-e^{2 (d+e x)}\right ) (e-b c \log (F))}{e^2}+\frac {b c F^{c (a+b x)} \log (F) \text {sech}(d+e x)}{2 e^2}+\frac {F^{c (a+b x)} \text {sech}(d+e x) \tanh (d+e x)}{2 e} \] Output:

exp(e*x+d)*F^(c*(b*x+a))*hypergeom([1, 1/2*(e+b*c*ln(F))/e],[3/2+1/2*b*c*l 
n(F)/e],-exp(2*e*x+2*d))*(e-b*c*ln(F))/e^2+1/2*b*c*F^(c*(b*x+a))*ln(F)*sec 
h(e*x+d)/e^2+1/2*F^(c*(b*x+a))*sech(e*x+d)*tanh(e*x+d)/e
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.77 \[ \int F^{c (a+b x)} \text {sech}^3(d+e x) \, dx=\frac {F^{c (a+b x)} \left (2 e^{d+e x} \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),-e^{2 (d+e x)}\right ) (e-b c \log (F))+\text {sech}(d+e x) (b c \log (F)+e \tanh (d+e x))\right )}{2 e^2} \] Input:

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

Output:

(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)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.12, 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.

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

\(\Big \downarrow \) 6013

\(\displaystyle \frac {1}{2} \left (1-\frac {b^2 c^2 \log ^2(F)}{e^2}\right ) \int F^{c (a+b x)} \text {sech}(d+e x)dx+\frac {b c \log (F) \text {sech}(d+e x) F^{c (a+b x)}}{2 e^2}+\frac {\tanh (d+e x) \text {sech}(d+e x) F^{c (a+b x)}}{2 e}\)

\(\Big \downarrow \) 6015

\(\displaystyle \frac {e^{d+e x} F^{c (a+b x)} \left (1-\frac {b^2 c^2 \log ^2(F)}{e^2}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (\frac {b c \log (F)}{e}+3\right ),-e^{2 (d+e x)}\right )}{b c \log (F)+e}+\frac {b c \log (F) \text {sech}(d+e x) F^{c (a+b x)}}{2 e^2}+\frac {\tanh (d+e x) \text {sech}(d+e x) F^{c (a+b x)}}{2 e}\)

Input:

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

Output:

(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))]*(1 - (b^2*c^2*Log[F]^2)/e^2))/(e 
 + b*c*Log[F]) + (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)
 

Defintions of rubi rules used

rule 6013

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]
 

rule 6015

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]
 

Maple [F]

\[\int F^{c \left (b x +a \right )} \operatorname {sech}\left (e x +d \right )^{3}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int F^{c (a+b x)} \text {sech}^3(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {sech}\left (e x + d\right )^{3} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int F^{c (a+b x)} \text {sech}^3(d+e x) \, dx=\int F^{c \left (a + b x\right )} \operatorname {sech}^{3}{\left (d + e x \right )}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int F^{c (a+b x)} \text {sech}^3(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {sech}\left (e x + d\right )^{3} \,d x } \] Input:

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

Output:

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*log(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*l 
og(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))
 

Giac [F]

\[ \int F^{c (a+b x)} \text {sech}^3(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {sech}\left (e x + d\right )^{3} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \text {sech}^3(d+e x) \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\mathrm {cosh}\left (d+e\,x\right )}^3} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int F^{c (a+b x)} \text {sech}^3(d+e x) \, dx=f^{a c} \left (\int f^{b c x} \mathrm {sech}\left (e x +d \right )^{3}d x \right ) \] Input:

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

Output:

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