\(\int F^{c (a+b x)} (f \text {sech}(d+\frac {b c x \log (F)}{-2+n}))^n \, dx\) [96]

Optimal result

Integrand size = 28, antiderivative size = 132 \[ \int F^{c (a+b x)} \left (f \text {sech}\left (d+\frac {b c x \log (F)}{-2+n}\right )\right )^n \, dx=\frac {f^2 F^{c (a+b x)} (2-n) \left (f \text {sech}\left (d-\frac {b c x \log (F)}{2-n}\right )\right )^{-2+n}}{b c (1-n) \log (F)}+\frac {f F^{c (a+b x)} (2-n) \left (f \text {sech}\left (d-\frac {b c x \log (F)}{2-n}\right )\right )^{-1+n} \sinh \left (d-\frac {b c x \log (F)}{2-n}\right )}{b c (1-n) \log (F)} \] Output:

f^2*F^(c*(b*x+a))*(2-n)*(f*sech(-d+b*c*x*ln(F)/(2-n)))^(-2+n)/b/c/(1-n)/ln 
(F)-f*F^(c*(b*x+a))*(2-n)*(f*sech(-d+b*c*x*ln(F)/(2-n)))^(-1+n)*sinh(-d+b* 
c*x*ln(F)/(2-n))/b/c/(1-n)/ln(F)
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.52 \[ \int F^{c (a+b x)} \left (f \text {sech}\left (d+\frac {b c x \log (F)}{-2+n}\right )\right )^n \, dx=\frac {F^{c (a+b x)} \left (1+e^{2 d} F^{\frac {2 b c x}{-2+n}}\right ) (-2+n) \left (f \text {sech}\left (d+\frac {b c x \log (F)}{-2+n}\right )\right )^n}{2 b c (-1+n) \log (F)} \] Input:

Integrate[F^(c*(a + b*x))*(f*Sech[d + (b*c*x*Log[F])/(-2 + n)])^n,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.27, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {7271, 6011}

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.

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

\(\Big \downarrow \) 7271

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

\(\Big \downarrow \) 6011

\(\displaystyle \text {sech}^{-n}\left (d-\frac {b c x \log (F)}{2-n}\right ) \left (f \text {sech}\left (d-\frac {b c x \log (F)}{2-n}\right )\right )^n \left (\frac {(2-n) F^{c (a+b x)} \text {sech}^{n-2}\left (d-\frac {b c x \log (F)}{2-n}\right )}{b c (1-n) \log (F)}+\frac {(2-n) F^{c (a+b x)} \sinh \left (d-\frac {b c x \log (F)}{2-n}\right ) \text {sech}^{n-1}\left (d-\frac {b c x \log (F)}{2-n}\right )}{b c (1-n) \log (F)}\right )\)

Input:

Int[F^(c*(a + b*x))*(f*Sech[d + (b*c*x*Log[F])/(-2 + n)])^n,x]
 

Output:

((f*Sech[d - (b*c*x*Log[F])/(2 - n)])^n*((F^(c*(a + b*x))*(2 - n)*Sech[d - 
 (b*c*x*Log[F])/(2 - n)]^(-2 + n))/(b*c*(1 - n)*Log[F]) + (F^(c*(a + b*x)) 
*(2 - n)*Sech[d - (b*c*x*Log[F])/(2 - n)]^(-1 + n)*Sinh[d - (b*c*x*Log[F]) 
/(2 - n)])/(b*c*(1 - n)*Log[F])))/Sech[d - (b*c*x*Log[F])/(2 - n)]^n
 

Defintions of rubi rules used

rule 6011

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] /; FreeQ[{F, a, b, c, d, e, n}, x] && EqQ[e^2*(n - 2)^2 - b 
^2*c^2*Log[F]^2, 0] && NeQ[n, 1] && NeQ[n, 2]
 

rule 7271

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*v^m)^ 
FracPart[p]/v^(m*FracPart[p]))   Int[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, 
x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&  !(Eq 
Q[v, x] && EqQ[m, 1])
 

Maple [F]

\[\int F^{c \left (b x +a \right )} \left (f \,\operatorname {sech}\left (d +\frac {b c x \ln \left (F \right )}{n -2}\right )\right )^{n}d x\]

Input:

int(F^(c*(b*x+a))*(f*sech(d+b*c*x*ln(F)/(n-2)))^n,x)
 

Output:

int(F^(c*(b*x+a))*(f*sech(d+b*c*x*ln(F)/(n-2)))^n,x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 500 vs. \(2 (125) = 250\).

Time = 0.10 (sec) , antiderivative size = 500, normalized size of antiderivative = 3.79 \[ \int F^{c (a+b x)} \left (f \text {sech}\left (d+\frac {b c x \log (F)}{-2+n}\right )\right )^n \, dx=\frac {{\left ({\left (n - 2\right )} \cosh \left ({\left (b c x + a c\right )} \log \left (F\right )\right ) \cosh \left (\frac {b c x \log \left (F\right ) + d n - 2 \, d}{n - 2}\right ) + {\left (n - 2\right )} \cosh \left (\frac {b c x \log \left (F\right ) + d n - 2 \, d}{n - 2}\right ) \sinh \left ({\left (b c x + a c\right )} \log \left (F\right )\right )\right )} \cosh \left (n \log \left (\frac {2 \, {\left (f \cosh \left (\frac {b c x \log \left (F\right ) + d n - 2 \, d}{n - 2}\right ) + f \sinh \left (\frac {b c x \log \left (F\right ) + d n - 2 \, d}{n - 2}\right )\right )}}{\cosh \left (\frac {b c x \log \left (F\right ) + d n - 2 \, d}{n - 2}\right )^{2} + 2 \, \cosh \left (\frac {b c x \log \left (F\right ) + d n - 2 \, d}{n - 2}\right ) \sinh \left (\frac {b c x \log \left (F\right ) + d n - 2 \, d}{n - 2}\right ) + \sinh \left (\frac {b c x \log \left (F\right ) + d n - 2 \, d}{n - 2}\right )^{2} + 1}\right )\right ) + {\left ({\left (n - 2\right )} \cosh \left ({\left (b c x + a c\right )} \log \left (F\right )\right ) \cosh \left (\frac {b c x \log \left (F\right ) + d n - 2 \, d}{n - 2}\right ) + {\left (n - 2\right )} \cosh \left (\frac {b c x \log \left (F\right ) + d n - 2 \, d}{n - 2}\right ) \sinh \left ({\left (b c x + a c\right )} \log \left (F\right )\right )\right )} \sinh \left (n \log \left (\frac {2 \, {\left (f \cosh \left (\frac {b c x \log \left (F\right ) + d n - 2 \, d}{n - 2}\right ) + f \sinh \left (\frac {b c x \log \left (F\right ) + d n - 2 \, d}{n - 2}\right )\right )}}{\cosh \left (\frac {b c x \log \left (F\right ) + d n - 2 \, d}{n - 2}\right )^{2} + 2 \, \cosh \left (\frac {b c x \log \left (F\right ) + d n - 2 \, d}{n - 2}\right ) \sinh \left (\frac {b c x \log \left (F\right ) + d n - 2 \, d}{n - 2}\right ) + \sinh \left (\frac {b c x \log \left (F\right ) + d n - 2 \, d}{n - 2}\right )^{2} + 1}\right )\right )}{{\left (b c n - b c\right )} \cosh \left (\frac {b c x \log \left (F\right ) + d n - 2 \, d}{n - 2}\right ) \log \left (F\right ) - {\left (b c n - b c\right )} \log \left (F\right ) \sinh \left (\frac {b c x \log \left (F\right ) + d n - 2 \, d}{n - 2}\right )} \] Input:

integrate(F^(c*(b*x+a))*(f*sech(d+b*c*x*log(F)/(-2+n)))^n,x, algorithm="fr 
icas")
 

Output:

(((n - 2)*cosh((b*c*x + a*c)*log(F))*cosh((b*c*x*log(F) + d*n - 2*d)/(n - 
2)) + (n - 2)*cosh((b*c*x*log(F) + d*n - 2*d)/(n - 2))*sinh((b*c*x + a*c)* 
log(F)))*cosh(n*log(2*(f*cosh((b*c*x*log(F) + d*n - 2*d)/(n - 2)) + f*sinh 
((b*c*x*log(F) + d*n - 2*d)/(n - 2)))/(cosh((b*c*x*log(F) + d*n - 2*d)/(n 
- 2))^2 + 2*cosh((b*c*x*log(F) + d*n - 2*d)/(n - 2))*sinh((b*c*x*log(F) + 
d*n - 2*d)/(n - 2)) + sinh((b*c*x*log(F) + d*n - 2*d)/(n - 2))^2 + 1))) + 
((n - 2)*cosh((b*c*x + a*c)*log(F))*cosh((b*c*x*log(F) + d*n - 2*d)/(n - 2 
)) + (n - 2)*cosh((b*c*x*log(F) + d*n - 2*d)/(n - 2))*sinh((b*c*x + a*c)*l 
og(F)))*sinh(n*log(2*(f*cosh((b*c*x*log(F) + d*n - 2*d)/(n - 2)) + f*sinh( 
(b*c*x*log(F) + d*n - 2*d)/(n - 2)))/(cosh((b*c*x*log(F) + d*n - 2*d)/(n - 
 2))^2 + 2*cosh((b*c*x*log(F) + d*n - 2*d)/(n - 2))*sinh((b*c*x*log(F) + d 
*n - 2*d)/(n - 2)) + sinh((b*c*x*log(F) + d*n - 2*d)/(n - 2))^2 + 1))))/(( 
b*c*n - b*c)*cosh((b*c*x*log(F) + d*n - 2*d)/(n - 2))*log(F) - (b*c*n - b* 
c)*log(F)*sinh((b*c*x*log(F) + d*n - 2*d)/(n - 2)))
 

Sympy [F]

\[ \int F^{c (a+b x)} \left (f \text {sech}\left (d+\frac {b c x \log (F)}{-2+n}\right )\right )^n \, dx=\int F^{c \left (a + b x\right )} \left (f \operatorname {sech}{\left (\frac {b c x \log {\left (F \right )}}{n - 2} + d \right )}\right )^{n}\, dx \] Input:

integrate(F**(c*(b*x+a))*(f*sech(d+b*c*x*ln(F)/(-2+n)))**n,x)
 

Output:

Integral(F**(c*(a + b*x))*(f*sech(b*c*x*log(F)/(n - 2) + d))**n, x)
 

Maxima [F]

\[ \int F^{c (a+b x)} \left (f \text {sech}\left (d+\frac {b c x \log (F)}{-2+n}\right )\right )^n \, dx=\int { \left (f \operatorname {sech}\left (\frac {b c x \log \left (F\right )}{n - 2} + d\right )\right )^{n} F^{{\left (b x + a\right )} c} \,d x } \] Input:

integrate(F^(c*(b*x+a))*(f*sech(d+b*c*x*log(F)/(-2+n)))^n,x, algorithm="ma 
xima")
 

Output:

integrate((f*sech(b*c*x*log(F)/(n - 2) + d))^n*F^((b*x + a)*c), x)
 

Giac [F]

\[ \int F^{c (a+b x)} \left (f \text {sech}\left (d+\frac {b c x \log (F)}{-2+n}\right )\right )^n \, dx=\int { \left (f \operatorname {sech}\left (\frac {b c x \log \left (F\right )}{n - 2} + d\right )\right )^{n} F^{{\left (b x + a\right )} c} \,d x } \] Input:

integrate(F^(c*(b*x+a))*(f*sech(d+b*c*x*log(F)/(-2+n)))^n,x, algorithm="gi 
ac")
 

Output:

integrate((f*sech(b*c*x*log(F)/(n - 2) + d))^n*F^((b*x + a)*c), x)
 

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \left (f \text {sech}\left (d+\frac {b c x \log (F)}{-2+n}\right )\right )^n \, dx=\int F^{c\,\left (a+b\,x\right )}\,{\left (\frac {f}{\mathrm {cosh}\left (d+\frac {b\,c\,x\,\ln \left (F\right )}{n-2}\right )}\right )}^n \,d x \] Input:

int(F^(c*(a + b*x))*(f/cosh(d + (b*c*x*log(F))/(n - 2)))^n,x)
 

Output:

int(F^(c*(a + b*x))*(f/cosh(d + (b*c*x*log(F))/(n - 2)))^n, x)
 

Reduce [F]

\[ \int F^{c (a+b x)} \left (f \text {sech}\left (d+\frac {b c x \log (F)}{-2+n}\right )\right )^n \, dx=f^{a c +n} \left (\int f^{b c x} \mathrm {sech}\left (\frac {\mathrm {log}\left (f \right ) b c x +d n -2 d}{n -2}\right )^{n}d x \right ) \] Input:

int(F^(c*(b*x+a))*(f*sech(d+b*c*x*log(F)/(-2+n)))^n,x)
 

Output:

f**(a*c + n)*int(f**(b*c*x)*sech((log(f)*b*c*x + d*n - 2*d)/(n - 2))**n,x)