\(\int F^{c (a+b x)} (f \sinh (d+e x))^n \, dx\) [19]

Optimal result

Integrand size = 20, antiderivative size = 98 \[ \int F^{c (a+b x)} (f \sinh (d+e x))^n \, dx=-\frac {\left (1-e^{2 d+2 e x}\right )^{-n} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (-n,\frac {1}{2} \left (-n+\frac {b c \log (F)}{e}\right ),\frac {1}{2} \left (2-n+\frac {b c \log (F)}{e}\right ),e^{2 d+2 e x}\right ) (f \sinh (d+e x))^n}{e n-b c \log (F)} \] Output:

-F^(c*(b*x+a))*hypergeom([-n, -1/2*n+1/2*b*c*ln(F)/e],[1-1/2*n+1/2*b*c*ln( 
F)/e],exp(2*e*x+2*d))*(f*sinh(e*x+d))^n/((1-exp(2*e*x+2*d))^n)/(e*n-b*c*ln 
(F))
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00 \[ \int F^{c (a+b x)} (f \sinh (d+e x))^n \, dx=\frac {\left (1-e^{2 (d+e x)}\right )^{-n} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (-n,\frac {-e n+b c \log (F)}{2 e},1+\frac {-e n+b c \log (F)}{2 e},e^{2 (d+e x)}\right ) (f \sinh (d+e x))^n}{-e n+b c \log (F)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {7271, 6005, 2689}

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.

\(\displaystyle \int F^{c (a+b x)} (f \sinh (d+e x))^n \, dx\)

\(\Big \downarrow \) 7271

\(\displaystyle \sinh ^{-n}(d+e x) (f \sinh (d+e x))^n \int F^{c (a+b x)} \sinh ^n(d+e x)dx\)

\(\Big \downarrow \) 6005

\(\displaystyle e^{n (d+e x)} \left (e^{2 (d+e x)}-1\right )^{-n} (f \sinh (d+e x))^n \int e^{-n (d+e x)} \left (-1+e^{2 (d+e x)}\right )^n F^{c (a+b x)}dx\)

\(\Big \downarrow \) 2689

\(\displaystyle -\frac {\left (1-e^{2 (d+e x)}\right )^{-n} F^{c (a+b x)} (f \sinh (d+e x))^n \operatorname {Hypergeometric2F1}\left (-n,-\frac {e n-b c \log (F)}{2 e},\frac {1}{2} \left (-n+\frac {b c \log (F)}{e}+2\right ),e^{2 (d+e x)}\right )}{e n-b c \log (F)}\)

Input:

Int[F^(c*(a + b*x))*(f*Sinh[d + e*x])^n,x]
 

Output:

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

Defintions of rubi rules used

rule 2689

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/((g*h*Log[G] + s*t*Lo 
g[H])*((a + b*F^(e*(c + d*x)))/a)^p))*Hypergeometric2F1[-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, Simpli 
fy[(-b/a)*F^(e*(c + d*x))]], x] /; FreeQ[{F, G, H, a, b, c, d, e, f, g, h, 
r, s, t, p}, x] &&  !IntegerQ[p]
 

rule 6005

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sinh[(d_.) + (e_.)*(x_)]^(n_), x_Symb 
ol] :> Simp[E^(n*(d + e*x))*(Sinh[d + e*x]^n/(-1 + E^(2*(d + e*x)))^n)   In 
t[F^(c*(a + b*x))*((-1 + E^(2*(d + e*x)))^n/E^(n*(d + e*x))), x], x] /; Fre 
eQ[{F, a, b, c, d, e, n}, x] &&  !IntegerQ[n]
 

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]

\[\int F^{c \left (b x +a \right )} \left (f \sinh \left (e x +d \right )\right )^{n}d x\]

Input:

int(F^(c*(b*x+a))*(f*sinh(e*x+d))^n,x)
 

Output:

int(F^(c*(b*x+a))*(f*sinh(e*x+d))^n,x)
 

Fricas [F]

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

integrate(F^(c*(b*x+a))*(f*sinh(e*x+d))^n,x, algorithm="fricas")
 

Output:

integral((f*sinh(e*x + d))^n*F^(b*c*x + a*c), x)
 

Sympy [F]

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

integrate(F**(c*(b*x+a))*(f*sinh(e*x+d))**n,x)
 

Output:

Integral(F**(c*(a + b*x))*(f*sinh(d + e*x))**n, x)
 

Maxima [F]

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

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

Output:

integrate((f*sinh(e*x + d))^n*F^((b*x + a)*c), x)
 

Giac [F]

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

integrate(F^(c*(b*x+a))*(f*sinh(e*x+d))^n,x, algorithm="giac")
 

Output:

integrate((f*sinh(e*x + d))^n*F^((b*x + a)*c), x)
 

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} (f \sinh (d+e x))^n \, dx=\int F^{c\,\left (a+b\,x\right )}\,{\left (f\,\mathrm {sinh}\left (d+e\,x\right )\right )}^n \,d x \] Input:

int(F^(c*(a + b*x))*(f*sinh(d + e*x))^n,x)
 

Output:

int(F^(c*(a + b*x))*(f*sinh(d + e*x))^n, x)
 

Reduce [F]

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

int(F^(c*(b*x+a))*(f*sinh(e*x+d))^n,x)
 

Output:

(f**(a*c + n)*(f**(b*c*x)*sinh(d + e*x)**n - int((f**(b*c*x)*sinh(d + e*x) 
**n*cosh(d + e*x))/sinh(d + e*x),x)*e*n))/(log(f)*b*c)