\(\int F^{c (a+b x)} (f \sinh (d+\frac {b c x \log (F)}{2+n}))^n \, dx\) [20]

Optimal result

Integrand size = 28, antiderivative size = 82 \[ \int F^{c (a+b x)} \left (f \sinh \left (d+\frac {b c x \log (F)}{2+n}\right )\right )^n \, dx=\frac {F^{c (a+b x)} (2+n) \left (\cosh \left (d+\frac {b c x \log (F)}{2+n}\right )-\sinh \left (d+\frac {b c x \log (F)}{2+n}\right )\right ) \left (f \sinh \left (d+\frac {b c x \log (F)}{2+n}\right )\right )^{1+n}}{b c f (1+n) \log (F)} \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.98 \[ \int F^{c (a+b x)} \left (f \sinh \left (d+\frac {b c x \log (F)}{2+n}\right )\right )^n \, dx=\frac {e^{-2 d} F^{c \left (a+\frac {b n x}{2+n}\right )} \left (-1+e^{2 d} F^{\frac {2 b c x}{2+n}}\right ) (2+n) \left (f \sinh \left (d+\frac {b c x \log (F)}{2+n}\right )\right )^n}{2 b c (1+n) \log (F)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.77, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {7271, 6001}

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.

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

\(\Big \downarrow \) 7271

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

\(\Big \downarrow \) 6001

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 6001

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

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

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (82) = 164\).

Time = 0.12 (sec) , antiderivative size = 267, normalized size of antiderivative = 3.26 \[ \int F^{c (a+b x)} \left (f \sinh \left (d+\frac {b c x \log (F)}{2+n}\right )\right )^n \, dx=\frac {{\left ({\left (n + 2\right )} \cosh \left ({\left (b c x + a c\right )} \log \left (F\right )\right ) \sinh \left (\frac {b c x \log \left (F\right ) + d n + 2 \, d}{n + 2}\right ) + {\left (n + 2\right )} \sinh \left ({\left (b c x + a c\right )} \log \left (F\right )\right ) \sinh \left (\frac {b c x \log \left (F\right ) + d n + 2 \, d}{n + 2}\right )\right )} \cosh \left (n \log \left (f \sinh \left (\frac {b c x \log \left (F\right ) + d n + 2 \, d}{n + 2}\right )\right )\right ) + {\left ({\left (n + 2\right )} \cosh \left ({\left (b c x + a c\right )} \log \left (F\right )\right ) \sinh \left (\frac {b c x \log \left (F\right ) + d n + 2 \, d}{n + 2}\right ) + {\left (n + 2\right )} \sinh \left ({\left (b c x + a c\right )} \log \left (F\right )\right ) \sinh \left (\frac {b c x \log \left (F\right ) + d n + 2 \, d}{n + 2}\right )\right )} \sinh \left (n \log \left (f \sinh \left (\frac {b c x \log \left (F\right ) + d n + 2 \, d}{n + 2}\right )\right )\right )}{{\left (b c n + b c\right )} \cosh \left (\frac {b c x \log \left (F\right ) + d n + 2 \, d}{n + 2}\right ) \log \left (F\right ) + {\left (b c n + b c\right )} \log \left (F\right ) \sinh \left (\frac {b c x \log \left (F\right ) + d n + 2 \, d}{n + 2}\right )} \] Input:

integrate(F^(c*(b*x+a))*(f*sinh(d+b*c*x*log(F)/(2+n)))^n,x, algorithm="fri 
cas")
 

Output:

(((n + 2)*cosh((b*c*x + a*c)*log(F))*sinh((b*c*x*log(F) + d*n + 2*d)/(n + 
2)) + (n + 2)*sinh((b*c*x + a*c)*log(F))*sinh((b*c*x*log(F) + d*n + 2*d)/( 
n + 2)))*cosh(n*log(f*sinh((b*c*x*log(F) + d*n + 2*d)/(n + 2)))) + ((n + 2 
)*cosh((b*c*x + a*c)*log(F))*sinh((b*c*x*log(F) + d*n + 2*d)/(n + 2)) + (n 
 + 2)*sinh((b*c*x + a*c)*log(F))*sinh((b*c*x*log(F) + d*n + 2*d)/(n + 2))) 
*sinh(n*log(f*sinh((b*c*x*log(F) + d*n + 2*d)/(n + 2)))))/((b*c*n + b*c)*c 
osh((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]

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

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

Output:

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

Maxima [F]

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

integrate(F^(c*(b*x+a))*(f*sinh(d+b*c*x*log(F)/(2+n)))^n,x, algorithm="max 
ima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 530 vs. \(2 (82) = 164\).

Time = 0.45 (sec) , antiderivative size = 530, normalized size of antiderivative = 6.46 \[ \int F^{c (a+b x)} \left (f \sinh \left (d+\frac {b c x \log (F)}{2+n}\right )\right )^n \, dx=\frac {F^{a c} n e^{\left (\frac {2 \, b c x \log \left (F\right ) - d n^{2} - n^{2} \log \left (2\right ) + n^{2} \log \left (f e^{\left (\frac {2 \, {\left (b c x \log \left (F\right ) + d n + 2 \, d\right )}}{n + 2}\right )} - f\right ) - 2 \, d n - 2 \, n \log \left (2\right ) + 2 \, n \log \left (f e^{\left (\frac {2 \, {\left (b c x \log \left (F\right ) + d n + 2 \, d\right )}}{n + 2}\right )} - f\right )}{n + 2} + \frac {2 \, {\left (b c x \log \left (F\right ) + d n + 2 \, d\right )}}{n + 2}\right )} - F^{a c} n e^{\left (\frac {2 \, b c x \log \left (F\right ) - d n^{2} - n^{2} \log \left (2\right ) + n^{2} \log \left (f e^{\left (\frac {2 \, {\left (b c x \log \left (F\right ) + d n + 2 \, d\right )}}{n + 2}\right )} - f\right ) - 2 \, d n - 2 \, n \log \left (2\right ) + 2 \, n \log \left (f e^{\left (\frac {2 \, {\left (b c x \log \left (F\right ) + d n + 2 \, d\right )}}{n + 2}\right )} - f\right )}{n + 2}\right )} + 2 \, F^{a c} e^{\left (\frac {2 \, b c x \log \left (F\right ) - d n^{2} - n^{2} \log \left (2\right ) + n^{2} \log \left (f e^{\left (\frac {2 \, {\left (b c x \log \left (F\right ) + d n + 2 \, d\right )}}{n + 2}\right )} - f\right ) - 2 \, d n - 2 \, n \log \left (2\right ) + 2 \, n \log \left (f e^{\left (\frac {2 \, {\left (b c x \log \left (F\right ) + d n + 2 \, d\right )}}{n + 2}\right )} - f\right )}{n + 2} + \frac {2 \, {\left (b c x \log \left (F\right ) + d n + 2 \, d\right )}}{n + 2}\right )} - 2 \, F^{a c} e^{\left (\frac {2 \, b c x \log \left (F\right ) - d n^{2} - n^{2} \log \left (2\right ) + n^{2} \log \left (f e^{\left (\frac {2 \, {\left (b c x \log \left (F\right ) + d n + 2 \, d\right )}}{n + 2}\right )} - f\right ) - 2 \, d n - 2 \, n \log \left (2\right ) + 2 \, n \log \left (f e^{\left (\frac {2 \, {\left (b c x \log \left (F\right ) + d n + 2 \, d\right )}}{n + 2}\right )} - f\right )}{n + 2}\right )}}{2 \, {\left (b c n e^{\left (\frac {2 \, {\left (b c x \log \left (F\right ) + d n + 2 \, d\right )}}{n + 2}\right )} \log \left (F\right ) + b c e^{\left (\frac {2 \, {\left (b c x \log \left (F\right ) + d n + 2 \, d\right )}}{n + 2}\right )} \log \left (F\right )\right )}} \] Input:

integrate(F^(c*(b*x+a))*(f*sinh(d+b*c*x*log(F)/(2+n)))^n,x, algorithm="gia 
c")
 

Output:

1/2*(F^(a*c)*n*e^((2*b*c*x*log(F) - d*n^2 - n^2*log(2) + n^2*log(f*e^(2*(b 
*c*x*log(F) + d*n + 2*d)/(n + 2)) - f) - 2*d*n - 2*n*log(2) + 2*n*log(f*e^ 
(2*(b*c*x*log(F) + d*n + 2*d)/(n + 2)) - f))/(n + 2) + 2*(b*c*x*log(F) + d 
*n + 2*d)/(n + 2)) - F^(a*c)*n*e^((2*b*c*x*log(F) - d*n^2 - n^2*log(2) + n 
^2*log(f*e^(2*(b*c*x*log(F) + d*n + 2*d)/(n + 2)) - f) - 2*d*n - 2*n*log(2 
) + 2*n*log(f*e^(2*(b*c*x*log(F) + d*n + 2*d)/(n + 2)) - f))/(n + 2)) + 2* 
F^(a*c)*e^((2*b*c*x*log(F) - d*n^2 - n^2*log(2) + n^2*log(f*e^(2*(b*c*x*lo 
g(F) + d*n + 2*d)/(n + 2)) - f) - 2*d*n - 2*n*log(2) + 2*n*log(f*e^(2*(b*c 
*x*log(F) + d*n + 2*d)/(n + 2)) - f))/(n + 2) + 2*(b*c*x*log(F) + d*n + 2* 
d)/(n + 2)) - 2*F^(a*c)*e^((2*b*c*x*log(F) - d*n^2 - n^2*log(2) + n^2*log( 
f*e^(2*(b*c*x*log(F) + d*n + 2*d)/(n + 2)) - f) - 2*d*n - 2*n*log(2) + 2*n 
*log(f*e^(2*(b*c*x*log(F) + d*n + 2*d)/(n + 2)) - f))/(n + 2)))/(b*c*n*e^( 
2*(b*c*x*log(F) + d*n + 2*d)/(n + 2))*log(F) + b*c*e^(2*(b*c*x*log(F) + d* 
n + 2*d)/(n + 2))*log(F))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.90 \[ \int F^{c (a+b x)} \left (f \sinh \left (d+\frac {b c x \log (F)}{2+n}\right )\right )^n \, dx=\frac {f^{a c +n} \left (-e^{\frac {2 \,\mathrm {log}\left (f \right ) b c x +2 d n +4 d}{n +2}} \left (e^{\frac {2 \,\mathrm {log}\left (f \right ) b c x +2 d n +4 d}{n +2}}-1\right )^{n} n -\left (e^{\frac {2 \,\mathrm {log}\left (f \right ) b c x +2 d n +4 d}{n +2}}-1\right )^{n} n -2 \left (e^{\frac {2 \,\mathrm {log}\left (f \right ) b c x +2 d n +4 d}{n +2}}-1\right )^{n}+2 f^{b c x} e^{d n +2 d} \sinh \left (\frac {\mathrm {log}\left (f \right ) b c x +d n +2 d}{n +2}\right )^{n} 2^{n} n +2 f^{b c x} e^{d n +2 d} \sinh \left (\frac {\mathrm {log}\left (f \right ) b c x +d n +2 d}{n +2}\right )^{n} 2^{n}\right )}{2 e^{d n +2 d} 2^{n} \mathrm {log}\left (f \right ) b c \left (n +1\right )} \] Input:

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

Output:

(f**(a*c + n)*( - e**((2*log(f)*b*c*x + 2*d*n + 4*d)/(n + 2))*(e**((2*log( 
f)*b*c*x + 2*d*n + 4*d)/(n + 2)) - 1)**n*n - (e**((2*log(f)*b*c*x + 2*d*n 
+ 4*d)/(n + 2)) - 1)**n*n - 2*(e**((2*log(f)*b*c*x + 2*d*n + 4*d)/(n + 2)) 
 - 1)**n + 2*f**(b*c*x)*e**(d*n + 2*d)*sinh((log(f)*b*c*x + d*n + 2*d)/(n 
+ 2))**n*2**n*n + 2*f**(b*c*x)*e**(d*n + 2*d)*sinh((log(f)*b*c*x + d*n + 2 
*d)/(n + 2))**n*2**n))/(2*e**(d*n + 2*d)*2**n*log(f)*b*c*(n + 1))