\(\int F^{c (a+b x)} \sinh (d+e x) \tanh (d+e x) \, dx\) [117]

Optimal result

Integrand size = 22, antiderivative size = 136 \[ \int F^{c (a+b x)} \sinh (d+e x) \tanh (d+e x) \, dx=\frac {3 e^{-d-e x} F^{c (a+b x)}}{2 (e-b c \log (F))}-\frac {2 e^{-d-e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (-1+\frac {b c \log (F)}{e}\right ),\frac {e+b c \log (F)}{2 e},-e^{2 d+2 e x}\right )}{e-b c \log (F)}+\frac {e^{d+e x} F^{c (a+b x)}}{2 (e+b c \log (F))} \] Output:

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

Mathematica [A] (verified)

Time = 1.07 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.79 \[ \int F^{c (a+b x)} \sinh (d+e x) \tanh (d+e x) \, dx=\frac {F^{c (a+b x)} \left (-b c \cosh (d+e x) \log (F)-2 e^{d+e x} \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),-e^{2 (d+e x)}\right ) (e-b c \log (F))+e \sinh (d+e x)\right )}{(e-b c \log (F)) (e+b c \log (F))} \] Input:

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

Output:

(F^(c*(a + b*x))*(-(b*c*Cosh[d + e*x]*Log[F]) - 2*E^(d + e*x)*Hypergeometr 
ic2F1[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*Sinh[d + e*x]))/((e - b*c*Log[F])*(e + b*c*Log[F]))
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.04, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6037, 2009}

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.

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

\(\Big \downarrow \) 6037

\(\displaystyle \int \left (-\frac {3}{2} e^{-d-e x} F^{a c+b c x}+\frac {1}{2} e^{2 (d+e x)-d-e x} F^{a c+b c x}+\frac {2 e^{-d-e x} F^{a c+b c x}}{e^{2 (d+e x)}+1}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6037

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*(G_)[(d_.) + (e_.)*(x_)]^(m_.)*(H_)[( 
d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigToExp[F^(c*(a + b*x)), 
 G[d + e*x]^m*H[d + e*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e}, x] && IGtQ[ 
m, 0] && IGtQ[n, 0] && HyperbolicQ[G] && HyperbolicQ[H]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

-4*F^(a*c)*e*integrate(F^(b*c*x)/((b*c*e^(5*d)*log(F) - 3*e*e^(5*d))*e^(5* 
e*x) + 2*(b*c*e^(3*d)*log(F) - 3*e*e^(3*d))*e^(3*e*x) + (b*c*e^d*log(F) - 
3*e*e^d)*e^(e*x)), x) + 1/2*(F^(a*c)*b^2*c^2*log(F)^2 + 6*F^(a*c)*b*c*e*lo 
g(F) + 5*F^(a*c)*e^2 + (F^(a*c)*b^2*c^2*e^(4*d)*log(F)^2 - 4*F^(a*c)*b*c*e 
*e^(4*d)*log(F) + 3*F^(a*c)*e^2*e^(4*d))*e^(4*e*x) - 2*(F^(a*c)*b^2*c^2*e^ 
(2*d)*log(F)^2 - F^(a*c)*b*c*e*e^(2*d)*log(F) - 6*F^(a*c)*e^2*e^(2*d))*e^( 
2*e*x))*F^(b*c*x)/((b^3*c^3*e^(3*d)*log(F)^3 - 3*b^2*c^2*e*e^(3*d)*log(F)^ 
2 - b*c*e^2*e^(3*d)*log(F) + 3*e^3*e^(3*d))*e^(3*e*x) + (b^3*c^3*e^d*log(F 
)^3 - 3*b^2*c^2*e*e^d*log(F)^2 - b*c*e^2*e^d*log(F) + 3*e^3*e^d)*e^(e*x))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

f**(a*c)*int(f**(b*c*x)*sinh(d + e*x)*tanh(d + e*x),x)