\(\int e^{2 (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x))^3 \, dx\) [9]

Optimal result

Integrand size = 29, antiderivative size = 110 \[ \int e^{2 (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x))^3 \, dx=\frac {e^{2 a-3 d-b x} (f-g)^3}{8 b}+\frac {3 e^{2 a-d+b x} (f-g)^2 (f+g)}{8 b}-\frac {e^{2 a+d+3 b x} (f-g) (f+g)^2}{8 b}+\frac {e^{2 a+3 d+5 b x} (f+g)^3}{40 b} \] Output:

1/8*exp(-b*x+2*a-3*d)*(f-g)^3/b+3/8*exp(b*x+2*a-d)*(f-g)^2*(f+g)/b-1/8*exp 
(3*b*x+2*a+d)*(f-g)*(f+g)^2/b+1/40*exp(5*b*x+2*a+3*d)*(f+g)^3/b
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.85 \[ \int e^{2 (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x))^3 \, dx=\frac {e^{2 a-2 d} \left (e^{-d-b x} (f-g)^3+3 e^{d+b x} (f-g)^2 (f+g)-e^{3 (d+b x)} (f-g) (f+g)^2+\frac {1}{5} e^{5 (d+b x)} (f+g)^3\right )}{8 b} \] Input:

Integrate[E^(2*(a + b*x))*(g*Cosh[d + b*x] + f*Sinh[d + b*x])^3,x]
 

Output:

(E^(2*a - 2*d)*(E^(-d - b*x)*(f - g)^3 + 3*E^(d + b*x)*(f - g)^2*(f + g) - 
 E^(3*(d + b*x))*(f - g)*(f + g)^2 + (E^(5*(d + b*x))*(f + g)^3)/5))/(8*b)
 

Rubi [A] (warning: unable to verify)

Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.70, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2720, 27, 244, 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.

\(\displaystyle \int e^{2 (a+b x)} (f \sinh (b x+d)+g \cosh (b x+d))^3 \, dx\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\int -\frac {1}{8} e^{2 a-2 b x} \left (f-g-e^{2 b x} (f+g)\right )^3de^{b x}}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {e^{2 a} \int e^{-2 b x} \left (f-g-e^{2 b x} (f+g)\right )^3de^{b x}}{8 b}\)

\(\Big \downarrow \) 244

\(\displaystyle -\frac {e^{2 a} \int \left (e^{-2 b x} (f-g)^3-3 (f+g) (f-g)^2+3 e^{2 b x} (f+g)^2 (f-g)-e^{4 b x} (f+g)^3\right )de^{b x}}{8 b}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {e^{2 a} \left (-e^{-b x} (f-g)^3-3 e^{b x} (f+g) (f-g)^2+e^{3 b x} (f+g)^2 (f-g)-\frac {1}{5} e^{5 b x} (f+g)^3\right )}{8 b}\)

Input:

Int[E^(2*(a + b*x))*(g*Cosh[d + b*x] + f*Sinh[d + b*x])^3,x]
 

Output:

-1/8*(E^(2*a)*(-((f - g)^3/E^(b*x)) - 3*E^(b*x)*(f - g)^2*(f + g) + E^(3*b 
*x)*(f - g)*(f + g)^2 - (E^(5*b*x)*(f + g)^3)/5))/b
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 244

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

rule 2009

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

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(312\) vs. \(2(98)=196\).

Time = 0.41 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.85

\[\frac {\left (-\frac {3}{8} f^{3}+\frac {3}{8} g^{3}+\frac {3}{8} f \,g^{2}-\frac {3}{8} f^{2} g \right ) \sinh \left (3 b x +2 a +d \right )}{3 b}+\frac {\left (-\frac {3}{8} f^{3}+\frac {3}{8} g^{3}+\frac {3}{8} f \,g^{2}-\frac {3}{8} f^{2} g \right ) \cosh \left (3 b x +2 a +d \right )}{3 b}-\frac {\left (-\frac {1}{8} f^{3}+\frac {1}{8} g^{3}-\frac {3}{8} f \,g^{2}+\frac {3}{8} f^{2} g \right ) \sinh \left (-b x +2 a -3 d \right )}{b}+\frac {\left (\frac {1}{8} f^{3}-\frac {3}{8} f^{2} g +\frac {3}{8} f \,g^{2}-\frac {1}{8} g^{3}\right ) \cosh \left (-b x +2 a -3 d \right )}{b}+\frac {\left (\frac {1}{8} f^{3}+\frac {1}{8} g^{3}+\frac {3}{8} f \,g^{2}+\frac {3}{8} f^{2} g \right ) \sinh \left (5 b x +2 a +3 d \right )}{5 b}+\frac {\left (\frac {1}{8} f^{3}+\frac {1}{8} g^{3}+\frac {3}{8} f \,g^{2}+\frac {3}{8} f^{2} g \right ) \cosh \left (5 b x +2 a +3 d \right )}{5 b}+\frac {\left (\frac {3}{8} f^{3}+\frac {3}{8} g^{3}-\frac {3}{8} f^{2} g -\frac {3}{8} f \,g^{2}\right ) \sinh \left (b x +2 a -d \right )}{b}+\frac {\left (\frac {3}{8} f^{3}+\frac {3}{8} g^{3}-\frac {3}{8} f^{2} g -\frac {3}{8} f \,g^{2}\right ) \cosh \left (b x +2 a -d \right )}{b}\]

Input:

int(exp(2*b*x+2*a)*(g*cosh(b*x+d)+f*sinh(b*x+d))^3,x)
 

Output:

1/3*(-3/8*f^3+3/8*g^3+3/8*f*g^2-3/8*f^2*g)/b*sinh(3*b*x+2*a+d)+1/3*(-3/8*f 
^3+3/8*g^3+3/8*f*g^2-3/8*f^2*g)*cosh(3*b*x+2*a+d)/b-(-1/8*f^3+1/8*g^3-3/8* 
f*g^2+3/8*f^2*g)/b*sinh(-b*x+2*a-3*d)+(1/8*f^3-3/8*f^2*g+3/8*f*g^2-1/8*g^3 
)*cosh(-b*x+2*a-3*d)/b+1/5*(1/8*f^3+1/8*g^3+3/8*f*g^2+3/8*f^2*g)/b*sinh(5* 
b*x+2*a+3*d)+1/5*(1/8*f^3+1/8*g^3+3/8*f*g^2+3/8*f^2*g)*cosh(5*b*x+2*a+3*d) 
/b+(3/8*f^3+3/8*g^3-3/8*f^2*g-3/8*f*g^2)/b*sinh(b*x+2*a-d)+(3/8*f^3+3/8*g^ 
3-3/8*f^2*g-3/8*f*g^2)*cosh(b*x+2*a-d)/b
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 501 vs. \(2 (98) = 196\).

Time = 0.09 (sec) , antiderivative size = 501, normalized size of antiderivative = 4.55 \[ \int e^{2 (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x))^3 \, dx=\frac {{\left (3 \, f^{3} - 6 \, f^{2} g + 9 \, f g^{2} - 2 \, g^{3}\right )} \cosh \left (b x + d\right )^{3} \cosh \left (-2 \, a + 2 \, d\right ) - {\left ({\left (2 \, f^{3} - 9 \, f^{2} g + 6 \, f g^{2} - 3 \, g^{3}\right )} \cosh \left (-2 \, a + 2 \, d\right ) - {\left (2 \, f^{3} - 9 \, f^{2} g + 6 \, f g^{2} - 3 \, g^{3}\right )} \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right )^{3} + 5 \, {\left (f^{3} - 2 \, f^{2} g - f g^{2} + 2 \, g^{3}\right )} \cosh \left (b x + d\right ) \cosh \left (-2 \, a + 2 \, d\right ) + 3 \, {\left ({\left (3 \, f^{3} - 6 \, f^{2} g + 9 \, f g^{2} - 2 \, g^{3}\right )} \cosh \left (b x + d\right ) \cosh \left (-2 \, a + 2 \, d\right ) - {\left (3 \, f^{3} - 6 \, f^{2} g + 9 \, f g^{2} - 2 \, g^{3}\right )} \cosh \left (b x + d\right ) \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right )^{2} - {\left (3 \, {\left (2 \, f^{3} - 9 \, f^{2} g + 6 \, f g^{2} - 3 \, g^{3}\right )} \cosh \left (b x + d\right )^{2} \cosh \left (-2 \, a + 2 \, d\right ) + 5 \, {\left (2 \, f^{3} - f^{2} g - 2 \, f g^{2} + g^{3}\right )} \cosh \left (-2 \, a + 2 \, d\right ) - {\left (10 \, f^{3} - 5 \, f^{2} g - 10 \, f g^{2} + 5 \, g^{3} + 3 \, {\left (2 \, f^{3} - 9 \, f^{2} g + 6 \, f g^{2} - 3 \, g^{3}\right )} \cosh \left (b x + d\right )^{2}\right )} \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right ) - {\left ({\left (3 \, f^{3} - 6 \, f^{2} g + 9 \, f g^{2} - 2 \, g^{3}\right )} \cosh \left (b x + d\right )^{3} + 5 \, {\left (f^{3} - 2 \, f^{2} g - f g^{2} + 2 \, g^{3}\right )} \cosh \left (b x + d\right )\right )} \sinh \left (-2 \, a + 2 \, d\right )}{20 \, {\left (b \cosh \left (b x + d\right )^{2} - 2 \, b \cosh \left (b x + d\right ) \sinh \left (b x + d\right ) + b \sinh \left (b x + d\right )^{2}\right )}} \] Input:

integrate(exp(2*b*x+2*a)*(g*cosh(b*x+d)+f*sinh(b*x+d))^3,x, algorithm="fri 
cas")
 

Output:

1/20*((3*f^3 - 6*f^2*g + 9*f*g^2 - 2*g^3)*cosh(b*x + d)^3*cosh(-2*a + 2*d) 
 - ((2*f^3 - 9*f^2*g + 6*f*g^2 - 3*g^3)*cosh(-2*a + 2*d) - (2*f^3 - 9*f^2* 
g + 6*f*g^2 - 3*g^3)*sinh(-2*a + 2*d))*sinh(b*x + d)^3 + 5*(f^3 - 2*f^2*g 
- f*g^2 + 2*g^3)*cosh(b*x + d)*cosh(-2*a + 2*d) + 3*((3*f^3 - 6*f^2*g + 9* 
f*g^2 - 2*g^3)*cosh(b*x + d)*cosh(-2*a + 2*d) - (3*f^3 - 6*f^2*g + 9*f*g^2 
 - 2*g^3)*cosh(b*x + d)*sinh(-2*a + 2*d))*sinh(b*x + d)^2 - (3*(2*f^3 - 9* 
f^2*g + 6*f*g^2 - 3*g^3)*cosh(b*x + d)^2*cosh(-2*a + 2*d) + 5*(2*f^3 - f^2 
*g - 2*f*g^2 + g^3)*cosh(-2*a + 2*d) - (10*f^3 - 5*f^2*g - 10*f*g^2 + 5*g^ 
3 + 3*(2*f^3 - 9*f^2*g + 6*f*g^2 - 3*g^3)*cosh(b*x + d)^2)*sinh(-2*a + 2*d 
))*sinh(b*x + d) - ((3*f^3 - 6*f^2*g + 9*f*g^2 - 2*g^3)*cosh(b*x + d)^3 + 
5*(f^3 - 2*f^2*g - f*g^2 + 2*g^3)*cosh(b*x + d))*sinh(-2*a + 2*d))/(b*cosh 
(b*x + d)^2 - 2*b*cosh(b*x + d)*sinh(b*x + d) + b*sinh(b*x + d)^2)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 541 vs. \(2 (88) = 176\).

Time = 1.02 (sec) , antiderivative size = 541, normalized size of antiderivative = 4.92 \[ \int e^{2 (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x))^3 \, dx=\begin {cases} \frac {2 f^{3} e^{2 a} e^{2 b x} \sinh ^{3}{\left (b x + d \right )}}{5 b} + \frac {f^{3} e^{2 a} e^{2 b x} \sinh ^{2}{\left (b x + d \right )} \cosh {\left (b x + d \right )}}{5 b} - \frac {4 f^{3} e^{2 a} e^{2 b x} \sinh {\left (b x + d \right )} \cosh ^{2}{\left (b x + d \right )}}{5 b} + \frac {2 f^{3} e^{2 a} e^{2 b x} \cosh ^{3}{\left (b x + d \right )}}{5 b} + \frac {f^{2} g e^{2 a} e^{2 b x} \sinh ^{3}{\left (b x + d \right )}}{5 b} - \frac {2 f^{2} g e^{2 a} e^{2 b x} \sinh ^{2}{\left (b x + d \right )} \cosh {\left (b x + d \right )}}{5 b} + \frac {8 f^{2} g e^{2 a} e^{2 b x} \sinh {\left (b x + d \right )} \cosh ^{2}{\left (b x + d \right )}}{5 b} - \frac {4 f^{2} g e^{2 a} e^{2 b x} \cosh ^{3}{\left (b x + d \right )}}{5 b} - \frac {4 f g^{2} e^{2 a} e^{2 b x} \sinh ^{3}{\left (b x + d \right )}}{5 b} + \frac {8 f g^{2} e^{2 a} e^{2 b x} \sinh ^{2}{\left (b x + d \right )} \cosh {\left (b x + d \right )}}{5 b} - \frac {2 f g^{2} e^{2 a} e^{2 b x} \sinh {\left (b x + d \right )} \cosh ^{2}{\left (b x + d \right )}}{5 b} + \frac {f g^{2} e^{2 a} e^{2 b x} \cosh ^{3}{\left (b x + d \right )}}{5 b} + \frac {2 g^{3} e^{2 a} e^{2 b x} \sinh ^{3}{\left (b x + d \right )}}{5 b} - \frac {4 g^{3} e^{2 a} e^{2 b x} \sinh ^{2}{\left (b x + d \right )} \cosh {\left (b x + d \right )}}{5 b} + \frac {g^{3} e^{2 a} e^{2 b x} \sinh {\left (b x + d \right )} \cosh ^{2}{\left (b x + d \right )}}{5 b} + \frac {2 g^{3} e^{2 a} e^{2 b x} \cosh ^{3}{\left (b x + d \right )}}{5 b} & \text {for}\: b \neq 0 \\x \left (f \sinh {\left (d \right )} + g \cosh {\left (d \right )}\right )^{3} e^{2 a} & \text {otherwise} \end {cases} \] Input:

integrate(exp(2*b*x+2*a)*(g*cosh(b*x+d)+f*sinh(b*x+d))**3,x)
 

Output:

Piecewise((2*f**3*exp(2*a)*exp(2*b*x)*sinh(b*x + d)**3/(5*b) + f**3*exp(2* 
a)*exp(2*b*x)*sinh(b*x + d)**2*cosh(b*x + d)/(5*b) - 4*f**3*exp(2*a)*exp(2 
*b*x)*sinh(b*x + d)*cosh(b*x + d)**2/(5*b) + 2*f**3*exp(2*a)*exp(2*b*x)*co 
sh(b*x + d)**3/(5*b) + f**2*g*exp(2*a)*exp(2*b*x)*sinh(b*x + d)**3/(5*b) - 
 2*f**2*g*exp(2*a)*exp(2*b*x)*sinh(b*x + d)**2*cosh(b*x + d)/(5*b) + 8*f** 
2*g*exp(2*a)*exp(2*b*x)*sinh(b*x + d)*cosh(b*x + d)**2/(5*b) - 4*f**2*g*ex 
p(2*a)*exp(2*b*x)*cosh(b*x + d)**3/(5*b) - 4*f*g**2*exp(2*a)*exp(2*b*x)*si 
nh(b*x + d)**3/(5*b) + 8*f*g**2*exp(2*a)*exp(2*b*x)*sinh(b*x + d)**2*cosh( 
b*x + d)/(5*b) - 2*f*g**2*exp(2*a)*exp(2*b*x)*sinh(b*x + d)*cosh(b*x + d)* 
*2/(5*b) + f*g**2*exp(2*a)*exp(2*b*x)*cosh(b*x + d)**3/(5*b) + 2*g**3*exp( 
2*a)*exp(2*b*x)*sinh(b*x + d)**3/(5*b) - 4*g**3*exp(2*a)*exp(2*b*x)*sinh(b 
*x + d)**2*cosh(b*x + d)/(5*b) + g**3*exp(2*a)*exp(2*b*x)*sinh(b*x + d)*co 
sh(b*x + d)**2/(5*b) + 2*g**3*exp(2*a)*exp(2*b*x)*cosh(b*x + d)**3/(5*b), 
Ne(b, 0)), (x*(f*sinh(d) + g*cosh(d))**3*exp(2*a), True))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (98) = 196\).

Time = 0.05 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.32 \[ \int e^{2 (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x))^3 \, dx=\frac {1}{40} \, g^{3} {\left (\frac {{\left (5 \, e^{\left (-2 \, b x - 2 \, d\right )} + 15 \, e^{\left (-4 \, b x - 4 \, d\right )} + 1\right )} e^{\left (5 \, b x + 2 \, a + 3 \, d\right )}}{b} - \frac {5 \, e^{\left (-b x + 2 \, a - 3 \, d\right )}}{b}\right )} - \frac {1}{40} \, f^{2} g {\left (\frac {{\left (5 \, e^{\left (-2 \, b x - 2 \, d\right )} + 15 \, e^{\left (-4 \, b x - 4 \, d\right )} - 3\right )} e^{\left (5 \, b x + 2 \, a + 3 \, d\right )}}{b} + \frac {15 \, e^{\left (-b x + 2 \, a - 3 \, d\right )}}{b}\right )} + \frac {1}{40} \, f g^{2} {\left (\frac {{\left (5 \, e^{\left (-2 \, b x - 2 \, d\right )} - 15 \, e^{\left (-4 \, b x - 4 \, d\right )} + 3\right )} e^{\left (5 \, b x + 2 \, a + 3 \, d\right )}}{b} + \frac {15 \, e^{\left (-b x + 2 \, a - 3 \, d\right )}}{b}\right )} - \frac {1}{40} \, f^{3} {\left (\frac {{\left (5 \, e^{\left (-2 \, b x - 2 \, d\right )} - 15 \, e^{\left (-4 \, b x - 4 \, d\right )} - 1\right )} e^{\left (5 \, b x + 2 \, a + 3 \, d\right )}}{b} - \frac {5 \, e^{\left (-b x + 2 \, a - 3 \, d\right )}}{b}\right )} \] Input:

integrate(exp(2*b*x+2*a)*(g*cosh(b*x+d)+f*sinh(b*x+d))^3,x, algorithm="max 
ima")
 

Output:

1/40*g^3*((5*e^(-2*b*x - 2*d) + 15*e^(-4*b*x - 4*d) + 1)*e^(5*b*x + 2*a + 
3*d)/b - 5*e^(-b*x + 2*a - 3*d)/b) - 1/40*f^2*g*((5*e^(-2*b*x - 2*d) + 15* 
e^(-4*b*x - 4*d) - 3)*e^(5*b*x + 2*a + 3*d)/b + 15*e^(-b*x + 2*a - 3*d)/b) 
 + 1/40*f*g^2*((5*e^(-2*b*x - 2*d) - 15*e^(-4*b*x - 4*d) + 3)*e^(5*b*x + 2 
*a + 3*d)/b + 15*e^(-b*x + 2*a - 3*d)/b) - 1/40*f^3*((5*e^(-2*b*x - 2*d) - 
 15*e^(-4*b*x - 4*d) - 1)*e^(5*b*x + 2*a + 3*d)/b - 5*e^(-b*x + 2*a - 3*d) 
/b)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (98) = 196\).

Time = 0.12 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.32 \[ \int e^{2 (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x))^3 \, dx=\frac {{\left (f^{3} e^{\left (5 \, b x + 2 \, a + 5 \, d\right )} + 3 \, f^{2} g e^{\left (5 \, b x + 2 \, a + 5 \, d\right )} + 3 \, f g^{2} e^{\left (5 \, b x + 2 \, a + 5 \, d\right )} + g^{3} e^{\left (5 \, b x + 2 \, a + 5 \, d\right )} - 5 \, f^{3} e^{\left (3 \, b x + 2 \, a + 3 \, d\right )} - 5 \, f^{2} g e^{\left (3 \, b x + 2 \, a + 3 \, d\right )} + 5 \, f g^{2} e^{\left (3 \, b x + 2 \, a + 3 \, d\right )} + 5 \, g^{3} e^{\left (3 \, b x + 2 \, a + 3 \, d\right )} + 15 \, f^{3} e^{\left (b x + 2 \, a + d\right )} - 15 \, f^{2} g e^{\left (b x + 2 \, a + d\right )} - 15 \, f g^{2} e^{\left (b x + 2 \, a + d\right )} + 15 \, g^{3} e^{\left (b x + 2 \, a + d\right )} + 5 \, {\left (f^{3} e^{\left (2 \, a\right )} - 3 \, f^{2} g e^{\left (2 \, a\right )} + 3 \, f g^{2} e^{\left (2 \, a\right )} - g^{3} e^{\left (2 \, a\right )}\right )} e^{\left (-b x - d\right )}\right )} e^{\left (-2 \, d\right )}}{40 \, b} \] Input:

integrate(exp(2*b*x+2*a)*(g*cosh(b*x+d)+f*sinh(b*x+d))^3,x, algorithm="gia 
c")
 

Output:

1/40*(f^3*e^(5*b*x + 2*a + 5*d) + 3*f^2*g*e^(5*b*x + 2*a + 5*d) + 3*f*g^2* 
e^(5*b*x + 2*a + 5*d) + g^3*e^(5*b*x + 2*a + 5*d) - 5*f^3*e^(3*b*x + 2*a + 
 3*d) - 5*f^2*g*e^(3*b*x + 2*a + 3*d) + 5*f*g^2*e^(3*b*x + 2*a + 3*d) + 5* 
g^3*e^(3*b*x + 2*a + 3*d) + 15*f^3*e^(b*x + 2*a + d) - 15*f^2*g*e^(b*x + 2 
*a + d) - 15*f*g^2*e^(b*x + 2*a + d) + 15*g^3*e^(b*x + 2*a + d) + 5*(f^3*e 
^(2*a) - 3*f^2*g*e^(2*a) + 3*f*g^2*e^(2*a) - g^3*e^(2*a))*e^(-b*x - d))*e^ 
(-2*d)/b
 

Mupad [B] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.72 \[ \int e^{2 (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x))^3 \, dx=\frac {{\mathrm {cosh}\left (d+b\,x\right )}^3\,{\mathrm {e}}^{2\,a+2\,b\,x}\,\left (2\,f^3-4\,f^2\,g+f\,g^2+2\,g^3\right )}{5\,b}+\frac {{\mathrm {e}}^{2\,a+2\,b\,x}\,{\mathrm {sinh}\left (d+b\,x\right )}^3\,\left (2\,f^3+f^2\,g-4\,f\,g^2+2\,g^3\right )}{5\,b}+\frac {\mathrm {cosh}\left (d+b\,x\right )\,{\mathrm {e}}^{2\,a+2\,b\,x}\,{\mathrm {sinh}\left (d+b\,x\right )}^2\,\left (f^3-2\,f^2\,g+8\,f\,g^2-4\,g^3\right )}{5\,b}-\frac {{\mathrm {cosh}\left (d+b\,x\right )}^2\,{\mathrm {e}}^{2\,a+2\,b\,x}\,\mathrm {sinh}\left (d+b\,x\right )\,\left (4\,f^3-8\,f^2\,g+2\,f\,g^2-g^3\right )}{5\,b} \] Input:

int(exp(2*a + 2*b*x)*(g*cosh(d + b*x) + f*sinh(d + b*x))^3,x)
 

Output:

(cosh(d + b*x)^3*exp(2*a + 2*b*x)*(f*g^2 - 4*f^2*g + 2*f^3 + 2*g^3))/(5*b) 
 + (exp(2*a + 2*b*x)*sinh(d + b*x)^3*(f^2*g - 4*f*g^2 + 2*f^3 + 2*g^3))/(5 
*b) + (cosh(d + b*x)*exp(2*a + 2*b*x)*sinh(d + b*x)^2*(8*f*g^2 - 2*f^2*g + 
 f^3 - 4*g^3))/(5*b) - (cosh(d + b*x)^2*exp(2*a + 2*b*x)*sinh(d + b*x)*(2* 
f*g^2 - 8*f^2*g + 4*f^3 - g^3))/(5*b)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.51 \[ \int e^{2 (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x))^3 \, dx=\frac {e^{2 b x +2 a} \left (2 \cosh \left (b x +d \right )^{3} f^{3}-4 \cosh \left (b x +d \right )^{3} f^{2} g +\cosh \left (b x +d \right )^{3} f \,g^{2}+2 \cosh \left (b x +d \right )^{3} g^{3}-4 \cosh \left (b x +d \right )^{2} \sinh \left (b x +d \right ) f^{3}+8 \cosh \left (b x +d \right )^{2} \sinh \left (b x +d \right ) f^{2} g -2 \cosh \left (b x +d \right )^{2} \sinh \left (b x +d \right ) f \,g^{2}+\cosh \left (b x +d \right )^{2} \sinh \left (b x +d \right ) g^{3}+\cosh \left (b x +d \right ) \sinh \left (b x +d \right )^{2} f^{3}-2 \cosh \left (b x +d \right ) \sinh \left (b x +d \right )^{2} f^{2} g +8 \cosh \left (b x +d \right ) \sinh \left (b x +d \right )^{2} f \,g^{2}-4 \cosh \left (b x +d \right ) \sinh \left (b x +d \right )^{2} g^{3}+2 \sinh \left (b x +d \right )^{3} f^{3}+\sinh \left (b x +d \right )^{3} f^{2} g -4 \sinh \left (b x +d \right )^{3} f \,g^{2}+2 \sinh \left (b x +d \right )^{3} g^{3}\right )}{5 b} \] Input:

int(exp(2*b*x+2*a)*(g*cosh(b*x+d)+f*sinh(b*x+d))^3,x)
 

Output:

(e**(2*a + 2*b*x)*(2*cosh(b*x + d)**3*f**3 - 4*cosh(b*x + d)**3*f**2*g + c 
osh(b*x + d)**3*f*g**2 + 2*cosh(b*x + d)**3*g**3 - 4*cosh(b*x + d)**2*sinh 
(b*x + d)*f**3 + 8*cosh(b*x + d)**2*sinh(b*x + d)*f**2*g - 2*cosh(b*x + d) 
**2*sinh(b*x + d)*f*g**2 + cosh(b*x + d)**2*sinh(b*x + d)*g**3 + cosh(b*x 
+ d)*sinh(b*x + d)**2*f**3 - 2*cosh(b*x + d)*sinh(b*x + d)**2*f**2*g + 8*c 
osh(b*x + d)*sinh(b*x + d)**2*f*g**2 - 4*cosh(b*x + d)*sinh(b*x + d)**2*g* 
*3 + 2*sinh(b*x + d)**3*f**3 + sinh(b*x + d)**3*f**2*g - 4*sinh(b*x + d)** 
3*f*g**2 + 2*sinh(b*x + d)**3*g**3))/(5*b)