\(\int \frac {e^{2 (a+b x)}}{(g \cosh (d+b x)+f \sinh (d+b x))^3} \, dx\) [14]

Optimal result

Integrand size = 29, antiderivative size = 143 \[ \int \frac {e^{2 (a+b x)}}{(g \cosh (d+b x)+f \sinh (d+b x))^3} \, dx=-\frac {2 e^{2 a+d+3 b x}}{b (f+g) \left (f-g-e^{2 d+2 b x} (f+g)\right )^2}+\frac {3 e^{2 a-d+b x}}{b (f+g)^2 \left (f-g-e^{2 d+2 b x} (f+g)\right )}-\frac {3 e^{2 a-2 d} \text {arctanh}\left (\frac {e^{d+b x} \sqrt {f+g}}{\sqrt {f-g}}\right )}{b \sqrt {f-g} (f+g)^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.92 \[ \int \frac {e^{2 (a+b x)}}{(g \cosh (d+b x)+f \sinh (d+b x))^3} \, dx=\frac {e^{2 a-2 d} \left (-\frac {e^{d+b x} \left (\left (-3+5 e^{2 (d+b x)}\right ) f+\left (3+5 e^{2 (d+b x)}\right ) g\right )}{(f+g)^2 \left (\left (-1+e^{2 (d+b x)}\right ) f+\left (1+e^{2 (d+b x)}\right ) g\right )^2}-\frac {3 \text {arctanh}\left (\frac {e^{d+b x} \sqrt {f+g}}{\sqrt {f-g}}\right )}{\sqrt {f-g} (f+g)^{5/2}}\right )}{b} \] 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)*((-3 + 5*E^(2*(d + b*x)))*f + (3 + 5*E^(2*( 
d + b*x)))*g))/((f + g)^2*((-1 + E^(2*(d + b*x)))*f + (1 + E^(2*(d + b*x)) 
)*g)^2)) - (3*ArcTanh[(E^(d + b*x)*Sqrt[f + g])/Sqrt[f - g]])/(Sqrt[f - g] 
*(f + g)^(5/2))))/b
 

Rubi [A] (warning: unable to verify)

Time = 0.31 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2720, 27, 252, 252, 221}

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.

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 252

\(\displaystyle -\frac {8 e^{2 a} \left (\frac {e^{3 b x}}{4 (f+g) \left (-e^{2 b x} (f+g)+f-g\right )^2}-\frac {3 \left (\frac {e^{b x}}{2 (f+g) \left (-e^{2 b x} (f+g)+f-g\right )}-\frac {\int \frac {1}{f-g-e^{2 b x} (f+g)}de^{b x}}{2 (f+g)}\right )}{4 (f+g)}\right )}{b}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {8 e^{2 a} \left (\frac {e^{3 b x}}{4 (f+g) \left (-e^{2 b x} (f+g)+f-g\right )^2}-\frac {3 \left (\frac {e^{b x}}{2 (f+g) \left (-e^{2 b x} (f+g)+f-g\right )}-\frac {\text {arctanh}\left (\frac {e^{b x} \sqrt {f+g}}{\sqrt {f-g}}\right )}{2 \sqrt {f-g} (f+g)^{3/2}}\right )}{4 (f+g)}\right )}{b}\)

Input:

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

Output:

(-8*E^(2*a)*(E^(3*b*x)/(4*(f + g)*(f - g - E^(2*b*x)*(f + g))^2) - (3*(E^( 
b*x)/(2*(f + g)*(f - g - E^(2*b*x)*(f + g))) - ArcTanh[(E^(b*x)*Sqrt[f + g 
])/Sqrt[f - g]]/(2*Sqrt[f - g]*(f + g)^(3/2))))/(4*(f + g))))/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 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

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 [A] (verified)

Time = 2.52 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.64

\[\frac {\left (-5 f \,{\mathrm e}^{2 b x +2 a +2 d}-5 g \,{\mathrm e}^{2 b x +2 a +2 d}+3 \,{\mathrm e}^{2 a} f -3 \,{\mathrm e}^{2 a} g \right ) {\mathrm e}^{b x +4 a -d}}{\left (-f \,{\mathrm e}^{2 b x +2 a +2 d}-g \,{\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a} f -{\mathrm e}^{2 a} g \right )^{2} \left (f +g \right )^{2} b}+\frac {3 \ln \left ({\mathrm e}^{b x +a}-\frac {\left (f -g \right ) {\mathrm e}^{a -d}}{\sqrt {f^{2}-g^{2}}}\right ) {\mathrm e}^{2 a -2 d}}{2 b \left (f +g \right )^{2} \sqrt {f^{2}-g^{2}}}-\frac {3 \ln \left ({\mathrm e}^{b x +a}+\frac {\left (f -g \right ) {\mathrm e}^{a -d}}{\sqrt {f^{2}-g^{2}}}\right ) {\mathrm e}^{2 a -2 d}}{2 b \left (f +g \right )^{2} \sqrt {f^{2}-g^{2}}}\]

Input:

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

Output:

1/(-f*exp(2*b*x+2*a+2*d)-g*exp(2*b*x+2*a+2*d)+exp(2*a)*f-exp(2*a)*g)^2/(f+ 
g)^2/b*(-5*f*exp(2*b*x+2*a+2*d)-5*g*exp(2*b*x+2*a+2*d)+3*exp(2*a)*f-3*exp( 
2*a)*g)*exp(b*x+4*a-d)+3/2*ln(exp(b*x+a)-1/(f^2-g^2)^(1/2)*(f-g)*exp(a-d)) 
/b/(f+g)^2/(f^2-g^2)^(1/2)*exp(2*a-2*d)-3/2*ln(exp(b*x+a)+1/(f^2-g^2)^(1/2 
)*(f-g)*exp(a-d))/b/(f+g)^2/(f^2-g^2)^(1/2)*exp(2*a-2*d)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1437 vs. \(2 (127) = 254\).

Time = 0.15 (sec) , antiderivative size = 2961, normalized size of antiderivative = 20.71 \[ \int \frac {e^{2 (a+b x)}}{(g \cosh (d+b x)+f \sinh (d+b x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/2*(10*(f^3 + f^2*g - f*g^2 - g^3)*cosh(b*x + d)^3*cosh(-2*a + 2*d) + 1 
0*((f^3 + f^2*g - f*g^2 - g^3)*cosh(-2*a + 2*d) - (f^3 + f^2*g - f*g^2 - g 
^3)*sinh(-2*a + 2*d))*sinh(b*x + d)^3 - 6*(f^3 - f^2*g - f*g^2 + g^3)*cosh 
(b*x + d)*cosh(-2*a + 2*d) + 30*((f^3 + f^2*g - f*g^2 - g^3)*cosh(b*x + d) 
*cosh(-2*a + 2*d) - (f^3 + f^2*g - f*g^2 - g^3)*cosh(b*x + d)*sinh(-2*a + 
2*d))*sinh(b*x + d)^2 - 3*((f^2 + 2*f*g + g^2)*cosh(b*x + d)^4*cosh(-2*a + 
 2*d) + ((f^2 + 2*f*g + g^2)*cosh(-2*a + 2*d) - (f^2 + 2*f*g + g^2)*sinh(- 
2*a + 2*d))*sinh(b*x + d)^4 - 2*(f^2 - g^2)*cosh(b*x + d)^2*cosh(-2*a + 2* 
d) + 4*((f^2 + 2*f*g + g^2)*cosh(b*x + d)*cosh(-2*a + 2*d) - (f^2 + 2*f*g 
+ g^2)*cosh(b*x + d)*sinh(-2*a + 2*d))*sinh(b*x + d)^3 + 2*(3*(f^2 + 2*f*g 
 + g^2)*cosh(b*x + d)^2*cosh(-2*a + 2*d) - (f^2 - g^2)*cosh(-2*a + 2*d) - 
(3*(f^2 + 2*f*g + g^2)*cosh(b*x + d)^2 - f^2 + g^2)*sinh(-2*a + 2*d))*sinh 
(b*x + d)^2 + (f^2 - 2*f*g + g^2)*cosh(-2*a + 2*d) + 4*((f^2 + 2*f*g + g^2 
)*cosh(b*x + d)^3*cosh(-2*a + 2*d) - (f^2 - g^2)*cosh(b*x + d)*cosh(-2*a + 
 2*d) - ((f^2 + 2*f*g + g^2)*cosh(b*x + d)^3 - (f^2 - g^2)*cosh(b*x + d))* 
sinh(-2*a + 2*d))*sinh(b*x + d) - ((f^2 + 2*f*g + g^2)*cosh(b*x + d)^4 - 2 
*(f^2 - g^2)*cosh(b*x + d)^2 + f^2 - 2*f*g + g^2)*sinh(-2*a + 2*d))*sqrt(f 
^2 - g^2)*log(((f + g)*cosh(b*x + d)^2 + 2*(f + g)*cosh(b*x + d)*sinh(b*x 
+ d) + (f + g)*sinh(b*x + d)^2 - 2*sqrt(f^2 - g^2)*(cosh(b*x + d) + sinh(b 
*x + d)) + f - g)/((f + g)*cosh(b*x + d)^2 + 2*(f + g)*cosh(b*x + d)*si...
 

Sympy [F]

\[ \int \frac {e^{2 (a+b x)}}{(g \cosh (d+b x)+f \sinh (d+b x))^3} \, dx=e^{2 a} \int \frac {e^{2 b x}}{f^{3} \sinh ^{3}{\left (b x + d \right )} + 3 f^{2} g \sinh ^{2}{\left (b x + d \right )} \cosh {\left (b x + d \right )} + 3 f g^{2} \sinh {\left (b x + d \right )} \cosh ^{2}{\left (b x + d \right )} + g^{3} \cosh ^{3}{\left (b x + d \right )}}\, dx \] Input:

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{2 (a+b x)}}{(g \cosh (d+b x)+f \sinh (d+b x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*f^2-4*g^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.26 \[ \int \frac {e^{2 (a+b x)}}{(g \cosh (d+b x)+f \sinh (d+b x))^3} \, dx=\frac {{\left (\frac {3 \, \arctan \left (\frac {f e^{\left (b x + d\right )} + g e^{\left (b x + d\right )}}{\sqrt {-f^{2} + g^{2}}}\right )}{{\left (f^{2} e^{\left (2 \, d\right )} + 2 \, f g e^{\left (2 \, d\right )} + g^{2} e^{\left (2 \, d\right )}\right )} \sqrt {-f^{2} + g^{2}}} - \frac {5 \, f e^{\left (3 \, b x + 3 \, d\right )} + 5 \, g e^{\left (3 \, b x + 3 \, d\right )} - 3 \, f e^{\left (b x + d\right )} + 3 \, g e^{\left (b x + d\right )}}{{\left (f^{2} e^{\left (2 \, d\right )} + 2 \, f g e^{\left (2 \, d\right )} + g^{2} e^{\left (2 \, d\right )}\right )} {\left (f e^{\left (2 \, b x + 2 \, d\right )} + g e^{\left (2 \, b x + 2 \, d\right )} - f + g\right )}^{2}}\right )} e^{\left (2 \, a\right )}}{b} \] Input:

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

Output:

(3*arctan((f*e^(b*x + d) + g*e^(b*x + d))/sqrt(-f^2 + g^2))/((f^2*e^(2*d) 
+ 2*f*g*e^(2*d) + g^2*e^(2*d))*sqrt(-f^2 + g^2)) - (5*f*e^(3*b*x + 3*d) + 
5*g*e^(3*b*x + 3*d) - 3*f*e^(b*x + d) + 3*g*e^(b*x + d))/((f^2*e^(2*d) + 2 
*f*g*e^(2*d) + g^2*e^(2*d))*(f*e^(2*b*x + 2*d) + g*e^(2*b*x + 2*d) - f + g 
)^2))*e^(2*a)/b
 

Mupad [B] (verification not implemented)

Time = 3.61 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.23 \[ \int \frac {e^{2 (a+b x)}}{(g \cosh (d+b x)+f \sinh (d+b x))^3} \, dx=-\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d}\,{\mathrm {e}}^{b\,x}\,\sqrt {-b^2\,f^6-4\,b^2\,f^5\,g-5\,b^2\,f^4\,g^2+5\,b^2\,f^2\,g^4+4\,b^2\,f\,g^5+b^2\,g^6}}{b\,f^3\,\sqrt {{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-4\,d}}-b\,g^3\,\sqrt {{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-4\,d}}-b\,f\,g^2\,\sqrt {{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-4\,d}}+b\,f^2\,g\,\sqrt {{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-4\,d}}}\right )\,\sqrt {{\mathrm {e}}^{4\,a-4\,d}}}{\sqrt {-b^2\,f^6-4\,b^2\,f^5\,g-5\,b^2\,f^4\,g^2+5\,b^2\,f^2\,g^4+4\,b^2\,f\,g^5+b^2\,g^6}}-\frac {2\,{\mathrm {e}}^{2\,a+d+3\,b\,x}}{b\,\left (f+g\right )\,\left ({\mathrm {e}}^{4\,d+4\,b\,x}\,{\left (f+g\right )}^2+{\left (f-g\right )}^2-2\,{\mathrm {e}}^{2\,d+2\,b\,x}\,\left (f+g\right )\,\left (f-g\right )\right )}-\frac {3\,{\mathrm {e}}^{2\,a-d+b\,x}}{b\,{\left (f+g\right )}^2\,\left (g-f+{\mathrm {e}}^{2\,d+2\,b\,x}\,\left (f+g\right )\right )} \] Input:

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

Output:

- (3*atan((exp(2*a)*exp(-d)*exp(b*x)*(b^2*g^6 - b^2*f^6 + 4*b^2*f*g^5 - 4* 
b^2*f^5*g + 5*b^2*f^2*g^4 - 5*b^2*f^4*g^2)^(1/2))/(b*f^3*(exp(4*a)*exp(-4* 
d))^(1/2) - b*g^3*(exp(4*a)*exp(-4*d))^(1/2) - b*f*g^2*(exp(4*a)*exp(-4*d) 
)^(1/2) + b*f^2*g*(exp(4*a)*exp(-4*d))^(1/2)))*exp(4*a - 4*d)^(1/2))/(b^2* 
g^6 - b^2*f^6 + 4*b^2*f*g^5 - 4*b^2*f^5*g + 5*b^2*f^2*g^4 - 5*b^2*f^4*g^2) 
^(1/2) - (2*exp(2*a + d + 3*b*x))/(b*(f + g)*(exp(4*d + 4*b*x)*(f + g)^2 + 
 (f - g)^2 - 2*exp(2*d + 2*b*x)*(f + g)*(f - g))) - (3*exp(2*a - d + b*x)) 
/(b*(f + g)^2*(g - f + exp(2*d + 2*b*x)*(f + g)))
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 801, normalized size of antiderivative = 5.60 \[ \int \frac {e^{2 (a+b x)}}{(g \cosh (d+b x)+f \sinh (d+b x))^3} \, dx=\frac {e^{2 a} \left (-3 e^{4 b x +4 d} \sqrt {-f^{2}+g^{2}}\, \mathit {atan} \left (\frac {e^{b x +d} f +e^{b x +d} g}{\sqrt {-f^{2}+g^{2}}}\right ) f^{2}-6 e^{4 b x +4 d} \sqrt {-f^{2}+g^{2}}\, \mathit {atan} \left (\frac {e^{b x +d} f +e^{b x +d} g}{\sqrt {-f^{2}+g^{2}}}\right ) f g -3 e^{4 b x +4 d} \sqrt {-f^{2}+g^{2}}\, \mathit {atan} \left (\frac {e^{b x +d} f +e^{b x +d} g}{\sqrt {-f^{2}+g^{2}}}\right ) g^{2}+6 e^{2 b x +2 d} \sqrt {-f^{2}+g^{2}}\, \mathit {atan} \left (\frac {e^{b x +d} f +e^{b x +d} g}{\sqrt {-f^{2}+g^{2}}}\right ) f^{2}-6 e^{2 b x +2 d} \sqrt {-f^{2}+g^{2}}\, \mathit {atan} \left (\frac {e^{b x +d} f +e^{b x +d} g}{\sqrt {-f^{2}+g^{2}}}\right ) g^{2}-3 \sqrt {-f^{2}+g^{2}}\, \mathit {atan} \left (\frac {e^{b x +d} f +e^{b x +d} g}{\sqrt {-f^{2}+g^{2}}}\right ) f^{2}+6 \sqrt {-f^{2}+g^{2}}\, \mathit {atan} \left (\frac {e^{b x +d} f +e^{b x +d} g}{\sqrt {-f^{2}+g^{2}}}\right ) f g -3 \sqrt {-f^{2}+g^{2}}\, \mathit {atan} \left (\frac {e^{b x +d} f +e^{b x +d} g}{\sqrt {-f^{2}+g^{2}}}\right ) g^{2}-5 e^{3 b x +3 d} f^{3}-5 e^{3 b x +3 d} f^{2} g +5 e^{3 b x +3 d} f \,g^{2}+5 e^{3 b x +3 d} g^{3}+3 e^{b x +d} f^{3}-3 e^{b x +d} f^{2} g -3 e^{b x +d} f \,g^{2}+3 e^{b x +d} g^{3}\right )}{e^{2 d} b \left (e^{4 b x +4 d} f^{6}+4 e^{4 b x +4 d} f^{5} g +5 e^{4 b x +4 d} f^{4} g^{2}-5 e^{4 b x +4 d} f^{2} g^{4}-4 e^{4 b x +4 d} f \,g^{5}-e^{4 b x +4 d} g^{6}-2 e^{2 b x +2 d} f^{6}-4 e^{2 b x +2 d} f^{5} g +2 e^{2 b x +2 d} f^{4} g^{2}+8 e^{2 b x +2 d} f^{3} g^{3}+2 e^{2 b x +2 d} f^{2} g^{4}-4 e^{2 b x +2 d} f \,g^{5}-2 e^{2 b x +2 d} g^{6}+f^{6}-3 f^{4} g^{2}+3 f^{2} g^{4}-g^{6}\right )} \] Input:

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

Output:

(e**(2*a)*( - 3*e**(4*b*x + 4*d)*sqrt( - f**2 + g**2)*atan((e**(b*x + d)*f 
 + e**(b*x + d)*g)/sqrt( - f**2 + g**2))*f**2 - 6*e**(4*b*x + 4*d)*sqrt( - 
 f**2 + g**2)*atan((e**(b*x + d)*f + e**(b*x + d)*g)/sqrt( - f**2 + g**2)) 
*f*g - 3*e**(4*b*x + 4*d)*sqrt( - f**2 + g**2)*atan((e**(b*x + d)*f + e**( 
b*x + d)*g)/sqrt( - f**2 + g**2))*g**2 + 6*e**(2*b*x + 2*d)*sqrt( - f**2 + 
 g**2)*atan((e**(b*x + d)*f + e**(b*x + d)*g)/sqrt( - f**2 + g**2))*f**2 - 
 6*e**(2*b*x + 2*d)*sqrt( - f**2 + g**2)*atan((e**(b*x + d)*f + e**(b*x + 
d)*g)/sqrt( - f**2 + g**2))*g**2 - 3*sqrt( - f**2 + g**2)*atan((e**(b*x + 
d)*f + e**(b*x + d)*g)/sqrt( - f**2 + g**2))*f**2 + 6*sqrt( - f**2 + g**2) 
*atan((e**(b*x + d)*f + e**(b*x + d)*g)/sqrt( - f**2 + g**2))*f*g - 3*sqrt 
( - f**2 + g**2)*atan((e**(b*x + d)*f + e**(b*x + d)*g)/sqrt( - f**2 + g** 
2))*g**2 - 5*e**(3*b*x + 3*d)*f**3 - 5*e**(3*b*x + 3*d)*f**2*g + 5*e**(3*b 
*x + 3*d)*f*g**2 + 5*e**(3*b*x + 3*d)*g**3 + 3*e**(b*x + d)*f**3 - 3*e**(b 
*x + d)*f**2*g - 3*e**(b*x + d)*f*g**2 + 3*e**(b*x + d)*g**3))/(e**(2*d)*b 
*(e**(4*b*x + 4*d)*f**6 + 4*e**(4*b*x + 4*d)*f**5*g + 5*e**(4*b*x + 4*d)*f 
**4*g**2 - 5*e**(4*b*x + 4*d)*f**2*g**4 - 4*e**(4*b*x + 4*d)*f*g**5 - e**( 
4*b*x + 4*d)*g**6 - 2*e**(2*b*x + 2*d)*f**6 - 4*e**(2*b*x + 2*d)*f**5*g + 
2*e**(2*b*x + 2*d)*f**4*g**2 + 8*e**(2*b*x + 2*d)*f**3*g**3 + 2*e**(2*b*x 
+ 2*d)*f**2*g**4 - 4*e**(2*b*x + 2*d)*f*g**5 - 2*e**(2*b*x + 2*d)*g**6 + f 
**6 - 3*f**4*g**2 + 3*f**2*g**4 - g**6))