\(\int \frac {1}{(a+a \cosh (x)+c \sinh (x))^3} \, dx\) [511]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F(-1)]
Maxima [B] (verification not implemented)
Giac [B] (verification not implemented)
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 12, antiderivative size = 89 \[ \int \frac {1}{(a+a \cosh (x)+c \sinh (x))^3} \, dx=\frac {\left (3 a^2-c^2\right ) \log \left (a+c \tanh \left (\frac {x}{2}\right )\right )}{2 c^5}-\frac {c \cosh (x)+a \sinh (x)}{2 c^2 (a+a \cosh (x)+c \sinh (x))^2}-\frac {3 \left (a c \cosh (x)+a^2 \sinh (x)\right )}{2 c^4 (a+a \cosh (x)+c \sinh (x))} \] Output:

1/2*(3*a^2-c^2)*ln(a+c*tanh(1/2*x))/c^5-1/2*(c*cosh(x)+a*sinh(x))/c^2/(a+a 
*cosh(x)+c*sinh(x))^2-3/2*(a*c*cosh(x)+a^2*sinh(x))/c^4/(a+a*cosh(x)+c*sin 
h(x))
 

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.66 \[ \int \frac {1}{(a+a \cosh (x)+c \sinh (x))^3} \, dx=\frac {4 \left (-3 a^2+c^2\right ) \log \left (\cosh \left (\frac {x}{2}\right )\right )+4 \left (3 a^2-c^2\right ) \log \left (a \cosh \left (\frac {x}{2}\right )+c \sinh \left (\frac {x}{2}\right )\right )-c^2 \text {sech}^2\left (\frac {x}{2}\right )+\frac {(a-c) c^2 (a+c)}{\left (a \cosh \left (\frac {x}{2}\right )+c \sinh \left (\frac {x}{2}\right )\right )^2}+\frac {6 c \left (-a^2+c^2\right ) \sinh \left (\frac {x}{2}\right )}{a \cosh \left (\frac {x}{2}\right )+c \sinh \left (\frac {x}{2}\right )}-6 a c \tanh \left (\frac {x}{2}\right )}{8 c^5} \] Input:

Integrate[(a + a*Cosh[x] + c*Sinh[x])^(-3),x]
 

Output:

(4*(-3*a^2 + c^2)*Log[Cosh[x/2]] + 4*(3*a^2 - c^2)*Log[a*Cosh[x/2] + c*Sin 
h[x/2]] - c^2*Sech[x/2]^2 + ((a - c)*c^2*(a + c))/(a*Cosh[x/2] + c*Sinh[x/ 
2])^2 + (6*c*(-a^2 + c^2)*Sinh[x/2])/(a*Cosh[x/2] + c*Sinh[x/2]) - 6*a*c*T 
anh[x/2])/(8*c^5)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3608, 25, 3042, 3632, 3042, 3603, 16}

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.

\(\displaystyle \int \frac {1}{(a \cosh (x)+a+c \sinh (x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a \cos (i x)+a-i c \sin (i x))^3}dx\)

\(\Big \downarrow \) 3608

\(\displaystyle -\frac {\int -\frac {-\cosh (x) a+2 a-c \sinh (x)}{(\cosh (x) a+a+c \sinh (x))^2}dx}{2 c^2}-\frac {a \sinh (x)+c \cosh (x)}{2 c^2 (a \cosh (x)+a+c \sinh (x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {-\cosh (x) a+2 a-c \sinh (x)}{(\cosh (x) a+a+c \sinh (x))^2}dx}{2 c^2}-\frac {a \sinh (x)+c \cosh (x)}{2 c^2 (a \cosh (x)+a+c \sinh (x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a \sinh (x)+c \cosh (x)}{2 c^2 (a \cosh (x)+a+c \sinh (x))^2}+\frac {\int \frac {-\cos (i x) a+2 a+i c \sin (i x)}{(\cos (i x) a+a-i c \sin (i x))^2}dx}{2 c^2}\)

\(\Big \downarrow \) 3632

\(\displaystyle \frac {-\left (1-\frac {3 a^2}{c^2}\right ) \int \frac {1}{\cosh (x) a+a+c \sinh (x)}dx-\frac {3 \left (a^2 \sinh (x)+a c \cosh (x)\right )}{c^2 (a \cosh (x)+a+c \sinh (x))}}{2 c^2}-\frac {a \sinh (x)+c \cosh (x)}{2 c^2 (a \cosh (x)+a+c \sinh (x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a \sinh (x)+c \cosh (x)}{2 c^2 (a \cosh (x)+a+c \sinh (x))^2}+\frac {-\frac {3 \left (a^2 \sinh (x)+a c \cosh (x)\right )}{c^2 (a \cosh (x)+a+c \sinh (x))}-\left (1-\frac {3 a^2}{c^2}\right ) \int \frac {1}{\cos (i x) a+a-i c \sin (i x)}dx}{2 c^2}\)

\(\Big \downarrow \) 3603

\(\displaystyle \frac {-2 \left (1-\frac {3 a^2}{c^2}\right ) \int \frac {1}{2 a+2 c \tanh \left (\frac {x}{2}\right )}d\tanh \left (\frac {x}{2}\right )-\frac {3 \left (a^2 \sinh (x)+a c \cosh (x)\right )}{c^2 (a \cosh (x)+a+c \sinh (x))}}{2 c^2}-\frac {a \sinh (x)+c \cosh (x)}{2 c^2 (a \cosh (x)+a+c \sinh (x))^2}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {-\frac {\left (1-\frac {3 a^2}{c^2}\right ) \log \left (a+c \tanh \left (\frac {x}{2}\right )\right )}{c}-\frac {3 \left (a^2 \sinh (x)+a c \cosh (x)\right )}{c^2 (a \cosh (x)+a+c \sinh (x))}}{2 c^2}-\frac {a \sinh (x)+c \cosh (x)}{2 c^2 (a \cosh (x)+a+c \sinh (x))^2}\)

Input:

Int[(a + a*Cosh[x] + c*Sinh[x])^(-3),x]
 

Output:

-1/2*(c*Cosh[x] + a*Sinh[x])/(c^2*(a + a*Cosh[x] + c*Sinh[x])^2) + (-(((1 
- (3*a^2)/c^2)*Log[a + c*Tanh[x/2]])/c) - (3*(a*c*Cosh[x] + a^2*Sinh[x]))/ 
(c^2*(a + a*Cosh[x] + c*Sinh[x])))/(2*c^2)
 

Defintions of rubi rules used

rule 16
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3603
Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ 
(-1), x_Symbol] :> Module[{f = FreeFactors[Tan[(d + e*x)/2], x]}, Simp[2*(f 
/e)   Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d + e*x) 
/2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]
 

rule 3608
Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ 
(n_), x_Symbol] :> Simp[((-c)*Cos[d + e*x] + b*Sin[d + e*x])*((a + b*Cos[d 
+ e*x] + c*Sin[d + e*x])^(n + 1)/(e*(n + 1)*(a^2 - b^2 - c^2))), x] + Simp[ 
1/((n + 1)*(a^2 - b^2 - c^2))   Int[(a*(n + 1) - b*(n + 2)*Cos[d + e*x] - c 
*(n + 2)*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1), x], x 
] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && LtQ[n, -1] && 
NeQ[n, -3/2]
 

rule 3632
Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)]) 
/((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)])^2, 
 x_Symbol] :> Simp[(c*B - b*C - (a*C - c*A)*Cos[d + e*x] + (a*B - b*A)*Sin[ 
d + e*x])/(e*(a^2 - b^2 - c^2)*(a + b*Cos[d + e*x] + c*Sin[d + e*x])), x] + 
 Simp[(a*A - b*B - c*C)/(a^2 - b^2 - c^2)   Int[1/(a + b*Cos[d + e*x] + c*S 
in[d + e*x]), x], x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[a^2 - b^2 
 - c^2, 0] && NeQ[a*A - b*B - c*C, 0]
 
Maple [A] (verified)

Time = 5.20 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.16

method result size
default \(-\frac {-\frac {c \tanh \left (\frac {x}{2}\right )^{2}}{2}+3 a \tanh \left (\frac {x}{2}\right )}{4 c^{4}}+\frac {a \left (a^{2}-c^{2}\right )}{c^{5} \left (a +c \tanh \left (\frac {x}{2}\right )\right )}+\frac {\left (6 a^{2}-2 c^{2}\right ) \ln \left (a +c \tanh \left (\frac {x}{2}\right )\right )}{4 c^{5}}-\frac {a^{4}-2 a^{2} c^{2}+c^{4}}{8 c^{5} \left (a +c \tanh \left (\frac {x}{2}\right )\right )^{2}}\) \(103\)
risch \(\frac {3 \,{\mathrm e}^{3 x} a^{3}+3 a^{2} c \,{\mathrm e}^{3 x}-{\mathrm e}^{3 x} a \,c^{2}-{\mathrm e}^{3 x} c^{3}+9 \,{\mathrm e}^{2 x} a^{3}-3 \,{\mathrm e}^{2 x} a \,c^{2}+9 a^{3} {\mathrm e}^{x}-9 \,{\mathrm e}^{x} a^{2} c +{\mathrm e}^{x} a \,c^{2}-{\mathrm e}^{x} c^{3}+3 a^{3}-6 a^{2} c +3 a \,c^{2}}{\left ({\mathrm e}^{2 x} a +c \,{\mathrm e}^{2 x}+2 a \,{\mathrm e}^{x}+a -c \right )^{2} c^{4}}+\frac {3 \ln \left ({\mathrm e}^{x}+\frac {a -c}{a +c}\right ) a^{2}}{2 c^{5}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {a -c}{a +c}\right )}{2 c^{3}}-\frac {3 \ln \left (1+{\mathrm e}^{x}\right ) a^{2}}{2 c^{5}}+\frac {\ln \left (1+{\mathrm e}^{x}\right )}{2 c^{3}}\) \(200\)

Input:

int(1/(a+a*cosh(x)+c*sinh(x))^3,x,method=_RETURNVERBOSE)
 

Output:

-1/4/c^4*(-1/2*c*tanh(1/2*x)^2+3*a*tanh(1/2*x))+a/c^5*(a^2-c^2)/(a+c*tanh( 
1/2*x))+1/4*(6*a^2-2*c^2)/c^5*ln(a+c*tanh(1/2*x))-1/8/c^5*(a^4-2*a^2*c^2+c 
^4)/(a+c*tanh(1/2*x))^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1504 vs. \(2 (81) = 162\).

Time = 0.10 (sec) , antiderivative size = 1504, normalized size of antiderivative = 16.90 \[ \int \frac {1}{(a+a \cosh (x)+c \sinh (x))^3} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+a*cosh(x)+c*sinh(x))^3,x, algorithm="fricas")
 

Output:

1/2*(6*a^3*c - 12*a^2*c^2 + 6*a*c^3 + 2*(3*a^3*c + 3*a^2*c^2 - a*c^3 - c^4 
)*cosh(x)^3 + 2*(3*a^3*c + 3*a^2*c^2 - a*c^3 - c^4)*sinh(x)^3 + 6*(3*a^3*c 
 - a*c^3)*cosh(x)^2 + 6*(3*a^3*c - a*c^3 + (3*a^3*c + 3*a^2*c^2 - a*c^3 - 
c^4)*cosh(x))*sinh(x)^2 + 2*(9*a^3*c - 9*a^2*c^2 + a*c^3 - c^4)*cosh(x) + 
((3*a^4 + 6*a^3*c + 2*a^2*c^2 - 2*a*c^3 - c^4)*cosh(x)^4 + (3*a^4 + 6*a^3* 
c + 2*a^2*c^2 - 2*a*c^3 - c^4)*sinh(x)^4 + 3*a^4 - 6*a^3*c + 2*a^2*c^2 + 2 
*a*c^3 - c^4 + 4*(3*a^4 + 3*a^3*c - a^2*c^2 - a*c^3)*cosh(x)^3 + 4*(3*a^4 
+ 3*a^3*c - a^2*c^2 - a*c^3 + (3*a^4 + 6*a^3*c + 2*a^2*c^2 - 2*a*c^3 - c^4 
)*cosh(x))*sinh(x)^3 + 2*(9*a^4 - 6*a^2*c^2 + c^4)*cosh(x)^2 + 2*(9*a^4 - 
6*a^2*c^2 + c^4 + 3*(3*a^4 + 6*a^3*c + 2*a^2*c^2 - 2*a*c^3 - c^4)*cosh(x)^ 
2 + 6*(3*a^4 + 3*a^3*c - a^2*c^2 - a*c^3)*cosh(x))*sinh(x)^2 + 4*(3*a^4 - 
3*a^3*c - a^2*c^2 + a*c^3)*cosh(x) + 4*(3*a^4 - 3*a^3*c - a^2*c^2 + a*c^3 
+ (3*a^4 + 6*a^3*c + 2*a^2*c^2 - 2*a*c^3 - c^4)*cosh(x)^3 + 3*(3*a^4 + 3*a 
^3*c - a^2*c^2 - a*c^3)*cosh(x)^2 + (9*a^4 - 6*a^2*c^2 + c^4)*cosh(x))*sin 
h(x))*log((a + c)*cosh(x) + (a + c)*sinh(x) + a - c) - ((3*a^4 + 6*a^3*c + 
 2*a^2*c^2 - 2*a*c^3 - c^4)*cosh(x)^4 + (3*a^4 + 6*a^3*c + 2*a^2*c^2 - 2*a 
*c^3 - c^4)*sinh(x)^4 + 3*a^4 - 6*a^3*c + 2*a^2*c^2 + 2*a*c^3 - c^4 + 4*(3 
*a^4 + 3*a^3*c - a^2*c^2 - a*c^3)*cosh(x)^3 + 4*(3*a^4 + 3*a^3*c - a^2*c^2 
 - a*c^3 + (3*a^4 + 6*a^3*c + 2*a^2*c^2 - 2*a*c^3 - c^4)*cosh(x))*sinh(x)^ 
3 + 2*(9*a^4 - 6*a^2*c^2 + c^4)*cosh(x)^2 + 2*(9*a^4 - 6*a^2*c^2 + c^4 ...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+a \cosh (x)+c \sinh (x))^3} \, dx=\text {Timed out} \] Input:

integrate(1/(a+a*cosh(x)+c*sinh(x))**3,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (81) = 162\).

Time = 0.05 (sec) , antiderivative size = 248, normalized size of antiderivative = 2.79 \[ \int \frac {1}{(a+a \cosh (x)+c \sinh (x))^3} \, dx=-\frac {3 \, a^{3} + 6 \, a^{2} c + 3 \, a c^{2} + {\left (9 \, a^{3} + 9 \, a^{2} c + a c^{2} + c^{3}\right )} e^{\left (-x\right )} + 3 \, {\left (3 \, a^{3} - a c^{2}\right )} e^{\left (-2 \, x\right )} + {\left (3 \, a^{3} - 3 \, a^{2} c - a c^{2} + c^{3}\right )} e^{\left (-3 \, x\right )}}{a^{2} c^{4} + 2 \, a c^{5} + c^{6} + 4 \, {\left (a^{2} c^{4} + a c^{5}\right )} e^{\left (-x\right )} + 2 \, {\left (3 \, a^{2} c^{4} - c^{6}\right )} e^{\left (-2 \, x\right )} + 4 \, {\left (a^{2} c^{4} - a c^{5}\right )} e^{\left (-3 \, x\right )} + {\left (a^{2} c^{4} - 2 \, a c^{5} + c^{6}\right )} e^{\left (-4 \, x\right )}} + \frac {{\left (3 \, a^{2} - c^{2}\right )} \log \left (-{\left (a - c\right )} e^{\left (-x\right )} - a - c\right )}{2 \, c^{5}} - \frac {{\left (3 \, a^{2} - c^{2}\right )} \log \left (e^{\left (-x\right )} + 1\right )}{2 \, c^{5}} \] Input:

integrate(1/(a+a*cosh(x)+c*sinh(x))^3,x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

-(3*a^3 + 6*a^2*c + 3*a*c^2 + (9*a^3 + 9*a^2*c + a*c^2 + c^3)*e^(-x) + 3*( 
3*a^3 - a*c^2)*e^(-2*x) + (3*a^3 - 3*a^2*c - a*c^2 + c^3)*e^(-3*x))/(a^2*c 
^4 + 2*a*c^5 + c^6 + 4*(a^2*c^4 + a*c^5)*e^(-x) + 2*(3*a^2*c^4 - c^6)*e^(- 
2*x) + 4*(a^2*c^4 - a*c^5)*e^(-3*x) + (a^2*c^4 - 2*a*c^5 + c^6)*e^(-4*x)) 
+ 1/2*(3*a^2 - c^2)*log(-(a - c)*e^(-x) - a - c)/c^5 - 1/2*(3*a^2 - c^2)*l 
og(e^(-x) + 1)/c^5
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (81) = 162\).

Time = 0.14 (sec) , antiderivative size = 205, normalized size of antiderivative = 2.30 \[ \int \frac {1}{(a+a \cosh (x)+c \sinh (x))^3} \, dx=\frac {{\left (3 \, a^{3} + 3 \, a^{2} c - a c^{2} - c^{3}\right )} \log \left ({\left | a e^{x} + c e^{x} + a - c \right |}\right )}{2 \, {\left (a c^{5} + c^{6}\right )}} - \frac {{\left (3 \, a^{2} - c^{2}\right )} \log \left (e^{x} + 1\right )}{2 \, c^{5}} + \frac {3 \, a^{3} e^{\left (3 \, x\right )} + 3 \, a^{2} c e^{\left (3 \, x\right )} - a c^{2} e^{\left (3 \, x\right )} - c^{3} e^{\left (3 \, x\right )} + 9 \, a^{3} e^{\left (2 \, x\right )} - 3 \, a c^{2} e^{\left (2 \, x\right )} + 9 \, a^{3} e^{x} - 9 \, a^{2} c e^{x} + a c^{2} e^{x} - c^{3} e^{x} + 3 \, a^{3} - 6 \, a^{2} c + 3 \, a c^{2}}{{\left (a e^{\left (2 \, x\right )} + c e^{\left (2 \, x\right )} + 2 \, a e^{x} + a - c\right )}^{2} c^{4}} \] Input:

integrate(1/(a+a*cosh(x)+c*sinh(x))^3,x, algorithm="giac")
 

Output:

1/2*(3*a^3 + 3*a^2*c - a*c^2 - c^3)*log(abs(a*e^x + c*e^x + a - c))/(a*c^5 
 + c^6) - 1/2*(3*a^2 - c^2)*log(e^x + 1)/c^5 + (3*a^3*e^(3*x) + 3*a^2*c*e^ 
(3*x) - a*c^2*e^(3*x) - c^3*e^(3*x) + 9*a^3*e^(2*x) - 3*a*c^2*e^(2*x) + 9* 
a^3*e^x - 9*a^2*c*e^x + a*c^2*e^x - c^3*e^x + 3*a^3 - 6*a^2*c + 3*a*c^2)/( 
(a*e^(2*x) + c*e^(2*x) + 2*a*e^x + a - c)^2*c^4)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+a \cosh (x)+c \sinh (x))^3} \, dx=\int \frac {1}{{\left (a+a\,\mathrm {cosh}\left (x\right )+c\,\mathrm {sinh}\left (x\right )\right )}^3} \,d x \] Input:

int(1/(a + a*cosh(x) + c*sinh(x))^3,x)
 

Output:

int(1/(a + a*cosh(x) + c*sinh(x))^3, x)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 1136, normalized size of antiderivative = 12.76 \[ \int \frac {1}{(a+a \cosh (x)+c \sinh (x))^3} \, dx =\text {Too large to display} \] Input:

int(1/(a+a*cosh(x)+c*sinh(x))^3,x)
 

Output:

( - 6*e**(4*x)*log(e**x + 1)*a**5 - 12*e**(4*x)*log(e**x + 1)*a**4*c - 4*e 
**(4*x)*log(e**x + 1)*a**3*c**2 + 4*e**(4*x)*log(e**x + 1)*a**2*c**3 + 2*e 
**(4*x)*log(e**x + 1)*a*c**4 + 6*e**(4*x)*log(e**x*a + e**x*c + a - c)*a** 
5 + 12*e**(4*x)*log(e**x*a + e**x*c + a - c)*a**4*c + 4*e**(4*x)*log(e**x* 
a + e**x*c + a - c)*a**3*c**2 - 4*e**(4*x)*log(e**x*a + e**x*c + a - c)*a* 
*2*c**3 - 2*e**(4*x)*log(e**x*a + e**x*c + a - c)*a*c**4 - 3*e**(4*x)*a**4 
*c - 6*e**(4*x)*a**3*c**2 - 2*e**(4*x)*a**2*c**3 + 2*e**(4*x)*a*c**4 + e** 
(4*x)*c**5 - 24*e**(3*x)*log(e**x + 1)*a**5 - 24*e**(3*x)*log(e**x + 1)*a* 
*4*c + 8*e**(3*x)*log(e**x + 1)*a**3*c**2 + 8*e**(3*x)*log(e**x + 1)*a**2* 
c**3 + 24*e**(3*x)*log(e**x*a + e**x*c + a - c)*a**5 + 24*e**(3*x)*log(e** 
x*a + e**x*c + a - c)*a**4*c - 8*e**(3*x)*log(e**x*a + e**x*c + a - c)*a** 
3*c**2 - 8*e**(3*x)*log(e**x*a + e**x*c + a - c)*a**2*c**3 - 36*e**(2*x)*l 
og(e**x + 1)*a**5 + 24*e**(2*x)*log(e**x + 1)*a**3*c**2 - 4*e**(2*x)*log(e 
**x + 1)*a*c**4 + 36*e**(2*x)*log(e**x*a + e**x*c + a - c)*a**5 - 24*e**(2 
*x)*log(e**x*a + e**x*c + a - c)*a**3*c**2 + 4*e**(2*x)*log(e**x*a + e**x* 
c + a - c)*a*c**4 + 18*e**(2*x)*a**4*c - 2*e**(2*x)*c**5 - 24*e**x*log(e** 
x + 1)*a**5 + 24*e**x*log(e**x + 1)*a**4*c + 8*e**x*log(e**x + 1)*a**3*c** 
2 - 8*e**x*log(e**x + 1)*a**2*c**3 + 24*e**x*log(e**x*a + e**x*c + a - c)* 
a**5 - 24*e**x*log(e**x*a + e**x*c + a - c)*a**4*c - 8*e**x*log(e**x*a + e 
**x*c + a - c)*a**3*c**2 + 8*e**x*log(e**x*a + e**x*c + a - c)*a**2*c**...