\(\int \frac {\cosh (x) \sinh ^2(x)}{a \cosh (x)+b \sinh (x)} \, dx\) [467]

Optimal result

Integrand size = 18, antiderivative size = 102 \[ \int \frac {\cosh (x) \sinh ^2(x)}{a \cosh (x)+b \sinh (x)} \, dx=-\frac {a b^2 x}{\left (a^2-b^2\right )^2}-\frac {a x}{2 \left (a^2-b^2\right )}+\frac {a^2 b \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^2}+\frac {a \cosh (x) \sinh (x)}{2 \left (a^2-b^2\right )}-\frac {b \sinh ^2(x)}{2 \left (a^2-b^2\right )} \] Output:

-a*b^2*x/(a^2-b^2)^2-a*x/(2*a^2-2*b^2)+a^2*b*ln(a*cosh(x)+b*sinh(x))/(a^2- 
b^2)^2+a*cosh(x)*sinh(x)/(2*a^2-2*b^2)-b*sinh(x)^2/(2*a^2-2*b^2)
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.72 \[ \int \frac {\cosh (x) \sinh ^2(x)}{a \cosh (x)+b \sinh (x)} \, dx=\frac {\left (-a^2 b+b^3\right ) \cosh (2 x)+a \left (-2 \left (a^2+b^2\right ) x+4 a b \log (a \cosh (x)+b \sinh (x))+\left (a^2-b^2\right ) \sinh (2 x)\right )}{4 (a-b)^2 (a+b)^2} \] Input:

Integrate[(Cosh[x]*Sinh[x]^2)/(a*Cosh[x] + b*Sinh[x]),x]
 

Output:

((-(a^2*b) + b^3)*Cosh[2*x] + a*(-2*(a^2 + b^2)*x + 4*a*b*Log[a*Cosh[x] + 
b*Sinh[x]] + (a^2 - b^2)*Sinh[2*x]))/(4*(a - b)^2*(a + b)^2)
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.65 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.08, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.889, Rules used = {3042, 25, 3588, 25, 26, 3042, 25, 26, 3044, 15, 3115, 24, 3576, 26, 3042, 3612}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {\sin (i x)^2 \cos (i x)}{a \cos (i x)-i b \sin (i x)}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {\cos (i x) \sin (i x)^2}{a \cos (i x)-i b \sin (i x)}dx\)

\(\Big \downarrow \) 3588

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3044

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

\(\Big \downarrow \) 15

\(\displaystyle -\frac {a \int \sin (i x)^2dx}{a^2-b^2}-\frac {i a b \int \frac {\sin (i x)}{a \cos (i x)-i b \sin (i x)}dx}{a^2-b^2}-\frac {b \sinh ^2(x)}{2 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 3576

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3612

\(\displaystyle -\frac {b \sinh ^2(x)}{2 \left (a^2-b^2\right )}-\frac {a \left (\frac {x}{2}-\frac {1}{2} \sinh (x) \cosh (x)\right )}{a^2-b^2}-\frac {i a b \left (\frac {i a \log (a \cosh (x)+b \sinh (x))}{a^2-b^2}-\frac {i b x}{a^2-b^2}\right )}{a^2-b^2}\)

Input:

Int[(Cosh[x]*Sinh[x]^2)/(a*Cosh[x] + b*Sinh[x]),x]
 

Output:

((-I)*a*b*(((-I)*b*x)/(a^2 - b^2) + (I*a*Log[a*Cosh[x] + b*Sinh[x]])/(a^2 
- b^2)))/(a^2 - b^2) - (b*Sinh[x]^2)/(2*(a^2 - b^2)) - (a*(x/2 - (Cosh[x]* 
Sinh[x])/2))/(a^2 - b^2)
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 25

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

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 3042

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

rule 3044

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ 
Symbol] :> Simp[1/(a*f)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a 
*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&  !(I 
ntegerQ[(m - 1)/2] && LtQ[0, m, n])
 

rule 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3576

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

rule 3588

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

rule 3612

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

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.96

method	result	size
risch	\(-\frac {x a}{2 \left (a +b \right )^{2}}+\frac {{\mathrm e}^{2 x}}{8 a +8 b}-\frac {{\mathrm e}^{-2 x}}{8 \left (a -b \right )}-\frac {2 a^{2} b x}{a^{4}-2 b^{2} a^{2}+b^{4}}+\frac {a^{2} b \ln \left ({\mathrm e}^{2 x}+\frac {a -b}{a +b}\right )}{a^{4}-2 b^{2} a^{2}+b^{4}}\)	\(98\)
default	\(\frac {4}{\left (8 a +8 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {8}{\left (16 a +16 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 \left (a +b \right )^{2}}+\frac {a^{2} b \ln \left (a +2 b \tanh \left (\frac {x}{2}\right )+a \tanh \left (\frac {x}{2}\right )^{2}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {4}{\left (8 a -8 b \right ) \left (1+\tanh \left (\frac {x}{2}\right )\right )^{2}}+\frac {8}{\left (16 a -16 b \right ) \left (1+\tanh \left (\frac {x}{2}\right )\right )}-\frac {a \ln \left (1+\tanh \left (\frac {x}{2}\right )\right )}{2 \left (a -b \right )^{2}}\)	\(145\)

Input:

int(cosh(x)*sinh(x)^2/(a*cosh(x)+b*sinh(x)),x,method=_RETURNVERBOSE)
 

Output:

-1/2*x/(a+b)^2*a+1/8/(a+b)*exp(2*x)-1/8/(a-b)*exp(-2*x)-2*a^2*b/(a^4-2*a^2 
*b^2+b^4)*x+a^2*b/(a^4-2*a^2*b^2+b^4)*ln(exp(2*x)+(a-b)/(a+b))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 334 vs. \(2 (96) = 192\).

Time = 0.11 (sec) , antiderivative size = 334, normalized size of antiderivative = 3.27 \[ \int \frac {\cosh (x) \sinh ^2(x)}{a \cosh (x)+b \sinh (x)} \, dx=\frac {{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right )^{4} + 4 \, {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{3} + {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \sinh \left (x\right )^{4} - 4 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} x \cosh \left (x\right )^{2} - a^{3} - a^{2} b + a b^{2} + b^{3} + 2 \, {\left (3 \, {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right )^{2} - 2 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} x\right )} \sinh \left (x\right )^{2} + 8 \, {\left (a^{2} b \cosh \left (x\right )^{2} + 2 \, a^{2} b \cosh \left (x\right ) \sinh \left (x\right ) + a^{2} b \sinh \left (x\right )^{2}\right )} \log \left (\frac {2 \, {\left (a \cosh \left (x\right ) + b \sinh \left (x\right )\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 4 \, {\left ({\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right )^{3} - 2 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} x \cosh \left (x\right )\right )} \sinh \left (x\right )}{8 \, {\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )^{2}\right )}} \] Input:

integrate(cosh(x)*sinh(x)^2/(a*cosh(x)+b*sinh(x)),x, algorithm="fricas")
 

Output:

1/8*((a^3 - a^2*b - a*b^2 + b^3)*cosh(x)^4 + 4*(a^3 - a^2*b - a*b^2 + b^3) 
*cosh(x)*sinh(x)^3 + (a^3 - a^2*b - a*b^2 + b^3)*sinh(x)^4 - 4*(a^3 + 2*a^ 
2*b + a*b^2)*x*cosh(x)^2 - a^3 - a^2*b + a*b^2 + b^3 + 2*(3*(a^3 - a^2*b - 
 a*b^2 + b^3)*cosh(x)^2 - 2*(a^3 + 2*a^2*b + a*b^2)*x)*sinh(x)^2 + 8*(a^2* 
b*cosh(x)^2 + 2*a^2*b*cosh(x)*sinh(x) + a^2*b*sinh(x)^2)*log(2*(a*cosh(x) 
+ b*sinh(x))/(cosh(x) - sinh(x))) + 4*((a^3 - a^2*b - a*b^2 + b^3)*cosh(x) 
^3 - 2*(a^3 + 2*a^2*b + a*b^2)*x*cosh(x))*sinh(x))/((a^4 - 2*a^2*b^2 + b^4 
)*cosh(x)^2 + 2*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)*sinh(x) + (a^4 - 2*a^2*b^2 
 + b^4)*sinh(x)^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cosh (x) \sinh ^2(x)}{a \cosh (x)+b \sinh (x)} \, dx=\text {Timed out} \] Input:

integrate(cosh(x)*sinh(x)**2/(a*cosh(x)+b*sinh(x)),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81 \[ \int \frac {\cosh (x) \sinh ^2(x)}{a \cosh (x)+b \sinh (x)} \, dx=\frac {a^{2} b \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {a x}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {e^{\left (2 \, x\right )}}{8 \, {\left (a + b\right )}} - \frac {e^{\left (-2 \, x\right )}}{8 \, {\left (a - b\right )}} \] Input:

integrate(cosh(x)*sinh(x)^2/(a*cosh(x)+b*sinh(x)),x, algorithm="maxima")
 

Output:

a^2*b*log(-(a - b)*e^(-2*x) - a - b)/(a^4 - 2*a^2*b^2 + b^4) - 1/2*a*x/(a^ 
2 + 2*a*b + b^2) + 1/8*e^(2*x)/(a + b) - 1/8*e^(-2*x)/(a - b)
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.99 \[ \int \frac {\cosh (x) \sinh ^2(x)}{a \cosh (x)+b \sinh (x)} \, dx=\frac {a^{2} b \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {a x}{2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac {{\left (2 \, a e^{\left (2 \, x\right )} - a + b\right )} e^{\left (-2 \, x\right )}}{8 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac {e^{\left (2 \, x\right )}}{8 \, {\left (a + b\right )}} \] Input:

integrate(cosh(x)*sinh(x)^2/(a*cosh(x)+b*sinh(x)),x, algorithm="giac")
 

Output:

a^2*b*log(abs(a*e^(2*x) + b*e^(2*x) + a - b))/(a^4 - 2*a^2*b^2 + b^4) - 1/ 
2*a*x/(a^2 - 2*a*b + b^2) + 1/8*(2*a*e^(2*x) - a + b)*e^(-2*x)/(a^2 - 2*a* 
b + b^2) + 1/8*e^(2*x)/(a + b)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 1.01 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.79 \[ \int \frac {\cosh (x) \sinh ^2(x)}{a \cosh (x)+b \sinh (x)} \, dx=\frac {{\mathrm {e}}^{2\,x}}{8\,a+8\,b}-\frac {{\mathrm {e}}^{-2\,x}}{8\,a-8\,b}-\frac {a\,x}{2\,{\left (a-b\right )}^2}+\frac {a^2\,b\,\ln \left (a-b+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a^4-2\,a^2\,b^2+b^4} \] Input:

int((cosh(x)*sinh(x)^2)/(a*cosh(x) + b*sinh(x)),x)
 

Output:

exp(2*x)/(8*a + 8*b) - exp(-2*x)/(8*a - 8*b) - (a*x)/(2*(a - b)^2) + (a^2* 
b*log(a - b + a*exp(2*x) + b*exp(2*x)))/(a^4 + b^4 - 2*a^2*b^2)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.49 \[ \int \frac {\cosh (x) \sinh ^2(x)}{a \cosh (x)+b \sinh (x)} \, dx=\frac {e^{4 x} a^{3}-e^{4 x} a^{2} b -e^{4 x} a \,b^{2}+e^{4 x} b^{3}+8 e^{2 x} \mathrm {log}\left (e^{2 x} a +e^{2 x} b +a -b \right ) a^{2} b -4 e^{2 x} a^{3} x -8 e^{2 x} a^{2} b x -4 e^{2 x} a \,b^{2} x -a^{3}-a^{2} b +a \,b^{2}+b^{3}}{8 e^{2 x} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )} \] Input:

int(cosh(x)*sinh(x)^2/(a*cosh(x)+b*sinh(x)),x)
 

Output:

(e**(4*x)*a**3 - e**(4*x)*a**2*b - e**(4*x)*a*b**2 + e**(4*x)*b**3 + 8*e** 
(2*x)*log(e**(2*x)*a + e**(2*x)*b + a - b)*a**2*b - 4*e**(2*x)*a**3*x - 8* 
e**(2*x)*a**2*b*x - 4*e**(2*x)*a*b**2*x - a**3 - a**2*b + a*b**2 + b**3)/( 
8*e**(2*x)*(a**4 - 2*a**2*b**2 + b**4))