Integrand size = 13, antiderivative size = 87 \[ \int \frac {\text {sech}^3(x)}{a+b \sinh (x)} \, dx=\frac {a \left (a^2+3 b^2\right ) \arctan (\sinh (x))}{2 \left (a^2+b^2\right )^2}-\frac {b^3 \log (\cosh (x))}{\left (a^2+b^2\right )^2}+\frac {b^3 \log (a+b \sinh (x))}{\left (a^2+b^2\right )^2}+\frac {\text {sech}^2(x) (b+a \sinh (x))}{2 \left (a^2+b^2\right )} \] Output:
1/2*a*(a^2+3*b^2)*arctan(sinh(x))/(a^2+b^2)^2-b^3*ln(cosh(x))/(a^2+b^2)^2+ b^3*ln(a+b*sinh(x))/(a^2+b^2)^2+sech(x)^2*(b+a*sinh(x))/(2*a^2+2*b^2)
Time = 0.13 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.89 \[ \int \frac {\text {sech}^3(x)}{a+b \sinh (x)} \, dx=\frac {2 a \left (a^2+3 b^2\right ) \arctan \left (\tanh \left (\frac {x}{2}\right )\right )+2 b^3 (-\log (\cosh (x))+\log (a+b \sinh (x)))+b \left (a^2+b^2\right ) \text {sech}^2(x)+a \left (a^2+b^2\right ) \text {sech}(x) \tanh (x)}{2 \left (a^2+b^2\right )^2} \] Input:
Integrate[Sech[x]^3/(a + b*Sinh[x]),x]
Output:
(2*a*(a^2 + 3*b^2)*ArcTan[Tanh[x/2]] + 2*b^3*(-Log[Cosh[x]] + Log[a + b*Si nh[x]]) + b*(a^2 + b^2)*Sech[x]^2 + a*(a^2 + b^2)*Sech[x]*Tanh[x])/(2*(a^2 + b^2)^2)
Time = 0.40 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.55, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3042, 3147, 496, 25, 657, 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.
| \(\displaystyle \int \frac {\text {sech}^3(x)}{a+b \sinh (x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (i x)^3 (a-i b \sin (i x))}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle b^3 \int \frac {1}{(a+b \sinh (x)) \left (\sinh ^2(x) b^2+b^2\right )^2}d(b \sinh (x))\) |
| \(\Big \downarrow \) 496 |
| \(\displaystyle b^3 \left (\frac {a b \sinh (x)+b^2}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \sinh ^2(x)+b^2\right )}-\frac {\int -\frac {a^2+b \sinh (x) a+2 b^2}{(a+b \sinh (x)) \left (\sinh ^2(x) b^2+b^2\right )}d(b \sinh (x))}{2 b^2 \left (a^2+b^2\right )}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle b^3 \left (\frac {\int \frac {a^2+b \sinh (x) a+2 b^2}{(a+b \sinh (x)) \left (\sinh ^2(x) b^2+b^2\right )}d(b \sinh (x))}{2 b^2 \left (a^2+b^2\right )}+\frac {a b \sinh (x)+b^2}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \sinh ^2(x)+b^2\right )}\right )\) |
| \(\Big \downarrow \) 657 |
| \(\displaystyle b^3 \left (\frac {\int \left (\frac {2 b^2}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {a^3+3 b^2 a-2 b^3 \sinh (x)}{\left (a^2+b^2\right ) \left (\sinh ^2(x) b^2+b^2\right )}\right )d(b \sinh (x))}{2 b^2 \left (a^2+b^2\right )}+\frac {a b \sinh (x)+b^2}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \sinh ^2(x)+b^2\right )}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle b^3 \left (\frac {\frac {a \left (a^2+3 b^2\right ) \arctan (\sinh (x))}{b \left (a^2+b^2\right )}-\frac {b^2 \log \left (b^2 \sinh ^2(x)+b^2\right )}{a^2+b^2}+\frac {2 b^2 \log (a+b \sinh (x))}{a^2+b^2}}{2 b^2 \left (a^2+b^2\right )}+\frac {a b \sinh (x)+b^2}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \sinh ^2(x)+b^2\right )}\right )\) |
Input:
Int[Sech[x]^3/(a + b*Sinh[x]),x]
Output:
b^3*(((a*(a^2 + 3*b^2)*ArcTan[Sinh[x]])/(b*(a^2 + b^2)) + (2*b^2*Log[a + b *Sinh[x]])/(a^2 + b^2) - (b^2*Log[b^2 + b^2*Sinh[x]^2])/(a^2 + b^2))/(2*b^ 2*(a^2 + b^2)) + (b^2 + a*b*Sinh[x])/(2*b^2*(a^2 + b^2)*(b^2 + b^2*Sinh[x] ^2)))
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-(a*d + b*c*x))*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1)*(b*c^2 + a*d^2))), x] + Simp[1/(2*a*(p + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*Simp[b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3) + b*c*d*(n + 2 *p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && IntQuad raticQ[a, 0, b, c, d, n, p, x]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 16.36 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.85
| method | result | size |
| default | \(\frac {b^{3} \ln \left (\tanh \left (\frac {x}{2}\right )^{2} a -2 b \tanh \left (\frac {x}{2}\right )-a \right )}{a^{4}+2 a^{2} b^{2}+b^{4}}+\frac {\frac {2 \left (\left (-\frac {1}{2} a^{3}-\frac {1}{2} a \,b^{2}\right ) \tanh \left (\frac {x}{2}\right )^{3}+\left (-a^{2} b -b^{3}\right ) \tanh \left (\frac {x}{2}\right )^{2}+\left (\frac {1}{2} a^{3}+\frac {1}{2} a \,b^{2}\right ) \tanh \left (\frac {x}{2}\right )\right )}{\left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )^{2}}-b^{3} \ln \left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )+\left (a^{3}+3 a \,b^{2}\right ) \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{a^{4}+2 a^{2} b^{2}+b^{4}}\) | \(161\) |
| risch | \(\frac {{\mathrm e}^{x} \left (a \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} b -a \right )}{\left ({\mathrm e}^{2 x}+1\right )^{2} \left (a^{2}+b^{2}\right )}+\frac {i \ln \left ({\mathrm e}^{x}+i\right ) a^{3}}{2 a^{4}+4 a^{2} b^{2}+2 b^{4}}+\frac {3 i \ln \left ({\mathrm e}^{x}+i\right ) a \,b^{2}}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {\ln \left ({\mathrm e}^{x}+i\right ) b^{3}}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {i \ln \left ({\mathrm e}^{x}-i\right ) a^{3}}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {3 i \ln \left ({\mathrm e}^{x}-i\right ) a \,b^{2}}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {\ln \left ({\mathrm e}^{x}-i\right ) b^{3}}{a^{4}+2 a^{2} b^{2}+b^{4}}+\frac {b^{3} \ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}-1\right )}{a^{4}+2 a^{2} b^{2}+b^{4}}\) | \(247\) |
Input:
int(sech(x)^3/(a+b*sinh(x)),x,method=_RETURNVERBOSE)
Output:
b^3/(a^4+2*a^2*b^2+b^4)*ln(tanh(1/2*x)^2*a-2*b*tanh(1/2*x)-a)+2/(a^4+2*a^2 *b^2+b^4)*(((-1/2*a^3-1/2*a*b^2)*tanh(1/2*x)^3+(-a^2*b-b^3)*tanh(1/2*x)^2+ (1/2*a^3+1/2*a*b^2)*tanh(1/2*x))/(1+tanh(1/2*x)^2)^2-1/2*b^3*ln(1+tanh(1/2 *x)^2)+1/2*(a^3+3*a*b^2)*arctan(tanh(1/2*x)))
Leaf count of result is larger than twice the leaf count of optimal. 652 vs. \(2 (83) = 166\).
Time = 0.11 (sec) , antiderivative size = 652, normalized size of antiderivative = 7.49 \[ \int \frac {\text {sech}^3(x)}{a+b \sinh (x)} \, dx =\text {Too large to display} \] Input:
integrate(sech(x)^3/(a+b*sinh(x)),x, algorithm="fricas")
Output:
((a^3 + a*b^2)*cosh(x)^3 + (a^3 + a*b^2)*sinh(x)^3 + 2*(a^2*b + b^3)*cosh( x)^2 + (2*a^2*b + 2*b^3 + 3*(a^3 + a*b^2)*cosh(x))*sinh(x)^2 + ((a^3 + 3*a *b^2)*cosh(x)^4 + 4*(a^3 + 3*a*b^2)*cosh(x)*sinh(x)^3 + (a^3 + 3*a*b^2)*si nh(x)^4 + a^3 + 3*a*b^2 + 2*(a^3 + 3*a*b^2)*cosh(x)^2 + 2*(a^3 + 3*a*b^2 + 3*(a^3 + 3*a*b^2)*cosh(x)^2)*sinh(x)^2 + 4*((a^3 + 3*a*b^2)*cosh(x)^3 + ( a^3 + 3*a*b^2)*cosh(x))*sinh(x))*arctan(cosh(x) + sinh(x)) - (a^3 + a*b^2) *cosh(x) + (b^3*cosh(x)^4 + 4*b^3*cosh(x)*sinh(x)^3 + b^3*sinh(x)^4 + 2*b^ 3*cosh(x)^2 + b^3 + 2*(3*b^3*cosh(x)^2 + b^3)*sinh(x)^2 + 4*(b^3*cosh(x)^3 + b^3*cosh(x))*sinh(x))*log(2*(b*sinh(x) + a)/(cosh(x) - sinh(x))) - (b^3 *cosh(x)^4 + 4*b^3*cosh(x)*sinh(x)^3 + b^3*sinh(x)^4 + 2*b^3*cosh(x)^2 + b ^3 + 2*(3*b^3*cosh(x)^2 + b^3)*sinh(x)^2 + 4*(b^3*cosh(x)^3 + b^3*cosh(x)) *sinh(x))*log(2*cosh(x)/(cosh(x) - sinh(x))) - (a^3 + a*b^2 - 3*(a^3 + a*b ^2)*cosh(x)^2 - 4*(a^2*b + b^3)*cosh(x))*sinh(x))/((a^4 + 2*a^2*b^2 + b^4) *cosh(x)^4 + 4*(a^4 + 2*a^2*b^2 + b^4)*cosh(x)*sinh(x)^3 + (a^4 + 2*a^2*b^ 2 + b^4)*sinh(x)^4 + a^4 + 2*a^2*b^2 + b^4 + 2*(a^4 + 2*a^2*b^2 + b^4)*cos h(x)^2 + 2*(a^4 + 2*a^2*b^2 + b^4 + 3*(a^4 + 2*a^2*b^2 + b^4)*cosh(x)^2)*s inh(x)^2 + 4*((a^4 + 2*a^2*b^2 + b^4)*cosh(x)^3 + (a^4 + 2*a^2*b^2 + b^4)* cosh(x))*sinh(x))
\[ \int \frac {\text {sech}^3(x)}{a+b \sinh (x)} \, dx=\int \frac {\operatorname {sech}^{3}{\left (x \right )}}{a + b \sinh {\left (x \right )}}\, dx \] Input:
integrate(sech(x)**3/(a+b*sinh(x)),x)
Output:
Integral(sech(x)**3/(a + b*sinh(x)), x)
Time = 0.14 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.83 \[ \int \frac {\text {sech}^3(x)}{a+b \sinh (x)} \, dx=\frac {b^{3} \log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {b^{3} \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (a^{3} + 3 \, a b^{2}\right )} \arctan \left (e^{\left (-x\right )}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {a e^{\left (-x\right )} + 2 \, b e^{\left (-2 \, x\right )} - a e^{\left (-3 \, x\right )}}{a^{2} + b^{2} + 2 \, {\left (a^{2} + b^{2}\right )} e^{\left (-2 \, x\right )} + {\left (a^{2} + b^{2}\right )} e^{\left (-4 \, x\right )}} \] Input:
integrate(sech(x)^3/(a+b*sinh(x)),x, algorithm="maxima")
Output:
b^3*log(-2*a*e^(-x) + b*e^(-2*x) - b)/(a^4 + 2*a^2*b^2 + b^4) - b^3*log(e^ (-2*x) + 1)/(a^4 + 2*a^2*b^2 + b^4) - (a^3 + 3*a*b^2)*arctan(e^(-x))/(a^4 + 2*a^2*b^2 + b^4) + (a*e^(-x) + 2*b*e^(-2*x) - a*e^(-3*x))/(a^2 + b^2 + 2 *(a^2 + b^2)*e^(-2*x) + (a^2 + b^2)*e^(-4*x))
Leaf count of result is larger than twice the leaf count of optimal. 214 vs. \(2 (83) = 166\).
Time = 0.14 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.46 \[ \int \frac {\text {sech}^3(x)}{a+b \sinh (x)} \, dx=\frac {b^{4} \log \left ({\left | -b {\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} - \frac {b^{3} \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} + \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} {\left (a^{3} + 3 \, a b^{2}\right )}}{4 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} + \frac {b^{3} {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} - 2 \, a^{3} {\left (e^{\left (-x\right )} - e^{x}\right )} - 2 \, a b^{2} {\left (e^{\left (-x\right )} - e^{x}\right )} + 4 \, a^{2} b + 8 \, b^{3}}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}} \] Input:
integrate(sech(x)^3/(a+b*sinh(x)),x, algorithm="giac")
Output:
b^4*log(abs(-b*(e^(-x) - e^x) + 2*a))/(a^4*b + 2*a^2*b^3 + b^5) - 1/2*b^3* log((e^(-x) - e^x)^2 + 4)/(a^4 + 2*a^2*b^2 + b^4) + 1/4*(pi + 2*arctan(1/2 *(e^(2*x) - 1)*e^(-x)))*(a^3 + 3*a*b^2)/(a^4 + 2*a^2*b^2 + b^4) + 1/2*(b^3 *(e^(-x) - e^x)^2 - 2*a^3*(e^(-x) - e^x) - 2*a*b^2*(e^(-x) - e^x) + 4*a^2* b + 8*b^3)/((a^4 + 2*a^2*b^2 + b^4)*((e^(-x) - e^x)^2 + 4))
Time = 3.18 (sec) , antiderivative size = 291, normalized size of antiderivative = 3.34 \[ \int \frac {\text {sech}^3(x)}{a+b \sinh (x)} \, dx=\frac {\frac {2\,\left (a^2\,b+b^3\right )}{{\left (a^2+b^2\right )}^2}+\frac {{\mathrm {e}}^x\,\left (a^3+a\,b^2\right )}{{\left (a^2+b^2\right )}^2}}{{\mathrm {e}}^{2\,x}+1}-\frac {\frac {2\,b}{a^2+b^2}+\frac {2\,a\,{\mathrm {e}}^x}{a^2+b^2}}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}-\frac {\ln \left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )\,\left (a+b\,2{}\mathrm {i}\right )}{2\,\left (-a^2\,1{}\mathrm {i}+2\,a\,b+b^2\,1{}\mathrm {i}\right )}+\frac {b^3\,\ln \left (16\,b^7\,{\mathrm {e}}^{2\,x}-a^6\,b-16\,b^7-9\,a^2\,b^5-6\,a^4\,b^3+2\,a^7\,{\mathrm {e}}^x+9\,a^2\,b^5\,{\mathrm {e}}^{2\,x}+6\,a^4\,b^3\,{\mathrm {e}}^{2\,x}+32\,a\,b^6\,{\mathrm {e}}^x+a^6\,b\,{\mathrm {e}}^{2\,x}+18\,a^3\,b^4\,{\mathrm {e}}^x+12\,a^5\,b^2\,{\mathrm {e}}^x\right )}{a^4+2\,a^2\,b^2+b^4}-\frac {\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,\left (2\,b+a\,1{}\mathrm {i}\right )}{2\,\left (-a^2+a\,b\,2{}\mathrm {i}+b^2\right )} \] Input:
int(1/(cosh(x)^3*(a + b*sinh(x))),x)
Output:
((2*(a^2*b + b^3))/(a^2 + b^2)^2 + (exp(x)*(a*b^2 + a^3))/(a^2 + b^2)^2)/( exp(2*x) + 1) - ((2*b)/(a^2 + b^2) + (2*a*exp(x))/(a^2 + b^2))/(2*exp(2*x) + exp(4*x) + 1) - (log(exp(x)*1i + 1)*(a + b*2i))/(2*(2*a*b - a^2*1i + b^ 2*1i)) + (b^3*log(16*b^7*exp(2*x) - a^6*b - 16*b^7 - 9*a^2*b^5 - 6*a^4*b^3 + 2*a^7*exp(x) + 9*a^2*b^5*exp(2*x) + 6*a^4*b^3*exp(2*x) + 32*a*b^6*exp(x ) + a^6*b*exp(2*x) + 18*a^3*b^4*exp(x) + 12*a^5*b^2*exp(x)))/(a^4 + b^4 + 2*a^2*b^2) - (log(exp(x) + 1i)*(a*1i + 2*b))/(2*(a*b*2i - a^2 + b^2))
Time = 0.16 (sec) , antiderivative size = 352, normalized size of antiderivative = 4.05 \[ \int \frac {\text {sech}^3(x)}{a+b \sinh (x)} \, dx=\frac {e^{4 x} \mathit {atan} \left (e^{x}\right ) a^{3}+3 e^{4 x} \mathit {atan} \left (e^{x}\right ) a \,b^{2}+2 e^{2 x} \mathit {atan} \left (e^{x}\right ) a^{3}+6 e^{2 x} \mathit {atan} \left (e^{x}\right ) a \,b^{2}+\mathit {atan} \left (e^{x}\right ) a^{3}+3 \mathit {atan} \left (e^{x}\right ) a \,b^{2}-e^{4 x} \mathrm {log}\left (e^{2 x}+1\right ) b^{3}+e^{4 x} \mathrm {log}\left (e^{2 x} b +2 e^{x} a -b \right ) b^{3}-e^{4 x} a^{2} b -e^{4 x} b^{3}+e^{3 x} a^{3}+e^{3 x} a \,b^{2}-2 e^{2 x} \mathrm {log}\left (e^{2 x}+1\right ) b^{3}+2 e^{2 x} \mathrm {log}\left (e^{2 x} b +2 e^{x} a -b \right ) b^{3}-e^{x} a^{3}-e^{x} a \,b^{2}-\mathrm {log}\left (e^{2 x}+1\right ) b^{3}+\mathrm {log}\left (e^{2 x} b +2 e^{x} a -b \right ) b^{3}-a^{2} b -b^{3}}{e^{4 x} a^{4}+2 e^{4 x} a^{2} b^{2}+e^{4 x} b^{4}+2 e^{2 x} a^{4}+4 e^{2 x} a^{2} b^{2}+2 e^{2 x} b^{4}+a^{4}+2 a^{2} b^{2}+b^{4}} \] Input:
int(sech(x)^3/(a+b*sinh(x)),x)
Output:
(e**(4*x)*atan(e**x)*a**3 + 3*e**(4*x)*atan(e**x)*a*b**2 + 2*e**(2*x)*atan (e**x)*a**3 + 6*e**(2*x)*atan(e**x)*a*b**2 + atan(e**x)*a**3 + 3*atan(e**x )*a*b**2 - e**(4*x)*log(e**(2*x) + 1)*b**3 + e**(4*x)*log(e**(2*x)*b + 2*e **x*a - b)*b**3 - e**(4*x)*a**2*b - e**(4*x)*b**3 + e**(3*x)*a**3 + e**(3* x)*a*b**2 - 2*e**(2*x)*log(e**(2*x) + 1)*b**3 + 2*e**(2*x)*log(e**(2*x)*b + 2*e**x*a - b)*b**3 - e**x*a**3 - e**x*a*b**2 - log(e**(2*x) + 1)*b**3 + log(e**(2*x)*b + 2*e**x*a - b)*b**3 - a**2*b - b**3)/(e**(4*x)*a**4 + 2*e* *(4*x)*a**2*b**2 + e**(4*x)*b**4 + 2*e**(2*x)*a**4 + 4*e**(2*x)*a**2*b**2 + 2*e**(2*x)*b**4 + a**4 + 2*a**2*b**2 + b**4)