Integrand size = 27, antiderivative size = 104 \[ \int \frac {\text {csch}^2(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {a \arctan (\sinh (c+d x))}{\left (a^2+b^2\right ) d}-\frac {\text {csch}(c+d x)}{a d}+\frac {b \log (\cosh (c+d x))}{\left (a^2+b^2\right ) d}-\frac {b \log (\sinh (c+d x))}{a^2 d}+\frac {b^3 \log (a+b \sinh (c+d x))}{a^2 \left (a^2+b^2\right ) d} \] Output:
-a*arctan(sinh(d*x+c))/(a^2+b^2)/d-csch(d*x+c)/a/d+b*ln(cosh(d*x+c))/(a^2+ b^2)/d-b*ln(sinh(d*x+c))/a^2/d+b^3*ln(a+b*sinh(d*x+c))/a^2/(a^2+b^2)/d
Time = 0.40 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.54 \[ \int \frac {\text {csch}^2(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {b^3 \left (\frac {\text {csch}(c+d x)}{a b^3}+\frac {\log (\sinh (c+d x))}{a^2 b^2}-\frac {\left (b^2+a \sqrt {-b^2}\right ) \log \left (\sqrt {-b^2}-b \sinh (c+d x)\right )}{2 b^4 \left (a^2+b^2\right )}-\frac {\log (a+b \sinh (c+d x))}{a^2 \left (a^2+b^2\right )}-\frac {\left (1+\frac {a}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}+b \sinh (c+d x)\right )}{2 b^2 \left (a^2+b^2\right )}\right )}{d} \] Input:
Integrate[(Csch[c + d*x]^2*Sech[c + d*x])/(a + b*Sinh[c + d*x]),x]
Output:
-((b^3*(Csch[c + d*x]/(a*b^3) + Log[Sinh[c + d*x]]/(a^2*b^2) - ((b^2 + a*S qrt[-b^2])*Log[Sqrt[-b^2] - b*Sinh[c + d*x]])/(2*b^4*(a^2 + b^2)) - Log[a + b*Sinh[c + d*x]]/(a^2*(a^2 + b^2)) - ((1 + a/Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Sinh[c + d*x]])/(2*b^2*(a^2 + b^2))))/d)
Time = 0.39 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3042, 25, 3316, 27, 615, 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.
| \(\displaystyle \int \frac {\text {csch}^2(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {1}{\sin (i c+i d x)^2 \cos (i c+i d x) (a-i b \sin (i c+i d x))}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {1}{\cos (i c+i d x) \sin (i c+i d x)^2 (a-i b \sin (i c+i d x))}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b \int \frac {\text {csch}^2(c+d x)}{(a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )}d(b \sinh (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^3 \int \frac {\text {csch}^2(c+d x)}{b^2 (a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )}d(b \sinh (c+d x))}{d}\) |
| \(\Big \downarrow \) 615 |
| \(\displaystyle \frac {b^3 \int \left (\frac {\text {csch}^2(c+d x)}{a b^4}-\frac {\text {csch}(c+d x)}{a^2 b^3}+\frac {1}{a^2 \left (a^2+b^2\right ) (a+b \sinh (c+d x))}+\frac {b \sinh (c+d x)-a}{b^2 \left (a^2+b^2\right ) \left (\sinh ^2(c+d x) b^2+b^2\right )}\right )d(b \sinh (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^3 \left (-\frac {a \arctan (\sinh (c+d x))}{b^3 \left (a^2+b^2\right )}+\frac {\log \left (b^2 \sinh ^2(c+d x)+b^2\right )}{2 b^2 \left (a^2+b^2\right )}-\frac {\log (b \sinh (c+d x))}{a^2 b^2}+\frac {\log (a+b \sinh (c+d x))}{a^2 \left (a^2+b^2\right )}-\frac {\text {csch}(c+d x)}{a b^3}\right )}{d}\) |
Input:
Int[(Csch[c + d*x]^2*Sech[c + d*x])/(a + b*Sinh[c + d*x]),x]
Output:
(b^3*(-((a*ArcTan[Sinh[c + d*x]])/(b^3*(a^2 + b^2))) - Csch[c + d*x]/(a*b^ 3) - Log[b*Sinh[c + d*x]]/(a^2*b^2) + Log[a + b*Sinh[c + d*x]]/(a^2*(a^2 + b^2)) + Log[b^2 + b^2*Sinh[c + d*x]^2]/(2*b^2*(a^2 + b^2))))/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 2.68 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.35
| method | result | size |
| derivativedivides | \(\frac {\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {1}{2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {2 b \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-4 a \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2}+2 b^{2}}+\frac {b^{3} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{a^{2} \left (a^{2}+b^{2}\right )}}{d}\) | \(140\) |
| default | \(\frac {\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {1}{2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {2 b \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-4 a \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2}+2 b^{2}}+\frac {b^{3} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{a^{2} \left (a^{2}+b^{2}\right )}}{d}\) | \(140\) |
| risch | \(-\frac {2 b \,d^{2} x}{a^{2} d^{2}+b^{2} d^{2}}-\frac {2 b d c}{a^{2} d^{2}+b^{2} d^{2}}-\frac {2 b^{3} x}{a^{2} \left (a^{2}+b^{2}\right )}-\frac {2 b^{3} c}{a^{2} d \left (a^{2}+b^{2}\right )}+\frac {2 b x}{a^{2}}+\frac {2 b c}{a^{2} d}-\frac {2 \,{\mathrm e}^{d x +c}}{d a \left ({\mathrm e}^{2 d x +2 c}-1\right )}+\frac {i \ln \left ({\mathrm e}^{d x +c}-i\right ) a}{\left (a^{2}+b^{2}\right ) d}+\frac {\ln \left ({\mathrm e}^{d x +c}-i\right ) b}{\left (a^{2}+b^{2}\right ) d}-\frac {i \ln \left ({\mathrm e}^{d x +c}+i\right ) a}{\left (a^{2}+b^{2}\right ) d}+\frac {\ln \left ({\mathrm e}^{d x +c}+i\right ) b}{\left (a^{2}+b^{2}\right ) d}+\frac {b^{3} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{a^{2} d \left (a^{2}+b^{2}\right )}-\frac {b \ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{a^{2} d}\) | \(295\) |
Input:
int(csch(d*x+c)^2*sech(d*x+c)/(a+b*sinh(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/2/a*tanh(1/2*d*x+1/2*c)-1/2/a/tanh(1/2*d*x+1/2*c)-1/a^2*b*ln(tanh(1 /2*d*x+1/2*c))+1/2/(a^2+b^2)*(2*b*ln(1+tanh(1/2*d*x+1/2*c)^2)-4*a*arctan(t anh(1/2*d*x+1/2*c)))+b^3/a^2/(a^2+b^2)*ln(tanh(1/2*d*x+1/2*c)^2*a-2*b*tanh (1/2*d*x+1/2*c)-a))
Leaf count of result is larger than twice the leaf count of optimal. 441 vs. \(2 (104) = 208\).
Time = 0.14 (sec) , antiderivative size = 441, normalized size of antiderivative = 4.24 \[ \int \frac {\text {csch}^2(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {2 \, {\left (a^{3} \cosh \left (d x + c\right )^{2} + 2 \, a^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{3} \sinh \left (d x + c\right )^{2} - a^{3}\right )} \arctan \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) + 2 \, {\left (a^{3} + a b^{2}\right )} \cosh \left (d x + c\right ) - {\left (b^{3} \cosh \left (d x + c\right )^{2} + 2 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{3} \sinh \left (d x + c\right )^{2} - b^{3}\right )} \log \left (\frac {2 \, {\left (b \sinh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) - {\left (a^{2} b \cosh \left (d x + c\right )^{2} + 2 \, a^{2} b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{2} b \sinh \left (d x + c\right )^{2} - a^{2} b\right )} \log \left (\frac {2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) - {\left (a^{2} b + b^{3} - {\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )^{2} - 2 \, {\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) - {\left (a^{2} b + b^{3}\right )} \sinh \left (d x + c\right )^{2}\right )} \log \left (\frac {2 \, \sinh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 2 \, {\left (a^{3} + a b^{2}\right )} \sinh \left (d x + c\right )}{{\left (a^{4} + a^{2} b^{2}\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{4} + a^{2} b^{2}\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{4} + a^{2} b^{2}\right )} d \sinh \left (d x + c\right )^{2} - {\left (a^{4} + a^{2} b^{2}\right )} d} \] Input:
integrate(csch(d*x+c)^2*sech(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="fricas ")
Output:
-(2*(a^3*cosh(d*x + c)^2 + 2*a^3*cosh(d*x + c)*sinh(d*x + c) + a^3*sinh(d* x + c)^2 - a^3)*arctan(cosh(d*x + c) + sinh(d*x + c)) + 2*(a^3 + a*b^2)*co sh(d*x + c) - (b^3*cosh(d*x + c)^2 + 2*b^3*cosh(d*x + c)*sinh(d*x + c) + b ^3*sinh(d*x + c)^2 - b^3)*log(2*(b*sinh(d*x + c) + a)/(cosh(d*x + c) - sin h(d*x + c))) - (a^2*b*cosh(d*x + c)^2 + 2*a^2*b*cosh(d*x + c)*sinh(d*x + c ) + a^2*b*sinh(d*x + c)^2 - a^2*b)*log(2*cosh(d*x + c)/(cosh(d*x + c) - si nh(d*x + c))) - (a^2*b + b^3 - (a^2*b + b^3)*cosh(d*x + c)^2 - 2*(a^2*b + b^3)*cosh(d*x + c)*sinh(d*x + c) - (a^2*b + b^3)*sinh(d*x + c)^2)*log(2*si nh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 2*(a^3 + a*b^2)*sinh(d*x + c))/((a^4 + a^2*b^2)*d*cosh(d*x + c)^2 + 2*(a^4 + a^2*b^2)*d*cosh(d*x + c) *sinh(d*x + c) + (a^4 + a^2*b^2)*d*sinh(d*x + c)^2 - (a^4 + a^2*b^2)*d)
Timed out. \[ \int \frac {\text {csch}^2(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(csch(d*x+c)**2*sech(d*x+c)/(a+b*sinh(d*x+c)),x)
Output:
Timed out
Time = 0.14 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.66 \[ \int \frac {\text {csch}^2(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {b^{3} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{4} + a^{2} b^{2}\right )} d} + \frac {2 \, a \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac {b \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac {2 \, e^{\left (-d x - c\right )}}{{\left (a e^{\left (-2 \, d x - 2 \, c\right )} - a\right )} d} - \frac {b \log \left (e^{\left (-d x - c\right )} + 1\right )}{a^{2} d} - \frac {b \log \left (e^{\left (-d x - c\right )} - 1\right )}{a^{2} d} \] Input:
integrate(csch(d*x+c)^2*sech(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="maxima ")
Output:
b^3*log(-2*a*e^(-d*x - c) + b*e^(-2*d*x - 2*c) - b)/((a^4 + a^2*b^2)*d) + 2*a*arctan(e^(-d*x - c))/((a^2 + b^2)*d) + b*log(e^(-2*d*x - 2*c) + 1)/((a ^2 + b^2)*d) + 2*e^(-d*x - c)/((a*e^(-2*d*x - 2*c) - a)*d) - b*log(e^(-d*x - c) + 1)/(a^2*d) - b*log(e^(-d*x - c) - 1)/(a^2*d)
Time = 0.13 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.92 \[ \int \frac {\text {csch}^2(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {2 \, b^{4} \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{a^{4} b + a^{2} b^{3}} - \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} a}{a^{2} + b^{2}} + \frac {b \log \left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}{a^{2} + b^{2}} - \frac {2 \, b \log \left ({\left | e^{\left (d x + c\right )} - e^{\left (-d x - c\right )} \right |}\right )}{a^{2}} + \frac {2 \, {\left (b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} - 2 \, a\right )}}{a^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}}{2 \, d} \] Input:
integrate(csch(d*x+c)^2*sech(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="giac")
Output:
1/2*(2*b^4*log(abs(b*(e^(d*x + c) - e^(-d*x - c)) + 2*a))/(a^4*b + a^2*b^3 ) - (pi + 2*arctan(1/2*(e^(2*d*x + 2*c) - 1)*e^(-d*x - c)))*a/(a^2 + b^2) + b*log((e^(d*x + c) - e^(-d*x - c))^2 + 4)/(a^2 + b^2) - 2*b*log(abs(e^(d *x + c) - e^(-d*x - c)))/a^2 + 2*(b*(e^(d*x + c) - e^(-d*x - c)) - 2*a)/(a ^2*(e^(d*x + c) - e^(-d*x - c))))/d
Time = 4.26 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.37 \[ \int \frac {\text {csch}^2(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\ln \left ({\mathrm {e}}^{c+d\,x}+1{}\mathrm {i}\right )}{b\,d+a\,d\,1{}\mathrm {i}}+\frac {b^3\,\ln \left (2\,a\,{\mathrm {e}}^{c+d\,x}-b+b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{d\,a^4+d\,a^2\,b^2}-\frac {2\,{\mathrm {e}}^{c+d\,x}}{a\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {b\,\ln \left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}{a^2\,d}+\frac {\ln \left (1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{a\,d+b\,d\,1{}\mathrm {i}} \] Input:
int(1/(cosh(c + d*x)*sinh(c + d*x)^2*(a + b*sinh(c + d*x))),x)
Output:
log(exp(c + d*x) + 1i)/(a*d*1i + b*d) + (log(exp(c + d*x)*1i + 1)*1i)/(a*d + b*d*1i) + (b^3*log(2*a*exp(c + d*x) - b + b*exp(2*c + 2*d*x)))/(a^4*d + a^2*b^2*d) - (2*exp(c + d*x))/(a*d*(exp(2*c + 2*d*x) - 1)) - (b*log(exp(2 *c + 2*d*x) - 1))/(a^2*d)
Time = 0.19 (sec) , antiderivative size = 390, normalized size of antiderivative = 3.75 \[ \int \frac {\text {csch}^2(c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {-2 e^{2 d x +2 c} \mathit {atan} \left (e^{d x +c}\right ) a^{3}+2 \mathit {atan} \left (e^{d x +c}\right ) a^{3}+e^{2 d x +2 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) a^{2} b -e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}-1\right ) a^{2} b -e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}-1\right ) b^{3}-e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}+1\right ) a^{2} b -e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}+1\right ) b^{3}+e^{2 d x +2 c} \mathrm {log}\left (e^{2 d x +2 c} b +2 e^{d x +c} a -b \right ) b^{3}-2 e^{d x +c} a^{3}-2 e^{d x +c} a \,b^{2}-\mathrm {log}\left (e^{2 d x +2 c}+1\right ) a^{2} b +\mathrm {log}\left (e^{d x +c}-1\right ) a^{2} b +\mathrm {log}\left (e^{d x +c}-1\right ) b^{3}+\mathrm {log}\left (e^{d x +c}+1\right ) a^{2} b +\mathrm {log}\left (e^{d x +c}+1\right ) b^{3}-\mathrm {log}\left (e^{2 d x +2 c} b +2 e^{d x +c} a -b \right ) b^{3}}{a^{2} d \left (e^{2 d x +2 c} a^{2}+e^{2 d x +2 c} b^{2}-a^{2}-b^{2}\right )} \] Input:
int(csch(d*x+c)^2*sech(d*x+c)/(a+b*sinh(d*x+c)),x)
Output:
( - 2*e**(2*c + 2*d*x)*atan(e**(c + d*x))*a**3 + 2*atan(e**(c + d*x))*a**3 + e**(2*c + 2*d*x)*log(e**(2*c + 2*d*x) + 1)*a**2*b - e**(2*c + 2*d*x)*lo g(e**(c + d*x) - 1)*a**2*b - e**(2*c + 2*d*x)*log(e**(c + d*x) - 1)*b**3 - e**(2*c + 2*d*x)*log(e**(c + d*x) + 1)*a**2*b - e**(2*c + 2*d*x)*log(e**( c + d*x) + 1)*b**3 + e**(2*c + 2*d*x)*log(e**(2*c + 2*d*x)*b + 2*e**(c + d *x)*a - b)*b**3 - 2*e**(c + d*x)*a**3 - 2*e**(c + d*x)*a*b**2 - log(e**(2* c + 2*d*x) + 1)*a**2*b + log(e**(c + d*x) - 1)*a**2*b + log(e**(c + d*x) - 1)*b**3 + log(e**(c + d*x) + 1)*a**2*b + log(e**(c + d*x) + 1)*b**3 - log (e**(2*c + 2*d*x)*b + 2*e**(c + d*x)*a - b)*b**3)/(a**2*d*(e**(2*c + 2*d*x )*a**2 + e**(2*c + 2*d*x)*b**2 - a**2 - b**2))