Integrand size = 15, antiderivative size = 228 \[ \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=-\frac {63 b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{4 (b x-\text {arctanh}(\tanh (a+b x)))^{11/2}}+\frac {63 b^2}{4 \sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))^5}+\frac {21 b}{4 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))^4}+\frac {63}{20 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))^3}+\frac {9}{4 b x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}+\frac {7}{4 b^2 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))}-\frac {1}{2 b x^{7/2} \text {arctanh}(\tanh (a+b x))^2}+\frac {7}{4 b^2 x^{9/2} \text {arctanh}(\tanh (a+b x))} \]
-63/4*b^(5/2)*arctanh(b^(1/2)*x^(1/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2))/(b *x-arctanh(tanh(b*x+a)))^(11/2)+21/4*b/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))^ 4+63/20/x^(5/2)/(b*x-arctanh(tanh(b*x+a)))^3+9/4/b/x^(7/2)/(b*x-arctanh(ta nh(b*x+a)))^2+7/4/b^2/x^(9/2)/(b*x-arctanh(tanh(b*x+a)))-1/2/b/x^(7/2)/arc tanh(tanh(b*x+a))^2+7/4/b^2/x^(9/2)/arctanh(tanh(b*x+a))+63/4*b^2/(b*x-arc tanh(tanh(b*x+a)))^5/x^(1/2)
Time = 0.24 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\frac {1}{20} \left (\frac {75 b^3 \sqrt {x}}{(b x-\text {arctanh}(\tanh (a+b x)))^5 \text {arctanh}(\tanh (a+b x))}-\frac {315 b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right )}{(-b x+\text {arctanh}(\tanh (a+b x)))^{11/2}}-\frac {10 b^3 \sqrt {x}}{\text {arctanh}(\tanh (a+b x))^2 (-b x+\text {arctanh}(\tanh (a+b x)))^4}+\frac {8 \left (36 b^2 x^2-7 b x \text {arctanh}(\tanh (a+b x))+\text {arctanh}(\tanh (a+b x))^2\right )}{x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))^5}\right ) \]
((75*b^3*Sqrt[x])/((b*x - ArcTanh[Tanh[a + b*x]])^5*ArcTanh[Tanh[a + b*x]] ) - (315*b^(5/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b *x]]]])/(-(b*x) + ArcTanh[Tanh[a + b*x]])^(11/2) - (10*b^3*Sqrt[x])/(ArcTa nh[Tanh[a + b*x]]^2*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4) + (8*(36*b^2*x^2 - 7*b*x*ArcTanh[Tanh[a + b*x]] + ArcTanh[Tanh[a + b*x]]^2))/(x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^5))/20
Time = 0.69 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.32, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {2599, 2599, 2594, 2594, 2594, 2594, 2594, 2593}
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.
| \(\displaystyle \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))^3} \, dx\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle -\frac {7 \int \frac {1}{x^{9/2} \text {arctanh}(\tanh (a+b x))^2}dx}{4 b}-\frac {1}{2 b x^{7/2} \text {arctanh}(\tanh (a+b x))^2}\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle -\frac {7 \left (-\frac {9 \int \frac {1}{x^{11/2} \text {arctanh}(\tanh (a+b x))}dx}{2 b}-\frac {1}{b x^{9/2} \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {1}{2 b x^{7/2} \text {arctanh}(\tanh (a+b x))^2}\) |
| \(\Big \downarrow \) 2594 |
| \(\displaystyle -\frac {7 \left (-\frac {9 \left (\frac {b \int \frac {1}{x^{9/2} \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{2 b}-\frac {1}{b x^{9/2} \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {1}{2 b x^{7/2} \text {arctanh}(\tanh (a+b x))^2}\) |
| \(\Big \downarrow \) 2594 |
| \(\displaystyle -\frac {7 \left (-\frac {9 \left (\frac {b \left (\frac {b \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{2 b}-\frac {1}{b x^{9/2} \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {1}{2 b x^{7/2} \text {arctanh}(\tanh (a+b x))^2}\) |
| \(\Big \downarrow \) 2594 |
| \(\displaystyle -\frac {7 \left (-\frac {9 \left (\frac {b \left (\frac {b \left (\frac {b \int \frac {1}{x^{5/2} \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{2 b}-\frac {1}{b x^{9/2} \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {1}{2 b x^{7/2} \text {arctanh}(\tanh (a+b x))^2}\) |
| \(\Big \downarrow \) 2594 |
| \(\displaystyle -\frac {7 \left (-\frac {9 \left (\frac {b \left (\frac {b \left (\frac {b \left (\frac {b \int \frac {1}{x^{3/2} \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{2 b}-\frac {1}{b x^{9/2} \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {1}{2 b x^{7/2} \text {arctanh}(\tanh (a+b x))^2}\) |
| \(\Big \downarrow \) 2594 |
| \(\displaystyle -\frac {7 \left (-\frac {9 \left (\frac {b \left (\frac {b \left (\frac {b \left (\frac {b \left (\frac {b \int \frac {1}{\sqrt {x} \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{2 b}-\frac {1}{b x^{9/2} \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {1}{2 b x^{7/2} \text {arctanh}(\tanh (a+b x))^2}\) |
| \(\Big \downarrow \) 2593 |
| \(\displaystyle -\frac {7 \left (-\frac {9 \left (\frac {b \left (\frac {b \left (\frac {b \left (\frac {2}{3 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {b \left (\frac {2}{\sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right )}{(b x-\text {arctanh}(\tanh (a+b x)))^{3/2}}\right )}{b x-\text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {2}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{2 b}-\frac {1}{b x^{9/2} \text {arctanh}(\tanh (a+b x))}\right )}{4 b}-\frac {1}{2 b x^{7/2} \text {arctanh}(\tanh (a+b x))^2}\) |
(-7*((-9*(2/(9*x^(9/2)*(b*x - ArcTanh[Tanh[a + b*x]])) + (b*(2/(7*x^(7/2)* (b*x - ArcTanh[Tanh[a + b*x]])) + (b*(2/(5*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])) + (b*(2/(3*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])) + (b*((-2*Sqrt [b]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^(3/2) + 2/(Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])) ))/(b*x - ArcTanh[Tanh[a + b*x]])))/(b*x - ArcTanh[Tanh[a + b*x]])))/(b*x - ArcTanh[Tanh[a + b*x]])))/(b*x - ArcTanh[Tanh[a + b*x]])))/(2*b) - 1/(b* x^(9/2)*ArcTanh[Tanh[a + b*x]])))/(4*b) - 1/(2*b*x^(7/2)*ArcTanh[Tanh[a + b*x]]^2)
3.3.14.3.1 Defintions of rubi rules used
Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simpli fy[D[v, x]]}, Simp[-2*(ArcTanh[Sqrt[v]/Rt[-(b*u - a*v)/a, 2]]/(a*Rt[-(b*u - a*v)/a, 2])), x] /; NeQ[b*u - a*v, 0] && NegQ[(b*u - a*v)/a]] /; Piecewise LinearQ[u, v, x]
Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[ D[v, x]]}, Simp[v^(n + 1)/((n + 1)*(b*u - a*v)), x] - Simp[a*((n + 1)/((n + 1)*(b*u - a*v))) Int[v^(n + 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; Piecew iseLinearQ[u, v, x] && LtQ[n, -1]
Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Sim plify[D[v, x]]}, Simp[u^(m + 1)*(v^n/(a*(m + 1))), x] - Simp[b*(n/(a*(m + 1 ))) Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n} , x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0 ] && !(ILtQ[m + n, -2] && (FractionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ [n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] && !IntegerQ[m]) || (ILt Q[m, 0] && !IntegerQ[n]))
Time = 0.95 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.69
| method | result | size |
| derivativedivides | \(-\frac {2 b^{3} \left (\frac {\frac {15 b \,x^{\frac {3}{2}}}{8}+\left (\frac {17 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{8}-\frac {17 b x}{8}\right ) \sqrt {x}}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}+\frac {63 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{8 \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{5}}-\frac {2}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} x^{\frac {5}{2}}}-\frac {12 b^{2}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{5} \sqrt {x}}+\frac {2 b}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{4} x^{\frac {3}{2}}}\) | \(158\) |
| default | \(-\frac {2 b^{3} \left (\frac {\frac {15 b \,x^{\frac {3}{2}}}{8}+\left (\frac {17 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{8}-\frac {17 b x}{8}\right ) \sqrt {x}}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}+\frac {63 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{8 \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{5}}-\frac {2}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3} x^{\frac {5}{2}}}-\frac {12 b^{2}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{5} \sqrt {x}}+\frac {2 b}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{4} x^{\frac {3}{2}}}\) | \(158\) |
| risch | \(\text {Expression too large to display}\) | \(36230\) |
-2/(arctanh(tanh(b*x+a))-b*x)^5*b^3*((15/8*b*x^(3/2)+(17/8*arctanh(tanh(b* x+a))-17/8*b*x)*x^(1/2))/arctanh(tanh(b*x+a))^2+63/8/((arctanh(tanh(b*x+a) )-b*x)*b)^(1/2)*arctan(b*x^(1/2)/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2)))-2/ 5/(arctanh(tanh(b*x+a))-b*x)^3/x^(5/2)-12/(arctanh(tanh(b*x+a))-b*x)^5*b^2 /x^(1/2)+2/(arctanh(tanh(b*x+a))-b*x)^4*b/x^(3/2)
Time = 0.28 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.21 \[ \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\left [\frac {315 \, {\left (b^{4} x^{5} + 2 \, a b^{3} x^{4} + a^{2} b^{2} x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) - 2 \, {\left (315 \, b^{4} x^{4} + 525 \, a b^{3} x^{3} + 168 \, a^{2} b^{2} x^{2} - 24 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt {x}}{40 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}}, \frac {315 \, {\left (b^{4} x^{5} + 2 \, a b^{3} x^{4} + a^{2} b^{2} x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (315 \, b^{4} x^{4} + 525 \, a b^{3} x^{3} + 168 \, a^{2} b^{2} x^{2} - 24 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt {x}}{20 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}}\right ] \]
[1/40*(315*(b^4*x^5 + 2*a*b^3*x^4 + a^2*b^2*x^3)*sqrt(-b/a)*log((b*x - 2*a *sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) - 2*(315*b^4*x^4 + 525*a*b^3*x^3 + 168 *a^2*b^2*x^2 - 24*a^3*b*x + 8*a^4)*sqrt(x))/(a^5*b^2*x^5 + 2*a^6*b*x^4 + a ^7*x^3), 1/20*(315*(b^4*x^5 + 2*a*b^3*x^4 + a^2*b^2*x^3)*sqrt(b/a)*arctan( a*sqrt(b/a)/(b*sqrt(x))) - (315*b^4*x^4 + 525*a*b^3*x^3 + 168*a^2*b^2*x^2 - 24*a^3*b*x + 8*a^4)*sqrt(x))/(a^5*b^2*x^5 + 2*a^6*b*x^4 + a^7*x^3)]
\[ \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\int \frac {1}{x^{\frac {7}{2}} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.43 \[ \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=-\frac {315 \, b^{4} x^{4} + 525 \, a b^{3} x^{3} + 168 \, a^{2} b^{2} x^{2} - 24 \, a^{3} b x + 8 \, a^{4}}{20 \, {\left (a^{5} b^{2} x^{\frac {9}{2}} + 2 \, a^{6} b x^{\frac {7}{2}} + a^{7} x^{\frac {5}{2}}\right )}} - \frac {63 \, b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{5}} \]
-1/20*(315*b^4*x^4 + 525*a*b^3*x^3 + 168*a^2*b^2*x^2 - 24*a^3*b*x + 8*a^4) /(a^5*b^2*x^(9/2) + 2*a^6*b*x^(7/2) + a^7*x^(5/2)) - 63/4*b^3*arctan(b*sqr t(x)/sqrt(a*b))/(sqrt(a*b)*a^5)
Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.35 \[ \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=-\frac {63 \, b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{5}} - \frac {15 \, b^{4} x^{\frac {3}{2}} + 17 \, a b^{3} \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{5}} - \frac {2 \, {\left (30 \, b^{2} x^{2} - 5 \, a b x + a^{2}\right )}}{5 \, a^{5} x^{\frac {5}{2}}} \]
-63/4*b^3*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5) - 1/4*(15*b^4*x^(3/2 ) + 17*a*b^3*sqrt(x))/((b*x + a)^2*a^5) - 2/5*(30*b^2*x^2 - 5*a*b*x + a^2) /(a^5*x^(5/2))
Time = 5.14 (sec) , antiderivative size = 2151, normalized size of antiderivative = 9.43 \[ \int \frac {1}{x^{7/2} \text {arctanh}(\tanh (a+b x))^3} \, dx=\text {Too large to display} \]
16/(5*x^(5/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b* x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3) - (x*((((448*b^4)/(3*(2*a*b - b *(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(ex p(2*a)*exp(2*b*x) + 1)) + 2*b*x))*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log( (2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3) - (16*b^3*( 2*b*(3*log(2/(exp(2*a)*exp(2*b*x) + 1)) - 3*log((2*exp(2*a)*exp(2*b*x))/(e xp(2*a)*exp(2*b*x) + 1)) + 6*b*x) - 14*b*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)))/(3*(2 *a*b - b*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + l og(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))*(log(2/(exp(2*a)*exp(2*b*x) + 1) ) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^4))*(l og(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*ex p(2*b*x) + 1)) + 2*b*x))/(2*b) - (112*b^3)/((2*a*b - b*(2*a - log((2*exp(2 *a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b* x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2) + (8*b^2*(2*b*(3*log(2/(exp(2*a )*exp(2*b*x) + 1)) - 3*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 6*b*x) - 14*b*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*ex p(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)))/((2*a*b - b*(2*a - log((2* exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2...