Integrand size = 17, antiderivative size = 84 \[ \int \frac {\tanh ^5(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{5/2}}-\frac {a^2}{3 b^2 (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}+\frac {a (a+2 b)}{b^2 (a+b)^2 \sqrt {a+b \tanh ^2(x)}} \]
arctanh((a+b*tanh(x)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)+a*(a+2*b)/b^2/(a+b) ^2/(a+b*tanh(x)^2)^(1/2)-1/3*a^2/b^2/(a+b)/(a+b*tanh(x)^2)^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.81 \[ \int \frac {\tanh ^5(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\frac {-b^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tanh ^2(x)}{a+b}\right )+(a+b) \left (2 a+b+3 b \tanh ^2(x)\right )}{3 b^2 (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}} \]
(-(b^2*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tanh[x]^2)/(a + b)]) + (a + b)*(2*a + b + 3*b*Tanh[x]^2))/(3*b^2*(a + b)*(a + b*Tanh[x]^2)^(3/2))
Time = 0.36 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {3042, 26, 4153, 26, 354, 98, 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.
| \(\displaystyle \int \frac {\tanh ^5(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {i \tan (i x)^5}{\left (a-b \tan (i x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle -i \int \frac {\tan (i x)^5}{\left (a-b \tan (i x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle -i \int \frac {i \tanh ^5(x)}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^2(x)+a\right )^{5/2}}d\tanh (x)\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \int \frac {\tanh ^5(x)}{\left (1-\tanh ^2(x)\right ) \left (a+b \tanh ^2(x)\right )^{5/2}}d\tanh (x)\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {1}{2} \int \frac {\tanh ^4(x)}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^2(x)+a\right )^{5/2}}d\tanh ^2(x)\) |
| \(\Big \downarrow \) 98 |
| \(\displaystyle \frac {1}{2} \int \left (\frac {a^2}{b (a+b) \left (b \tanh ^2(x)+a\right )^{5/2}}-\frac {(a+2 b) a}{b (a+b)^2 \left (b \tanh ^2(x)+a\right )^{3/2}}-\frac {1}{(a+b)^2 \left (\tanh ^2(x)-1\right ) \sqrt {b \tanh ^2(x)+a}}\right )d\tanh ^2(x)\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (-\frac {2 a^2}{3 b^2 (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{5/2}}+\frac {2 a (a+2 b)}{b^2 (a+b)^2 \sqrt {a+b \tanh ^2(x)}}\right )\) |
((2*ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a + b]])/(a + b)^(5/2) - (2*a^2)/(3 *b^2*(a + b)*(a + b*Tanh[x]^2)^(3/2)) + (2*a*(a + 2*b))/(b^2*(a + b)^2*Sqr t[a + b*Tanh[x]^2]))/2
3.3.47.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x _)), x_] :> Int[ExpandIntegrand[(e + f*x)^FractionalPart[p], (c + d*x)^n*(( e + f*x)^IntegerPart[p]/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio nalQ[n]))
Leaf count of result is larger than twice the leaf count of optimal. \(468\) vs. \(2(72)=144\).
Time = 0.10 (sec) , antiderivative size = 469, normalized size of antiderivative = 5.58
| method | result | size |
| derivativedivides | \(\frac {1}{3 b \left (a +b \tanh \left (x \right )^{2}\right )^{\frac {3}{2}}}+\frac {\tanh \left (x \right )^{2}}{b \left (a +b \tanh \left (x \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 a}{3 b^{2} \left (a +b \tanh \left (x \right )^{2}\right )^{\frac {3}{2}}}-\frac {1}{6 \left (a +b \right ) \left (b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b \right )^{\frac {3}{2}}}+\frac {b \tanh \left (x \right )}{6 \left (a +b \right ) a \left (b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b \right )^{\frac {3}{2}}}+\frac {b \tanh \left (x \right )}{3 \left (a +b \right ) a^{2} \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}-\frac {1}{2 \left (a +b \right )^{2} \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {b \tanh \left (x \right )}{2 \left (a +b \right )^{2} a \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2 \left (a +b \right )^{\frac {5}{2}}}-\frac {1}{6 \left (a +b \right ) \left (b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b \right )^{\frac {3}{2}}}-\frac {b \tanh \left (x \right )}{6 \left (a +b \right ) a \left (b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b \right )^{\frac {3}{2}}}-\frac {b \tanh \left (x \right )}{3 \left (a +b \right ) a^{2} \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}-\frac {1}{2 \left (a +b \right )^{2} \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}-\frac {b \tanh \left (x \right )}{2 \left (a +b \right )^{2} a \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b -2 b \left (1+\tanh \left (x \right )\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}{1+\tanh \left (x \right )}\right )}{2 \left (a +b \right )^{\frac {5}{2}}}\) | \(469\) |
| default | \(\frac {1}{3 b \left (a +b \tanh \left (x \right )^{2}\right )^{\frac {3}{2}}}+\frac {\tanh \left (x \right )^{2}}{b \left (a +b \tanh \left (x \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 a}{3 b^{2} \left (a +b \tanh \left (x \right )^{2}\right )^{\frac {3}{2}}}-\frac {1}{6 \left (a +b \right ) \left (b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b \right )^{\frac {3}{2}}}+\frac {b \tanh \left (x \right )}{6 \left (a +b \right ) a \left (b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b \right )^{\frac {3}{2}}}+\frac {b \tanh \left (x \right )}{3 \left (a +b \right ) a^{2} \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}-\frac {1}{2 \left (a +b \right )^{2} \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {b \tanh \left (x \right )}{2 \left (a +b \right )^{2} a \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2 \left (a +b \right )^{\frac {5}{2}}}-\frac {1}{6 \left (a +b \right ) \left (b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b \right )^{\frac {3}{2}}}-\frac {b \tanh \left (x \right )}{6 \left (a +b \right ) a \left (b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b \right )^{\frac {3}{2}}}-\frac {b \tanh \left (x \right )}{3 \left (a +b \right ) a^{2} \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}-\frac {1}{2 \left (a +b \right )^{2} \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}-\frac {b \tanh \left (x \right )}{2 \left (a +b \right )^{2} a \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b -2 b \left (1+\tanh \left (x \right )\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}{1+\tanh \left (x \right )}\right )}{2 \left (a +b \right )^{\frac {5}{2}}}\) | \(469\) |
1/3/b/(a+b*tanh(x)^2)^(3/2)+tanh(x)^2/b/(a+b*tanh(x)^2)^(3/2)+2/3*a/b^2/(a +b*tanh(x)^2)^(3/2)-1/6/(a+b)/(b*(tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(3/2)+ 1/6*b/(a+b)/a/(b*(tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(3/2)*tanh(x)+1/3*b/(a +b)/a^2/(b*(tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(1/2)*tanh(x)-1/2/(a+b)^2/(b *(tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(1/2)+1/2/(a+b)^2/a/(b*(tanh(x)-1)^2+2 *b*(tanh(x)-1)+a+b)^(1/2)*b*tanh(x)+1/2/(a+b)^(5/2)*ln((2*a+2*b+2*b*(tanh( x)-1)+2*(a+b)^(1/2)*(b*(tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(1/2))/(tanh(x)- 1))-1/6/(a+b)/(b*(1+tanh(x))^2-2*b*(1+tanh(x))+a+b)^(3/2)-1/6*b/(a+b)/a/(b *(1+tanh(x))^2-2*b*(1+tanh(x))+a+b)^(3/2)*tanh(x)-1/3*b/(a+b)/a^2/(b*(1+ta nh(x))^2-2*b*(1+tanh(x))+a+b)^(1/2)*tanh(x)-1/2/(a+b)^2/(b*(1+tanh(x))^2-2 *b*(1+tanh(x))+a+b)^(1/2)-1/2/(a+b)^2/a/(b*(1+tanh(x))^2-2*b*(1+tanh(x))+a +b)^(1/2)*b*tanh(x)+1/2/(a+b)^(5/2)*ln((2*a+2*b-2*b*(1+tanh(x))+2*(a+b)^(1 /2)*(b*(1+tanh(x))^2-2*b*(1+tanh(x))+a+b)^(1/2))/(1+tanh(x)))
Leaf count of result is larger than twice the leaf count of optimal. 3234 vs. \(2 (72) = 144\).
Time = 0.67 (sec) , antiderivative size = 7033, normalized size of antiderivative = 83.73 \[ \int \frac {\tanh ^5(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {\tanh ^5(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\int \frac {\tanh ^{5}{\left (x \right )}}{\left (a + b \tanh ^{2}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\tanh ^5(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\int { \frac {\tanh \left (x\right )^{5}}{{\left (b \tanh \left (x\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 711 vs. \(2 (72) = 144\).
Time = 0.53 (sec) , antiderivative size = 711, normalized size of antiderivative = 8.46 \[ \int \frac {\tanh ^5(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\frac {2 \, {\left ({\left ({\left (\frac {{\left (a^{9} + 8 \, a^{8} b + 25 \, a^{7} b^{2} + 40 \, a^{6} b^{3} + 35 \, a^{5} b^{4} + 16 \, a^{4} b^{5} + 3 \, a^{3} b^{6}\right )} e^{\left (2 \, x\right )}}{a^{8} b^{2} + 6 \, a^{7} b^{3} + 15 \, a^{6} b^{4} + 20 \, a^{5} b^{5} + 15 \, a^{4} b^{6} + 6 \, a^{3} b^{7} + a^{2} b^{8}} + \frac {3 \, {\left (a^{9} + 6 \, a^{8} b + 13 \, a^{7} b^{2} + 12 \, a^{6} b^{3} + 3 \, a^{5} b^{4} - 2 \, a^{4} b^{5} - a^{3} b^{6}\right )}}{a^{8} b^{2} + 6 \, a^{7} b^{3} + 15 \, a^{6} b^{4} + 20 \, a^{5} b^{5} + 15 \, a^{4} b^{6} + 6 \, a^{3} b^{7} + a^{2} b^{8}}\right )} e^{\left (2 \, x\right )} + \frac {3 \, {\left (a^{9} + 6 \, a^{8} b + 13 \, a^{7} b^{2} + 12 \, a^{6} b^{3} + 3 \, a^{5} b^{4} - 2 \, a^{4} b^{5} - a^{3} b^{6}\right )}}{a^{8} b^{2} + 6 \, a^{7} b^{3} + 15 \, a^{6} b^{4} + 20 \, a^{5} b^{5} + 15 \, a^{4} b^{6} + 6 \, a^{3} b^{7} + a^{2} b^{8}}\right )} e^{\left (2 \, x\right )} + \frac {a^{9} + 8 \, a^{8} b + 25 \, a^{7} b^{2} + 40 \, a^{6} b^{3} + 35 \, a^{5} b^{4} + 16 \, a^{4} b^{5} + 3 \, a^{3} b^{6}}{a^{8} b^{2} + 6 \, a^{7} b^{3} + 15 \, a^{6} b^{4} + 20 \, a^{5} b^{5} + 15 \, a^{4} b^{6} + 6 \, a^{3} b^{7} + a^{2} b^{8}}\right )}}{3 \, {\left (a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b\right )}^{\frac {3}{2}}} - \frac {\log \left ({\left | -{\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )} {\left (a + b\right )} - \sqrt {a + b} {\left (a - b\right )} \right |}\right )}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a + b}} + \frac {\log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} + \sqrt {a + b} \right |}\right )}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a + b}} - \frac {\log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} - \sqrt {a + b} \right |}\right )}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a + b}} \]
2/3*((((a^9 + 8*a^8*b + 25*a^7*b^2 + 40*a^6*b^3 + 35*a^5*b^4 + 16*a^4*b^5 + 3*a^3*b^6)*e^(2*x)/(a^8*b^2 + 6*a^7*b^3 + 15*a^6*b^4 + 20*a^5*b^5 + 15*a ^4*b^6 + 6*a^3*b^7 + a^2*b^8) + 3*(a^9 + 6*a^8*b + 13*a^7*b^2 + 12*a^6*b^3 + 3*a^5*b^4 - 2*a^4*b^5 - a^3*b^6)/(a^8*b^2 + 6*a^7*b^3 + 15*a^6*b^4 + 20 *a^5*b^5 + 15*a^4*b^6 + 6*a^3*b^7 + a^2*b^8))*e^(2*x) + 3*(a^9 + 6*a^8*b + 13*a^7*b^2 + 12*a^6*b^3 + 3*a^5*b^4 - 2*a^4*b^5 - a^3*b^6)/(a^8*b^2 + 6*a ^7*b^3 + 15*a^6*b^4 + 20*a^5*b^5 + 15*a^4*b^6 + 6*a^3*b^7 + a^2*b^8))*e^(2 *x) + (a^9 + 8*a^8*b + 25*a^7*b^2 + 40*a^6*b^3 + 35*a^5*b^4 + 16*a^4*b^5 + 3*a^3*b^6)/(a^8*b^2 + 6*a^7*b^3 + 15*a^6*b^4 + 20*a^5*b^5 + 15*a^4*b^6 + 6*a^3*b^7 + a^2*b^8))/(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b)^(3/2) - 1/2*log(abs(-(sqrt(a + b)*e^(2*x) - sqrt(a*e^(4*x) + b*e^( 4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b))*(a + b) - sqrt(a + b)*(a - b))) /((a^2 + 2*a*b + b^2)*sqrt(a + b)) + 1/2*log(abs(-sqrt(a + b)*e^(2*x) + sq rt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b) + sqrt(a + b )))/((a^2 + 2*a*b + b^2)*sqrt(a + b)) - 1/2*log(abs(-sqrt(a + b)*e^(2*x) + sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b) - sqrt(a + b)))/((a^2 + 2*a*b + b^2)*sqrt(a + b))
Time = 4.41 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.10 \[ \int \frac {\tanh ^5(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a}\,\left (2\,a^2+4\,a\,b+2\,b^2\right )}{2\,{\left (a+b\right )}^{5/2}}\right )}{{\left (a+b\right )}^{5/2}}-\frac {\frac {a^2}{3\,\left (a+b\right )}-\frac {\left (a^2+2\,b\,a\right )\,\left (b\,{\mathrm {tanh}\left (x\right )}^2+a\right )}{{\left (a+b\right )}^2}}{b^2\,{\left (b\,{\mathrm {tanh}\left (x\right )}^2+a\right )}^{3/2}} \]