Integrand size = 17, antiderivative size = 88 \[ \int \frac {\tanh ^2(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b} \tanh (x)}{\sqrt {a+b \tanh ^2(x)}}\right )}{(a+b)^{5/2}}-\frac {\tanh (x)}{3 (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}-\frac {(2 a-b) \tanh (x)}{3 a (a+b)^2 \sqrt {a+b \tanh ^2(x)}} \]
arctanh((a+b)^(1/2)*tanh(x)/(a+b*tanh(x)^2)^(1/2))/(a+b)^(5/2)-1/3*(2*a-b) *tanh(x)/a/(a+b)^2/(a+b*tanh(x)^2)^(1/2)-1/3*tanh(x)/(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 = 7.11 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.45 \[ \int \frac {\tanh ^2(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\frac {\sinh ^2(x) \tanh (x) \left (-\frac {4 (a+b) \operatorname {Hypergeometric2F1}\left (2,2,\frac {9}{2},-\frac {(a+b) \sinh ^2(x)}{a}\right ) \sinh ^2(x) \left (a+b \tanh ^2(x)\right )}{35 a^2}-\frac {\coth ^4(x) \left (-5 a-2 b \tanh ^2(x)\right ) \left (3 \arcsin \left (\sqrt {-\frac {(a+b) \sinh ^2(x)}{a}}\right ) \left (a+b \tanh ^2(x)\right )^2+a \text {sech}^2(x) \sqrt {-\frac {(a+b) \cosh ^2(x) \sinh ^2(x) \left (a+b \tanh ^2(x)\right )}{a^2}} \left (-4 b \tanh ^2(x)+a \left (-3-\tanh ^2(x)\right )\right )\right )}{3 a (a+b)^2 \sqrt {-\frac {(a+b) \cosh ^2(x) \sinh ^2(x) \left (a+b \tanh ^2(x)\right )}{a^2}}}\right )}{3 a^2 \sqrt {a+b \tanh ^2(x)} \left (1+\frac {b \tanh ^2(x)}{a}\right )} \]
(Sinh[x]^2*Tanh[x]*((-4*(a + b)*Hypergeometric2F1[2, 2, 9/2, -(((a + b)*Si nh[x]^2)/a)]*Sinh[x]^2*(a + b*Tanh[x]^2))/(35*a^2) - (Coth[x]^4*(-5*a - 2* b*Tanh[x]^2)*(3*ArcSin[Sqrt[-(((a + b)*Sinh[x]^2)/a)]]*(a + b*Tanh[x]^2)^2 + a*Sech[x]^2*Sqrt[-(((a + b)*Cosh[x]^2*Sinh[x]^2*(a + b*Tanh[x]^2))/a^2) ]*(-4*b*Tanh[x]^2 + a*(-3 - Tanh[x]^2))))/(3*a*(a + b)^2*Sqrt[-(((a + b)*C osh[x]^2*Sinh[x]^2*(a + b*Tanh[x]^2))/a^2)])))/(3*a^2*Sqrt[a + b*Tanh[x]^2 ]*(1 + (b*Tanh[x]^2)/a))
Time = 0.34 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {3042, 25, 4153, 25, 373, 402, 27, 291, 219}
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.
| \(\displaystyle \int \frac {\tanh ^2(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\tan (i x)^2}{\left (a-b \tan (i x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\tan (i x)^2}{\left (a-b \tan (i x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle -\int -\frac {\tanh ^2(x)}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^2(x)+a\right )^{5/2}}d\tanh (x)\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {\tanh ^2(x)}{\left (1-\tanh ^2(x)\right ) \left (a+b \tanh ^2(x)\right )^{5/2}}d\tanh (x)\) |
| \(\Big \downarrow \) 373 |
| \(\displaystyle \frac {\int \frac {2 \tanh ^2(x)+1}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^2(x)+a\right )^{3/2}}d\tanh (x)}{3 (a+b)}-\frac {\tanh (x)}{3 (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {-\frac {\int -\frac {3 a}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh (x)}{a (a+b)}-\frac {(2 a-b) \tanh (x)}{a (a+b) \sqrt {a+b \tanh ^2(x)}}}{3 (a+b)}-\frac {\tanh (x)}{3 (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3 \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh (x)}{a+b}-\frac {(2 a-b) \tanh (x)}{a (a+b) \sqrt {a+b \tanh ^2(x)}}}{3 (a+b)}-\frac {\tanh (x)}{3 (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {3 \int \frac {1}{1-\frac {(a+b) \tanh ^2(x)}{b \tanh ^2(x)+a}}d\frac {\tanh (x)}{\sqrt {b \tanh ^2(x)+a}}}{a+b}-\frac {(2 a-b) \tanh (x)}{a (a+b) \sqrt {a+b \tanh ^2(x)}}}{3 (a+b)}-\frac {\tanh (x)}{3 (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {3 \text {arctanh}\left (\frac {\sqrt {a+b} \tanh (x)}{\sqrt {a+b \tanh ^2(x)}}\right )}{(a+b)^{3/2}}-\frac {(2 a-b) \tanh (x)}{a (a+b) \sqrt {a+b \tanh ^2(x)}}}{3 (a+b)}-\frac {\tanh (x)}{3 (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}\) |
-1/3*Tanh[x]/((a + b)*(a + b*Tanh[x]^2)^(3/2)) + ((3*ArcTanh[(Sqrt[a + b]* Tanh[x])/Sqrt[a + b*Tanh[x]^2]])/(a + b)^(3/2) - ((2*a - b)*Tanh[x])/(a*(a + b)*Sqrt[a + b*Tanh[x]^2]))/(3*(a + b))
3.3.50.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1)) Int[(e *x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
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. \(453\) vs. \(2(74)=148\).
Time = 0.08 (sec) , antiderivative size = 454, normalized size of antiderivative = 5.16
| method | result | size |
| derivativedivides | \(-\frac {\tanh \left (x \right )}{3 a \left (a +b \tanh \left (x \right )^{2}\right )^{\frac {3}{2}}}-\frac {2 \tanh \left (x \right )}{3 a^{2} \sqrt {a +b \tanh \left (x \right )^{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}}}\) | \(454\) |
| default | \(-\frac {\tanh \left (x \right )}{3 a \left (a +b \tanh \left (x \right )^{2}\right )^{\frac {3}{2}}}-\frac {2 \tanh \left (x \right )}{3 a^{2} \sqrt {a +b \tanh \left (x \right )^{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}}}\) | \(454\) |
-1/3*tanh(x)/a/(a+b*tanh(x)^2)^(3/2)-2/3/a^2*tanh(x)/(a+b*tanh(x)^2)^(1/2) -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*(t anh(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+tanh(x))^2-2*b*(1+tan h(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. 2939 vs. \(2 (74) = 148\).
Time = 0.66 (sec) , antiderivative size = 6507, normalized size of antiderivative = 73.94 \[ \int \frac {\tanh ^2(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {\tanh ^2(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\int \frac {\tanh ^{2}{\left (x \right )}}{\left (a + b \tanh ^{2}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\tanh ^2(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\int { \frac {\tanh \left (x\right )^{2}}{{\left (b \tanh \left (x\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 728 vs. \(2 (74) = 148\).
Time = 0.52 (sec) , antiderivative size = 728, normalized size of antiderivative = 8.27 \[ \int \frac {\tanh ^2(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=-\frac {{\left ({\left (\frac {{\left (3 \, a^{7} b^{2} + 14 \, a^{6} b^{3} + 25 \, a^{5} b^{4} + 20 \, a^{4} b^{5} + 5 \, a^{3} b^{6} - 2 \, a^{2} b^{7} - a b^{8}\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^{7} b^{2} + 2 \, a^{6} b^{3} - a^{5} b^{4} - 4 \, a^{4} b^{5} - a^{3} b^{6} + 2 \, a^{2} b^{7} + a b^{8}\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^{7} b^{2} + 2 \, a^{6} b^{3} - a^{5} b^{4} - 4 \, a^{4} b^{5} - a^{3} b^{6} + 2 \, a^{2} b^{7} + a b^{8}\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 \, a^{7} b^{2} + 14 \, a^{6} b^{3} + 25 \, a^{5} b^{4} + 20 \, a^{4} b^{5} + 5 \, a^{3} b^{6} - 2 \, a^{2} b^{7} - a b^{8}}{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 \, {\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}} \]
-1/3*((((3*a^7*b^2 + 14*a^6*b^3 + 25*a^5*b^4 + 20*a^4*b^5 + 5*a^3*b^6 - 2* a^2*b^7 - a*b^8)*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^7*b^2 + 2*a^6*b^3 - a^5*b^4 - 4*a ^4*b^5 - a^3*b^6 + 2*a^2*b^7 + a*b^8)/(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^7*b^2 + 2*a ^6*b^3 - a^5*b^4 - 4*a^4*b^5 - a^3*b^6 + 2*a^2*b^7 + a*b^8)/(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^7*b^2 + 14*a^6*b^3 + 25*a^5*b^4 + 20*a^4*b^5 + 5*a^3*b^6 - 2*a^ 2*b^7 - a*b^8)/(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) + 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)) + 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))
Timed out. \[ \int \frac {\tanh ^2(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\int \frac {{\mathrm {tanh}\left (x\right )}^2}{{\left (b\,{\mathrm {tanh}\left (x\right )}^2+a\right )}^{5/2}} \,d x \]