\(\int \frac {\tanh (x)}{(a+b \tanh ^2(x))^{5/2}} \, dx\) [251]

Optimal result

Integrand size = 15, antiderivative size = 70 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{5/2}}-\frac {1}{3 (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}-\frac {1}{(a+b)^2 \sqrt {a+b \tanh ^2(x)}} \] Output:

arctanh((a+b*tanh(x)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)-1/3/(a+b)/(a+b*tanh 
(x)^2)^(3/2)-1/(a+b)^2/(a+b*tanh(x)^2)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.61 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tanh ^2(x)}{a+b}\right )}{3 (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}} \] Input:

Integrate[Tanh[x]/(a + b*Tanh[x]^2)^(5/2),x]
 

Output:

-1/3*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tanh[x]^2)/(a + b)]/((a + b)* 
(a + b*Tanh[x]^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.17, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 26, 4153, 26, 353, 61, 61, 73, 221}

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.

\(\displaystyle \int \frac {\tanh (x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {i \tan (i x)}{\left (a-b \tan (i x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 26

\(\displaystyle -i \int \frac {\tan (i x)}{\left (a-b \tan (i x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 4153

\(\displaystyle -i \int \frac {i \tanh (x)}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^2(x)+a\right )^{5/2}}d\tanh (x)\)

\(\Big \downarrow \) 26

\(\displaystyle \int \frac {\tanh (x)}{\left (1-\tanh ^2(x)\right ) \left (a+b \tanh ^2(x)\right )^{5/2}}d\tanh (x)\)

\(\Big \downarrow \) 353

\(\displaystyle \frac {1}{2} \int \frac {1}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^2(x)+a\right )^{5/2}}d\tanh ^2(x)\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {1}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^2(x)+a\right )^{3/2}}d\tanh ^2(x)}{a+b}-\frac {2}{3 (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}\right )\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {1}{2} \left (\frac {\frac {\int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)}{a+b}-\frac {2}{(a+b) \sqrt {a+b \tanh ^2(x)}}}{a+b}-\frac {2}{3 (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{2} \left (\frac {\frac {2 \int \frac {1}{\frac {a+b}{b}-\frac {\tanh ^4(x)}{b}}d\sqrt {b \tanh ^2(x)+a}}{b (a+b)}-\frac {2}{(a+b) \sqrt {a+b \tanh ^2(x)}}}{a+b}-\frac {2}{3 (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{2} \left (\frac {\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {2}{(a+b) \sqrt {a+b \tanh ^2(x)}}}{a+b}-\frac {2}{3 (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}\right )\)

Input:

Int[Tanh[x]/(a + b*Tanh[x]^2)^(5/2),x]
 

Output:

(-2/(3*(a + b)*(a + b*Tanh[x]^2)^(3/2)) + ((2*ArcTanh[Sqrt[a + b*Tanh[x]^2 
]/Sqrt[a + b]])/(a + b)^(3/2) - 2/((a + b)*Sqrt[a + b*Tanh[x]^2]))/(a + b) 
)/2
 

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 353

Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] 
 :> Simp[1/2   Subst[Int[(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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4153

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]))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(419\) vs. \(2(58)=116\).

Time = 0.05 (sec) , antiderivative size = 420, normalized size of antiderivative = 6.00

method	result	size
derivativedivides	\(-\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 (\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}}}\)	\(420\)
default	\(-\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 (\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}}}\)	\(420\)

Input:

int(tanh(x)/(a+b*tanh(x)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-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*(t 
anh(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))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2607 vs. \(2 (58) = 116\).

Time = 0.54 (sec) , antiderivative size = 5779, normalized size of antiderivative = 82.56 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(tanh(x)/(a+b*tanh(x)^2)^(5/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [A] (verification not implemented)

Time = 11.11 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.63 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=- \begin {cases} \frac {2 \left (\frac {b}{6 \left (a + b\right ) \left (a + b \tanh ^{2}{\left (x \right )}\right )^{\frac {3}{2}}} + \frac {b}{2 \left (a + b\right )^{2} \sqrt {a + b \tanh ^{2}{\left (x \right )}}} + \frac {b \operatorname {atan}{\left (\frac {\sqrt {a + b \tanh ^{2}{\left (x \right )}}}{\sqrt {- a - b}} \right )}}{2 \sqrt {- a - b} \left (a + b\right )^{2}}\right )}{b} & \text {for}\: b \neq 0 \\\begin {cases} \tilde {\infty } \tanh ^{2}{\left (x \right )} & \text {for}\: a^{\frac {5}{2}} = 0 \\\frac {\log {\left (2 a^{\frac {5}{2}} \tanh ^{2}{\left (x \right )} - 2 a^{\frac {5}{2}} \right )}}{2 a^{\frac {5}{2}}} & \text {otherwise} \end {cases} & \text {otherwise} \end {cases} \] Input:

integrate(tanh(x)/(a+b*tanh(x)**2)**(5/2),x)
 

Output:

-Piecewise((2*(b/(6*(a + b)*(a + b*tanh(x)**2)**(3/2)) + b/(2*(a + b)**2*s 
qrt(a + b*tanh(x)**2)) + b*atan(sqrt(a + b*tanh(x)**2)/sqrt(-a - b))/(2*sq 
rt(-a - b)*(a + b)**2))/b, Ne(b, 0)), (Piecewise((zoo*tanh(x)**2, Eq(a**(5 
/2), 0)), (log(2*a**(5/2)*tanh(x)**2 - 2*a**(5/2))/(2*a**(5/2)), True)), T 
rue))
 

Maxima [F]

\[ \int \frac {\tanh (x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\int { \frac {\tanh \left (x\right )}{{\left (b \tanh \left (x\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(tanh(x)/(a+b*tanh(x)^2)^(5/2),x, algorithm="maxima")
 

Output:

integrate(tanh(x)/(b*tanh(x)^2 + a)^(5/2), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 683 vs. \(2 (58) = 116\).

Time = 0.32 (sec) , antiderivative size = 683, normalized size of antiderivative = 9.76 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(tanh(x)/(a+b*tanh(x)^2)^(5/2),x, algorithm="giac")
 

Output:

-4/3*((((a^7*b^2 + 5*a^6*b^3 + 10*a^5*b^4 + 10*a^4*b^5 + 5*a^3*b^6 + a^2*b 
^7)*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 + 4*a^6*b^3 + 6*a^5*b^4 + 4*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^7*b^2 + 4*a^6*b^3 + 6*a^5*b^4 + 4*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^7*b^2 + 5*a^6*b^3 + 10*a^5*b^4 + 10*a^4* 
b^5 + 5*a^3*b^6 + a^2*b^7)/(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))
 

Mupad [B] (verification not implemented)

Time = 4.82 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.09 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a}\,\left (2\,a^2+4\,a\,b+2\,b^2\right )}{2\,{\left (a+b\right )}^{5/2}}\right )}{{\left (a+b\right )}^{5/2}}-\frac {\frac {1}{3\,\left (a+b\right )}+\frac {b\,{\mathrm {tanh}\left (x\right )}^2+a}{{\left (a+b\right )}^2}}{{\left (b\,{\mathrm {tanh}\left (x\right )}^2+a\right )}^{3/2}} \] Input:

int(tanh(x)/(a + b*tanh(x)^2)^(5/2),x)
 

Output:

atanh(((a + b*tanh(x)^2)^(1/2)*(4*a*b + 2*a^2 + 2*b^2))/(2*(a + b)^(5/2))) 
/(a + b)^(5/2) - (1/(3*(a + b)) + (a + b*tanh(x)^2)/(a + b)^2)/(a + b*tanh 
(x)^2)^(3/2)
 

Reduce [F]

\[ \int \frac {\tanh (x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\tanh \left (x \right )^{2} b +a}\, \tanh \left (x \right )}{\tanh \left (x \right )^{6} b^{3}+3 \tanh \left (x \right )^{4} a \,b^{2}+3 \tanh \left (x \right )^{2} a^{2} b +a^{3}}d x \] Input:

int(tanh(x)/(a+b*tanh(x)^2)^(5/2),x)
 

Output:

int((sqrt(tanh(x)**2*b + a)*tanh(x))/(tanh(x)**6*b**3 + 3*tanh(x)**4*a*b** 
2 + 3*tanh(x)**2*a**2*b + a**3),x)