\(\int \frac {1}{(a+b \tanh ^2(x))^{5/2}} \, dx\) [252]

Optimal result

Integrand size = 12, antiderivative size = 93 \[ \int \frac {1}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b} \tanh (x)}{\sqrt {a+b \tanh ^2(x)}}\right )}{(a+b)^{5/2}}+\frac {b \tanh (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}+\frac {b (5 a+2 b) \tanh (x)}{3 a^2 (a+b)^2 \sqrt {a+b \tanh ^2(x)}} \] Output:

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

Mathematica [C] (warning: unable to verify)

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

Time = 5.94 (sec) , antiderivative size = 943, normalized size of antiderivative = 10.14 \[ \int \frac {1}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(Cosh[x]*Sinh[x]*(1575*ArcSin[Sqrt[-(((a + b)*Sinh[x]^2)/a)]] + (3150*(a + 
 b)*ArcSin[Sqrt[-(((a + b)*Sinh[x]^2)/a)]]*Sinh[x]^2)/a + (1575*(a + b)^2* 
ArcSin[Sqrt[-(((a + b)*Sinh[x]^2)/a)]]*Sinh[x]^4)/a^2 + 2100*(-(((a + b)*S 
inh[x]^2)/a))^(3/2)*Sqrt[Cosh[x]^2 + (b*Sinh[x]^2)/a] + 96*Hypergeometric2 
F1[2, 2, 9/2, -(((a + b)*Sinh[x]^2)/a)]*(-(((a + b)*Sinh[x]^2)/a))^(7/2)*S 
qrt[Cosh[x]^2 + (b*Sinh[x]^2)/a] + 24*HypergeometricPFQ[{2, 2, 2}, {1, 9/2 
}, -(((a + b)*Sinh[x]^2)/a)]*(-(((a + b)*Sinh[x]^2)/a))^(7/2)*Sqrt[Cosh[x] 
^2 + (b*Sinh[x]^2)/a] + (2100*b*ArcSin[Sqrt[-(((a + b)*Sinh[x]^2)/a)]]*Tan 
h[x]^2)/a + (4200*b*(a + b)*ArcSin[Sqrt[-(((a + b)*Sinh[x]^2)/a)]]*Sinh[x] 
^2*Tanh[x]^2)/a^2 + (2100*b*(a + b)^2*ArcSin[Sqrt[-(((a + b)*Sinh[x]^2)/a) 
]]*Sinh[x]^4*Tanh[x]^2)/a^3 + (2800*b*(-(((a + b)*Sinh[x]^2)/a))^(3/2)*Sqr 
t[Cosh[x]^2 + (b*Sinh[x]^2)/a]*Tanh[x]^2)/a + (168*b*Hypergeometric2F1[2, 
2, 9/2, -(((a + b)*Sinh[x]^2)/a)]*(-(((a + b)*Sinh[x]^2)/a))^(7/2)*Sqrt[Co 
sh[x]^2 + (b*Sinh[x]^2)/a]*Tanh[x]^2)/a + (48*b*HypergeometricPFQ[{2, 2, 2 
}, {1, 9/2}, -(((a + b)*Sinh[x]^2)/a)]*(-(((a + b)*Sinh[x]^2)/a))^(7/2)*Sq 
rt[Cosh[x]^2 + (b*Sinh[x]^2)/a]*Tanh[x]^2)/a + (840*b^2*ArcSin[Sqrt[-(((a 
+ b)*Sinh[x]^2)/a)]]*Tanh[x]^4)/a^2 + (1680*b^2*(a + b)*ArcSin[Sqrt[-(((a 
+ b)*Sinh[x]^2)/a)]]*Sinh[x]^2*Tanh[x]^4)/a^3 + (840*b^2*(a + b)^2*ArcSin[ 
Sqrt[-(((a + b)*Sinh[x]^2)/a)]]*Sinh[x]^4*Tanh[x]^4)/a^4 + (1120*b^2*(-((( 
a + b)*Sinh[x]^2)/a))^(3/2)*Sqrt[Cosh[x]^2 + (b*Sinh[x]^2)/a]*Tanh[x]^4...
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {3042, 4144, 316, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4144

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 402

\(\displaystyle \frac {b \tanh (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}-\frac {-\frac {\int \frac {3 a^2}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh (x)}{a (a+b)}-\frac {b (5 a+2 b) \tanh (x)}{a (a+b) \sqrt {a+b \tanh ^2(x)}}}{3 a (a+b)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b \tanh (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}-\frac {-\frac {3 a \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh (x)}{a+b}-\frac {b (5 a+2 b) \tanh (x)}{a (a+b) \sqrt {a+b \tanh ^2(x)}}}{3 a (a+b)}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {b \tanh (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}-\frac {-\frac {3 a \int \frac {1}{1-\frac {(a+b) \tanh ^2(x)}{b \tanh ^2(x)+a}}d\frac {\tanh (x)}{\sqrt {b \tanh ^2(x)+a}}}{a+b}-\frac {b (5 a+2 b) \tanh (x)}{a (a+b) \sqrt {a+b \tanh ^2(x)}}}{3 a (a+b)}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {b \tanh (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}-\frac {-\frac {3 a \text {arctanh}\left (\frac {\sqrt {a+b} \tanh (x)}{\sqrt {a+b \tanh ^2(x)}}\right )}{(a+b)^{3/2}}-\frac {b (5 a+2 b) \tanh (x)}{a (a+b) \sqrt {a+b \tanh ^2(x)}}}{3 a (a+b)}\)

Input:

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

Output:

(b*Tanh[x])/(3*a*(a + b)*(a + b*Tanh[x]^2)^(3/2)) - ((-3*a*ArcTanh[(Sqrt[a 
 + b]*Tanh[x])/Sqrt[a + b*Tanh[x]^2]])/(a + b)^(3/2) - (b*(5*a + 2*b)*Tanh 
[x])/(a*(a + b)*Sqrt[a + b*Tanh[x]^2]))/(3*a*(a + b))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

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])
 

rule 291

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]
 

rule 316

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, 
 p, q, x]
 

rule 402

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]
 

rule 3042

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

rule 4144

Int[((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[(a + b* 
(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, 
 b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || 
EqQ[n^2, 16])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(419\) vs. \(2(79)=158\).

Time = 0.04 (sec) , antiderivative size = 420, normalized size of antiderivative = 4.52

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(1/(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. 3152 vs. \(2 (79) = 158\).

Time = 0.66 (sec) , antiderivative size = 6933, normalized size of antiderivative = 74.55 \[ \int \frac {1}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*tanh(x)^2)^(5/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {1}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\int \frac {1}{\left (a + b \tanh ^{2}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(a+b*tanh(x)**2)**(5/2),x)
 

Output:

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

Maxima [F]

\[ \int \frac {1}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b \tanh \left (x\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(a+b*tanh(x)^2)^(5/2),x, algorithm="maxima")
 

Output:

integrate((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. 714 vs. \(2 (79) = 158\).

Time = 0.32 (sec) , antiderivative size = 714, normalized size of antiderivative = 7.68 \[ \int \frac {1}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(1/(a+b*tanh(x)^2)^(5/2),x, algorithm="giac")
 

Output:

2/3*((((3*a^6*b^3 + 16*a^5*b^4 + 35*a^4*b^5 + 40*a^3*b^6 + 25*a^2*b^7 + 8* 
a*b^8 + b^9)*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^6*b^3 + 2*a^5*b^4 - 3*a^4*b^5 - 12*a^ 
3*b^6 - 13*a^2*b^7 - 6*a*b^8 - b^9)/(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^6*b^3 + 2*a^5 
*b^4 - 3*a^4*b^5 - 12*a^3*b^6 - 13*a^2*b^7 - 6*a*b^8 - b^9)/(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^6*b^3 + 16*a^5*b^4 + 35*a^4*b^5 + 40*a^3*b^6 + 25*a^2*b^7 + 8*a 
*b^8 + b^9)/(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))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (b\,{\mathrm {tanh}\left (x\right )}^2+a\right )}^{5/2}} \,d x \] Input:

int(1/(a + b*tanh(x)^2)^(5/2),x)
 

Output:

int(1/(a + b*tanh(x)^2)^(5/2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\tanh \left (x \right )^{2} b +a}}{\tanh \left (x \right )^{6} b^{3}+3 \tanh \left (x \right )^{4} a \,b^{2}+3 \tanh \left (x \right )^{2} a^{2} b +a^{3}}d x \] Input:

int(1/(a+b*tanh(x)^2)^(5/2),x)
 

Output:

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