\(\int \frac {\tanh ^4(x)}{(a+b \tanh ^2(x))^{3/2}} \, dx\) [239]

Optimal result

Integrand size = 17, antiderivative size = 84 \[ \int \frac {\tanh ^4(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a+b \tanh ^2(x)}}\right )}{b^{3/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b} \tanh (x)}{\sqrt {a+b \tanh ^2(x)}}\right )}{(a+b)^{3/2}}+\frac {a \tanh (x)}{b (a+b) \sqrt {a+b \tanh ^2(x)}} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 1.79 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.24 \[ \int \frac {\tanh ^4(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=-\frac {a \left (-2 a-2 b+\sqrt {2} (a+b) \sqrt {\frac {(a-b+(a+b) \cosh (2 x)) \text {csch}^2(x)}{b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a-b+(a+b) \cosh (2 x)) \text {csch}^2(x)}{b}}}{\sqrt {2}}\right ),1\right )+\sqrt {2} b \sqrt {\frac {(a-b+(a+b) \cosh (2 x)) \text {csch}^2(x)}{b}} \operatorname {EllipticPi}\left (\frac {b}{a+b},\arcsin \left (\frac {\sqrt {\frac {(a-b+(a+b) \cosh (2 x)) \text {csch}^2(x)}{b}}}{\sqrt {2}}\right ),1\right )\right ) \tanh (x)}{\sqrt {2} b (a+b)^2 \sqrt {(a-b+(a+b) \cosh (2 x)) \text {sech}^2(x)}} \] Input:

Integrate[Tanh[x]^4/(a + b*Tanh[x]^2)^(3/2),x]
 

Output:

-((a*(-2*a - 2*b + Sqrt[2]*(a + b)*Sqrt[((a - b + (a + b)*Cosh[2*x])*Csch[ 
x]^2)/b]*EllipticF[ArcSin[Sqrt[((a - b + (a + b)*Cosh[2*x])*Csch[x]^2)/b]/ 
Sqrt[2]], 1] + Sqrt[2]*b*Sqrt[((a - b + (a + b)*Cosh[2*x])*Csch[x]^2)/b]*E 
llipticPi[b/(a + b), ArcSin[Sqrt[((a - b + (a + b)*Cosh[2*x])*Csch[x]^2)/b 
]/Sqrt[2]], 1])*Tanh[x])/(Sqrt[2]*b*(a + b)^2*Sqrt[(a - b + (a + b)*Cosh[2 
*x])*Sech[x]^2]))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.18, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {3042, 4153, 372, 398, 224, 219, 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.

\(\displaystyle \int \frac {\tanh ^4(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (i x)^4}{\left (a-b \tan (i x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

Int[Tanh[x]^4/(a + b*Tanh[x]^2)^(3/2),x]
 

Output:

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

Defintions of rubi rules used

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 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 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 372

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 
)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 
))   Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + 
 (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, 
e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a 
, b, c, d, e, m, 2, p, q, x]
 

rule 398

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) 
, x_Symbol] :> Simp[f/b   Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ 
b   Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} 
, x]
 

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. \(327\) vs. \(2(70)=140\).

Time = 0.05 (sec) , antiderivative size = 328, normalized size of antiderivative = 3.90

method	result	size
derivativedivides	\(-\frac {\tanh \left (x \right )}{a \sqrt {a +b \tanh \left (x \right )^{2}}}+\frac {\tanh \left (x \right )}{b \sqrt {a +b \tanh \left (x \right )^{2}}}-\frac {\ln \left (\sqrt {b}\, \tanh \left (x \right )+\sqrt {a +b \tanh \left (x \right )^{2}}\right )}{b^{\frac {3}{2}}}-\frac {1}{2 \left (a +b \right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {b \left (2 b \left (\tanh \left (x \right )-1\right )+2 b \right )}{\left (a +b \right ) \left (4 b \left (a +b \right )-4 b^{2}\right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}+\frac {1}{2 \left (a +b \right ) \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}+\frac {b \left (2 b \left (\tanh \left (x \right )+1\right )-2 b \right )}{\left (a +b \right ) \left (4 b \left (a +b \right )-4 b^{2}\right ) \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}-\frac {\ln \left (\frac {2 a +2 b -2 b \left (\tanh \left (x \right )+1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}{\tanh \left (x \right )+1}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}\)	\(328\)
default	\(-\frac {\tanh \left (x \right )}{a \sqrt {a +b \tanh \left (x \right )^{2}}}+\frac {\tanh \left (x \right )}{b \sqrt {a +b \tanh \left (x \right )^{2}}}-\frac {\ln \left (\sqrt {b}\, \tanh \left (x \right )+\sqrt {a +b \tanh \left (x \right )^{2}}\right )}{b^{\frac {3}{2}}}-\frac {1}{2 \left (a +b \right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {b \left (2 b \left (\tanh \left (x \right )-1\right )+2 b \right )}{\left (a +b \right ) \left (4 b \left (a +b \right )-4 b^{2}\right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}+\frac {1}{2 \left (a +b \right ) \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}+\frac {b \left (2 b \left (\tanh \left (x \right )+1\right )-2 b \right )}{\left (a +b \right ) \left (4 b \left (a +b \right )-4 b^{2}\right ) \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}-\frac {\ln \left (\frac {2 a +2 b -2 b \left (\tanh \left (x \right )+1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}{\tanh \left (x \right )+1}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}\)	\(328\)

Input:

int(tanh(x)^4/(a+b*tanh(x)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-tanh(x)/a/(a+b*tanh(x)^2)^(1/2)+tanh(x)/b/(a+b*tanh(x)^2)^(1/2)-1/b^(3/2) 
*ln(b^(1/2)*tanh(x)+(a+b*tanh(x)^2)^(1/2))-1/2/(a+b)/(b*(tanh(x)-1)^2+2*b* 
(tanh(x)-1)+a+b)^(1/2)+b/(a+b)*(2*b*(tanh(x)-1)+2*b)/(4*b*(a+b)-4*b^2)/(b* 
(tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(1/2)+1/2/(a+b)^(3/2)*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))/(tan 
h(x)-1))+1/2/(a+b)/(b*(tanh(x)+1)^2-2*b*(tanh(x)+1)+a+b)^(1/2)+b/(a+b)*(2* 
b*(tanh(x)+1)-2*b)/(4*b*(a+b)-4*b^2)/(b*(tanh(x)+1)^2-2*b*(tanh(x)+1)+a+b) 
^(1/2)-1/2/(a+b)^(3/2)*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. 1364 vs. \(2 (70) = 140\).

Time = 0.37 (sec) , antiderivative size = 6833, normalized size of antiderivative = 81.35 \[ \int \frac {\tanh ^4(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(tanh(x)^4/(a+b*tanh(x)^2)^(3/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

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

integrate(tanh(x)**4/(a+b*tanh(x)**2)**(3/2),x)
 

Output:

Integral(tanh(x)**4/(a + b*tanh(x)**2)**(3/2), x)
 

Maxima [F]

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

integrate(tanh(x)^4/(a+b*tanh(x)^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate(tanh(x)^4/(b*tanh(x)^2 + a)^(3/2), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 371 vs. \(2 (70) = 140\).

Time = 0.32 (sec) , antiderivative size = 371, normalized size of antiderivative = 4.42 \[ \int \frac {\tanh ^4(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\frac {\frac {{\left (a^{3} b^{2} + a^{2} b^{3}\right )} e^{\left (2 \, x\right )}}{a^{3} b^{3} + 2 \, a^{2} b^{4} + a b^{5}} - \frac {a^{3} b^{2} + a^{2} b^{3}}{a^{3} b^{3} + 2 \, a^{2} b^{4} + a b^{5}}}{\sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}} - \frac {2 \, \arctan \left (-\frac {\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} + \sqrt {a + b}}{2 \, \sqrt {-b}}\right )}{\sqrt {-b} b} - \frac {\log \left ({\left | -{\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )} {\left (a + b\right )} - \sqrt {a + b} {\left (a - b\right )} \right |}\right )}{2 \, {\left (a + b\right )}^{\frac {3}{2}}} - \frac {\log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} + \sqrt {a + b} \right |}\right )}{2 \, {\left (a + b\right )}^{\frac {3}{2}}} + \frac {\log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} - \sqrt {a + b} \right |}\right )}{2 \, {\left (a + b\right )}^{\frac {3}{2}}} \] Input:

integrate(tanh(x)^4/(a+b*tanh(x)^2)^(3/2),x, algorithm="giac")
 

Output:

((a^3*b^2 + a^2*b^3)*e^(2*x)/(a^3*b^3 + 2*a^2*b^4 + a*b^5) - (a^3*b^2 + a^ 
2*b^3)/(a^3*b^3 + 2*a^2*b^4 + a*b^5))/sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^( 
2*x) - 2*b*e^(2*x) + a + b) - 2*arctan(-1/2*(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))/sq 
rt(-b))/(sqrt(-b)*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))*(a + b) - sqrt(a + b)*(a 
 - b)))/(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) + sqrt(a + b)))/(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) - sqrt(a + b)))/(a + b)^(3/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tanh ^4(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {tanh}\left (x\right )}^4}{{\left (b\,{\mathrm {tanh}\left (x\right )}^2+a\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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