\(\int \frac {1}{(a+b \tanh ^2(x))^{3/2}} \, dx\) [243]

Optimal result

Integrand size = 12, antiderivative size = 56 \[ \int \frac {1}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b} \tanh (x)}{\sqrt {a+b \tanh ^2(x)}}\right )}{(a+b)^{3/2}}+\frac {b \tanh (x)}{a (a+b) \sqrt {a+b \tanh ^2(x)}} \] Output:

arctanh((a+b)^(1/2)*tanh(x)/(a+b*tanh(x)^2)^(1/2))/(a+b)^(3/2)+b*tanh(x)/a 
/(a+b)/(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 = 2.96 (sec) , antiderivative size = 223, normalized size of antiderivative = 3.98 \[ \int \frac {1}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=-\frac {\sinh ^2(x) \left (\frac {15}{4} a (3 a-2 b+(3 a+2 b) \cosh (2 x)) \text {csch}(x) \text {sech}(x) \left ((a-b) \arcsin \left (\sqrt {-\frac {(a+b) \sinh ^2(x)}{a}}\right )+(a+b) \arcsin \left (\sqrt {-\frac {(a+b) \sinh ^2(x)}{a}}\right ) \cosh (2 x)-2 a \sqrt {-\frac {(a+b) \left (b+a \coth ^2(x)\right ) \sinh ^4(x)}{a^2}}\right )+\sqrt {2} a^2 (a+b) \operatorname {Hypergeometric2F1}\left (2,2,\frac {7}{2},-\frac {(a+b) \sinh ^2(x)}{a}\right ) \left (-\frac {(a+b) (a-b+(a+b) \cosh (2 x)) \sinh ^2(x)}{a^2}\right )^{3/2} \tanh (x)\right )}{15 a^4 \left (-\frac {(a+b) \sinh ^2(x)}{a}\right )^{3/2} \sqrt {\cosh ^2(x)+\frac {b \sinh ^2(x)}{a}} \sqrt {a+b \tanh ^2(x)}} \] Input:

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

Output:

-1/15*(Sinh[x]^2*((15*a*(3*a - 2*b + (3*a + 2*b)*Cosh[2*x])*Csch[x]*Sech[x 
]*((a - b)*ArcSin[Sqrt[-(((a + b)*Sinh[x]^2)/a)]] + (a + b)*ArcSin[Sqrt[-( 
((a + b)*Sinh[x]^2)/a)]]*Cosh[2*x] - 2*a*Sqrt[-(((a + b)*(b + a*Coth[x]^2) 
*Sinh[x]^4)/a^2)]))/4 + Sqrt[2]*a^2*(a + b)*Hypergeometric2F1[2, 2, 7/2, - 
(((a + b)*Sinh[x]^2)/a)]*(-(((a + b)*(a - b + (a + b)*Cosh[2*x])*Sinh[x]^2 
)/a^2))^(3/2)*Tanh[x]))/(a^4*(-(((a + b)*Sinh[x]^2)/a))^(3/2)*Sqrt[Cosh[x] 
^2 + (b*Sinh[x]^2)/a]*Sqrt[a + b*Tanh[x]^2])
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3042, 4144, 296, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a-b \tan (i x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 4144

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

\(\Big \downarrow \) 296

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

ArcTanh[(Sqrt[a + b]*Tanh[x])/Sqrt[a + b*Tanh[x]^2]]/(a + b)^(3/2) + (b*Ta 
nh[x])/(a*(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 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 296

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[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d))   Int[ 
(a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N 
eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1 
]) && NeQ[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. \(271\) vs. \(2(48)=96\).

Time = 0.04 (sec) , antiderivative size = 272, normalized size of antiderivative = 4.86

method	result	size
derivativedivides	\(-\frac {1}{2 \left (a +b \right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {b \left (2 b \left (\tanh \left (x \right )-1\right )+2 b \right )}{\left (a +b \right ) \left (4 b \left (a +b \right )-4 b^{2}\right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}+\frac {1}{2 \left (a +b \right ) \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}+\frac {b \left (2 b \left (\tanh \left (x \right )+1\right )-2 b \right )}{\left (a +b \right ) \left (4 b \left (a +b \right )-4 b^{2}\right ) \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}-\frac {\ln \left (\frac {2 a +2 b -2 b \left (\tanh \left (x \right )+1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}{\tanh \left (x \right )+1}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}\)	\(272\)
default	\(-\frac {1}{2 \left (a +b \right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {b \left (2 b \left (\tanh \left (x \right )-1\right )+2 b \right )}{\left (a +b \right ) \left (4 b \left (a +b \right )-4 b^{2}\right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}+\frac {1}{2 \left (a +b \right ) \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}+\frac {b \left (2 b \left (\tanh \left (x \right )+1\right )-2 b \right )}{\left (a +b \right ) \left (4 b \left (a +b \right )-4 b^{2}\right ) \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}-\frac {\ln \left (\frac {2 a +2 b -2 b \left (\tanh \left (x \right )+1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}{\tanh \left (x \right )+1}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}\)	\(272\)

Input:

int(1/(a+b*tanh(x)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-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))+1/2/(a+b)/(b*(tanh(x)+1)^2-2*b*(t 
anh(x)+1)+a+b)^(1/2)+b/(a+b)*(2*b*(tanh(x)+1)-2*b)/(4*b*(a+b)-4*b^2)/(b*(t 
anh(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*(ta 
nh(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. 905 vs. \(2 (48) = 96\).

Time = 0.20 (sec) , antiderivative size = 2439, normalized size of antiderivative = 43.55 \[ \int \frac {1}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*tanh(x)^2)^(3/2),x, algorithm="fricas")
 

Output:

[1/4*(((a^2 + a*b)*cosh(x)^4 + 4*(a^2 + a*b)*cosh(x)*sinh(x)^3 + (a^2 + a* 
b)*sinh(x)^4 + 2*(a^2 - a*b)*cosh(x)^2 + 2*(3*(a^2 + a*b)*cosh(x)^2 + a^2 
- a*b)*sinh(x)^2 + a^2 + a*b + 4*((a^2 + a*b)*cosh(x)^3 + (a^2 - a*b)*cosh 
(x))*sinh(x))*sqrt(a + b)*log(-((a*b^2 + b^3)*cosh(x)^8 + 8*(a*b^2 + b^3)* 
cosh(x)*sinh(x)^7 + (a*b^2 + b^3)*sinh(x)^8 - 2*(a*b^2 + 2*b^3)*cosh(x)^6 
- 2*(a*b^2 + 2*b^3 - 14*(a*b^2 + b^3)*cosh(x)^2)*sinh(x)^6 + 4*(14*(a*b^2 
+ b^3)*cosh(x)^3 - 3*(a*b^2 + 2*b^3)*cosh(x))*sinh(x)^5 + (a^3 - a^2*b + 4 
*a*b^2 + 6*b^3)*cosh(x)^4 + (70*(a*b^2 + b^3)*cosh(x)^4 + a^3 - a^2*b + 4* 
a*b^2 + 6*b^3 - 30*(a*b^2 + 2*b^3)*cosh(x)^2)*sinh(x)^4 + 4*(14*(a*b^2 + b 
^3)*cosh(x)^5 - 10*(a*b^2 + 2*b^3)*cosh(x)^3 + (a^3 - a^2*b + 4*a*b^2 + 6* 
b^3)*cosh(x))*sinh(x)^3 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 + 2*(a^3 - 3*a*b^2 
 - 2*b^3)*cosh(x)^2 + 2*(14*(a*b^2 + b^3)*cosh(x)^6 - 15*(a*b^2 + 2*b^3)*c 
osh(x)^4 + a^3 - 3*a*b^2 - 2*b^3 + 3*(a^3 - a^2*b + 4*a*b^2 + 6*b^3)*cosh( 
x)^2)*sinh(x)^2 + sqrt(2)*(b^2*cosh(x)^6 + 6*b^2*cosh(x)*sinh(x)^5 + b^2*s 
inh(x)^6 - 3*b^2*cosh(x)^4 + 3*(5*b^2*cosh(x)^2 - b^2)*sinh(x)^4 + 4*(5*b^ 
2*cosh(x)^3 - 3*b^2*cosh(x))*sinh(x)^3 - (a^2 - 2*a*b - 3*b^2)*cosh(x)^2 + 
 (15*b^2*cosh(x)^4 - 18*b^2*cosh(x)^2 - a^2 + 2*a*b + 3*b^2)*sinh(x)^2 - a 
^2 - 2*a*b - b^2 + 2*(3*b^2*cosh(x)^5 - 6*b^2*cosh(x)^3 - (a^2 - 2*a*b - 3 
*b^2)*cosh(x))*sinh(x))*sqrt(a + b)*sqrt(((a + b)*cosh(x)^2 + (a + b)*sinh 
(x)^2 + a - b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4*(2*(a*b...
 

Sympy [F]

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

integrate(1/(a+b*tanh(x)**2)**(3/2),x)
 

Output:

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

Maxima [F]

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

integrate(1/(a+b*tanh(x)^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate((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. 288 vs. \(2 (48) = 96\).

Time = 0.27 (sec) , antiderivative size = 288, normalized size of antiderivative = 5.14 \[ \int \frac {1}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\frac {\frac {{\left (a b^{2} + b^{3}\right )} e^{\left (2 \, x\right )}}{a^{3} b + 2 \, a^{2} b^{2} + a b^{3}} - \frac {a b^{2} + b^{3}}{a^{3} b + 2 \, a^{2} b^{2} + a b^{3}}}{\sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}} - \frac {\log \left ({\left | -{\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )} {\left (a + b\right )} - \sqrt {a + b} {\left (a - b\right )} \right |}\right )}{2 \, {\left (a + b\right )}^{\frac {3}{2}}} - \frac {\log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} + \sqrt {a + b} \right |}\right )}{2 \, {\left (a + b\right )}^{\frac {3}{2}}} + \frac {\log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} - \sqrt {a + b} \right |}\right )}{2 \, {\left (a + b\right )}^{\frac {3}{2}}} \] Input:

integrate(1/(a+b*tanh(x)^2)^(3/2),x, algorithm="giac")
 

Output:

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

int(1/(a + b*tanh(x)^2)^(3/2),x)
 

Output:

int(1/(a + b*tanh(x)^2)^(3/2), x)
 

Reduce [F]

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

int(1/(a+b*tanh(x)^2)^(3/2),x)
 

Output:

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