\(\int \frac {\coth ^2(x)}{(a+b \tanh ^2(x))^{3/2}} \, dx\) [245]

Optimal result

Integrand size = 17, antiderivative size = 85 \[ \int \frac {\coth ^2(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b} \tanh (x)}{\sqrt {a+b \tanh ^2(x)}}\right )}{(a+b)^{3/2}}+\frac {b \coth (x)}{a (a+b) \sqrt {a+b \tanh ^2(x)}}-\frac {(a+2 b) \coth (x) \sqrt {a+b \tanh ^2(x)}}{a^2 (a+b)} \] Output:

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

Mathematica [C] (warning: unable to verify)

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

Time = 10.65 (sec) , antiderivative size = 263, normalized size of antiderivative = 3.09 \[ \int \frac {\coth ^2(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=-\frac {\cosh ^2(x) \coth (x) \left (\frac {8 (a+b) \cosh ^2(x) \, _3F_2\left (2,2,2;1,\frac {7}{2};-\frac {(a+b) \sinh ^2(x)}{a}\right ) \left (i a \tanh (x)+i b \tanh ^3(x)\right )^2}{15 a^3}-\frac {8 (a+b) \operatorname {Hypergeometric2F1}\left (2,2,\frac {7}{2},-\frac {(a+b) \sinh ^2(x)}{a}\right ) \sinh ^2(x) \left (2 a^2+5 a b \tanh ^2(x)+3 b^2 \tanh ^4(x)\right )}{15 a^3}-\frac {\coth ^2(x) \left (3 a^2+12 a b \tanh ^2(x)+8 b^2 \tanh ^4(x)\right ) \left (\arcsin \left (\sqrt {-\frac {(a+b) \sinh ^2(x)}{a}}\right ) \left (-a-b \tanh ^2(x)\right )+a \text {sech}^2(x) \sqrt {-\frac {(a+b) \cosh ^2(x) \sinh ^2(x) \left (a+b \tanh ^2(x)\right )}{a^2}}\right )}{a^2 (a+b) \sqrt {-\frac {(a+b) \cosh ^2(x) \sinh ^2(x) \left (a+b \tanh ^2(x)\right )}{a^2}}}\right )}{a \sqrt {a+b \tanh ^2(x)}} \] Input:

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

Output:

-((Cosh[x]^2*Coth[x]*((8*(a + b)*Cosh[x]^2*HypergeometricPFQ[{2, 2, 2}, {1 
, 7/2}, -(((a + b)*Sinh[x]^2)/a)]*(I*a*Tanh[x] + I*b*Tanh[x]^3)^2)/(15*a^3 
) - (8*(a + b)*Hypergeometric2F1[2, 2, 7/2, -(((a + b)*Sinh[x]^2)/a)]*Sinh 
[x]^2*(2*a^2 + 5*a*b*Tanh[x]^2 + 3*b^2*Tanh[x]^4))/(15*a^3) - (Coth[x]^2*( 
3*a^2 + 12*a*b*Tanh[x]^2 + 8*b^2*Tanh[x]^4)*(ArcSin[Sqrt[-(((a + b)*Sinh[x 
]^2)/a)]]*(-a - b*Tanh[x]^2) + a*Sech[x]^2*Sqrt[-(((a + b)*Cosh[x]^2*Sinh[ 
x]^2*(a + b*Tanh[x]^2))/a^2)]))/(a^2*(a + b)*Sqrt[-(((a + b)*Cosh[x]^2*Sin 
h[x]^2*(a + b*Tanh[x]^2))/a^2)])))/(a*Sqrt[a + b*Tanh[x]^2]))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.647, Rules used = {3042, 25, 4153, 25, 374, 25, 445, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 374

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 445

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 374

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

rule 445

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

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 [F]

\[\int \frac {\coth \left (x \right )^{2}}{\left (a +b \tanh \left (x \right )^{2}\right )^{\frac {3}{2}}}d x\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1615 vs. \(2 (75) = 150\).

Time = 0.33 (sec) , antiderivative size = 3859, normalized size of antiderivative = 45.40 \[ \int \frac {\coth ^2(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate(coth(x)^2/(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. 459 vs. \(2 (75) = 150\).

Time = 0.38 (sec) , antiderivative size = 459, normalized size of antiderivative = 5.40 \[ \int \frac {\coth ^2(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=-\frac {\frac {{\left (a^{2} b^{3} + a b^{4}\right )} e^{\left (2 \, x\right )}}{a^{5} b + 2 \, a^{4} b^{2} + a^{3} b^{3}} - \frac {a^{2} b^{3} + a b^{4}}{a^{5} b + 2 \, a^{4} b^{2} + a^{3} 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}}} + \frac {4 \, {\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 )}}{{\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 )}^{2} - 2 \, {\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 )} \sqrt {a + b} - 3 \, a + b\right )} a} \] Input:

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

Output:

-((a^2*b^3 + a*b^4)*e^(2*x)/(a^5*b + 2*a^4*b^2 + a^3*b^3) - (a^2*b^3 + a*b 
^4)/(a^5*b + 2*a^4*b^2 + a^3*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) + 4*(sqr 
t(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))/(((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))^2 - 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) - 
3*a + b)*a)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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