\(\int \frac {\coth (x)}{(a+b \tanh ^2(x))^{5/2}} \, dx\) [253]

Optimal result

Integrand size = 15, antiderivative size = 108 \[ \int \frac {\coth (x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{5/2}}+\frac {b}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}+\frac {b (2 a+b)}{a^2 (a+b)^2 \sqrt {a+b \tanh ^2(x)}} \] Output:

-arctanh((a+b*tanh(x)^2)^(1/2)/a^(1/2))/a^(5/2)+arctanh((a+b*tanh(x)^2)^(1 
/2)/(a+b)^(1/2))/(a+b)^(5/2)+1/3*b/a/(a+b)/(a+b*tanh(x)^2)^(3/2)+b*(2*a+b) 
/a^2/(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 = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.68 \[ \int \frac {\coth (x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\frac {-a \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tanh ^2(x)}{a+b}\right )+(a+b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},1+\frac {b \tanh ^2(x)}{a}\right )}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.31, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {3042, 26, 4153, 26, 354, 96, 25, 169, 27, 174, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 96

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

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 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 96

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Simp[f*((e + f*x)^(p + 1)/((p + 1)*(b*e - a*f)*(d*e - c*f))), x] + S 
imp[1/((b*e - a*f)*(d*e - c*f))   Int[(b*d*e - b*c*f - a*d*f - b*d*f*x)*((e 
 + f*x)^(p + 1)/((a + b*x)*(c + d*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, 
 x] && LtQ[p, -1]
 

rule 169

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n 
*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n, 2*p]
 

rule 174

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(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] && IntegerQ 
[(m - 1)/2]
 

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]

\[\int \frac {\coth \left (x \right )}{\left (a +b \tanh \left (x \right )^{2}\right )^{\frac {5}{2}}}d x\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4433 vs. \(2 (90) = 180\).

Time = 1.45 (sec) , antiderivative size = 19151, normalized size of antiderivative = 177.32 \[ \int \frac {\coth (x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate(coth(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. 808 vs. \(2 (90) = 180\).

Time = 0.45 (sec) , antiderivative size = 808, normalized size of antiderivative = 7.48 \[ \int \frac {\coth (x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/3*((((7*a^14*b^3 + 38*a^13*b^4 + 85*a^12*b^5 + 100*a^11*b^6 + 65*a^10*b^ 
7 + 22*a^9*b^8 + 3*a^8*b^9)*e^(2*x)/(a^16*b^2 + 6*a^15*b^3 + 15*a^14*b^4 + 
 20*a^13*b^5 + 15*a^12*b^6 + 6*a^11*b^7 + a^10*b^8) + 3*(7*a^14*b^3 + 30*a 
^13*b^4 + 49*a^12*b^5 + 36*a^11*b^6 + 9*a^10*b^7 - 2*a^9*b^8 - a^8*b^9)/(a 
^16*b^2 + 6*a^15*b^3 + 15*a^14*b^4 + 20*a^13*b^5 + 15*a^12*b^6 + 6*a^11*b^ 
7 + a^10*b^8))*e^(2*x) + 3*(7*a^14*b^3 + 30*a^13*b^4 + 49*a^12*b^5 + 36*a^ 
11*b^6 + 9*a^10*b^7 - 2*a^9*b^8 - a^8*b^9)/(a^16*b^2 + 6*a^15*b^3 + 15*a^1 
4*b^4 + 20*a^13*b^5 + 15*a^12*b^6 + 6*a^11*b^7 + a^10*b^8))*e^(2*x) + (7*a 
^14*b^3 + 38*a^13*b^4 + 85*a^12*b^5 + 100*a^11*b^6 + 65*a^10*b^7 + 22*a^9* 
b^8 + 3*a^8*b^9)/(a^16*b^2 + 6*a^15*b^3 + 15*a^14*b^4 + 20*a^13*b^5 + 15*a 
^12*b^6 + 6*a^11*b^7 + a^10*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)) + 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))/sqrt(-a))/(sqrt(-a)*a^2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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