\(\int \frac {\coth ^2(x)}{(a+b \text {sech}^2(x))^{5/2}} \, dx\) [219]

Optimal result

Integrand size = 17, antiderivative size = 133 \[ \int \frac {\coth ^2(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{a^{5/2}}-\frac {b \coth (x)}{3 a (a+b) \left (a+b-b \tanh ^2(x)\right )^{3/2}}-\frac {b (7 a+3 b) \coth (x)}{3 a^2 (a+b)^2 \sqrt {a+b-b \tanh ^2(x)}}-\frac {(a-3 b) (3 a+b) \coth (x) \sqrt {a+b-b \tanh ^2(x)}}{3 a^2 (a+b)^3} \] Output:

arctanh(a^(1/2)*tanh(x)/(a+b-b*tanh(x)^2)^(1/2))/a^(5/2)-1/3*b*coth(x)/a/( 
a+b)/(a+b-b*tanh(x)^2)^(3/2)-1/3*b*(7*a+3*b)*coth(x)/a^2/(a+b)^2/(a+b-b*ta 
nh(x)^2)^(1/2)-1/3*(a-3*b)*(3*a+b)*coth(x)*(a+b-b*tanh(x)^2)^(1/2)/a^2/(a+ 
b)^3
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.17 \[ \int \frac {\coth ^2(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\frac {\text {sech}^5(x) \left (\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sinh (x)}{\sqrt {a+2 b+a \cosh (2 x)}}\right ) (a+2 b+a \cosh (2 x))^{5/2}}{a^{5/2}}-\frac {(a+2 b+a \cosh (2 x)) \left (3 a^2 (a+2 b+a \cosh (2 x))^2 \text {csch}(x)-4 b^3 (a+b) \sinh (x)+2 b^2 (9 a+4 b) (a+2 b+a \cosh (2 x)) \sinh (x)\right )}{3 a^2 (a+b)^3}\right )}{8 \left (a+b \text {sech}^2(x)\right )^{5/2}} \] Input:

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

Output:

(Sech[x]^5*((Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[a]*Sinh[x])/Sqrt[a + 2*b + a*Co 
sh[2*x]]]*(a + 2*b + a*Cosh[2*x])^(5/2))/a^(5/2) - ((a + 2*b + a*Cosh[2*x] 
)*(3*a^2*(a + 2*b + a*Cosh[2*x])^2*Csch[x] - 4*b^3*(a + b)*Sinh[x] + 2*b^2 
*(9*a + 4*b)*(a + 2*b + a*Cosh[2*x])*Sinh[x]))/(3*a^2*(a + b)^3)))/(8*(a + 
 b*Sech[x]^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.17, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.765, Rules used = {3042, 25, 4629, 25, 2075, 374, 25, 441, 25, 445, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2075

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

\(\Big \downarrow \) 374

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 441

\(\displaystyle \frac {-\frac {\int -\frac {\coth ^2(x) \left (2 b (7 a+3 b) \tanh ^2(x)+(a-3 b) (3 a+b)\right )}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)}{a (a+b)}-\frac {b (7 a+3 b) \coth (x)}{a (a+b) \sqrt {a-b \tanh ^2(x)+b}}}{3 a (a+b)}-\frac {b \coth (x)}{3 a (a+b) \left (a-b \tanh ^2(x)+b\right )^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {\coth ^2(x) \left (2 b (7 a+3 b) \tanh ^2(x)+(a-3 b) (3 a+b)\right )}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)}{a (a+b)}-\frac {b (7 a+3 b) \coth (x)}{a (a+b) \sqrt {a-b \tanh ^2(x)+b}}}{3 a (a+b)}-\frac {b \coth (x)}{3 a (a+b) \left (a-b \tanh ^2(x)+b\right )^{3/2}}\)

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {-\frac {\int -\frac {3 (a+b)^3}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)}{a+b}-\frac {(a-3 b) (3 a+b) \coth (x) \sqrt {a-b \tanh ^2(x)+b}}{a+b}}{a (a+b)}-\frac {b (7 a+3 b) \coth (x)}{a (a+b) \sqrt {a-b \tanh ^2(x)+b}}}{3 a (a+b)}-\frac {b \coth (x)}{3 a (a+b) \left (a-b \tanh ^2(x)+b\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 (a+b)^2 \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {-b \tanh ^2(x)+a+b}}d\tanh (x)-\frac {(a-3 b) (3 a+b) \coth (x) \sqrt {a-b \tanh ^2(x)+b}}{a+b}}{a (a+b)}-\frac {b (7 a+3 b) \coth (x)}{a (a+b) \sqrt {a-b \tanh ^2(x)+b}}}{3 a (a+b)}-\frac {b \coth (x)}{3 a (a+b) \left (a-b \tanh ^2(x)+b\right )^{3/2}}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {3 (a+b)^2 \int \frac {1}{1-\frac {a \tanh ^2(x)}{-b \tanh ^2(x)+a+b}}d\frac {\tanh (x)}{\sqrt {-b \tanh ^2(x)+a+b}}-\frac {(a-3 b) (3 a+b) \coth (x) \sqrt {a-b \tanh ^2(x)+b}}{a+b}}{a (a+b)}-\frac {b (7 a+3 b) \coth (x)}{a (a+b) \sqrt {a-b \tanh ^2(x)+b}}}{3 a (a+b)}-\frac {b \coth (x)}{3 a (a+b) \left (a-b \tanh ^2(x)+b\right )^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {3 (a+b)^2 \text {arctanh}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )}{\sqrt {a}}-\frac {(a-3 b) (3 a+b) \coth (x) \sqrt {a-b \tanh ^2(x)+b}}{a+b}}{a (a+b)}-\frac {b (7 a+3 b) \coth (x)}{a (a+b) \sqrt {a-b \tanh ^2(x)+b}}}{3 a (a+b)}-\frac {b \coth (x)}{3 a (a+b) \left (a-b \tanh ^2(x)+b\right )^{3/2}}\)

Input:

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

Output:

-1/3*(b*Coth[x])/(a*(a + b)*(a + b - b*Tanh[x]^2)^(3/2)) + (-((b*(7*a + 3* 
b)*Coth[x])/(a*(a + b)*Sqrt[a + b - b*Tanh[x]^2])) + ((3*(a + b)^2*ArcTanh 
[(Sqrt[a]*Tanh[x])/Sqrt[a + b - b*Tanh[x]^2]])/Sqrt[a] - ((a - 3*b)*(3*a + 
 b)*Coth[x]*Sqrt[a + b - b*Tanh[x]^2])/(a + b))/(a*(a + b)))/(3*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 441

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

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 2075

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa 
ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi 
alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] &&  ! 
BinomialMatchQ[{u, v}, x]
 

rule 3042

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

rule 4629

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f 
_.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim 
p[ff/f   Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 
)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte 
gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5323 vs. \(2 (115) = 230\).

Time = 0.85 (sec) , antiderivative size = 11205, normalized size of antiderivative = 84.25 \[ \int \frac {\coth ^2(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 854 vs. \(2 (115) = 230\).

Time = 0.39 (sec) , antiderivative size = 854, normalized size of antiderivative = 6.42 \[ \int \frac {\coth ^2(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/3*((((9*a^13*b^4 + 67*a^12*b^5 + 217*a^11*b^6 + 399*a^10*b^7 + 455*a^9* 
b^8 + 329*a^8*b^9 + 147*a^7*b^10 + 37*a^6*b^11 + 4*a^5*b^12)*e^(2*x)/(a^16 
*b^2 + 10*a^15*b^3 + 45*a^14*b^4 + 120*a^13*b^5 + 210*a^12*b^6 + 252*a^11* 
b^7 + 210*a^10*b^8 + 120*a^9*b^9 + 45*a^8*b^10 + 10*a^7*b^11 + a^6*b^12) + 
 3*(3*a^13*b^4 + 33*a^12*b^5 + 151*a^11*b^6 + 385*a^10*b^7 + 609*a^9*b^8 + 
 623*a^8*b^9 + 413*a^7*b^10 + 171*a^6*b^11 + 40*a^5*b^12 + 4*a^4*b^13)/(a^ 
16*b^2 + 10*a^15*b^3 + 45*a^14*b^4 + 120*a^13*b^5 + 210*a^12*b^6 + 252*a^1 
1*b^7 + 210*a^10*b^8 + 120*a^9*b^9 + 45*a^8*b^10 + 10*a^7*b^11 + a^6*b^12) 
)*e^(2*x) - 3*(3*a^13*b^4 + 33*a^12*b^5 + 151*a^11*b^6 + 385*a^10*b^7 + 60 
9*a^9*b^8 + 623*a^8*b^9 + 413*a^7*b^10 + 171*a^6*b^11 + 40*a^5*b^12 + 4*a^ 
4*b^13)/(a^16*b^2 + 10*a^15*b^3 + 45*a^14*b^4 + 120*a^13*b^5 + 210*a^12*b^ 
6 + 252*a^11*b^7 + 210*a^10*b^8 + 120*a^9*b^9 + 45*a^8*b^10 + 10*a^7*b^11 
+ a^6*b^12))*e^(2*x) - (9*a^13*b^4 + 67*a^12*b^5 + 217*a^11*b^6 + 399*a^10 
*b^7 + 455*a^9*b^8 + 329*a^8*b^9 + 147*a^7*b^10 + 37*a^6*b^11 + 4*a^5*b^12 
)/(a^16*b^2 + 10*a^15*b^3 + 45*a^14*b^4 + 120*a^13*b^5 + 210*a^12*b^6 + 25 
2*a^11*b^7 + 210*a^10*b^8 + 120*a^9*b^9 + 45*a^8*b^10 + 10*a^7*b^11 + a^6* 
b^12))/(a*e^(4*x) + 2*a*e^(2*x) + 4*b*e^(2*x) + a)^(3/2) + 4*(sqrt(a)*e^(2 
*x) - sqrt(a*e^(4*x) + 2*a*e^(2*x) + 4*b*e^(2*x) + a) + sqrt(a))/(((sqrt(a 
)*e^(2*x) - sqrt(a*e^(4*x) + 2*a*e^(2*x) + 4*b*e^(2*x) + a))^2 - 2*(sqrt(a 
)*e^(2*x) - sqrt(a*e^(4*x) + 2*a*e^(2*x) + 4*b*e^(2*x) + a))*sqrt(a) - ...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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