\(\int \coth ^4(x) \sqrt {a+b \text {sech}^2(x)} \, dx\) [185]

Optimal result

Integrand size = 17, antiderivative size = 84 \[ \int \coth ^4(x) \sqrt {a+b \text {sech}^2(x)} \, dx=\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )-\frac {(3 a+2 b) \coth (x) \sqrt {a+b-b \tanh ^2(x)}}{3 (a+b)}-\frac {1}{3} \coth ^3(x) \sqrt {a+b-b \tanh ^2(x)} \] Output:

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

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.77 \[ \int \coth ^4(x) \sqrt {a+b \text {sech}^2(x)} \, dx=\frac {\sqrt {2} \cosh (x) \sqrt {a+b \text {sech}^2(x)} \sqrt {a+b+a \sinh ^2(x)} \left (3 \sqrt {a} (a+b) \text {arcsinh}\left (\frac {\sqrt {a} \sinh (x)}{\sqrt {a+b}}\right )-\sqrt {a+b} \text {csch}(x) \left (4 a+3 b+(a+b) \text {csch}^2(x)\right ) \sqrt {\frac {a+b+a \sinh ^2(x)}{a+b}}\right )}{3 (a+b)^{3/2} \sqrt {a+2 b+a \cosh (2 x)} \sqrt {\frac {a+b+a \sinh ^2(x)}{a+b}}} \] Input:

Integrate[Coth[x]^4*Sqrt[a + b*Sech[x]^2],x]
 

Output:

(Sqrt[2]*Cosh[x]*Sqrt[a + b*Sech[x]^2]*Sqrt[a + b + a*Sinh[x]^2]*(3*Sqrt[a 
]*(a + b)*ArcSinh[(Sqrt[a]*Sinh[x])/Sqrt[a + b]] - Sqrt[a + b]*Csch[x]*(4* 
a + 3*b + (a + b)*Csch[x]^2)*Sqrt[(a + b + a*Sinh[x]^2)/(a + b)]))/(3*(a + 
 b)^(3/2)*Sqrt[a + 2*b + a*Cosh[2*x]]*Sqrt[(a + b + a*Sinh[x]^2)/(a + b)])
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {3042, 4629, 2075, 377, 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.

\(\displaystyle \int \coth ^4(x) \sqrt {a+b \text {sech}^2(x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a+b \sec (i x)^2}}{\tan (i x)^4}dx\)

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 2075

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

\(\Big \downarrow \) 377

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

\(\Big \downarrow \) 445

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

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

Input:

Int[Coth[x]^4*Sqrt[a + b*Sech[x]^2],x]
 

Output:

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

Defintions of rubi rules used

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 377

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

\[\int \coth \left (x \right )^{4} \sqrt {a +\operatorname {sech}\left (x \right )^{2} b}d x\]

Input:

int(coth(x)^4*(a+sech(x)^2*b)^(1/2),x)
 

Output:

int(coth(x)^4*(a+sech(x)^2*b)^(1/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 858 vs. \(2 (70) = 140\).

Time = 0.18 (sec) , antiderivative size = 2275, normalized size of antiderivative = 27.08 \[ \int \coth ^4(x) \sqrt {a+b \text {sech}^2(x)} \, dx=\text {Too large to display} \] Input:

integrate(coth(x)^4*(a+b*sech(x)^2)^(1/2),x, algorithm="fricas")
 

Output:

[1/12*(3*((a + b)*cosh(x)^6 + 6*(a + b)*cosh(x)*sinh(x)^5 + (a + b)*sinh(x 
)^6 - 3*(a + b)*cosh(x)^4 + 3*(5*(a + b)*cosh(x)^2 - a - b)*sinh(x)^4 + 4* 
(5*(a + b)*cosh(x)^3 - 3*(a + b)*cosh(x))*sinh(x)^3 + 3*(a + b)*cosh(x)^2 
+ 3*(5*(a + b)*cosh(x)^4 - 6*(a + b)*cosh(x)^2 + a + b)*sinh(x)^2 + 6*((a 
+ b)*cosh(x)^5 - 2*(a + b)*cosh(x)^3 + (a + b)*cosh(x))*sinh(x) - a - b)*s 
qrt(a)*log((a*b^2*cosh(x)^8 + 8*a*b^2*cosh(x)*sinh(x)^7 + a*b^2*sinh(x)^8 
- 2*(a*b^2 - b^3)*cosh(x)^6 + 2*(14*a*b^2*cosh(x)^2 - a*b^2 + b^3)*sinh(x) 
^6 + 4*(14*a*b^2*cosh(x)^3 - 3*(a*b^2 - b^3)*cosh(x))*sinh(x)^5 + (a^3 + 4 
*a^2*b + 9*a*b^2)*cosh(x)^4 + (70*a*b^2*cosh(x)^4 + a^3 + 4*a^2*b + 9*a*b^ 
2 - 30*(a*b^2 - b^3)*cosh(x)^2)*sinh(x)^4 + 4*(14*a*b^2*cosh(x)^5 - 10*(a* 
b^2 - b^3)*cosh(x)^3 + (a^3 + 4*a^2*b + 9*a*b^2)*cosh(x))*sinh(x)^3 + a^3 
+ 2*(a^3 + 3*a^2*b)*cosh(x)^2 + 2*(14*a*b^2*cosh(x)^6 - 15*(a*b^2 - b^3)*c 
osh(x)^4 + a^3 + 3*a^2*b + 3*(a^3 + 4*a^2*b + 9*a*b^2)*cosh(x)^2)*sinh(x)^ 
2 + sqrt(2)*(b^2*cosh(x)^6 + 6*b^2*cosh(x)*sinh(x)^5 + b^2*sinh(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 + 4*a*b)*cosh(x)^2 + (15*b^2*cosh(x)^4 - 1 
8*b^2*cosh(x)^2 - a^2 - 4*a*b)*sinh(x)^2 - a^2 + 2*(3*b^2*cosh(x)^5 - 6*b^ 
2*cosh(x)^3 - (a^2 + 4*a*b)*cosh(x))*sinh(x))*sqrt(a)*sqrt((a*cosh(x)^2 + 
a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4*(2 
*a*b^2*cosh(x)^7 - 3*(a*b^2 - b^3)*cosh(x)^5 + (a^3 + 4*a^2*b + 9*a*b^2...
 

Sympy [F]

\[ \int \coth ^4(x) \sqrt {a+b \text {sech}^2(x)} \, dx=\int \sqrt {a + b \operatorname {sech}^{2}{\left (x \right )}} \coth ^{4}{\left (x \right )}\, dx \] Input:

integrate(coth(x)**4*(a+b*sech(x)**2)**(1/2),x)
 

Output:

Integral(sqrt(a + b*sech(x)**2)*coth(x)**4, x)
 

Maxima [F]

\[ \int \coth ^4(x) \sqrt {a+b \text {sech}^2(x)} \, dx=\int { \sqrt {b \operatorname {sech}\left (x\right )^{2} + a} \coth \left (x\right )^{4} \,d x } \] Input:

integrate(coth(x)^4*(a+b*sech(x)^2)^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*sech(x)^2 + a)*coth(x)^4, x)
 

Giac [F(-2)]

Exception generated. \[ \int \coth ^4(x) \sqrt {a+b \text {sech}^2(x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(coth(x)^4*(a+b*sech(x)^2)^(1/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
 

Mupad [F(-1)]

Timed out. \[ \int \coth ^4(x) \sqrt {a+b \text {sech}^2(x)} \, dx=\int {\mathrm {coth}\left (x\right )}^4\,\sqrt {a+\frac {b}{{\mathrm {cosh}\left (x\right )}^2}} \,d x \] Input:

int(coth(x)^4*(a + b/cosh(x)^2)^(1/2),x)
 

Output:

int(coth(x)^4*(a + b/cosh(x)^2)^(1/2), x)
                                                                                    
                                                                                    
 

Reduce [F]

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

int(coth(x)^4*(a+b*sech(x)^2)^(1/2),x)
 

Output:

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