\(\int \frac {\coth ^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx\) [147]

Optimal result

Integrand size = 23, antiderivative size = 87 \[ \int \frac {\coth ^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {x}{a}-\frac {b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a (a+b)^{5/2} d}-\frac {(a+2 b) \coth (c+d x)}{(a+b)^2 d}-\frac {\coth ^3(c+d x)}{3 (a+b) d} \] Output:

x/a-b^(5/2)*arctanh(b^(1/2)*tanh(d*x+c)/(a+b)^(1/2))/a/(a+b)^(5/2)/d-(a+2* 
b)*coth(d*x+c)/(a+b)^2/d-1/3*coth(d*x+c)^3/(a+b)/d
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(380\) vs. \(2(87)=174\).

Time = 2.69 (sec) , antiderivative size = 380, normalized size of antiderivative = 4.37 \[ \int \frac {\coth ^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {(a+2 b+a \cosh (2 (c+d x))) \text {sech}^2(c+d x) \left (3 b^3 \text {arctanh}\left (\frac {\text {sech}(d x) (\cosh (2 c)-\sinh (2 c)) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right ) (-\cosh (2 c)+\sinh (2 c))+\frac {1}{8} \sqrt {a+b} \text {csch}(c) \text {csch}^3(c+d x) \sqrt {b (\cosh (c)-\sinh (c))^4} \left (9 (a+b)^2 d x \cosh (d x)-9 (a+b)^2 d x \cosh (2 c+d x)-3 a^2 d x \cosh (2 c+3 d x)-6 a b d x \cosh (2 c+3 d x)-3 b^2 d x \cosh (2 c+3 d x)+3 a^2 d x \cosh (4 c+3 d x)+6 a b d x \cosh (4 c+3 d x)+3 b^2 d x \cosh (4 c+3 d x)-12 a^2 \sinh (d x)-24 a b \sinh (d x)-12 a^2 \sinh (2 c+d x)-18 a b \sinh (2 c+d x)+8 a^2 \sinh (2 c+3 d x)+14 a b \sinh (2 c+3 d x)\right )\right )}{6 a (a+b)^{5/2} d \left (a+b \text {sech}^2(c+d x)\right ) \sqrt {b (\cosh (c)-\sinh (c))^4}} \] Input:

Integrate[Coth[c + d*x]^4/(a + b*Sech[c + d*x]^2),x]
 

Output:

((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^2*(3*b^3*ArcTanh[(Sech[d*x] 
*(Cosh[2*c] - Sinh[2*c])*((a + 2*b)*Sinh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqr 
t[a + b]*Sqrt[b*(Cosh[c] - Sinh[c])^4])]*(-Cosh[2*c] + Sinh[2*c]) + (Sqrt[ 
a + b]*Csch[c]*Csch[c + d*x]^3*Sqrt[b*(Cosh[c] - Sinh[c])^4]*(9*(a + b)^2* 
d*x*Cosh[d*x] - 9*(a + b)^2*d*x*Cosh[2*c + d*x] - 3*a^2*d*x*Cosh[2*c + 3*d 
*x] - 6*a*b*d*x*Cosh[2*c + 3*d*x] - 3*b^2*d*x*Cosh[2*c + 3*d*x] + 3*a^2*d* 
x*Cosh[4*c + 3*d*x] + 6*a*b*d*x*Cosh[4*c + 3*d*x] + 3*b^2*d*x*Cosh[4*c + 3 
*d*x] - 12*a^2*Sinh[d*x] - 24*a*b*Sinh[d*x] - 12*a^2*Sinh[2*c + d*x] - 18* 
a*b*Sinh[2*c + d*x] + 8*a^2*Sinh[2*c + 3*d*x] + 14*a*b*Sinh[2*c + 3*d*x])) 
/8))/(6*a*(a + b)^(5/2)*d*(a + b*Sech[c + d*x]^2)*Sqrt[b*(Cosh[c] - Sinh[c 
])^4])
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.23, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 4629, 2075, 382, 27, 445, 25, 397, 219, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (i c+i d x)^4 \left (a+b \sec (i c+i d x)^2\right )}dx\)

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 2075

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

\(\Big \downarrow \) 382

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 445

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 397

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {\frac {(a+b)^2 \text {arctanh}(\tanh (c+d x))}{a}-\frac {b^3 \int \frac {1}{-b \tanh ^2(c+d x)+a+b}d\tanh (c+d x)}{a}}{a+b}-\frac {(a+2 b) \coth (c+d x)}{a+b}}{a+b}-\frac {\coth ^3(c+d x)}{3 (a+b)}}{d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {\frac {(a+b)^2 \text {arctanh}(\tanh (c+d x))}{a}-\frac {b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}}{a+b}-\frac {(a+2 b) \coth (c+d x)}{a+b}}{a+b}-\frac {\coth ^3(c+d x)}{3 (a+b)}}{d}\)

Input:

Int[Coth[c + d*x]^4/(a + b*Sech[c + d*x]^2),x]
 

Output:

(-1/3*Coth[c + d*x]^3/(a + b) + ((((a + b)^2*ArcTanh[Tanh[c + d*x]])/a - ( 
b^(5/2)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(a*Sqrt[a + b]))/(a 
+ b) - ((a + 2*b)*Coth[c + d*x])/(a + b))/(a + b))/d
 

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 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 382

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 + 1)/ 
(a*c*e*(m + 1))), x] - Simp[1/(a*c*e^2*(m + 1))   Int[(e*x)^(m + 2)*(a + b* 
x^2)^p*(c + d*x^2)^q*Simp[(b*c + a*d)*(m + 3) + 2*(b*c*p + a*d*q) + b*d*(m 
+ 2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[ 
b*c - a*d, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
 

rule 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(191\) vs. \(2(77)=154\).

Time = 4.30 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.21

method	result	size
risch	\(\frac {x}{a}-\frac {2 \left (6 \,{\mathrm e}^{4 d x +4 c} a +9 \,{\mathrm e}^{4 d x +4 c} b -6 a \,{\mathrm e}^{2 d x +2 c}-12 b \,{\mathrm e}^{2 d x +2 c}+4 a +7 b \right )}{3 d \left (a +b \right )^{2} \left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}+\frac {\sqrt {b \left (a +b \right )}\, b^{2} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a +2 \sqrt {b \left (a +b \right )}+2 b}{a}\right )}{2 \left (a +b \right )^{3} d a}-\frac {\sqrt {b \left (a +b \right )}\, b^{2} \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {-a +2 \sqrt {b \left (a +b \right )}-2 b}{a}\right )}{2 \left (a +b \right )^{3} d a}\)	\(192\)
derivativedivides	\(\frac {-\frac {\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a}{3}+\frac {b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+5 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+9 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 \left (a +b \right )^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a}-\frac {1}{24 \left (a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {5 a +9 b}{8 \left (a +b \right )^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 b^{3} \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{a \left (a +b \right )^{2}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a}}{d}\)	\(252\)
default	\(\frac {-\frac {\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a}{3}+\frac {b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+5 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+9 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 \left (a +b \right )^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a}-\frac {1}{24 \left (a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {5 a +9 b}{8 \left (a +b \right )^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 b^{3} \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{a \left (a +b \right )^{2}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a}}{d}\)	\(252\)

Input:

int(coth(d*x+c)^4/(a+b*sech(d*x+c)^2),x,method=_RETURNVERBOSE)
 

Output:

x/a-2/3*(6*exp(4*d*x+4*c)*a+9*exp(4*d*x+4*c)*b-6*a*exp(2*d*x+2*c)-12*b*exp 
(2*d*x+2*c)+4*a+7*b)/d/(a+b)^2/(exp(2*d*x+2*c)-1)^3+1/2*(b*(a+b))^(1/2)/(a 
+b)^3*b^2/d/a*ln(exp(2*d*x+2*c)+(a+2*(b*(a+b))^(1/2)+2*b)/a)-1/2*(b*(a+b)) 
^(1/2)/(a+b)^3*b^2/d/a*ln(exp(2*d*x+2*c)-(-a+2*(b*(a+b))^(1/2)-2*b)/a)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1214 vs. \(2 (77) = 154\).

Time = 0.42 (sec) , antiderivative size = 2705, normalized size of antiderivative = 31.09 \[ \int \frac {\coth ^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/6*(6*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^6 + 36*(a^2 + 2*a*b + b^2)*d 
*x*cosh(d*x + c)*sinh(d*x + c)^5 + 6*(a^2 + 2*a*b + b^2)*d*x*sinh(d*x + c) 
^6 - 6*(3*(a^2 + 2*a*b + b^2)*d*x + 4*a^2 + 6*a*b)*cosh(d*x + c)^4 + 6*(15 
*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^2 - 3*(a^2 + 2*a*b + b^2)*d*x - 4*a 
^2 - 6*a*b)*sinh(d*x + c)^4 + 24*(5*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^ 
3 - (3*(a^2 + 2*a*b + b^2)*d*x + 4*a^2 + 6*a*b)*cosh(d*x + c))*sinh(d*x + 
c)^3 - 6*(a^2 + 2*a*b + b^2)*d*x + 6*(3*(a^2 + 2*a*b + b^2)*d*x + 4*a^2 + 
8*a*b)*cosh(d*x + c)^2 + 6*(15*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^4 + 3 
*(a^2 + 2*a*b + b^2)*d*x - 6*(3*(a^2 + 2*a*b + b^2)*d*x + 4*a^2 + 6*a*b)*c 
osh(d*x + c)^2 + 4*a^2 + 8*a*b)*sinh(d*x + c)^2 + 3*(b^2*cosh(d*x + c)^6 + 
 6*b^2*cosh(d*x + c)*sinh(d*x + c)^5 + b^2*sinh(d*x + c)^6 - 3*b^2*cosh(d* 
x + c)^4 + 3*(5*b^2*cosh(d*x + c)^2 - b^2)*sinh(d*x + c)^4 + 3*b^2*cosh(d* 
x + c)^2 + 4*(5*b^2*cosh(d*x + c)^3 - 3*b^2*cosh(d*x + c))*sinh(d*x + c)^3 
 + 3*(5*b^2*cosh(d*x + c)^4 - 6*b^2*cosh(d*x + c)^2 + b^2)*sinh(d*x + c)^2 
 - b^2 + 6*(b^2*cosh(d*x + c)^5 - 2*b^2*cosh(d*x + c)^3 + b^2*cosh(d*x + c 
))*sinh(d*x + c))*sqrt(b/(a + b))*log((a^2*cosh(d*x + c)^4 + 4*a^2*cosh(d* 
x + c)*sinh(d*x + c)^3 + a^2*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b)*cosh(d*x + 
c)^2 + 2*(3*a^2*cosh(d*x + c)^2 + a^2 + 2*a*b)*sinh(d*x + c)^2 + a^2 + 8*a 
*b + 8*b^2 + 4*(a^2*cosh(d*x + c)^3 + (a^2 + 2*a*b)*cosh(d*x + c))*sinh(d* 
x + c) + 4*((a^2 + a*b)*cosh(d*x + c)^2 + 2*(a^2 + a*b)*cosh(d*x + c)*s...
 

Sympy [F]

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

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

Output:

Integral(coth(c + d*x)**4/(a + b*sech(c + d*x)**2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1435 vs. \(2 (77) = 154\).

Time = 0.27 (sec) , antiderivative size = 1435, normalized size of antiderivative = 16.49 \[ \int \frac {\coth ^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

3/16*a*b*log((a*e^(-2*d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^(-2*d 
*x - 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/((a^2 + 2*a*b + b^2)*sqrt((a + b 
)*b)*d) + 1/8*(a*b + 2*b^2)*log(a*e^(4*d*x + 4*c) + 2*(a + 2*b)*e^(2*d*x + 
 2*c) + a)/((a^3 + 2*a^2*b + a*b^2)*d) - 1/4*b*log(a*e^(4*d*x + 4*c) + 2*( 
a + 2*b)*e^(2*d*x + 2*c) + a)/((a^2 + 2*a*b + b^2)*d) - 1/8*(a*b + 2*b^2)* 
log(2*(a + 2*b)*e^(-2*d*x - 2*c) + a*e^(-4*d*x - 4*c) + a)/((a^3 + 2*a^2*b 
 + a*b^2)*d) + 1/4*b*log(2*(a + 2*b)*e^(-2*d*x - 2*c) + a*e^(-4*d*x - 4*c) 
 + a)/((a^2 + 2*a*b + b^2)*d) + 1/4*(2*a + 3*b)*log(e^(2*d*x + 2*c) - 1)/( 
(a^2 + 2*a*b + b^2)*d) + 1/2*b*log(e^(2*d*x + 2*c) - 1)/((a^2 + 2*a*b + b^ 
2)*d) - 1/4*(2*a + 3*b)*log(e^(-2*d*x - 2*c) - 1)/((a^2 + 2*a*b + b^2)*d) 
- 1/2*b*log(e^(-2*d*x - 2*c) - 1)/((a^2 + 2*a*b + b^2)*d) - 1/32*(a^2*b + 
8*a*b^2 + 8*b^3)*log((a*e^(2*d*x + 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a* 
e^(2*d*x + 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/((a^3 + 2*a^2*b + a*b^2)*s 
qrt((a + b)*b)*d) + 1/8*(a*b + 2*b^2)*log((a*e^(2*d*x + 2*c) + a + 2*b - 2 
*sqrt((a + b)*b))/(a*e^(2*d*x + 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/((a^2 
 + 2*a*b + b^2)*sqrt((a + b)*b)*d) + 1/32*(a^2*b + 8*a*b^2 + 8*b^3)*log((a 
*e^(-2*d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^(-2*d*x - 2*c) + a + 
 2*b + 2*sqrt((a + b)*b)))/((a^3 + 2*a^2*b + a*b^2)*sqrt((a + b)*b)*d) - 1 
/8*(a*b + 2*b^2)*log((a*e^(-2*d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a 
*e^(-2*d*x - 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/((a^2 + 2*a*b + b^2)*...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (77) = 154\).

Time = 0.53 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.89 \[ \int \frac {\coth ^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=-\frac {\frac {3 \, b^{3} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right )}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt {-a b - b^{2}}} - \frac {3 \, {\left (d x + c\right )}}{a} + \frac {2 \, {\left (6 \, a e^{\left (4 \, d x + 4 \, c\right )} + 9 \, b e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a e^{\left (2 \, d x + 2 \, c\right )} - 12 \, b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a + 7 \, b\right )}}{{\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{3 \, d} \] Input:

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

Output:

-1/3*(3*b^3*arctan(1/2*(a*e^(2*d*x + 2*c) + a + 2*b)/sqrt(-a*b - b^2))/((a 
^3 + 2*a^2*b + a*b^2)*sqrt(-a*b - b^2)) - 3*(d*x + c)/a + 2*(6*a*e^(4*d*x 
+ 4*c) + 9*b*e^(4*d*x + 4*c) - 6*a*e^(2*d*x + 2*c) - 12*b*e^(2*d*x + 2*c) 
+ 4*a + 7*b)/((a^2 + 2*a*b + b^2)*(e^(2*d*x + 2*c) - 1)^3))/d
 

Mupad [B] (verification not implemented)

Time = 4.36 (sec) , antiderivative size = 779, normalized size of antiderivative = 8.95 \[ \int \frac {\coth ^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {x}{a}-\frac {8}{3\,\left (a\,d+b\,d\right )\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1\right )}-\frac {\mathrm {atan}\left (\left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (\frac {4\,b^3}{a^3\,d\,{\left (a+b\right )}^2\,\sqrt {b^5}\,\left (a^3+2\,a^2\,b+a\,b^2\right )}+\frac {\left (a+2\,b\right )\,\left (a^4\,d\,\sqrt {b^5}+2\,a\,b^3\,d\,\sqrt {b^5}+4\,a^3\,b\,d\,\sqrt {b^5}+5\,a^2\,b^2\,d\,\sqrt {b^5}\right )}{a^2\,b^3\,\left (a^3+2\,a^2\,b+a\,b^2\right )\,\sqrt {-a^2\,d^2\,{\left (a+b\right )}^5}\,\sqrt {-a^7\,d^2-5\,a^6\,b\,d^2-10\,a^5\,b^2\,d^2-10\,a^4\,b^3\,d^2-5\,a^3\,b^4\,d^2-a^2\,b^5\,d^2}}\right )+\frac {\left (a+2\,b\right )\,\left (a^4\,d\,\sqrt {b^5}+2\,a^3\,b\,d\,\sqrt {b^5}+a^2\,b^2\,d\,\sqrt {b^5}\right )}{a^2\,b^3\,\left (a^3+2\,a^2\,b+a\,b^2\right )\,\sqrt {-a^2\,d^2\,{\left (a+b\right )}^5}\,\sqrt {-a^7\,d^2-5\,a^6\,b\,d^2-10\,a^5\,b^2\,d^2-10\,a^4\,b^3\,d^2-5\,a^3\,b^4\,d^2-a^2\,b^5\,d^2}}\right )\,\left (\frac {a^4\,\sqrt {-a^7\,d^2-5\,a^6\,b\,d^2-10\,a^5\,b^2\,d^2-10\,a^4\,b^3\,d^2-5\,a^3\,b^4\,d^2-a^2\,b^5\,d^2}}{2}+\frac {a^2\,b^2\,\sqrt {-a^7\,d^2-5\,a^6\,b\,d^2-10\,a^5\,b^2\,d^2-10\,a^4\,b^3\,d^2-5\,a^3\,b^4\,d^2-a^2\,b^5\,d^2}}{2}+a^3\,b\,\sqrt {-a^7\,d^2-5\,a^6\,b\,d^2-10\,a^5\,b^2\,d^2-10\,a^4\,b^3\,d^2-5\,a^3\,b^4\,d^2-a^2\,b^5\,d^2}\right )\right )\,\sqrt {b^5}}{\sqrt {-a^7\,d^2-5\,a^6\,b\,d^2-10\,a^5\,b^2\,d^2-10\,a^4\,b^3\,d^2-5\,a^3\,b^4\,d^2-a^2\,b^5\,d^2}}-\frac {4\,\left (a^2+b\,a\right )}{a\,\left (a+b\right )\,\left (a\,d+b\,d\right )\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {2\,\left (2\,a^2+3\,b\,a\right )}{a\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )\,\left (a+b\right )\,\left (a\,d+b\,d\right )} \] Input:

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

Output:

x/a - 8/(3*(a*d + b*d)*(3*exp(2*c + 2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c 
+ 6*d*x) - 1)) - (atan((exp(2*c)*exp(2*d*x)*((4*b^3)/(a^3*d*(a + b)^2*(b^5 
)^(1/2)*(a*b^2 + 2*a^2*b + a^3)) + ((a + 2*b)*(a^4*d*(b^5)^(1/2) + 2*a*b^3 
*d*(b^5)^(1/2) + 4*a^3*b*d*(b^5)^(1/2) + 5*a^2*b^2*d*(b^5)^(1/2)))/(a^2*b^ 
3*(a*b^2 + 2*a^2*b + a^3)*(-a^2*d^2*(a + b)^5)^(1/2)*(- a^7*d^2 - 5*a^6*b* 
d^2 - a^2*b^5*d^2 - 5*a^3*b^4*d^2 - 10*a^4*b^3*d^2 - 10*a^5*b^2*d^2)^(1/2) 
)) + ((a + 2*b)*(a^4*d*(b^5)^(1/2) + 2*a^3*b*d*(b^5)^(1/2) + a^2*b^2*d*(b^ 
5)^(1/2)))/(a^2*b^3*(a*b^2 + 2*a^2*b + a^3)*(-a^2*d^2*(a + b)^5)^(1/2)*(- 
a^7*d^2 - 5*a^6*b*d^2 - a^2*b^5*d^2 - 5*a^3*b^4*d^2 - 10*a^4*b^3*d^2 - 10* 
a^5*b^2*d^2)^(1/2)))*((a^4*(- a^7*d^2 - 5*a^6*b*d^2 - a^2*b^5*d^2 - 5*a^3* 
b^4*d^2 - 10*a^4*b^3*d^2 - 10*a^5*b^2*d^2)^(1/2))/2 + (a^2*b^2*(- a^7*d^2 
- 5*a^6*b*d^2 - a^2*b^5*d^2 - 5*a^3*b^4*d^2 - 10*a^4*b^3*d^2 - 10*a^5*b^2* 
d^2)^(1/2))/2 + a^3*b*(- a^7*d^2 - 5*a^6*b*d^2 - a^2*b^5*d^2 - 5*a^3*b^4*d 
^2 - 10*a^4*b^3*d^2 - 10*a^5*b^2*d^2)^(1/2)))*(b^5)^(1/2))/(- a^7*d^2 - 5* 
a^6*b*d^2 - a^2*b^5*d^2 - 5*a^3*b^4*d^2 - 10*a^4*b^3*d^2 - 10*a^5*b^2*d^2) 
^(1/2) - (4*(a*b + a^2))/(a*(a + b)*(a*d + b*d)*(exp(4*c + 4*d*x) - 2*exp( 
2*c + 2*d*x) + 1)) - (2*(3*a*b + 2*a^2))/(a*(exp(2*c + 2*d*x) - 1)*(a + b) 
*(a*d + b*d))
 

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 1112, normalized size of antiderivative = 12.78 \[ \int \frac {\coth ^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 3*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a + b)*log( - sqrt(2*sqrt(b)*sqrt(a + 
b) - a - 2*b) + e**(c + d*x)*sqrt(a))*b**2 - 3*e**(6*c + 6*d*x)*sqrt(b)*sq 
rt(a + b)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a) 
)*b**2 + 3*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a + b)*log(2*sqrt(b)*sqrt(a + b) 
+ e**(2*c + 2*d*x)*a + a + 2*b)*b**2 + 9*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a + 
 b)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*b 
**2 + 9*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a + b)*log(sqrt(2*sqrt(b)*sqrt(a + b 
) - a - 2*b) + e**(c + d*x)*sqrt(a))*b**2 - 9*e**(4*c + 4*d*x)*sqrt(b)*sqr 
t(a + b)*log(2*sqrt(b)*sqrt(a + b) + e**(2*c + 2*d*x)*a + a + 2*b)*b**2 - 
9*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a + b)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - 
 a - 2*b) + e**(c + d*x)*sqrt(a))*b**2 - 9*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a 
 + b)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*b* 
*2 + 9*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a + b)*log(2*sqrt(b)*sqrt(a + b) + e* 
*(2*c + 2*d*x)*a + a + 2*b)*b**2 + 3*sqrt(b)*sqrt(a + b)*log( - sqrt(2*sqr 
t(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*b**2 + 3*sqrt(b)*sqrt( 
a + b)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*b 
**2 - 3*sqrt(b)*sqrt(a + b)*log(2*sqrt(b)*sqrt(a + b) + e**(2*c + 2*d*x)*a 
 + a + 2*b)*b**2 + 6*e**(6*c + 6*d*x)*a**3*d*x - 8*e**(6*c + 6*d*x)*a**3 + 
 18*e**(6*c + 6*d*x)*a**2*b*d*x - 20*e**(6*c + 6*d*x)*a**2*b + 18*e**(6*c 
+ 6*d*x)*a*b**2*d*x - 12*e**(6*c + 6*d*x)*a*b**2 + 6*e**(6*c + 6*d*x)*b...