\(\int \frac {\coth ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx\) [145]

Optimal result

Integrand size = 23, antiderivative size = 62 \[ \int \frac {\coth ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {x}{a}-\frac {b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a (a+b)^{3/2} d}-\frac {\coth (c+d x)}{(a+b) d} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(193\) vs. \(2(62)=124\).

Time = 1.25 (sec) , antiderivative size = 193, normalized size of antiderivative = 3.11 \[ \int \frac {\coth ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {(a+2 b+a \cosh (2 (c+d x))) \text {sech}^2(c+d x) \left (b^2 \text {arctanh}\left (\frac {\text {sech}(d x) (\cosh (2 c)-\sinh (2 c)) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right ) (-\cosh (2 c)+\sinh (2 c))+\sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4} ((a+b) d x+a \text {csch}(c) \text {csch}(c+d x) \sinh (d x))\right )}{2 a (a+b)^{3/2} d \left (a+b \text {sech}^2(c+d x)\right ) \sqrt {b (\cosh (c)-\sinh (c))^4}} \] Input:

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

Output:

((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^2*(b^2*ArcTanh[(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])]*(-Cosh[2*c] + Sinh[2*c]) + Sqrt[a + 
 b]*Sqrt[b*(Cosh[c] - Sinh[c])^4]*((a + b)*d*x + a*Csch[c]*Csch[c + d*x]*S 
inh[d*x])))/(2*a*(a + b)^(3/2)*d*(a + b*Sech[c + d*x]^2)*Sqrt[b*(Cosh[c] - 
 Sinh[c])^4])
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.24, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 25, 4629, 25, 2075, 382, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2075

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

\(\Big \downarrow \) 382

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

\(\Big \downarrow \) 397

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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. \(132\) vs. \(2(54)=108\).

Time = 1.70 (sec) , antiderivative size = 133, normalized size of antiderivative = 2.15

method	result	size
risch	\(\frac {x}{a}-\frac {2}{d \left (a +b \right ) \left ({\mathrm e}^{2 d x +2 c}-1\right )}+\frac {\sqrt {b \left (a +b \right )}\, b \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a +2 \sqrt {b \left (a +b \right )}+2 b}{a}\right )}{2 \left (a +b \right )^{2} d a}-\frac {\sqrt {b \left (a +b \right )}\, b \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {-a +2 \sqrt {b \left (a +b \right )}-2 b}{a}\right )}{2 \left (a +b \right )^{2} d a}\)	\(133\)
derivativedivides	\(\frac {-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right )}-\frac {1}{2 \left (a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a}+\frac {2 b^{2} \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{a \left (a +b \right )}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a}}{d}\)	\(183\)
default	\(\frac {-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right )}-\frac {1}{2 \left (a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a}+\frac {2 b^{2} \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{a \left (a +b \right )}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a}}{d}\)	\(183\)

Input:

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

Output:

x/a-2/d/(a+b)/(exp(2*d*x+2*c)-1)+1/2*(b*(a+b))^(1/2)/(a+b)^2*b/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)^2*b/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. 235 vs. \(2 (54) = 108\).

Time = 0.44 (sec) , antiderivative size = 749, normalized size of antiderivative = 12.08 \[ \int \frac {\coth ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/2*(2*(a + b)*d*x*cosh(d*x + c)^2 + 4*(a + b)*d*x*cosh(d*x + c)*sinh(d*x 
 + c) + 2*(a + b)*d*x*sinh(d*x + c)^2 - 2*(a + b)*d*x + (b*cosh(d*x + c)^2 
 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 - b)*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*sin 
h(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)*sinh(d*x + c) + (a^2 + a*b)*sinh(d*x 
 + c)^2 + a^2 + 3*a*b + 2*b^2)*sqrt(b/(a + b)))/(a*cosh(d*x + c)^4 + 4*a*c 
osh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + 
c)^2 + 2*(3*a*cosh(d*x + c)^2 + a + 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + 
 c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + a)) - 4*a)/((a^2 + a*b)*d 
*cosh(d*x + c)^2 + 2*(a^2 + a*b)*d*cosh(d*x + c)*sinh(d*x + c) + (a^2 + a* 
b)*d*sinh(d*x + c)^2 - (a^2 + a*b)*d), ((a + b)*d*x*cosh(d*x + c)^2 + 2*(a 
 + b)*d*x*cosh(d*x + c)*sinh(d*x + c) + (a + b)*d*x*sinh(d*x + c)^2 - (a + 
 b)*d*x - (b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d* 
x + c)^2 - b)*sqrt(-b/(a + b))*arctan(1/2*(a*cosh(d*x + c)^2 + 2*a*cosh(d* 
x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2 + a + 2*b)*sqrt(-b/(a + b))/b) - 
2*a)/((a^2 + a*b)*d*cosh(d*x + c)^2 + 2*(a^2 + a*b)*d*cosh(d*x + c)*sinh(d 
*x + c) + (a^2 + a*b)*d*sinh(d*x + c)^2 - (a^2 + a*b)*d)]
 

Sympy [F]

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

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

Output:

Integral(coth(c + d*x)**2/(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. 429 vs. \(2 (54) = 108\).

Time = 0.15 (sec) , antiderivative size = 429, normalized size of antiderivative = 6.92 \[ \int \frac {\coth ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {b \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (a + 2 \, b\right )} e^{\left (2 \, d x + 2 \, c\right )} + a\right )}{4 \, {\left (a^{2} + a b\right )} d} - \frac {b \log \left (2 \, {\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{4 \, {\left (a^{2} + a b\right )} d} - \frac {{\left (a b + 2 \, b^{2}\right )} \log \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{8 \, {\left (a^{2} + a b\right )} \sqrt {{\left (a + b\right )} b} d} + \frac {{\left (a b + 2 \, b^{2}\right )} \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{8 \, {\left (a^{2} + a b\right )} \sqrt {{\left (a + b\right )} b} d} - \frac {b \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{4 \, \sqrt {{\left (a + b\right )} b} {\left (a + b\right )} d} + \frac {\log \left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}{2 \, {\left (a + b\right )} d} - \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}{2 \, {\left (a + b\right )} d} - \frac {1}{2 \, {\left ({\left (a + b\right )} e^{\left (2 \, d x + 2 \, c\right )} - a - b\right )} d} + \frac {3}{2 \, {\left ({\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} - a - b\right )} d} \] Input:

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

Output:

1/4*b*log(a*e^(4*d*x + 4*c) + 2*(a + 2*b)*e^(2*d*x + 2*c) + a)/((a^2 + a*b 
)*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 + a*b)*d) - 1/8*(a*b + 2*b^2)*log((a*e^(2*d*x + 2*c) + a + 2*b - 2*sqr 
t((a + b)*b))/(a*e^(2*d*x + 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/((a^2 + a 
*b)*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 + a*b)*sqrt((a + b)*b)*d) - 1/4*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))) 
/(sqrt((a + b)*b)*(a + b)*d) + 1/2*log(e^(2*d*x + 2*c) - 1)/((a + b)*d) - 
1/2*log(e^(-2*d*x - 2*c) - 1)/((a + b)*d) - 1/2/(((a + b)*e^(2*d*x + 2*c) 
- a - b)*d) + 3/2/(((a + b)*e^(-2*d*x - 2*c) - a - b)*d)
 

Giac [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.48 \[ \int \frac {\coth ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=-\frac {\frac {b^{2} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right )}{{\left (a^{2} + a b\right )} \sqrt {-a b - b^{2}}} - \frac {d x + c}{a} + \frac {2}{{\left (a + b\right )} {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}}}{d} \] Input:

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

Output:

-(b^2*arctan(1/2*(a*e^(2*d*x + 2*c) + a + 2*b)/sqrt(-a*b - b^2))/((a^2 + a 
*b)*sqrt(-a*b - b^2)) - (d*x + c)/a + 2/((a + b)*(e^(2*d*x + 2*c) - 1)))/d
 

Mupad [B] (verification not implemented)

Time = 4.03 (sec) , antiderivative size = 977, normalized size of antiderivative = 15.76 \[ \int \frac {\coth ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {x}{a}-\frac {2}{\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )\,\left (a\,d+b\,d\right )}+\frac {\mathrm {atan}\left (\frac {\left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (\frac {8\,\left (a+2\,b\right )\,\left (4\,d\,a^4\,b^2+16\,d\,a^3\,b^3+20\,d\,a^2\,b^4+8\,d\,a\,b^5\right )}{a^6\,\left (a+b\right )\,\left (a^3+2\,a^2\,b+a\,b^2\right )\,\sqrt {-a^2\,d^2\,{\left (a+b\right )}^3}\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}}+\frac {2\,\sqrt {b^3}\,\left (a^2+8\,a\,b+8\,b^2\right )\,\left (a^2\,\sqrt {b^3}\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}+8\,b^2\,\sqrt {b^3}\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}+8\,a\,b\,\sqrt {b^3}\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}\right )}{a^7\,b^2\,d\,{\left (a+b\right )}^3\,\left (a^3+2\,a^2\,b+a\,b^2\right )\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}}\right )+\frac {8\,\left (a+2\,b\right )\,\left (2\,d\,a^4\,b^2+4\,d\,a^3\,b^3+2\,d\,a^2\,b^4\right )}{a^6\,\left (a+b\right )\,\left (a^3+2\,a^2\,b+a\,b^2\right )\,\sqrt {-a^2\,d^2\,{\left (a+b\right )}^3}\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}}+\frac {2\,\left (a^2\,\sqrt {b^3}\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}+2\,a\,b\,\sqrt {b^3}\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}\right )\,\sqrt {b^3}\,\left (a^2+8\,a\,b+8\,b^2\right )}{a^7\,b^2\,d\,{\left (a+b\right )}^3\,\left (a^3+2\,a^2\,b+a\,b^2\right )\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}}\right )\,\left (a^7\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}+a^4\,b^3\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}+3\,a^5\,b^2\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}+3\,a^6\,b\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}\right )}{4\,\sqrt {b^3}}\right )\,\sqrt {b^3}}{\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}} \] Input:

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

Output:

x/a - 2/((exp(2*c + 2*d*x) - 1)*(a*d + b*d)) + (atan(((exp(2*c)*exp(2*d*x) 
*((8*(a + 2*b)*(20*a^2*b^4*d + 16*a^3*b^3*d + 4*a^4*b^2*d + 8*a*b^5*d))/(a 
^6*(a + b)*(a*b^2 + 2*a^2*b + a^3)*(-a^2*d^2*(a + b)^3)^(1/2)*(- a^5*d^2 - 
 3*a^4*b*d^2 - a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2)) + (2*(b^3)^(1/2)*(8*a*b 
 + a^2 + 8*b^2)*(a^2*(b^3)^(1/2)*(- a^5*d^2 - 3*a^4*b*d^2 - a^2*b^3*d^2 - 
3*a^3*b^2*d^2)^(1/2) + 8*b^2*(b^3)^(1/2)*(- a^5*d^2 - 3*a^4*b*d^2 - a^2*b^ 
3*d^2 - 3*a^3*b^2*d^2)^(1/2) + 8*a*b*(b^3)^(1/2)*(- a^5*d^2 - 3*a^4*b*d^2 
- a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2)))/(a^7*b^2*d*(a + b)^3*(a*b^2 + 2*a^2 
*b + a^3)*(- a^5*d^2 - 3*a^4*b*d^2 - a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2))) 
+ (8*(a + 2*b)*(2*a^2*b^4*d + 4*a^3*b^3*d + 2*a^4*b^2*d))/(a^6*(a + b)*(a* 
b^2 + 2*a^2*b + a^3)*(-a^2*d^2*(a + b)^3)^(1/2)*(- a^5*d^2 - 3*a^4*b*d^2 - 
 a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2)) + (2*(a^2*(b^3)^(1/2)*(- a^5*d^2 - 3* 
a^4*b*d^2 - a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2) + 2*a*b*(b^3)^(1/2)*(- a^5* 
d^2 - 3*a^4*b*d^2 - a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2))*(b^3)^(1/2)*(8*a*b 
 + a^2 + 8*b^2))/(a^7*b^2*d*(a + b)^3*(a*b^2 + 2*a^2*b + a^3)*(- a^5*d^2 - 
 3*a^4*b*d^2 - a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2)))*(a^7*(- a^5*d^2 - 3*a^ 
4*b*d^2 - a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2) + a^4*b^3*(- a^5*d^2 - 3*a^4* 
b*d^2 - a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2) + 3*a^5*b^2*(- a^5*d^2 - 3*a^4* 
b*d^2 - a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2) + 3*a^6*b*(- a^5*d^2 - 3*a^4*b* 
d^2 - a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2)))/(4*(b^3)^(1/2)))*(b^3)^(1/2)...
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 416, normalized size of antiderivative = 6.71 \[ \int \frac {\coth ^2(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {-e^{2 d x +2 c} \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (-\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+e^{d x +c} \sqrt {a}\right ) b -e^{2 d x +2 c} \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+e^{d x +c} \sqrt {a}\right ) b +e^{2 d x +2 c} \sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (2 \sqrt {b}\, \sqrt {a +b}+e^{2 d x +2 c} a +a +2 b \right ) b +\sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (-\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+e^{d x +c} \sqrt {a}\right ) b +\sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+e^{d x +c} \sqrt {a}\right ) b -\sqrt {b}\, \sqrt {a +b}\, \mathrm {log}\left (2 \sqrt {b}\, \sqrt {a +b}+e^{2 d x +2 c} a +a +2 b \right ) b +2 e^{2 d x +2 c} a^{2} d x -4 e^{2 d x +2 c} a^{2}+4 e^{2 d x +2 c} a b d x -4 e^{2 d x +2 c} a b +2 e^{2 d x +2 c} b^{2} d x -2 a^{2} d x -4 a b d x -2 b^{2} d x}{2 a d \left (e^{2 d x +2 c} a^{2}+2 e^{2 d x +2 c} a b +e^{2 d x +2 c} b^{2}-a^{2}-2 a b -b^{2}\right )} \] Input:

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

Output:

( - 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 - 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 + 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 + sqrt(b)*sqrt(a + b)*log( - sqrt(2*sqrt(b)*sqrt(a + 
 b) - a - 2*b) + e**(c + d*x)*sqrt(a))*b + sqrt(b)*sqrt(a + b)*log(sqrt(2* 
sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*b - sqrt(b)*sqrt(a 
+ b)*log(2*sqrt(b)*sqrt(a + b) + e**(2*c + 2*d*x)*a + a + 2*b)*b + 2*e**(2 
*c + 2*d*x)*a**2*d*x - 4*e**(2*c + 2*d*x)*a**2 + 4*e**(2*c + 2*d*x)*a*b*d* 
x - 4*e**(2*c + 2*d*x)*a*b + 2*e**(2*c + 2*d*x)*b**2*d*x - 2*a**2*d*x - 4* 
a*b*d*x - 2*b**2*d*x)/(2*a*d*(e**(2*c + 2*d*x)*a**2 + 2*e**(2*c + 2*d*x)*a 
*b + e**(2*c + 2*d*x)*b**2 - a**2 - 2*a*b - b**2))