\(\int \frac {\coth (c+d x)}{(a+b \text {sech}^2(c+d x))^2} \, dx\) [154]

Optimal result

Integrand size = 21, antiderivative size = 83 \[ \int \frac {\coth (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\frac {b^2}{2 a^2 (a+b) d \left (b+a \cosh ^2(c+d x)\right )}+\frac {b (2 a+b) \log \left (b+a \cosh ^2(c+d x)\right )}{2 a^2 (a+b)^2 d}+\frac {\log (\sinh (c+d x))}{(a+b)^2 d} \] Output:

1/2*b^2/a^2/(a+b)/d/(b+a*cosh(d*x+c)^2)+1/2*b*(2*a+b)*ln(b+a*cosh(d*x+c)^2 
)/a^2/(a+b)^2/d+ln(sinh(d*x+c))/(a+b)^2/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.39 \[ \int \frac {\coth (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\frac {(a+b) \left (2 a^2 \log (\sinh (c+d x))+b \left (b+(2 a+b) \log \left (a+b+a \sinh ^2(c+d x)\right )\right )\right )+a \left (2 a^2 \log (\sinh (c+d x))+b (2 a+b) \log \left (a+b+a \sinh ^2(c+d x)\right )\right ) \sinh ^2(c+d x)}{a^2 (a+b)^2 d (a+2 b+a \cosh (2 (c+d x)))} \] Input:

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

Output:

((a + b)*(2*a^2*Log[Sinh[c + d*x]] + b*(b + (2*a + b)*Log[a + b + a*Sinh[c 
 + d*x]^2])) + a*(2*a^2*Log[Sinh[c + d*x]] + b*(2*a + b)*Log[a + b + a*Sin 
h[c + d*x]^2])*Sinh[c + d*x]^2)/(a^2*(a + b)^2*d*(a + 2*b + a*Cosh[2*(c + 
d*x)]))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 26, 4626, 354, 99, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 4626

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {-\frac {b^2}{a^2 (a+b) \left (a \cosh ^2(c+d x)+b\right )}-\frac {b (2 a+b) \log \left (a \cosh ^2(c+d x)+b\right )}{a^2 (a+b)^2}-\frac {\log \left (1-\cosh ^2(c+d x)\right )}{(a+b)^2}}{2 d}\)

Input:

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

Output:

-1/2*(-(b^2/(a^2*(a + b)*(b + a*Cosh[c + d*x]^2))) - Log[1 - Cosh[c + d*x] 
^2]/(a + b)^2 - (b*(2*a + b)*Log[b + a*Cosh[c + d*x]^2])/(a^2*(a + b)^2))/ 
d
 

Defintions of rubi rules used

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 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4626

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_ 
)]^(m_.), x_Symbol] :> Module[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-(f 
*ff^(m + n*p - 1))^(-1)   Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff* 
x)^n)^p/x^(m + n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n} 
, x] && IntegerQ[(m - 1)/2] && IntegerQ[n] && IntegerQ[p]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(205\) vs. \(2(79)=158\).

Time = 15.53 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.48

method	result	size
derivativedivides	\(\frac {\frac {b \left (-\frac {2 a b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}+\frac {\left (2 a +b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}{2}\right )}{a^{2} \left (a +b \right )^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a +b \right )^{2}}}{d}\)	\(206\)
default	\(\frac {\frac {b \left (-\frac {2 a b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}+\frac {\left (2 a +b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}{2}\right )}{a^{2} \left (a +b \right )^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a +b \right )^{2}}}{d}\)	\(206\)
risch	\(\frac {x}{a^{2}}-\frac {2 x}{a^{2}+2 a b +b^{2}}-\frac {2 c}{d \left (a^{2}+2 a b +b^{2}\right )}-\frac {4 b x}{a \left (a^{2}+2 a b +b^{2}\right )}-\frac {4 b c}{a d \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 b^{2} x}{a^{2} \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 b^{2} c}{a^{2} d \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 b^{2} {\mathrm e}^{2 d x +2 c}}{a^{2} \left (a +b \right ) d \left ({\mathrm e}^{4 d x +4 c} a +2 a \,{\mathrm e}^{2 d x +2 c}+4 b \,{\mathrm e}^{2 d x +2 c}+a \right )}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{d \left (a^{2}+2 a b +b^{2}\right )}+\frac {b \ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 d x +2 c}}{a}+1\right )}{a d \left (a^{2}+2 a b +b^{2}\right )}+\frac {b^{2} \ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 d x +2 c}}{a}+1\right )}{2 a^{2} d \left (a^{2}+2 a b +b^{2}\right )}\)	\(332\)

Input:

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

Output:

1/d*(b/a^2/(a+b)^2*(-2*a*b*tanh(1/2*d*x+1/2*c)^2/(tanh(1/2*d*x+1/2*c)^4*a+ 
tanh(1/2*d*x+1/2*c)^4*b+2*tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)^2* 
b+a+b)+1/2*(2*a+b)*ln(tanh(1/2*d*x+1/2*c)^4*a+tanh(1/2*d*x+1/2*c)^4*b+2*ta 
nh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)^2*b+a+b))-1/a^2*ln(tanh(1/2*d* 
x+1/2*c)+1)-1/a^2*ln(tanh(1/2*d*x+1/2*c)-1)+1/(a+b)^2*ln(tanh(1/2*d*x+1/2* 
c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1031 vs. \(2 (79) = 158\).

Time = 0.41 (sec) , antiderivative size = 1031, normalized size of antiderivative = 12.42 \[ \int \frac {\coth (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(2*(a^3 + 2*a^2*b + a*b^2)*d*x*cosh(d*x + c)^4 + 8*(a^3 + 2*a^2*b + a 
*b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*(a^3 + 2*a^2*b + a*b^2)*d*x*si 
nh(d*x + c)^4 + 2*(a^3 + 2*a^2*b + a*b^2)*d*x - 4*(a*b^2 + b^3 - (a^3 + 4* 
a^2*b + 5*a*b^2 + 2*b^3)*d*x)*cosh(d*x + c)^2 + 4*(3*(a^3 + 2*a^2*b + a*b^ 
2)*d*x*cosh(d*x + c)^2 - a*b^2 - b^3 + (a^3 + 4*a^2*b + 5*a*b^2 + 2*b^3)*d 
*x)*sinh(d*x + c)^2 - ((2*a^2*b + a*b^2)*cosh(d*x + c)^4 + 4*(2*a^2*b + a* 
b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (2*a^2*b + a*b^2)*sinh(d*x + c)^4 + 2 
*a^2*b + a*b^2 + 2*(2*a^2*b + 5*a*b^2 + 2*b^3)*cosh(d*x + c)^2 + 2*(2*a^2* 
b + 5*a*b^2 + 2*b^3 + 3*(2*a^2*b + a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^2 
 + 4*((2*a^2*b + a*b^2)*cosh(d*x + c)^3 + (2*a^2*b + 5*a*b^2 + 2*b^3)*cosh 
(d*x + c))*sinh(d*x + c))*log(2*(a*cosh(d*x + c)^2 + a*sinh(d*x + c)^2 + a 
 + 2*b)/(cosh(d*x + c)^2 - 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 
)) - 2*(a^3*cosh(d*x + c)^4 + 4*a^3*cosh(d*x + c)*sinh(d*x + c)^3 + a^3*si 
nh(d*x + c)^4 + a^3 + 2*(a^3 + 2*a^2*b)*cosh(d*x + c)^2 + 2*(3*a^3*cosh(d* 
x + c)^2 + a^3 + 2*a^2*b)*sinh(d*x + c)^2 + 4*(a^3*cosh(d*x + c)^3 + (a^3 
+ 2*a^2*b)*cosh(d*x + c))*sinh(d*x + c))*log(2*sinh(d*x + c)/(cosh(d*x + c 
) - sinh(d*x + c))) + 8*((a^3 + 2*a^2*b + a*b^2)*d*x*cosh(d*x + c)^3 - (a* 
b^2 + b^3 - (a^3 + 4*a^2*b + 5*a*b^2 + 2*b^3)*d*x)*cosh(d*x + c))*sinh(d*x 
 + c))/((a^5 + 2*a^4*b + a^3*b^2)*d*cosh(d*x + c)^4 + 4*(a^5 + 2*a^4*b + a 
^3*b^2)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^5 + 2*a^4*b + a^3*b^2)*d*s...
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (79) = 158\).

Time = 0.05 (sec) , antiderivative size = 209, normalized size of antiderivative = 2.52 \[ \int \frac {\coth (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\frac {2 \, b^{2} e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (a^{4} + a^{3} b + 2 \, {\left (a^{4} + 3 \, a^{3} b + 2 \, a^{2} b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{4} + a^{3} b\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} + \frac {{\left (2 \, a b + b^{2}\right )} \log \left (2 \, {\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{2 \, {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {d x + c}{a^{2} d} \] Input:

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

Output:

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

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable 
to make series expansion Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 1990, normalized size of antiderivative = 23.98 \[ \int \frac {\coth (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*e**(4*c + 4*d*x)*log(e**(c + d*x) - 1)*a**4 + 4*e**(4*c + 4*d*x)*log(e* 
*(c + d*x) - 1)*a**3*b + 2*e**(4*c + 4*d*x)*log(e**(c + d*x) + 1)*a**4 + 4 
*e**(4*c + 4*d*x)*log(e**(c + d*x) + 1)*a**3*b + 2*e**(4*c + 4*d*x)*log( - 
 sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*a**3*b + 5* 
e**(4*c + 4*d*x)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d* 
x)*sqrt(a))*a**2*b**2 + 2*e**(4*c + 4*d*x)*log( - sqrt(2*sqrt(b)*sqrt(a + 
b) - a - 2*b) + e**(c + d*x)*sqrt(a))*a*b**3 + 2*e**(4*c + 4*d*x)*log(sqrt 
(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*a**3*b + 5*e**(4 
*c + 4*d*x)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt( 
a))*a**2*b**2 + 2*e**(4*c + 4*d*x)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2* 
b) + e**(c + d*x)*sqrt(a))*a*b**3 + 2*e**(4*c + 4*d*x)*log(2*sqrt(b)*sqrt( 
a + b) + e**(2*c + 2*d*x)*a + a + 2*b)*a**3*b + 5*e**(4*c + 4*d*x)*log(2*s 
qrt(b)*sqrt(a + b) + e**(2*c + 2*d*x)*a + a + 2*b)*a**2*b**2 + 2*e**(4*c + 
 4*d*x)*log(2*sqrt(b)*sqrt(a + b) + e**(2*c + 2*d*x)*a + a + 2*b)*a*b**3 - 
 2*e**(4*c + 4*d*x)*a**4*d*x - 8*e**(4*c + 4*d*x)*a**3*b*d*x - 10*e**(4*c 
+ 4*d*x)*a**2*b**2*d*x - 2*e**(4*c + 4*d*x)*a**2*b**2 - 4*e**(4*c + 4*d*x) 
*a*b**3*d*x - 2*e**(4*c + 4*d*x)*a*b**3 + 4*e**(2*c + 2*d*x)*log(e**(c + d 
*x) - 1)*a**4 + 16*e**(2*c + 2*d*x)*log(e**(c + d*x) - 1)*a**3*b + 16*e**( 
2*c + 2*d*x)*log(e**(c + d*x) - 1)*a**2*b**2 + 4*e**(2*c + 2*d*x)*log(e**( 
c + d*x) + 1)*a**4 + 16*e**(2*c + 2*d*x)*log(e**(c + d*x) + 1)*a**3*b +...