\(\int \frac {\cosh (c+d x) \coth (c+d x)}{a+b \sinh (c+d x)} \, dx\) [428]

Optimal result

Integrand size = 25, antiderivative size = 71 \[ \int \frac {\cosh (c+d x) \coth (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {x}{b}-\frac {\text {arctanh}(\cosh (c+d x))}{a d}+\frac {2 \sqrt {a^2+b^2} \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a b d} \] Output:

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

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.32 \[ \int \frac {\cosh (c+d x) \coth (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {a c+a d x+2 \sqrt {-a^2-b^2} \arctan \left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )-b \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )\right )+b \log \left (\sinh \left (\frac {1}{2} (c+d x)\right )\right )}{a b d} \] Input:

Integrate[(Cosh[c + d*x]*Coth[c + d*x])/(a + b*Sinh[c + d*x]),x]
 

Output:

(a*c + a*d*x + 2*Sqrt[-a^2 - b^2]*ArcTan[(b - a*Tanh[(c + d*x)/2])/Sqrt[-a 
^2 - b^2]] - b*Log[Cosh[(c + d*x)/2]] + b*Log[Sinh[(c + d*x)/2]])/(a*b*d)
 

Rubi [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 0.58 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.13, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3042, 26, 3368, 26, 3042, 26, 3537, 26, 3042, 26, 3139, 1083, 217, 4257}

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.

\(\displaystyle \int \frac {\cosh (c+d x) \coth (c+d x)}{a+b \sinh (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {i \cos (i c+i d x)^2}{\sin (i c+i d x) (a-i b \sin (i c+i d x))}dx\)

\(\Big \downarrow \) 26

\(\displaystyle i \int \frac {\cos (i c+i d x)^2}{\sin (i c+i d x) (a-i b \sin (i c+i d x))}dx\)

\(\Big \downarrow \) 3368

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

\(\Big \downarrow \) 26

\(\displaystyle \int \frac {\left (\sinh ^2(c+d x)+1\right ) \text {csch}(c+d x)}{a+b \sinh (c+d x)}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3537

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

\(\displaystyle i \left (\frac {\int \csc (i c+i d x)dx}{a}+\frac {2 i \sqrt {a^2+b^2} \text {arctanh}\left (\frac {\tanh \left (\frac {1}{2} (c+d x)\right )}{2 \sqrt {a^2+b^2}}\right )}{a b d}-\frac {i x}{b}\right )\)

\(\Big \downarrow \) 4257

\(\displaystyle i \left (\frac {2 i \sqrt {a^2+b^2} \text {arctanh}\left (\frac {\tanh \left (\frac {1}{2} (c+d x)\right )}{2 \sqrt {a^2+b^2}}\right )}{a b d}+\frac {i \text {arctanh}(\cosh (c+d x))}{a d}-\frac {i x}{b}\right )\)

Input:

Int[(Cosh[c + d*x]*Coth[c + d*x])/(a + b*Sinh[c + d*x]),x]
 

Output:

I*(((-I)*x)/b + (I*ArcTanh[Cosh[c + d*x]])/(a*d) + ((2*I)*Sqrt[a^2 + b^2]* 
ArcTanh[Tanh[(c + d*x)/2]/(2*Sqrt[a^2 + b^2])])/(a*b*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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

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

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ 
[a^2 - b^2, 0]
 

rule 3368

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + 
(b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a 
 + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, m, n 
}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])
 

rule 3537

Int[((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_) + (b_.)*sin[(e_.) + (f 
_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C*(x 
/(b*d)), x] + (Simp[(A*b^2 + a^2*C)/(b*(b*c - a*d))   Int[1/(a + b*Sin[e + 
f*x]), x], x] - Simp[(c^2*C + A*d^2)/(d*(b*c - a*d))   Int[1/(c + d*Sin[e + 
 f*x]), x], x]) /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] 
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 0.99 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.54

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}-\frac {\left (2 a^{2}+2 b^{2}\right ) \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a b \sqrt {a^{2}+b^{2}}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{d}\)	\(109\)
default	\(\frac {\frac {\ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}-\frac {\left (2 a^{2}+2 b^{2}\right ) \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a b \sqrt {a^{2}+b^{2}}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{d}\)	\(109\)
risch	\(\frac {x}{b}+\frac {\ln \left ({\mathrm e}^{d x +c}-1\right )}{d a}-\frac {\ln \left ({\mathrm e}^{d x +c}+1\right )}{d a}+\frac {\sqrt {a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{d x +c}+\frac {a +\sqrt {a^{2}+b^{2}}}{b}\right )}{d b a}-\frac {\sqrt {a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{d x +c}-\frac {-a +\sqrt {a^{2}+b^{2}}}{b}\right )}{d b a}\)	\(128\)

Input:

int(cosh(d*x+c)*coth(d*x+c)/(a+b*sinh(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/d*(1/b*ln(1+tanh(1/2*d*x+1/2*c))-1/b*ln(tanh(1/2*d*x+1/2*c)-1)-(2*a^2+2* 
b^2)/a/b/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*d*x+1/2*c)-2*b)/(a^2+b^ 
2)^(1/2))+1/a*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. 209 vs. \(2 (68) = 136\).

Time = 0.10 (sec) , antiderivative size = 209, normalized size of antiderivative = 2.94 \[ \int \frac {\cosh (c+d x) \coth (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {a d x - b \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) + 1\right ) + b \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) - 1\right ) + \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \, {\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right )}{a b d} \] Input:

integrate(cosh(d*x+c)*coth(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="fricas")
 

Output:

(a*d*x - b*log(cosh(d*x + c) + sinh(d*x + c) + 1) + b*log(cosh(d*x + c) + 
sinh(d*x + c) - 1) + sqrt(a^2 + b^2)*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d 
*x + c)^2 + 2*a*b*cosh(d*x + c) + 2*a^2 + b^2 + 2*(b^2*cosh(d*x + c) + a*b 
)*sinh(d*x + c) + 2*sqrt(a^2 + b^2)*(b*cosh(d*x + c) + b*sinh(d*x + c) + a 
))/(b*cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh( 
d*x + c) + a)*sinh(d*x + c) - b)))/(a*b*d)
 

Sympy [F]

\[ \int \frac {\cosh (c+d x) \coth (c+d x)}{a+b \sinh (c+d x)} \, dx=\int \frac {\cosh {\left (c + d x \right )} \coth {\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \] Input:

integrate(cosh(d*x+c)*coth(d*x+c)/(a+b*sinh(d*x+c)),x)
 

Output:

Integral(cosh(c + d*x)*coth(c + d*x)/(a + b*sinh(c + d*x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.77 \[ \int \frac {\cosh (c+d x) \coth (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {d x + c}{b d} - \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d} - \frac {\sqrt {a^{2} + b^{2}} \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{a b d} \] Input:

integrate(cosh(d*x+c)*coth(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.59 \[ \int \frac {\cosh (c+d x) \coth (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {d x + c}{b} - \frac {\log \left (e^{\left (d x + c\right )} + 1\right )}{a} + \frac {\log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a} - \frac {\sqrt {a^{2} + b^{2}} \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{a b}}{d} \] Input:

integrate(cosh(d*x+c)*coth(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 1.55 (sec) , antiderivative size = 384, normalized size of antiderivative = 5.41 \[ \int \frac {\cosh (c+d x) \coth (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {x}{b}+\frac {\ln \left (32\,a\,b^2+32\,a^3-32\,a^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-32\,a\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{a\,d}-\frac {\ln \left (32\,a\,b^2+32\,a^3+32\,a^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+32\,a\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{a\,d}-\frac {\ln \left (128\,a^5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-64\,a^2\,b^3-64\,a^4\,b-32\,a\,b^3\,\sqrt {a^2+b^2}-64\,a^3\,b\,\sqrt {a^2+b^2}+160\,a^3\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+128\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}+32\,a\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+96\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{a\,b\,d}+\frac {\ln \left (128\,a^5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-64\,a^2\,b^3-64\,a^4\,b+32\,a\,b^3\,\sqrt {a^2+b^2}+64\,a^3\,b\,\sqrt {a^2+b^2}+160\,a^3\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-128\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}+32\,a\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-96\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{a\,b\,d} \] Input:

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

Output:

x/b + log(32*a*b^2 + 32*a^3 - 32*a^3*exp(d*x)*exp(c) - 32*a*b^2*exp(d*x)*e 
xp(c))/(a*d) - log(32*a*b^2 + 32*a^3 + 32*a^3*exp(d*x)*exp(c) + 32*a*b^2*e 
xp(d*x)*exp(c))/(a*d) - (log(128*a^5*exp(d*x)*exp(c) - 64*a^2*b^3 - 64*a^4 
*b - 32*a*b^3*(a^2 + b^2)^(1/2) - 64*a^3*b*(a^2 + b^2)^(1/2) + 160*a^3*b^2 
*exp(d*x)*exp(c) + 128*a^4*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2) + 32*a*b^4*ex 
p(d*x)*exp(c) + 96*a^2*b^2*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2))*(a^2 + b^2)^ 
(1/2))/(a*b*d) + (log(128*a^5*exp(d*x)*exp(c) - 64*a^2*b^3 - 64*a^4*b + 32 
*a*b^3*(a^2 + b^2)^(1/2) + 64*a^3*b*(a^2 + b^2)^(1/2) + 160*a^3*b^2*exp(d* 
x)*exp(c) - 128*a^4*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2) + 32*a*b^4*exp(d*x)* 
exp(c) - 96*a^2*b^2*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2))*(a^2 + b^2)^(1/2))/ 
(a*b*d)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.08 \[ \int \frac {\cosh (c+d x) \coth (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {-2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{d x +c} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) i +\mathrm {log}\left (e^{d x +c}-1\right ) b -\mathrm {log}\left (e^{d x +c}+1\right ) b +a d x}{a b d} \] Input:

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

Output:

( - 2*sqrt(a**2 + b**2)*atan((e**(c + d*x)*b*i + a*i)/sqrt(a**2 + b**2))*i 
 + log(e**(c + d*x) - 1)*b - log(e**(c + d*x) + 1)*b + a*d*x)/(a*b*d)