3.3.31 \(\int \frac {\tanh (x)}{a+b \sinh (x)} \, dx\) [231]

3.3.31.1 Optimal result

Integrand size = 11, antiderivative size = 48 \[ \int \frac {\tanh (x)}{a+b \sinh (x)} \, dx=\frac {b \arctan (\sinh (x))}{a^2+b^2}+\frac {a \log (\cosh (x))}{a^2+b^2}-\frac {a \log (a+b \sinh (x))}{a^2+b^2} \]

output

b*arctan(sinh(x))/(a^2+b^2)+a*ln(cosh(x))/(a^2+b^2)-a*ln(a+b*sinh(x))/(a^2 
+b^2)
 

3.3.31.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.17 \[ \int \frac {\tanh (x)}{a+b \sinh (x)} \, dx=\frac {(a-i b) \log (i-\sinh (x))+(a+i b) \log (i+\sinh (x))-2 a \log (a+b \sinh (x))}{2 \left (a^2+b^2\right )} \]

input

Integrate[Tanh[x]/(a + b*Sinh[x]),x]
 

output

((a - I*b)*Log[I - Sinh[x]] + (a + I*b)*Log[I + Sinh[x]] - 2*a*Log[a + b*S 
inh[x]])/(2*(a^2 + b^2))
 

3.3.31.3 Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.818, Rules used = {3042, 26, 3200, 25, 587, 16, 452, 216, 240}

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.

\(\displaystyle \int \frac {\tanh (x)}{a+b \sinh (x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {i \tan (i x)}{a-i b \sin (i x)}dx\)

\(\Big \downarrow \) 26

\(\displaystyle -i \int \frac {\tan (i x)}{a-i b \sin (i x)}dx\)

\(\Big \downarrow \) 3200

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

\(\Big \downarrow \) 25

\(\displaystyle \int \frac {b \sinh (x)}{\left (b^2 \sinh ^2(x)+b^2\right ) (a+b \sinh (x))}d(b \sinh (x))\)

\(\Big \downarrow \) 587

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

\(\Big \downarrow \) 16

\(\displaystyle \frac {\int \frac {b^2+a \sinh (x) b}{\sinh ^2(x) b^2+b^2}d(b \sinh (x))}{a^2+b^2}-\frac {a \log (a+b \sinh (x))}{a^2+b^2}\)

\(\Big \downarrow \) 452

\(\displaystyle \frac {a \int \frac {b \sinh (x)}{\sinh ^2(x) b^2+b^2}d(b \sinh (x))+b^2 \int \frac {1}{\sinh ^2(x) b^2+b^2}d(b \sinh (x))}{a^2+b^2}-\frac {a \log (a+b \sinh (x))}{a^2+b^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {a \int \frac {b \sinh (x)}{\sinh ^2(x) b^2+b^2}d(b \sinh (x))+b \arctan (\sinh (x))}{a^2+b^2}-\frac {a \log (a+b \sinh (x))}{a^2+b^2}\)

\(\Big \downarrow \) 240

\(\displaystyle \frac {\frac {1}{2} a \log \left (b^2 \sinh ^2(x)+b^2\right )+b \arctan (\sinh (x))}{a^2+b^2}-\frac {a \log (a+b \sinh (x))}{a^2+b^2}\)

input

Int[Tanh[x]/(a + b*Sinh[x]),x]
 

output

-((a*Log[a + b*Sinh[x]])/(a^2 + b^2)) + (b*ArcTan[Sinh[x]] + (a*Log[b^2 + 
b^2*Sinh[x]^2])/2)/(a^2 + b^2)
 

3.3.31.3.1 Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 216

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

rule 240

Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x 
^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
 

rule 452

Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[c   Int[1/ 
(a + b*x^2), x], x] + Simp[d   Int[x/(a + b*x^2), x], x] /; FreeQ[{a, b, c, 
 d}, x] && NeQ[b*c^2 + a*d^2, 0]
 

rule 587

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

rule 3042

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

rule 3200

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p 
_.), x_Symbol] :> Simp[1/f   Subst[Int[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1) 
/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b 
^2, 0] && IntegerQ[(p + 1)/2]
 

3.3.31.4 Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.52

method	result	size
default	\(-\frac {2 a \ln \left (\tanh \left (\frac {x}{2}\right )^{2} a -2 b \tanh \left (\frac {x}{2}\right )-a \right )}{2 a^{2}+2 b^{2}}+\frac {2 a \ln \left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )+4 b \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{2 a^{2}+2 b^{2}}\)	\(73\)
risch	\(\frac {i \ln \left ({\mathrm e}^{x}+i\right ) b}{a^{2}+b^{2}}+\frac {\ln \left ({\mathrm e}^{x}+i\right ) a}{a^{2}+b^{2}}-\frac {i \ln \left ({\mathrm e}^{x}-i\right ) b}{a^{2}+b^{2}}+\frac {\ln \left ({\mathrm e}^{x}-i\right ) a}{a^{2}+b^{2}}-\frac {a \ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}-1\right )}{a^{2}+b^{2}}\)	\(101\)

input

int(tanh(x)/(a+b*sinh(x)),x,method=_RETURNVERBOSE)
 

output

-2*a/(2*a^2+2*b^2)*ln(tanh(1/2*x)^2*a-2*b*tanh(1/2*x)-a)+4/(2*a^2+2*b^2)*( 
1/2*a*ln(1+tanh(1/2*x)^2)+b*arctan(tanh(1/2*x)))
 

3.3.31.5 Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.19 \[ \int \frac {\tanh (x)}{a+b \sinh (x)} \, dx=\frac {2 \, b \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - a \log \left (\frac {2 \, {\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + a \log \left (\frac {2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{2} + b^{2}} \]

input

integrate(tanh(x)/(a+b*sinh(x)),x, algorithm="fricas")
 

output

(2*b*arctan(cosh(x) + sinh(x)) - a*log(2*(b*sinh(x) + a)/(cosh(x) - sinh(x 
))) + a*log(2*cosh(x)/(cosh(x) - sinh(x))))/(a^2 + b^2)
 

3.3.31.6 Sympy [F]

\[ \int \frac {\tanh (x)}{a+b \sinh (x)} \, dx=\int \frac {\tanh {\left (x \right )}}{a + b \sinh {\left (x \right )}}\, dx \]

input

integrate(tanh(x)/(a+b*sinh(x)),x)
 

output

Integral(tanh(x)/(a + b*sinh(x)), x)
 

3.3.31.7 Maxima [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.38 \[ \int \frac {\tanh (x)}{a+b \sinh (x)} \, dx=-\frac {2 \, b \arctan \left (e^{\left (-x\right )}\right )}{a^{2} + b^{2}} - \frac {a \log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{a^{2} + b^{2}} + \frac {a \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{2} + b^{2}} \]

input

integrate(tanh(x)/(a+b*sinh(x)),x, algorithm="maxima")
 

output

-2*b*arctan(e^(-x))/(a^2 + b^2) - a*log(-2*a*e^(-x) + b*e^(-2*x) - b)/(a^2 
 + b^2) + a*log(e^(-2*x) + 1)/(a^2 + b^2)
 

3.3.31.8 Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.85 \[ \int \frac {\tanh (x)}{a+b \sinh (x)} \, dx=-\frac {a b \log \left ({\left | -b {\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{a^{2} b + b^{3}} + \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} b}{2 \, {\left (a^{2} + b^{2}\right )}} + \frac {a \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}{2 \, {\left (a^{2} + b^{2}\right )}} \]

input

integrate(tanh(x)/(a+b*sinh(x)),x, algorithm="giac")
 

output

-a*b*log(abs(-b*(e^(-x) - e^x) + 2*a))/(a^2*b + b^3) + 1/2*(pi + 2*arctan( 
1/2*(e^(2*x) - 1)*e^(-x)))*b/(a^2 + b^2) + 1/2*a*log((e^(-x) - e^x)^2 + 4) 
/(a^2 + b^2)
 

3.3.31.9 Mupad [B] (verification not implemented)

Time = 2.14 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.98 \[ \int \frac {\tanh (x)}{a+b \sinh (x)} \, dx=\frac {\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )}{a-b\,1{}\mathrm {i}}-\frac {a\,\ln \left (b^3\,{\mathrm {e}}^{2\,x}-4\,a^2\,b-b^3+8\,a^3\,{\mathrm {e}}^x+2\,a\,b^2\,{\mathrm {e}}^x+4\,a^2\,b\,{\mathrm {e}}^{2\,x}\right )}{a^2+b^2}+\frac {\ln \left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{-b+a\,1{}\mathrm {i}} \]

input

int(tanh(x)/(a + b*sinh(x)),x)
 

output

(log(exp(x)*1i + 1)*1i)/(a*1i - b) + log(exp(x) + 1i)/(a - b*1i) - (a*log( 
b^3*exp(2*x) - 4*a^2*b - b^3 + 8*a^3*exp(x) + 2*a*b^2*exp(x) + 4*a^2*b*exp 
(2*x)))/(a^2 + b^2)