\(\int \frac {\tanh (x)}{a+b \coth (x)} \, dx\) [143]

Optimal result

Integrand size = 11, antiderivative size = 51 \[ \int \frac {\tanh (x)}{a+b \coth (x)} \, dx=-\frac {b x}{a^2-b^2}+\frac {\log (\cosh (x))}{a}+\frac {b^2 \log (b \cosh (x)+a \sinh (x))}{a \left (a^2-b^2\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3652, 3611, 3556} \[ \int \frac {\tanh (x)}{a+b \coth (x)} \, dx=-\frac {b x}{a^2-b^2}+\frac {b^2 \log (a \sinh (x)+b \cosh (x))}{a \left (a^2-b^2\right )}+\frac {\log (\cosh (x))}{a} \]

[In]

Int[Tanh[x]/(a + b*Coth[x]),x]

[Out]

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

Rule 3556

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

Rule 3611

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c/(b*f))
*Log[RemoveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3652

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

Rubi steps \begin{align*} \text {integral}& = -\frac {b x}{a^2-b^2}+\frac {\int \tanh (x) \, dx}{a}+\frac {\left (i b^2\right ) \int \frac {-i b-i a \coth (x)}{a+b \coth (x)} \, dx}{a \left (a^2-b^2\right )} \\ & = -\frac {b x}{a^2-b^2}+\frac {\log (\cosh (x))}{a}+\frac {b^2 \log (b \cosh (x)+a \sinh (x))}{a \left (a^2-b^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.29 \[ \int \frac {\tanh (x)}{a+b \coth (x)} \, dx=-\frac {\log (1-\coth (x))}{2 (a+b)}-\frac {\log (1+\coth (x))}{2 (a-b)}+\frac {b^2 \log (a+b \coth (x))}{a \left (a^2-b^2\right )}-\frac {\log (\tanh (x))}{a} \]

[In]

Integrate[Tanh[x]/(a + b*Coth[x]),x]

[Out]

-1/2*Log[1 - Coth[x]]/(a + b) - Log[1 + Coth[x]]/(2*(a - b)) + (b^2*Log[a + b*Coth[x]])/(a*(a^2 - b^2)) - Log[
Tanh[x]]/a

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.86

method	result	size
parallelrisch	\(\frac {b^{2} \ln \left (b +a \tanh \left (x \right )\right )-\left (a \ln \left (1-\tanh \left (x \right )\right )+\left (a +b \right ) x \right ) a}{a^{3}-a \,b^{2}}\)	\(44\)
derivativedivides	\(-\frac {\ln \left (1+\coth \left (x \right )\right )}{2 a -2 b}+\frac {b^{2} \ln \left (a +b \coth \left (x \right )\right )}{a \left (a +b \right ) \left (a -b \right )}-\frac {\ln \left (\coth \left (x \right )-1\right )}{2 a +2 b}+\frac {\ln \left (\coth \left (x \right )\right )}{a}\)	\(67\)
default	\(-\frac {\ln \left (1+\coth \left (x \right )\right )}{2 a -2 b}+\frac {b^{2} \ln \left (a +b \coth \left (x \right )\right )}{a \left (a +b \right ) \left (a -b \right )}-\frac {\ln \left (\coth \left (x \right )-1\right )}{2 a +2 b}+\frac {\ln \left (\coth \left (x \right )\right )}{a}\)	\(67\)
risch	\(\frac {x}{a +b}-\frac {2 x}{a}-\frac {2 x \,b^{2}}{a \left (a^{2}-b^{2}\right )}+\frac {\ln \left (1+{\mathrm e}^{2 x}\right )}{a}+\frac {b^{2} \ln \left ({\mathrm e}^{2 x}-\frac {a -b}{a +b}\right )}{a \left (a^{2}-b^{2}\right )}\)	\(82\)

[In]

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

[Out]

(b^2*ln(b+a*tanh(x))-(a*ln(1-tanh(x))+(a+b)*x)*a)/(a^3-a*b^2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.43 \[ \int \frac {\tanh (x)}{a+b \coth (x)} \, dx=\frac {b^{2} \log \left (\frac {2 \, {\left (b \cosh \left (x\right ) + a \sinh \left (x\right )\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - {\left (a^{2} + a b\right )} x + {\left (a^{2} - b^{2}\right )} \log \left (\frac {2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{3} - a b^{2}} \]

[In]

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

[Out]

(b^2*log(2*(b*cosh(x) + a*sinh(x))/(cosh(x) - sinh(x))) - (a^2 + a*b)*x + (a^2 - b^2)*log(2*cosh(x)/(cosh(x) -
sinh(x))))/(a^3 - a*b^2)

Sympy [F]

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

[In]

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

[Out]

Integral(tanh(x)/(a + b*coth(x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.98 \[ \int \frac {\tanh (x)}{a+b \coth (x)} \, dx=\frac {b^{2} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} + a + b\right )}{a^{3} - a b^{2}} + \frac {x}{a + b} + \frac {\log \left (e^{\left (-2 \, x\right )} + 1\right )}{a} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.12 \[ \int \frac {\tanh (x)}{a+b \coth (x)} \, dx=\frac {b^{2} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} - a + b \right |}\right )}{a^{3} - a b^{2}} - \frac {x}{a - b} + \frac {\log \left (e^{\left (2 \, x\right )} + 1\right )}{a} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.14 \[ \int \frac {\tanh (x)}{a+b \coth (x)} \, dx=\frac {\ln \left ({\mathrm {e}}^{2\,x}+1\right )}{a}-\frac {x}{a-b}-\frac {b^2\,\ln \left (b-a+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a\,b^2-a^3} \]

[In]

int(tanh(x)/(a + b*coth(x)),x)

[Out]

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