\(\int \frac {\coth (x)}{a+b \tanh (x)} \, dx\) [139]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.06 (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} \[ \int \frac {\coth (x)}{a+b \tanh (x)} \, dx=-\frac {b x}{a^2-b^2}+\frac {b^2 \log (a \cosh (x)+b \sinh (x))}{a \left (a^2-b^2\right )}+\frac {\log (\sinh (x))}{a} \]

[In]

Int[Coth[x]/(a + b*Tanh[x]),x]

[Out]

-((b*x)/(a^2 - b^2)) + Log[Sinh[x]]/a + (b^2*Log[a*Cosh[x] + b*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 \coth (x) \, dx}{a}+\frac {\left (i b^2\right ) \int \frac {-i b-i a \tanh (x)}{a+b \tanh (x)} \, dx}{a \left (a^2-b^2\right )} \\ & = -\frac {b x}{a^2-b^2}+\frac {\log (\sinh (x))}{a}+\frac {b^2 \log (a \cosh (x)+b \sinh (x))}{a \left (a^2-b^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.27 \[ \int \frac {\coth (x)}{a+b \tanh (x)} \, dx=-\frac {\log (1-\tanh (x))}{2 (a+b)}+\frac {\log (\tanh (x))}{a}-\frac {\log (1+\tanh (x))}{2 (a-b)}+\frac {b^2 \log (a+b \tanh (x))}{a \left (a^2-b^2\right )} \]

[In]

Integrate[Coth[x]/(a + b*Tanh[x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.10

method	result	size
parallelrisch	\(\frac {b^{2} \ln \left (a +b \tanh \left (x \right )\right )-a^{2} \ln \left (1-\tanh \left (x \right )\right )-\left (a +b \right ) \left (\left (-a +b \right ) \ln \left (\tanh \left (x \right )\right )+a x \right )}{a^{3}-a \,b^{2}}\)	\(56\)
derivativedivides	\(-\frac {\ln \left (\tanh \left (x \right )-1\right )}{2 a +2 b}+\frac {b^{2} \ln \left (a +b \tanh \left (x \right )\right )}{a \left (a +b \right ) \left (a -b \right )}+\frac {\ln \left (\tanh \left (x \right )\right )}{a}-\frac {\ln \left (1+\tanh \left (x \right )\right )}{2 a -2 b}\)	\(67\)
default	\(-\frac {\ln \left (\tanh \left (x \right )-1\right )}{2 a +2 b}+\frac {b^{2} \ln \left (a +b \tanh \left (x \right )\right )}{a \left (a +b \right ) \left (a -b \right )}+\frac {\ln \left (\tanh \left (x \right )\right )}{a}-\frac {\ln \left (1+\tanh \left (x \right )\right )}{2 a -2 b}\)	\(67\)
risch	\(\frac {x}{a +b}-\frac {2 x}{a}-\frac {2 x \,b^{2}}{a \left (a^{2}-b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{2 x}-1\right )}{a}+\frac {b^{2} \ln \left ({\mathrm e}^{2 x}+\frac {a -b}{a +b}\right )}{a \left (a^{2}-b^{2}\right )}\)	\(81\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.43 \[ \int \frac {\coth (x)}{a+b \tanh (x)} \, dx=\frac {b^{2} \log \left (\frac {2 \, {\left (a \cosh \left (x\right ) + b \sinh \left (x\right )\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - {\left (a^{2} + a b\right )} x + {\left (a^{2} - b^{2}\right )} \log \left (\frac {2 \, \sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{3} - a b^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(coth(x)/(a + b*tanh(x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.27 \[ \int \frac {\coth (x)}{a+b \tanh (x)} \, dx=\frac {b^{2} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{3} - a b^{2}} + \frac {x}{a + b} + \frac {\log \left (e^{\left (-x\right )} + 1\right )}{a} + \frac {\log \left (e^{\left (-x\right )} - 1\right )}{a} \]

[In]

integrate(coth(x)/(a+b*tanh(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^(-x) + 1)/a + log(e^(-x) - 1)/a

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.14 \[ \int \frac {\coth (x)}{a+b \tanh (x)} \, dx=\frac {b^{2} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{3} - a b^{2}} - \frac {x}{a - b} + \frac {\log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right )}{a} \]

[In]

integrate(coth(x)/(a+b*tanh(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(abs(e^(2*x) - 1))/a

Mupad [B] (verification not implemented)

Time = 1.99 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.14 \[ \int \frac {\coth (x)}{a+b \tanh (x)} \, dx=\frac {\ln \left ({\mathrm {e}}^{2\,x}-1\right )}{a}-\frac {x}{a-b}-\frac {b^2\,\ln \left (a-b+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a\,b^2-a^3} \]

[In]

int(coth(x)/(a + b*tanh(x)),x)

[Out]

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