\(\int \frac {\tanh (x)}{a+a \cosh (x)} \, dx\) [192]

Optimal result

Integrand size = 11, antiderivative size = 18 \[ \int \frac {\tanh (x)}{a+a \cosh (x)} \, dx=\frac {\log (\cosh (x))}{a}-\frac {\log (1+\cosh (x))}{a} \]

[Out]

ln(cosh(x))/a-ln(1+cosh(x))/a

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2786, 36, 29, 31} \[ \int \frac {\tanh (x)}{a+a \cosh (x)} \, dx=\frac {\log (\cosh (x))}{a}-\frac {\log (\cosh (x)+1)}{a} \]

[In]

Int[Tanh[x]/(a + a*Cosh[x]),x]

[Out]

Log[Cosh[x]]/a - Log[1 + Cosh[x]]/a

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

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

Rule 2786

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

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{x (a+x)} \, dx,x,a \cosh (x)\right ) \\ & = \frac {\text {Subst}\left (\int \frac {1}{x} \, dx,x,a \cosh (x)\right )}{a}-\frac {\text {Subst}\left (\int \frac {1}{a+x} \, dx,x,a \cosh (x)\right )}{a} \\ & = \frac {\log (\cosh (x))}{a}-\frac {\log (1+\cosh (x))}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {\tanh (x)}{a+a \cosh (x)} \, dx=-\frac {2 \text {arctanh}(1+2 \cosh (x))}{a} \]

[In]

Integrate[Tanh[x]/(a + a*Cosh[x]),x]

[Out]

(-2*ArcTanh[1 + 2*Cosh[x]])/a

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78

method	result	size
default	\(\frac {\ln \left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )}{a}\)	\(14\)
risch	\(-\frac {2 \ln \left ({\mathrm e}^{x}+1\right )}{a}+\frac {\ln \left (1+{\mathrm e}^{2 x}\right )}{a}\)	\(23\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.56 \[ \int \frac {\tanh (x)}{a+a \cosh (x)} \, dx=\frac {\log \left (\frac {2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - 2 \, \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right )}{a} \]

[In]

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

[Out]

(log(2*cosh(x)/(cosh(x) - sinh(x))) - 2*log(cosh(x) + sinh(x) + 1))/a

Sympy [F]

\[ \int \frac {\tanh (x)}{a+a \cosh (x)} \, dx=\frac {\int \frac {\tanh {\left (x \right )}}{\cosh {\left (x \right )} + 1}\, dx}{a} \]

[In]

integrate(tanh(x)/(a+a*cosh(x)),x)

[Out]

Integral(tanh(x)/(cosh(x) + 1), x)/a

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \frac {\tanh (x)}{a+a \cosh (x)} \, dx=-\frac {2 \, \log \left (e^{\left (-x\right )} + 1\right )}{a} + \frac {\log \left (e^{\left (-2 \, x\right )} + 1\right )}{a} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {\tanh (x)}{a+a \cosh (x)} \, dx=\frac {\log \left (e^{\left (2 \, x\right )} + 1\right )}{a} - \frac {2 \, \log \left (e^{x} + 1\right )}{a} \]

[In]

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

[Out]

log(e^(2*x) + 1)/a - 2*log(e^x + 1)/a

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {\tanh (x)}{a+a \cosh (x)} \, dx=-\frac {2\,\ln \left (36\,{\mathrm {e}}^x+36\right )-\ln \left (3\,{\mathrm {e}}^{2\,x}+3\right )}{a} \]

[In]

int(tanh(x)/(a + a*cosh(x)),x)

[Out]

-(2*log(36*exp(x) + 36) - log(3*exp(2*x) + 3))/a