\(\int \frac {\text {sech}^2(x)}{a+b \tanh (x)} \, dx\) [105]

Optimal result

Integrand size = 13, antiderivative size = 11 \[ \int \frac {\text {sech}^2(x)}{a+b \tanh (x)} \, dx=\frac {\log (a+b \tanh (x))}{b} \]

[Out]

ln(a+b*tanh(x))/b

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3587, 31} \[ \int \frac {\text {sech}^2(x)}{a+b \tanh (x)} \, dx=\frac {\log (a+b \tanh (x))}{b} \]

[In]

Int[Sech[x]^2/(a + b*Tanh[x]),x]

[Out]

Log[a + b*Tanh[x]]/b

Rule 31

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

Rule 3587

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(b*f), Subst
[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b
^2, 0] && IntegerQ[m/2]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {\text {sech}^2(x)}{a+b \tanh (x)} \, dx=\frac {\log (a+b \tanh (x))}{b} \]

[In]

Integrate[Sech[x]^2/(a + b*Tanh[x]),x]

[Out]

Log[a + b*Tanh[x]]/b

Maple [A] (verified)

Time = 1.59 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {\ln \left (a +b \tanh \left (x \right )\right )}{b}\)	\(12\)
default	\(\frac {\ln \left (a +b \tanh \left (x \right )\right )}{b}\)	\(12\)
risch	\(-\frac {\ln \left (1+{\mathrm e}^{2 x}\right )}{b}+\frac {\ln \left ({\mathrm e}^{2 x}+\frac {a -b}{a +b}\right )}{b}\)	\(35\)

[In]

int(sech(x)^2/(a+b*tanh(x)),x,method=_RETURNVERBOSE)

[Out]

ln(a+b*tanh(x))/b

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (11) = 22\).

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 3.82 \[ \int \frac {\text {sech}^2(x)}{a+b \tanh (x)} \, dx=\frac {\log \left (\frac {2 \, {\left (a \cosh \left (x\right ) + b \sinh \left (x\right )\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - \log \left (\frac {2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{b} \]

[In]

integrate(sech(x)^2/(a+b*tanh(x)),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\text {sech}^2(x)}{a+b \tanh (x)} \, dx=\int \frac {\operatorname {sech}^{2}{\left (x \right )}}{a + b \tanh {\left (x \right )}}\, dx \]

[In]

integrate(sech(x)**2/(a+b*tanh(x)),x)

[Out]

Integral(sech(x)**2/(a + b*tanh(x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {\text {sech}^2(x)}{a+b \tanh (x)} \, dx=\frac {\log \left (b \tanh \left (x\right ) + a\right )}{b} \]

[In]

integrate(sech(x)^2/(a+b*tanh(x)),x, algorithm="maxima")

[Out]

log(b*tanh(x) + a)/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (11) = 22\).

Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 4.09 \[ \int \frac {\text {sech}^2(x)}{a+b \tanh (x)} \, dx=\frac {{\left (a + b\right )} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a b + b^{2}} - \frac {\log \left (e^{\left (2 \, x\right )} + 1\right )}{b} \]

[In]

integrate(sech(x)^2/(a+b*tanh(x)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 50, normalized size of antiderivative = 4.55 \[ \int \frac {\text {sech}^2(x)}{a+b \tanh (x)} \, dx=-\frac {2\,\mathrm {atan}\left (\frac {a\,\sqrt {-b^2}+a\,{\mathrm {e}}^{2\,x}\,\sqrt {-b^2}+b\,{\mathrm {e}}^{2\,x}\,\sqrt {-b^2}}{b^2}\right )}{\sqrt {-b^2}} \]

[In]

int(1/(cosh(x)^2*(a + b*tanh(x))),x)

[Out]

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