3.119 \(\int \frac {\tanh (x)}{a+b \text {sech}(x)} \, dx\)

Optimal. Leaf size=19 \[ \frac {\log (a+b \text {sech}(x))}{a}+\frac {\log (\cosh (x))}{a} \]

[Out]

ln(cosh(x))/a+ln(a+b*sech(x))/a

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Rubi [A]  time = 0.03, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3885, 36, 29, 31} \[ \frac {\log (a+b \text {sech}(x))}{a}+\frac {\log (\cosh (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]/(a + b*Sech[x]),x]

[Out]

Log[Cosh[x]]/a + Log[a + b*Sech[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 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\tanh (x)}{a+b \text {sech}(x)} \, dx &=-\operatorname {Subst}\left (\int \frac {1}{x (a+x)} \, dx,x,b \text {sech}(x)\right )\\ &=-\frac {\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,b \text {sech}(x)\right )}{a}+\frac {\operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \text {sech}(x)\right )}{a}\\ &=\frac {\log (\cosh (x))}{a}+\frac {\log (a+b \text {sech}(x))}{a}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 11, normalized size = 0.58 \[ \frac {\log (a \cosh (x)+b)}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]/(a + b*Sech[x]),x]

[Out]

Log[b + a*Cosh[x]]/a

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fricas [A]  time = 0.43, size = 27, normalized size = 1.42 \[ -\frac {x - \log \left (\frac {2 \, {\left (a \cosh \relax (x) + b\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.13, size = 19, normalized size = 1.00 \[ \frac {\log \left ({\left | a {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, b \right |}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(a*(e^(-x) + e^x) + 2*b))/a

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maple [A]  time = 0.10, size = 21, normalized size = 1.11 \[ \frac {\ln \left (a +b \,\mathrm {sech}\relax (x )\right )}{a}-\frac {\ln \left (\mathrm {sech}\relax (x )\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(a+b*sech(x))/a-1/a*ln(sech(x))

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maxima [A]  time = 0.31, size = 26, normalized size = 1.37 \[ \frac {x}{a} + \frac {\log \left (2 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} + a\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.11, size = 23, normalized size = 1.21 \[ -\frac {x-\ln \left (a+2\,b\,{\mathrm {e}}^x+a\,{\mathrm {e}}^{2\,x}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)/(a + b/cosh(x)),x)

[Out]

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

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sympy [A]  time = 0.46, size = 41, normalized size = 2.16 \[ \begin {cases} \frac {\tilde {\infty }}{\operatorname {sech}{\relax (x )}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {1}{b \operatorname {sech}{\relax (x )}} & \text {for}\: a = 0 \\\frac {x - \log {\left (\tanh {\relax (x )} + 1 \right )}}{a} & \text {for}\: b = 0 \\\frac {x}{a} + \frac {\log {\left (\frac {a}{b} + \operatorname {sech}{\relax (x )} \right )}}{a} - \frac {\log {\left (\tanh {\relax (x )} + 1 \right )}}{a} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo/sech(x), Eq(a, 0) & Eq(b, 0)), (1/(b*sech(x)), Eq(a, 0)), ((x - log(tanh(x) + 1))/a, Eq(b, 0)),
(x/a + log(a/b + sech(x))/a - log(tanh(x) + 1)/a, True))

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