### 3.619 $$\int \frac{1}{a \text{sech}(x)+b \tanh (x)} \, dx$$

Optimal. Leaf size=11 $\frac{\log (a+b \sinh (x))}{b}$

[Out]

Log[a + b*Sinh[x]]/b

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Rubi [A]  time = 0.0385105, antiderivative size = 11, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.273, Rules used = {3159, 2668, 31} $\frac{\log (a+b \sinh (x))}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a*Sech[x] + b*Tanh[x])^(-1),x]

[Out]

Log[a + b*Sinh[x]]/b

Rule 3159

Int[((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Int[Cos[d + e*x
]/(b + a*Cos[d + e*x] + c*Sin[d + e*x]), x] /; FreeQ[{a, b, c, d, e}, x]

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 31

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

Rubi steps

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

Mathematica [A]  time = 0.0058458, size = 11, normalized size = 1. $\frac{\log (a+b \sinh (x))}{b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a*Sech[x] + b*Tanh[x])^(-1),x]

[Out]

Log[a + b*Sinh[x]]/b

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Maple [B]  time = 0.046, size = 50, normalized size = 4.6 \begin{align*} -{\frac{1}{b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{1}{b}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }-{\frac{1}{b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sech(x)+b*tanh(x)),x)

[Out]

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

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Maxima [B]  time = 1.02182, size = 38, normalized size = 3.45 \begin{align*} \frac{x}{b} + \frac{\log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sech(x)+b*tanh(x)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.28765, size = 72, normalized size = 6.55 \begin{align*} -\frac{x - \log \left (\frac{2 \,{\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sech(x)+b*tanh(x)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.654301, size = 32, normalized size = 2.91 \begin{align*} \begin{cases} \frac{x}{b} + \frac{\log{\left (\frac{a \operatorname{sech}{\left (x \right )}}{b} + \tanh{\left (x \right )} \right )}}{b} - \frac{\log{\left (\tanh{\left (x \right )} + 1 \right )}}{b} & \text{for}\: b \neq 0 \\\frac{\tanh{\left (x \right )}}{a \operatorname{sech}{\left (x \right )}} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sech(x)+b*tanh(x)),x)

[Out]

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

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Giac [A]  time = 1.15566, size = 30, normalized size = 2.73 \begin{align*} \frac{\log \left ({\left | -b{\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sech(x)+b*tanh(x)),x, algorithm="giac")

[Out]

log(abs(-b*(e^(-x) - e^x) + 2*a))/b