### 3.687 $$\int \frac{1}{\sinh (x)-\tanh (x)} \, dx$$

Optimal. Leaf size=20 $\frac{1}{2 (1-\cosh (x))}-\frac{1}{2} \tanh ^{-1}(\cosh (x))$

[Out]

-ArcTanh[Cosh[x]]/2 + 1/(2*(1 - Cosh[x]))

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Rubi [A]  time = 0.0732545, antiderivative size = 24, normalized size of antiderivative = 1.2, number of steps used = 6, number of rules used = 6, integrand size = 9, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.667, Rules used = {4397, 2706, 2606, 30, 2611, 3770} $-\frac{1}{2} \text{csch}^2(x)-\frac{1}{2} \tanh ^{-1}(\cosh (x))-\frac{1}{2} \coth (x) \text{csch}(x)$

Antiderivative was successfully veriﬁed.

[In]

Int[(Sinh[x] - Tanh[x])^(-1),x]

[Out]

-ArcTanh[Cosh[x]]/2 - (Coth[x]*Csch[x])/2 - Csch[x]^2/2

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rule 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rule 2606

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2611

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{1}{\sinh (x)-\tanh (x)} \, dx &=-\left (i \int \frac{\coth (x)}{i-i \cosh (x)} \, dx\right )\\ &=\int \coth ^2(x) \text{csch}(x) \, dx+\int \coth (x) \text{csch}^2(x) \, dx\\ &=-\frac{1}{2} \coth (x) \text{csch}(x)+\frac{1}{2} \int \text{csch}(x) \, dx+\operatorname{Subst}(\int x \, dx,x,-i \text{csch}(x))\\ &=-\frac{1}{2} \tanh ^{-1}(\cosh (x))-\frac{1}{2} \coth (x) \text{csch}(x)-\frac{\text{csch}^2(x)}{2}\\ \end{align*}

Mathematica [B]  time = 0.0431123, size = 50, normalized size = 2.5 $-\frac{1}{4} \text{csch}^2\left (\frac{x}{2}\right ) \left (\log \left (\sinh \left (\frac{x}{2}\right )\right )-\log \left (\cosh \left (\frac{x}{2}\right )\right )+\cosh (x) \left (\log \left (\cosh \left (\frac{x}{2}\right )\right )-\log \left (\sinh \left (\frac{x}{2}\right )\right )\right )+1\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Sinh[x] - Tanh[x])^(-1),x]

[Out]

-(Csch[x/2]^2*(1 - Log[Cosh[x/2]] + Cosh[x]*(Log[Cosh[x/2]] - Log[Sinh[x/2]]) + Log[Sinh[x/2]]))/4

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Maple [A]  time = 0.032, size = 17, normalized size = 0.9 \begin{align*} -{\frac{1}{4} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}}+{\frac{1}{2}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)-tanh(x)),x)

[Out]

-1/4/tanh(1/2*x)^2+1/2*ln(tanh(1/2*x))

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Maxima [B]  time = 1.22083, size = 54, normalized size = 2.7 \begin{align*} \frac{e^{\left (-x\right )}}{2 \, e^{\left (-x\right )} - e^{\left (-2 \, x\right )} - 1} - \frac{1}{2} \, \log \left (e^{\left (-x\right )} + 1\right ) + \frac{1}{2} \, \log \left (e^{\left (-x\right )} - 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sinh(x)-tanh(x)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.69748, size = 386, normalized size = 19.3 \begin{align*} -\frac{{\left (\cosh \left (x\right )^{2} + 2 \,{\left (\cosh \left (x\right ) - 1\right )} \sinh \left (x\right ) + \sinh \left (x\right )^{2} - 2 \, \cosh \left (x\right ) + 1\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) -{\left (\cosh \left (x\right )^{2} + 2 \,{\left (\cosh \left (x\right ) - 1\right )} \sinh \left (x\right ) + \sinh \left (x\right )^{2} - 2 \, \cosh \left (x\right ) + 1\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right ) + 2 \, \cosh \left (x\right ) + 2 \, \sinh \left (x\right )}{2 \,{\left (\cosh \left (x\right )^{2} + 2 \,{\left (\cosh \left (x\right ) - 1\right )} \sinh \left (x\right ) + \sinh \left (x\right )^{2} - 2 \, \cosh \left (x\right ) + 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sinh(x)-tanh(x)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sinh{\left (x \right )} - \tanh{\left (x \right )}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sinh(x)-tanh(x)),x)

[Out]

Integral(1/(sinh(x) - tanh(x)), x)

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Giac [B]  time = 1.11335, size = 58, normalized size = 2.9 \begin{align*} -\frac{e^{\left (-x\right )} + e^{x} + 2}{4 \,{\left (e^{\left (-x\right )} + e^{x} - 2\right )}} - \frac{1}{4} \, \log \left (e^{\left (-x\right )} + e^{x} + 2\right ) + \frac{1}{4} \, \log \left (e^{\left (-x\right )} + e^{x} - 2\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sinh(x)-tanh(x)),x, algorithm="giac")

[Out]

-1/4*(e^(-x) + e^x + 2)/(e^(-x) + e^x - 2) - 1/4*log(e^(-x) + e^x + 2) + 1/4*log(e^(-x) + e^x - 2)