### 3.477 $$\int \text{csch}(a+b x) \text{sech}^2(a+b x) \, dx$$

Optimal. Leaf size=23 $\frac{\text{sech}(a+b x)}{b}-\frac{\tanh ^{-1}(\cosh (a+b x))}{b}$

[Out]

-(ArcTanh[Cosh[a + b*x]]/b) + Sech[a + b*x]/b

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Rubi [A]  time = 0.0264866, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {2622, 321, 207} $\frac{\text{sech}(a+b x)}{b}-\frac{\tanh ^{-1}(\cosh (a+b x))}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Csch[a + b*x]*Sech[a + b*x]^2,x]

[Out]

-(ArcTanh[Cosh[a + b*x]]/b) + Sech[a + b*x]/b

Rule 2622

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0287162, size = 26, normalized size = 1.13 $\frac{\text{sech}(a+b x)}{b}+\frac{\log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Csch[a + b*x]*Sech[a + b*x]^2,x]

[Out]

Log[Tanh[(a + b*x)/2]]/b + Sech[a + b*x]/b

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Maple [A]  time = 0.013, size = 23, normalized size = 1. \begin{align*}{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{-1}-2\,{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) }{b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(csch(b*x+a)*sech(b*x+a)^2,x)

[Out]

1/b*(1/cosh(b*x+a)-2*arctanh(exp(b*x+a)))

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Maxima [B]  time = 1.11271, size = 82, normalized size = 3.57 \begin{align*} -\frac{\log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac{\log \left (e^{\left (-b x - a\right )} - 1\right )}{b} + \frac{2 \, e^{\left (-b x - a\right )}}{b{\left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.31909, size = 462, normalized size = 20.09 \begin{align*} -\frac{{\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) -{\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) - 2 \, \cosh \left (b x + a\right ) - 2 \, \sinh \left (b x + a\right )}{b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2} + b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{csch}{\left (a + b x \right )} \operatorname{sech}^{2}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)*sech(b*x+a)**2,x)

[Out]

Integral(csch(a + b*x)*sech(a + b*x)**2, x)

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Giac [B]  time = 1.19063, size = 95, normalized size = 4.13 \begin{align*} -\frac{\log \left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} + 2\right )}{2 \, b} + \frac{\log \left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} - 2\right )}{2 \, b} + \frac{2}{b{\left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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