### 3.876 $$\int e^{2 (a+b x)} \text{csch}^2(a+b x) \, dx$$

Optimal. Leaf size=42 $\frac{2}{b \left (1-e^{2 a+2 b x}\right )}+\frac{2 \log \left (1-e^{2 a+2 b x}\right )}{b}$

[Out]

2/(b*(1 - E^(2*a + 2*b*x))) + (2*Log[1 - E^(2*a + 2*b*x)])/b

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Rubi [A]  time = 0.0427926, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.222, Rules used = {2282, 12, 266, 43} $\frac{2}{b \left (1-e^{2 a+2 b x}\right )}+\frac{2 \log \left (1-e^{2 a+2 b x}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(2*(a + b*x))*Csch[a + b*x]^2,x]

[Out]

2/(b*(1 - E^(2*a + 2*b*x))) + (2*Log[1 - E^(2*a + 2*b*x)])/b

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int e^{2 (a+b x)} \text{csch}^2(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{4 x^3}{\left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{4 \operatorname{Subst}\left (\int \frac{x^3}{\left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{x}{(1-x)^2} \, dx,x,e^{2 a+2 b x}\right )}{b}\\ &=\frac{2 \operatorname{Subst}\left (\int \left (\frac{1}{(-1+x)^2}+\frac{1}{-1+x}\right ) \, dx,x,e^{2 a+2 b x}\right )}{b}\\ &=\frac{2}{b \left (1-e^{2 a+2 b x}\right )}+\frac{2 \log \left (1-e^{2 a+2 b x}\right )}{b}\\ \end{align*}

Mathematica [A]  time = 0.0489636, size = 37, normalized size = 0.88 $\frac{2 \left (\frac{1}{1-e^{2 a+2 b x}}+\log \left (1-e^{2 a+2 b x}\right )\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^(2*(a + b*x))*Csch[a + b*x]^2,x]

[Out]

(2*((1 - E^(2*a + 2*b*x))^(-1) + Log[1 - E^(2*a + 2*b*x)]))/b

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Maple [A]  time = 0.032, size = 43, normalized size = 1. \begin{align*} -4\,{\frac{a}{b}}-2\,{\frac{1}{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }}+2\,{\frac{\ln \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }{b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*a/b-2/b/(exp(2*b*x+2*a)-1)+2/b*ln(exp(2*b*x+2*a)-1)

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Maxima [A]  time = 1.45041, size = 84, normalized size = 2. \begin{align*} 4 \, x + \frac{4 \, a}{b} + \frac{2 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac{2 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{b} + \frac{2}{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(exp(2*b*x+2*a)*csch(b*x+a)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.77485, size = 286, normalized size = 6.81 \begin{align*} \frac{2 \,{\left ({\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 (\frac{2 \, \sinh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right ) - 1\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(exp(2*b*x+2*a)*csch(b*x+a)^2,x, algorithm="fricas")

[Out]

2*((cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 - 1)*log(2*sinh(b*x + a)/(cosh(b*x + a)
- sinh(b*x + a))) - 1)/(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*} e^{2 a} \int e^{2 b x} \operatorname{csch}^{2}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15745, size = 65, normalized size = 1.55 \begin{align*} \frac{2 \,{\left (e^{\left (-2 \, a\right )} \log \left ({\left | e^{\left (2 \, b x + 2 \, a\right )} - 1 \right |}\right ) - \frac{e^{\left (2 \, b x\right )}}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right )} e^{\left (2 \, a\right )}}{b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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