### 3.922 $$\int e^{2 (a+b x)} \coth (a+b x) \, dx$$

Optimal. Leaf size=37 $\frac{e^{2 a+2 b x}}{2 b}+\frac{\log \left (1-e^{2 a+2 b x}\right )}{b}$

[Out]

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

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Rubi [A]  time = 0.032322, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.188, Rules used = {2282, 444, 43} $\frac{e^{2 a+2 b x}}{2 b}+\frac{\log \left (1-e^{2 a+2 b x}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(2*(a + b*x))*Coth[a + b*x],x]

[Out]

E^(2*a + 2*b*x)/(2*b) + 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 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

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

Mathematica [A]  time = 0.0245513, size = 35, normalized size = 0.95 $\frac{e^{2 a+2 b x}+2 \log \left (1-e^{2 a+2 b x}\right )}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.034, size = 38, normalized size = 1. \begin{align*}{\frac{{{\rm e}^{2\,bx+2\,a}}}{2\,b}}-2\,{\frac{a}{b}}+{\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)*cosh(b*x+a)*csch(b*x+a),x)

[Out]

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

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Maxima [A]  time = 1.0186, size = 77, normalized size = 2.08 \begin{align*} \frac{2 \,{\left (b x + a\right )}}{b} + \frac{e^{\left (2 \, b x + 2 \, a\right )}}{2 \, b} + \frac{\log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac{\log \left (e^{\left (-b x - a\right )} - 1\right )}{b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*cosh(b*x+a)*csch(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.85139, size = 178, normalized size = 4.81 \begin{align*} \frac{\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} + 2 \, \log \left (\frac{2 \, \sinh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right )}{2 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*cosh(b*x+a)*csch(b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.16356, size = 41, normalized size = 1.11 \begin{align*} \frac{e^{\left (2 \, b x + 2 \, a\right )} + 2 \, \log \left ({\left | e^{\left (2 \, b x + 2 \, a\right )} - 1 \right |}\right )}{2 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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