### 3.905 $$\int e^{a+b x} \coth (a+b x) \text{csch}(a+b x) \, dx$$

Optimal. Leaf size=41 $\frac{2}{b \left (1-e^{2 a+2 b x}\right )}+\frac{\log \left (1-e^{2 a+2 b x}\right )}{b}$

[Out]

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

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Rubi [A]  time = 0.0448145, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {2282, 12, 444, 43} $\frac{2}{b \left (1-e^{2 a+2 b x}\right )}+\frac{\log \left (1-e^{2 a+2 b x}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0466463, size = 34, normalized size = 0.83 $\frac{\log \left (1-e^{2 (a+b x)}\right )-\frac{2}{e^{2 (a+b x)}-1}}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.017, size = 30, normalized size = 0.7 \begin{align*} x+{\frac{\ln \left ( \sinh \left ( bx+a \right ) \right ) }{b}}-{\frac{{\rm coth} \left (bx+a\right )}{b}}+{\frac{a}{b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x+ln(sinh(b*x+a))/b-coth(b*x+a)/b+a/b

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Maxima [A]  time = 1.02518, size = 61, normalized size = 1.49 \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}{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(b*x+a)*cosh(b*x+a)*csch(b*x+a)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.47199, size = 284, normalized size = 6.93 \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 (\frac{2 \, \sinh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right ) - 2}{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(b*x+a)*cosh(b*x+a)*csch(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(2*sinh(b*x + a)/(cosh(b*x + a) -
sinh(b*x + a))) - 2)/(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(-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(b*x+a)*cosh(b*x+a)*csch(b*x+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.2552, size = 63, normalized size = 1.54 \begin{align*} \frac{\log \left ({\left | e^{\left (2 \, b x + 2 \, a\right )} - 1 \right |}\right )}{b} - \frac{e^{\left (2 \, b x + 2 \, a\right )} + 1}{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(b*x+a)*cosh(b*x+a)*csch(b*x+a)^2,x, algorithm="giac")

[Out]

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