### 3.923 $$\int e^{2 (a+b x)} \coth (a+b x) \text{csch}(a+b x) \, dx$$

Optimal. Leaf size=54 $\frac{2 e^{a+b x}}{b}+\frac{2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{4 \tanh ^{-1}\left (e^{a+b x}\right )}{b}$

[Out]

(2*E^(a + b*x))/b + (2*E^(a + b*x))/(b*(1 - E^(2*a + 2*b*x))) - (4*ArcTanh[E^(a + b*x)])/b

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Rubi [A]  time = 0.0434778, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.227, Rules used = {2282, 12, 455, 388, 206} $\frac{2 e^{a+b x}}{b}+\frac{2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{4 \tanh ^{-1}\left (e^{a+b x}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*E^(a + b*x))/b + (2*E^(a + b*x))/(b*(1 - E^(2*a + 2*b*x))) - (4*ArcTanh[E^(a + 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 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 388

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

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 0.116034, size = 62, normalized size = 1.15 $\frac{2 \left (\frac{e^{a+b x} \left (e^{2 (a+b x)}-2\right )}{e^{2 (a+b x)}-1}+\log \left (1-e^{a+b x}\right )-\log \left (e^{a+b x}+1\right )\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.039, size = 65, normalized size = 1.2 \begin{align*} 2\,{\frac{{{\rm e}^{bx+a}}}{b}}-2\,{\frac{{{\rm e}^{bx+a}}}{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }}-2\,{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{b}}+2\,{\frac{\ln \left ({{\rm e}^{bx+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)^2,x)

[Out]

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

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Maxima [A]  time = 1.05406, size = 103, normalized size = 1.91 \begin{align*} -\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 \,{\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} - 1\right )}}{b{\left (e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )}\right )}} \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)^2,x, algorithm="maxima")

[Out]

-2*log(e^(-b*x - a) + 1)/b + 2*log(e^(-b*x - a) - 1)/b - 2*(2*e^(-2*b*x - 2*a) - 1)/(b*(e^(-b*x - a) - e^(-3*b
*x - 3*a)))

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Fricas [B]  time = 1.85101, size = 587, normalized size = 10.87 \begin{align*} \frac{2 \,{\left (\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3} -{\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 ) +{\left (3 \, \cosh \left (b x + a\right )^{2} - 2\right )} \sinh \left (b x + a\right ) - 2 \, \cosh \left (b x + a\right )\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)*cosh(b*x+a)*csch(b*x+a)^2,x, algorithm="fricas")

[Out]

2*(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 + sinh(b*x + a)^3 - (cosh(b*x + a)^2 + 2*cosh(b*x + a)*si
nh(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) + (3*cosh(b*x + a)^2 - 2)*sinh(b*
x + a) - 2*cosh(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(-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)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.15969, size = 74, normalized size = 1.37 \begin{align*} -\frac{2 \,{\left (\frac{e^{\left (b x + a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} - 1} - e^{\left (b x + a\right )} + \log \left (e^{\left (b x + a\right )} + 1\right ) - \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )\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)^2,x, algorithm="giac")

[Out]

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