### 3.906 $$\int e^{a+b x} \coth (a+b x) \text{csch}^2(a+b x) \, dx$$

Optimal. Leaf size=70 $\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}-\frac{\tanh ^{-1}\left (e^{a+b x}\right )}{b}$

[Out]

(-2*E^(a + b*x))/(b*(1 - E^(2*a + 2*b*x))^2) + (3*E^(a + b*x))/(b*(1 - E^(2*a + 2*b*x))) - ArcTanh[E^(a + b*x)
]/b

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Rubi [A]  time = 0.0602406, antiderivative size = 70, 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, 385, 206} $\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}-\frac{\tanh ^{-1}\left (e^{a+b x}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*E^(a + b*x))/(b*(1 - E^(2*a + 2*b*x))^2) + (3*E^(a + b*x))/(b*(1 - E^(2*a + 2*b*x))) - 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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 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^{a+b x} \coth (a+b x) \text{csch}^2(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{4 x^2 \left (-1-x^2\right )}{\left (1-x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{4 \operatorname{Subst}\left (\int \frac{x^2 \left (-1-x^2\right )}{\left (1-x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=-\frac{2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}-\frac{\operatorname{Subst}\left (\int \frac{-2-4 x^2}{\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 )^2}+\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{\operatorname{Subst}\left (\int \frac{1}{1-x^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 )^2}+\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{\tanh ^{-1}\left (e^{a+b x}\right )}{b}\\ \end{align*}

Mathematica [A]  time = 0.0802675, size = 59, normalized size = 0.84 $\frac{e^{a+b x}-3 e^{3 (a+b x)}+\left (e^{2 (a+b x)}-1\right )^2 \left (-\tanh ^{-1}\left (e^{a+b x}\right )\right )}{b \left (e^{2 (a+b x)}-1\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(E^(a + b*x) - 3*E^(3*(a + b*x)) - (-1 + E^(2*(a + b*x)))^2*ArcTanh[E^(a + b*x)])/(b*(-1 + E^(2*(a + b*x)))^2)

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Maple [A]  time = 0.017, size = 69, normalized size = 1. \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{\sinh \left ( bx+a \right ) }}+\sinh \left ( bx+a \right ) -{\frac{\cosh \left ( bx+a \right ) }{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}}+{\frac{{\rm csch} \left (bx+a\right ){\rm coth} \left (bx+a\right )}{2}}-{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) \right ) } \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)^3,x)

[Out]

1/b*(-1/sinh(b*x+a)*cosh(b*x+a)^2+sinh(b*x+a)-1/sinh(b*x+a)^2*cosh(b*x+a)+1/2*csch(b*x+a)*coth(b*x+a)-arctanh(
exp(b*x+a)))

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Maxima [A]  time = 1.01084, size = 105, normalized size = 1.5 \begin{align*} -\frac{\log \left (e^{\left (b x + a\right )} + 1\right )}{2 \, b} + \frac{\log \left (e^{\left (b x + a\right )} - 1\right )}{2 \, b} - \frac{3 \, e^{\left (3 \, b x + 3 \, a\right )} - e^{\left (b x + a\right )}}{b{\left (e^{\left (4 \, b x + 4 \, a\right )} - 2 \, 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)^3,x, algorithm="maxima")

[Out]

-1/2*log(e^(b*x + a) + 1)/b + 1/2*log(e^(b*x + a) - 1)/b - (3*e^(3*b*x + 3*a) - e^(b*x + a))/(b*(e^(4*b*x + 4*
a) - 2*e^(2*b*x + 2*a) + 1))

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Fricas [B]  time = 1.5487, size = 1089, normalized size = 15.56 \begin{align*} -\frac{6 \, \cosh \left (b x + a\right )^{3} + 18 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + 6 \, \sinh \left (b x + a\right )^{3} +{\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \,{\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 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 )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \,{\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 2 \,{\left (9 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) - 2 \, \cosh \left (b x + a\right )}{2 \,{\left (b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} - 2 \, b \cosh \left (b x + a\right )^{2} + 2 \,{\left (3 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{2} + 4 \,{\left (b \cosh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b\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)^3,x, algorithm="fricas")

[Out]

-1/2*(6*cosh(b*x + a)^3 + 18*cosh(b*x + a)*sinh(b*x + a)^2 + 6*sinh(b*x + a)^3 + (cosh(b*x + a)^4 + 4*cosh(b*x
+ a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4*(c
osh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + sinh(b*x + a) + 1) - (cosh(b*x + a)^4 +
4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x +
a)^2 + 4*(cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + sinh(b*x + a) - 1) + 2*(9*co
sh(b*x + a)^2 - 1)*sinh(b*x + a) - 2*cosh(b*x + a))/(b*cosh(b*x + a)^4 + 4*b*cosh(b*x + a)*sinh(b*x + a)^3 + b
*sinh(b*x + a)^4 - 2*b*cosh(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^2 + 4*(b*cosh(b*x + a)^3 -
b*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(b*x+a)*cosh(b*x+a)*csch(b*x+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.37044, size = 92, normalized size = 1.31 \begin{align*} -\frac{\log \left (e^{\left (b x + a\right )} + 1\right )}{2 \, b} + \frac{\log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{2 \, b} - \frac{3 \, e^{\left (3 \, b x + 3 \, a\right )} - e^{\left (b x + a\right )}}{b{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}} \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)^3,x, algorithm="giac")

[Out]

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