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

Optimal. Leaf size=63 \[ \frac {6}{b \left (1-e^{2 a+2 b x}\right )}-\frac {2}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {2 \log \left (1-e^{2 a+2 b x}\right )}{b} \]

[Out]

-2/b/(1-exp(2*b*x+2*a))^2+6/b/(1-exp(2*b*x+2*a))+2*ln(1-exp(2*b*x+2*a))/b

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Rubi [A]  time = 0.07, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2282, 12, 446, 77} \[ \frac {6}{b \left (1-e^{2 a+2 b x}\right )}-\frac {2}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {2 \log \left (1-e^{2 a+2 b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2/(b*(1 - E^(2*a + 2*b*x))^2) + 6/(b*(1 - E^(2*a + 2*b*x))) + (2*Log[1 - E^(2*a + 2*b*x)])/b

Rule 12

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

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]]]

Rubi steps

\begin {align*} \int e^{2 (a+b x)} \coth (a+b x) \text {csch}^2(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {4 x^3 \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^3 \left (-1-x^2\right )}{\left (1-x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {2 \operatorname {Subst}\left (\int \frac {(-1-x) x}{(1-x)^3} \, dx,x,e^{2 a+2 b x}\right )}{b}\\ &=\frac {2 \operatorname {Subst}\left (\int \left (\frac {2}{(-1+x)^3}+\frac {3}{(-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 )^2}+\frac {6}{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*}

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Mathematica [A]  time = 0.08, size = 47, normalized size = 0.75 \[ \frac {2 \left (\frac {2-3 e^{2 (a+b x)}}{\left (e^{2 (a+b x)}-1\right )^2}+\log \left (1-e^{2 (a+b x)}\right )\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.52, size = 262, normalized size = 4.16 \[ -\frac {2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\frac {2 \, \sinh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right ) + 6 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 3 \, \sinh \left (b x + a\right )^{2} - 2\right )}}{b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} - 2 \, b \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b \cosh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b} \]

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)^3,x, algorithm="fricas")

[Out]

-2*(3*cosh(b*x + a)^2 - (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(2*
sinh(b*x + a)/(cosh(b*x + a) - sinh(b*x + a))) + 6*cosh(b*x + a)*sinh(b*x + a) + 3*sinh(b*x + a)^2 - 2)/(b*cos
h(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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giac [A]  time = 0.15, size = 48, normalized size = 0.76 \[ -\frac {\frac {3 \, e^{\left (4 \, b x + 4 \, a\right )} - 1}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}} - 2 \, \log \left ({\left | e^{\left (2 \, b x + 2 \, a\right )} - 1 \right |}\right )}{b} \]

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)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.22, size = 56, normalized size = 0.89 \[ -\frac {4 a}{b}-\frac {2 \left (3 \,{\mathrm e}^{2 b x +2 a}-2\right )}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )^{2}}+\frac {2 \ln \left ({\mathrm e}^{2 b x +2 a}-1\right )}{b} \]

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)^3,x)

[Out]

-4*a/b-2*(3*exp(2*b*x+2*a)-2)/b/(exp(2*b*x+2*a)-1)^2+2/b*ln(exp(2*b*x+2*a)-1)

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maxima [A]  time = 0.32, size = 86, normalized size = 1.37 \[ 4 \, x + \frac {4 \, a}{b} + \frac {2 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac {2 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{b} - \frac {2 \, {\left (e^{\left (-2 \, b x - 2 \, a\right )} - 2\right )}}{b {\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )}} \]

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

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mupad [B]  time = 1.86, size = 66, normalized size = 1.05 \[ \frac {2\,\ln \left ({\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1\right )}{b}-\frac {6}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )}-\frac {2}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(a + b*x)*exp(2*a + 2*b*x))/sinh(a + b*x)^3,x)

[Out]

(2*log(exp(2*a)*exp(2*b*x) - 1))/b - 6/(b*(exp(2*a + 2*b*x) - 1)) - 2/(b*(exp(4*a + 4*b*x) - 2*exp(2*a + 2*b*x
) + 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**3,x)

[Out]

Timed out

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