3.928 \(\int e^{2 (a+b x)} \cosh (a+b x) \coth (a+b x) \, dx\)

Optimal. Leaf size=45 \[ \frac{3 e^{a+b x}}{2 b}+\frac{e^{3 a+3 b x}}{6 b}-\frac{2 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

[Out]

(3*E^(a + b*x))/(2*b) + E^(3*a + 3*b*x)/(6*b) - (2*ArcTanh[E^(a + b*x)])/b

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Rubi [A]  time = 0.0341525, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2282, 12, 390, 207} \[ \frac{3 e^{a+b x}}{2 b}+\frac{e^{3 a+3 b x}}{6 b}-\frac{2 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*E^(a + b*x))/(2*b) + E^(3*a + 3*b*x)/(6*b) - (2*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 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 207

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

Rubi steps

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

Mathematica [A]  time = 0.262552, size = 58, normalized size = 1.29 \[ -\frac{e^{a+b x} \left (-\frac{1}{3} e^{2 (a+b x)}+\frac{4 \tanh ^{-1}\left (\sqrt{e^{2 (a+b x)}}\right )}{\sqrt{e^{2 (a+b x)}}}-3\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^(a + b*x)*(-3 - E^(2*(a + b*x))/3 + (4*ArcTanh[Sqrt[E^(2*(a + b*x))]])/Sqrt[E^(2*(a + b*x))]))/(2*b)

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Maple [A]  time = 0.073, size = 54, normalized size = 1.2 \begin{align*}{\frac{{{\rm e}^{3\,bx+3\,a}}}{6\,b}}+{\frac{3\,{{\rm e}^{bx+a}}}{2\,b}}-{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{b}}+{\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)^2*csch(b*x+a),x)

[Out]

1/6*exp(3*b*x+3*a)/b+3/2*exp(b*x+a)/b-1/b*ln(1+exp(b*x+a))+1/b*ln(exp(b*x+a)-1)

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Maxima [A]  time = 1.00814, size = 82, normalized size = 1.82 \begin{align*} \frac{{\left (9 \, e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )} e^{\left (3 \, b x + 3 \, a\right )}}{6 \, 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)^2*csch(b*x+a),x, algorithm="maxima")

[Out]

1/6*(9*e^(-2*b*x - 2*a) + 1)*e^(3*b*x + 3*a)/b - log(e^(-b*x - a) + 1)/b + log(e^(-b*x - a) - 1)/b

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Fricas [B]  time = 1.84004, size = 298, normalized size = 6.62 \begin{align*} \frac{\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} + 3 \,{\left (\cosh \left (b x + a\right )^{2} + 3\right )} \sinh \left (b x + a\right ) + 9 \, \cosh \left (b x + a\right ) - 6 \, \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 6 \, \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right )}{6 \, 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)^2*csch(b*x+a),x, algorithm="fricas")

[Out]

1/6*(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 + sinh(b*x + a)^3 + 3*(cosh(b*x + a)^2 + 3)*sinh(b*x +
a) + 9*cosh(b*x + a) - 6*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 6*log(cosh(b*x + a) + sinh(b*x + a) - 1))/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)**2*csch(b*x+a),x)

[Out]

Timed out

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Giac [A]  time = 1.15038, size = 73, normalized size = 1.62 \begin{align*} \frac{{\left (e^{\left (3 \, b x + 9 \, a\right )} + 9 \, e^{\left (b x + 7 \, a\right )}\right )} e^{\left (-6 \, a\right )} - 6 \, \log \left (e^{\left (b x + a\right )} + 1\right ) + 6 \, \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{6 \, 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)^2*csch(b*x+a),x, algorithm="giac")

[Out]

1/6*((e^(3*b*x + 9*a) + 9*e^(b*x + 7*a))*e^(-6*a) - 6*log(e^(b*x + a) + 1) + 6*log(abs(e^(b*x + a) - 1)))/b