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

Optimal. Leaf size=37 \[ \frac {e^{2 a+2 b x}}{2 b}+\frac {\log \left (1-e^{2 a+2 b x}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2282, 444, 43} \[ \frac {e^{2 a+2 b x}}{2 b}+\frac {\log \left (1-e^{2 a+2 b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*(a + b*x))*Coth[a + b*x],x]

[Out]

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

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])

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

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Mathematica [A]  time = 0.03, size = 35, normalized size = 0.95 \[ \frac {e^{2 a+2 b x}+2 \log \left (1-e^{2 a+2 b x}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 64, normalized size = 1.73 \[ \frac {\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 2 \, \log \left (\frac {2 \, \sinh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right )}{2 \, b} \]

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),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.11, size = 30, normalized size = 0.81 \[ \frac {e^{\left (2 \, b x + 2 \, a\right )} + 2 \, \log \left ({\left | e^{\left (2 \, b x + 2 \, a\right )} - 1 \right |}\right )}{2 \, b} \]

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),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.26, size = 38, normalized size = 1.03 \[ \frac {{\mathrm e}^{2 b x +2 a}}{2 b}-\frac {2 a}{b}+\frac {\ln \left ({\mathrm e}^{2 b x +2 a}-1\right )}{b} \]

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),x)

[Out]

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

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maxima [A]  time = 0.32, size = 57, normalized size = 1.54 \[ \frac {2 \, {\left (b x + a\right )}}{b} + \frac {e^{\left (2 \, b x + 2 \, a\right )}}{2 \, b} + \frac {\log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac {\log \left (e^{\left (-b x - a\right )} - 1\right )}{b} \]

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),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.06, size = 30, normalized size = 0.81 \[ \frac {{\mathrm {e}}^{2\,a+2\,b\,x}+2\,\ln \left ({\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1\right )}{2\,b} \]

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),x)

[Out]

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

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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),x)

[Out]

Timed out

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