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

Optimal. Leaf size=42 \[ \frac {e^{2 a+2 b x}}{4 b}+\frac {\log \left (1-e^{2 a+2 b x}\right )}{b}-\frac {x}{2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2282, 12, 446, 72} \[ \frac {e^{2 a+2 b x}}{4 b}+\frac {\log \left (1-e^{2 a+2 b x}\right )}{b}-\frac {x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

E^(2*a + 2*b*x)/(4*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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 39, normalized size = 0.93 \[ \frac {e^{2 a+2 b x}+4 \log \left (1-e^{2 a+2 b x}\right )-2 b x}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.48, size = 72, normalized size = 1.71 \[ -\frac {2 \, b x - \cosh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - \sinh \left (b x + a\right )^{2} - 4 \, \log \left (\frac {2 \, \sinh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)^2*csch(b*x+a),x, algorithm="fricas")

[Out]

-1/4*(2*b*x - cosh(b*x + a)^2 - 2*cosh(b*x + a)*sinh(b*x + a) - sinh(b*x + a)^2 - 4*log(2*sinh(b*x + a)/(cosh(
b*x + a) - sinh(b*x + a))))/b

________________________________________________________________________________________

giac [A]  time = 0.14, size = 39, normalized size = 0.93 \[ -\frac {2 \, b x + 2 \, a - e^{\left (2 \, b x + 2 \, a\right )} - 4 \, \log \left ({\left | e^{\left (2 \, b x + 2 \, a\right )} - 1 \right |}\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)^2*csch(b*x+a),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.20, size = 52, normalized size = 1.24 \[ \frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{2 b}+\frac {x}{2}+\frac {a}{2 b}+\frac {\cosh ^{2}\left (b x +a \right )}{2 b}+\frac {\ln \left (\sinh \left (b x +a \right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*cosh(b*x+a)^2*csch(b*x+a),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.33, size = 50, normalized size = 1.19 \[ -\frac {1}{2} \, x - \frac {a}{2 \, b} + \frac {e^{\left (2 \, b x + 2 \, a\right )}}{4 \, b} + \frac {\log \left (e^{\left (b x + a\right )} + 1\right )}{b} + \frac {\log \left (e^{\left (b x + a\right )} - 1\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)^2*csch(b*x+a),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.71, size = 35, normalized size = 0.83 \[ \frac {\ln \left ({\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1\right )}{b}-\frac {x}{2}+\frac {{\mathrm {e}}^{2\,a+2\,b\,x}}{4\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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(b*x+a)*cosh(b*x+a)**2*csch(b*x+a),x)

[Out]

Timed out

________________________________________________________________________________________