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]

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

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Rubi [A]  time = 0.0433302, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, 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 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 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 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]

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 = 0.0293295, 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)

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Maple [A]  time = 0.018, size = 52, normalized size = 1.2 \begin{align*}{\frac{\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{2\,b}}+{\frac{x}{2}}+{\frac{a}{2\,b}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{2\,b}}+{\frac{\ln \left ( \sinh \left ( bx+a \right ) \right ) }{b}} \end{align*}

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

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Maxima [A]  time = 1.02406, size = 68, normalized size = 1.62 \begin{align*} -\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} \end{align*}

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

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Fricas [A]  time = 1.65233, size = 190, normalized size = 4.52 \begin{align*} -\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} \end{align*}

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

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

[Out]

Timed out

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Giac [A]  time = 1.17022, size = 57, normalized size = 1.36 \begin{align*} -\frac{b x + a}{2 \, b} + \frac{e^{\left (2 \, b x + 2 \, a\right )}}{4 \, b} + \frac{\log \left ({\left | e^{\left (2 \, b x + 2 \, a\right )} - 1 \right |}\right )}{b} \end{align*}

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]

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