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

Optimal. Leaf size=63 \[ \frac{e^{2 a+2 b x}}{4 b}+\frac{2}{b \left (1-e^{2 a+2 b x}\right )}+\frac{\log \left (1-e^{2 a+2 b x}\right )}{b}+\frac{x}{2} \]

[Out]

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

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Rubi [A]  time = 0.0645684, antiderivative size = 63, 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, 446, 88} \[ \frac{e^{2 a+2 b x}}{4 b}+\frac{2}{b \left (1-e^{2 a+2 b x}\right )}+\frac{\log \left (1-e^{2 a+2 b x}\right )}{b}+\frac{x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

E^(2*a + 2*b*x)/(4*b) + 2/(b*(1 - E^(2*a + 2*b*x))) + 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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int e^{a+b x} \cosh (a+b x) \coth ^2(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^3}{2 x \left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^3}{x \left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{2 b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(1+x)^3}{(1-x)^2 x} \, dx,x,e^{2 a+2 b x}\right )}{4 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (1+\frac{8}{(-1+x)^2}+\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{2}{b \left (1-e^{2 a+2 b x}\right )}+\frac{x}{2}+\frac{\log \left (1-e^{2 a+2 b x}\right )}{b}\\ \end{align*}

Mathematica [A]  time = 0.081858, size = 52, normalized size = 0.83 \[ \frac{e^{2 (a+b x)}-\frac{8}{e^{2 (a+b x)}-1}+4 \log \left (1-e^{2 (a+b x)}\right )+2 b x}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(2*(a + b*x)) - 8/(-1 + E^(2*(a + b*x))) + 2*b*x + 4*Log[1 - E^(2*(a + b*x))])/(4*b)

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Maple [A]  time = 0.02, size = 67, normalized size = 1.1 \begin{align*}{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{2\,b}}+{\frac{\ln \left ( \sinh \left ( bx+a \right ) \right ) }{b}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{2\,b\sinh \left ( bx+a \right ) }}+{\frac{3\,x}{2}}+{\frac{3\,a}{2\,b}}-{\frac{3\,{\rm coth} \left (bx+a\right )}{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*cosh(b*x+a)^2/b+ln(sinh(b*x+a))/b+1/2/b*cosh(b*x+a)^3/sinh(b*x+a)+3/2*x+3/2*a/b-3/2*coth(b*x+a)/b

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Maxima [A]  time = 1.02303, size = 92, normalized size = 1.46 \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} - \frac{2}{b{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)^3*csch(b*x+a)^2,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 - 2/(b*(e^(2*b*x + 2
*a) - 1))

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Fricas [B]  time = 1.92155, size = 594, normalized size = 9.43 \begin{align*} \frac{\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} +{\left (2 \, b x - 1\right )} \cosh \left (b x + a\right )^{2} +{\left (2 \, b x + 6 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, b x + 4 \,{\left (\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} - 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 ) + 2 \,{\left (2 \, \cosh \left (b x + a\right )^{3} +{\left (2 \, b x - 1\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - 8}{4 \,{\left (b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2} - b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)^3*csch(b*x+a)^2,x, algorithm="fricas")

[Out]

1/4*(cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + (2*b*x - 1)*cosh(b*x + a)^2 + (2*b*
x + 6*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*b*x + 4*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh
(b*x + a)^2 - 1)*log(2*sinh(b*x + a)/(cosh(b*x + a) - sinh(b*x + a))) + 2*(2*cosh(b*x + a)^3 + (2*b*x - 1)*cos
h(b*x + a))*sinh(b*x + a) - 8)/(b*cosh(b*x + a)^2 + 2*b*cosh(b*x + a)*sinh(b*x + a) + b*sinh(b*x + a)^2 - 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)**3*csch(b*x+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.15194, size = 96, normalized size = 1.52 \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} - \frac{e^{\left (2 \, b x + 2 \, a\right )} + 1}{b{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)^3*csch(b*x+a)^2,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 - (e^(2*b*x + 2*a) + 1)/(b*(e^(2*b*x
+ 2*a) - 1))