### 3.934 $$\int e^{2 (a+b x)} \cosh ^2(a+b x) \coth (a+b x) \, dx$$

Optimal. Leaf size=59 $\frac{e^{2 a+2 b x}}{2 b}+\frac{e^{4 a+4 b x}}{16 b}+\frac{\log \left (1-e^{2 a+2 b x}\right )}{b}-\frac{x}{4}$

[Out]

E^(2*a + 2*b*x)/(2*b) + E^(4*a + 4*b*x)/(16*b) - x/4 + Log[1 - E^(2*a + 2*b*x)]/b

________________________________________________________________________________________

Rubi [A]  time = 0.0603323, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {2282, 12, 446, 72} $\frac{e^{2 a+2 b x}}{2 b}+\frac{e^{4 a+4 b x}}{16 b}+\frac{\log \left (1-e^{2 a+2 b x}\right )}{b}-\frac{x}{4}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

E^(2*a + 2*b*x)/(2*b) + E^(4*a + 4*b*x)/(16*b) - x/4 + 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^{2 (a+b x)} \cosh ^2(a+b x) \coth (a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^3}{4 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 )^3}{x \left (-1+x^2\right )} \, dx,x,e^{a+b x}\right )}{4 b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(1+x)^3}{(-1+x) x} \, dx,x,e^{2 a+2 b x}\right )}{8 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (4+\frac{8}{-1+x}-\frac{1}{x}+x\right ) \, dx,x,e^{2 a+2 b x}\right )}{8 b}\\ &=\frac{e^{2 a+2 b x}}{2 b}+\frac{e^{4 a+4 b x}}{16 b}-\frac{x}{4}+\frac{\log \left (1-e^{2 a+2 b x}\right )}{b}\\ \end{align*}

Mathematica [A]  time = 0.0424042, size = 48, normalized size = 0.81 $\frac{8 e^{2 (a+b x)}+e^{4 (a+b x)}+16 \log \left (1-e^{2 (a+b x)}\right )-4 b x}{16 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(8*E^(2*(a + b*x)) + E^(4*(a + b*x)) - 4*b*x + 16*Log[1 - E^(2*(a + b*x))])/(16*b)

________________________________________________________________________________________

Maple [A]  time = 0.115, size = 55, normalized size = 0.9 \begin{align*} -{\frac{x}{4}}+{\frac{{{\rm e}^{4\,bx+4\,a}}}{16\,b}}+{\frac{{{\rm e}^{2\,bx+2\,a}}}{2\,b}}-2\,{\frac{a}{b}}+{\frac{\ln \left ({{\rm e}^{2\,bx+2\,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)^3*csch(b*x+a),x)

[Out]

-1/4*x+1/16*exp(4*b*x+4*a)/b+1/2*exp(2*b*x+2*a)/b-2*a/b+1/b*ln(exp(2*b*x+2*a)-1)

________________________________________________________________________________________

Maxima [A]  time = 1.0337, size = 95, normalized size = 1.61 \begin{align*} \frac{{\left (8 \, e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )} e^{\left (4 \, b x + 4 \, a\right )}}{16 \, b} + \frac{7 \,{\left (b x + 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(2*b*x+2*a)*cosh(b*x+a)^3*csch(b*x+a),x, algorithm="maxima")

[Out]

1/16*(8*e^(-2*b*x - 2*a) + 1)*e^(4*b*x + 4*a)/b + 7/4*(b*x + a)/b + log(e^(-b*x - a) + 1)/b + log(e^(-b*x - a)
- 1)/b

________________________________________________________________________________________

Fricas [B]  time = 1.94596, size = 354, normalized size = 6. \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} + 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} + 4\right )} \sinh \left (b x + a\right )^{2} - 4 \, b x + 8 \, \cosh \left (b x + a\right )^{2} + 4 \,{\left (\cosh \left (b x + a\right )^{3} + 4 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 16 \, \log \left (\frac{2 \, \sinh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right )}{16 \, 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)^3*csch(b*x+a),x, algorithm="fricas")

[Out]

1/16*(cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 4)*sinh(b*x
+ a)^2 - 4*b*x + 8*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 + 4*cosh(b*x + a))*sinh(b*x + a) + 16*log(2*sinh(b*x
+ a)/(cosh(b*x + a) - sinh(b*x + a))))/b

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.21171, size = 70, normalized size = 1.19 \begin{align*} -\frac{4 \, b x -{\left (e^{\left (4 \, b x + 8 \, a\right )} + 8 \, e^{\left (2 \, b x + 6 \, a\right )}\right )} e^{\left (-4 \, a\right )} - 16 \, \log \left ({\left | e^{\left (2 \, b x + 2 \, a\right )} - 1 \right |}\right )}{16 \, 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)^3*csch(b*x+a),x, algorithm="giac")

[Out]

-1/16*(4*b*x - (e^(4*b*x + 8*a) + 8*e^(2*b*x + 6*a))*e^(-4*a) - 16*log(abs(e^(2*b*x + 2*a) - 1)))/b