### 3.911 $$\int e^{a+b x} \coth ^2(a+b x) \, dx$$

Optimal. Leaf size=53 $\frac{e^{a+b x}}{b}+\frac{2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}$

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0381733, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {2282, 390, 288, 206} $\frac{e^{a+b x}}{b}+\frac{2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(a + b*x)*Coth[a + b*x]^2,x]

[Out]

E^(a + b*x)/b + (2*E^(a + b*x))/(b*(1 - E^(2*a + 2*b*x))) - (2*ArcTanh[E^(a + 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 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 2.54259, size = 179, normalized size = 3.38 $\frac{e^{a+b x} \left (\frac{4}{105} \left (e^{a+b x}+e^{3 (a+b x)}\right )^2 \, _4F_3\left (\frac{3}{2},2,2,2;1,1,\frac{9}{2};e^{2 (a+b x)}\right )+\frac{1}{48} e^{-4 (a+b x)} \left (-713 e^{2 (a+b x)}-181 e^{4 (a+b x)}+61 e^{6 (a+b x)}+\frac{3 \left (196 e^{2 (a+b x)}-14 e^{4 (a+b x)}-52 e^{6 (a+b x)}+e^{8 (a+b x)}+125\right ) \tanh ^{-1}\left (\sqrt{e^{2 (a+b x)}}\right )}{\sqrt{e^{2 (a+b x)}}}-375\right )\right )}{b}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(E^(a + b*x)*((-375 - 713*E^(2*(a + b*x)) - 181*E^(4*(a + b*x)) + 61*E^(6*(a + b*x)) + (3*(125 + 196*E^(2*(a +
b*x)) - 14*E^(4*(a + b*x)) - 52*E^(6*(a + b*x)) + E^(8*(a + b*x)))*ArcTanh[Sqrt[E^(2*(a + b*x))]])/Sqrt[E^(2*
(a + b*x))])/(48*E^(4*(a + b*x))) + (4*(E^(a + b*x) + E^(3*(a + b*x)))^2*HypergeometricPFQ[{3/2, 2, 2, 2}, {1,
1, 9/2}, E^(2*(a + b*x))])/105))/b

________________________________________________________________________________________

Maple [A]  time = 0.019, size = 47, normalized size = 0.9 \begin{align*}{\frac{1}{b} \left ( \cosh \left ( bx+a \right ) -2\,{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) -{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{\sinh \left ( bx+a \right ) }}+2\,\sinh \left ( bx+a \right ) \right ) } \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)^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.999177, size = 84, normalized size = 1.58 \begin{align*} \frac{e^{\left (b x + a\right )}}{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 \, e^{\left (b x + a\right )}}{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)^2*csch(b*x+a)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 1.59806, size = 585, normalized size = 11.04 \begin{align*} \frac{\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3} -{\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 (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) +{\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 (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 3 \,{\left (\cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) - 3 \, \cosh \left (b x + a\right )}{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} \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)^2,x, algorithm="fricas")

[Out]

(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 + sinh(b*x + a)^3 - (cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh
(b*x + a) + sinh(b*x + a)^2 - 1)*log(cosh(b*x + a) + sinh(b*x + a) + 1) + (cosh(b*x + a)^2 + 2*cosh(b*x + a)*s
inh(b*x + a) + sinh(b*x + a)^2 - 1)*log(cosh(b*x + a) + sinh(b*x + a) - 1) + 3*(cosh(b*x + a)^2 - 1)*sinh(b*x
+ a) - 3*cosh(b*x + a))/(b*cosh(b*x + a)^2 + 2*b*cosh(b*x + a)*sinh(b*x + a) + b*sinh(b*x + a)^2 - 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(b*x+a)*cosh(b*x+a)**2*csch(b*x+a)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.1461, size = 85, normalized size = 1.6 \begin{align*} \frac{e^{\left (b x + a\right )}}{b} - \frac{\log \left (e^{\left (b x + a\right )} + 1\right )}{b} + \frac{\log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{b} - \frac{2 \, e^{\left (b x + a\right )}}{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)^2*csch(b*x+a)^2,x, algorithm="giac")

[Out]

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