### 3.916 $$\int e^{a+b x} \cosh ^2(a+b x) \coth (a+b x) \, dx$$

Optimal. Leaf size=59 $\frac{e^{-a-b x}}{4 b}+\frac{e^{a+b x}}{b}+\frac{e^{3 a+3 b x}}{12 b}-\frac{2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}$

[Out]

E^(-a - b*x)/(4*b) + E^(a + b*x)/b + E^(3*a + 3*b*x)/(12*b) - (2*ArcTanh[E^(a + b*x)])/b

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Rubi [A]  time = 0.0537115, antiderivative size = 59, 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, 461, 207} $\frac{e^{-a-b x}}{4 b}+\frac{e^{a+b x}}{b}+\frac{e^{3 a+3 b x}}{12 b}-\frac{2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

E^(-a - b*x)/(4*b) + E^(a + b*x)/b + E^(3*a + 3*b*x)/(12*b) - (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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 207

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

Rubi steps

\begin{align*} \int e^{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^2 \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^2 \left (-1+x^2\right )} \, dx,x,e^{a+b x}\right )}{4 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (4-\frac{1}{x^2}+x^2+\frac{8}{-1+x^2}\right ) \, dx,x,e^{a+b x}\right )}{4 b}\\ &=\frac{e^{-a-b x}}{4 b}+\frac{e^{a+b x}}{b}+\frac{e^{3 a+3 b x}}{12 b}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{e^{-a-b x}}{4 b}+\frac{e^{a+b x}}{b}+\frac{e^{3 a+3 b x}}{12 b}-\frac{2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}\\ \end{align*}

Mathematica [A]  time = 0.347232, size = 68, normalized size = 1.15 $\frac{e^{-a-b x} \left (12 e^{2 (a+b x)}+e^{4 (a+b x)}-24 \sqrt{e^{2 (a+b x)}} \tanh ^{-1}\left (\sqrt{e^{2 (a+b x)}}\right )+3\right )}{12 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(E^(-a - b*x)*(3 + 12*E^(2*(a + b*x)) + E^(4*(a + b*x)) - 24*Sqrt[E^(2*(a + b*x))]*ArcTanh[Sqrt[E^(2*(a + b*x)
)]]))/(12*b)

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Maple [A]  time = 0.02, size = 50, normalized size = 0.9 \begin{align*}{\frac{1}{b} \left ( \left ({\frac{2}{3}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{3}} \right ) \sinh \left ( bx+a \right ) +{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{3}}+\cosh \left ( bx+a \right ) -2\,{\it Artanh} \left ({{\rm e}^{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)^3*csch(b*x+a),x)

[Out]

1/b*((2/3+1/3*cosh(b*x+a)^2)*sinh(b*x+a)+1/3*cosh(b*x+a)^3+cosh(b*x+a)-2*arctanh(exp(b*x+a)))

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Maxima [A]  time = 0.984571, size = 88, normalized size = 1.49 \begin{align*} \frac{e^{\left (3 \, b x + 3 \, a\right )} + 12 \, e^{\left (b x + a\right )}}{12 \, b} + \frac{e^{\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(b*x+a)*cosh(b*x+a)^3*csch(b*x+a),x, algorithm="maxima")

[Out]

1/12*(e^(3*b*x + 3*a) + 12*e^(b*x + a))/b + 1/4*e^(-b*x - a)/b - log(e^(b*x + a) + 1)/b + log(e^(b*x + a) - 1)
/b

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Fricas [B]  time = 1.83499, size = 520, normalized size = 8.81 \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} + 6 \,{\left (\cosh \left (b x + a\right )^{2} + 2\right )} \sinh \left (b x + a\right )^{2} + 12 \, \cosh \left (b x + a\right )^{2} - 12 \,{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 12 \,{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 4 \,{\left (\cosh \left (b x + a\right )^{3} + 6 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 3}{12 \,{\left (b \cosh \left (b x + a\right ) + b \sinh \left (b x + a\right )\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),x, algorithm="fricas")

[Out]

1/12*(cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 6*(cosh(b*x + a)^2 + 2)*sinh(b*x +
a)^2 + 12*cosh(b*x + a)^2 - 12*(cosh(b*x + a) + sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 12*(c
osh(b*x + a) + sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) - 1) + 4*(cosh(b*x + a)^3 + 6*cosh(b*x + a))*s
inh(b*x + a) + 3)/(b*cosh(b*x + a) + b*sinh(b*x + a))

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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),x)

[Out]

Timed out

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Giac [A]  time = 1.17613, size = 99, normalized size = 1.68 \begin{align*} \frac{e^{\left (-b x - a\right )}}{4 \, 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{b^{2} e^{\left (3 \, b x + 3 \, a\right )} + 12 \, b^{2} e^{\left (b x + a\right )}}{12 \, b^{3}} \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),x, algorithm="giac")

[Out]

1/4*e^(-b*x - a)/b - log(e^(b*x + a) + 1)/b + log(abs(e^(b*x + a) - 1))/b + 1/12*(b^2*e^(3*b*x + 3*a) + 12*b^2
*e^(b*x + a))/b^3