### 3.935 $$\int e^{2 (a+b x)} \cosh (a+b x) \coth ^2(a+b x) \, dx$$

Optimal. Leaf size=73 $\frac{5 e^{a+b x}}{2 b}+\frac{e^{3 a+3 b x}}{6 b}+\frac{2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{4 \tanh ^{-1}\left (e^{a+b x}\right )}{b}$

[Out]

(5*E^(a + b*x))/(2*b) + E^(3*a + 3*b*x)/(6*b) + (2*E^(a + b*x))/(b*(1 - E^(2*a + 2*b*x))) - (4*ArcTanh[E^(a +
b*x)])/b

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Rubi [A]  time = 0.0537078, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.208, Rules used = {2282, 12, 390, 385, 206} $\frac{5 e^{a+b x}}{2 b}+\frac{e^{3 a+3 b x}}{6 b}+\frac{2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{4 \tanh ^{-1}\left (e^{a+b x}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(5*E^(a + b*x))/(2*b) + E^(3*a + 3*b*x)/(6*b) + (2*E^(a + b*x))/(b*(1 - E^(2*a + 2*b*x))) - (4*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 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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

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^{2 (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 \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}{\left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{2 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (5+x^2-\frac{4 \left (1-3 x^2\right )}{\left (1-x^2\right )^2}\right ) \, dx,x,e^{a+b x}\right )}{2 b}\\ &=\frac{5 e^{a+b x}}{2 b}+\frac{e^{3 a+3 b x}}{6 b}-\frac{2 \operatorname{Subst}\left (\int \frac{1-3 x^2}{\left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{5 e^{a+b x}}{2 b}+\frac{e^{3 a+3 b x}}{6 b}+\frac{2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{4 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{5 e^{a+b x}}{2 b}+\frac{e^{3 a+3 b x}}{6 b}+\frac{2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{4 \tanh ^{-1}\left (e^{a+b x}\right )}{b}\\ \end{align*}

Mathematica [C]  time = 2.07236, size = 220, normalized size = 3.01 $\frac{e^{-5 (a+b x)} \left (256 e^{8 (a+b x)} \left (e^{2 (a+b x)}+1\right )^3 \, _5F_4\left (\frac{3}{2},2,2,2,2;1,1,1,\frac{11}{2};e^{2 (a+b x)}\right )-21 \left (91925 e^{2 (a+b x)}+61158 e^{4 (a+b x)}-20166 e^{6 (a+b x)}-15061 e^{8 (a+b x)}+753 e^{10 (a+b x)}+36015\right )-\frac{315 \left (-5328 e^{2 (a+b x)}-1821 e^{4 (a+b x)}+3264 e^{6 (a+b x)}+1149 e^{8 (a+b x)}-240 e^{10 (a+b x)}+e^{12 (a+b x)}-2401\right ) \tanh ^{-1}\left (\sqrt{e^{2 (a+b x)}}\right )}{\sqrt{e^{2 (a+b x)}}}\right )}{60480 b}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-21*(36015 + 91925*E^(2*(a + b*x)) + 61158*E^(4*(a + b*x)) - 20166*E^(6*(a + b*x)) - 15061*E^(8*(a + b*x)) +
753*E^(10*(a + b*x))) - (315*(-2401 - 5328*E^(2*(a + b*x)) - 1821*E^(4*(a + b*x)) + 3264*E^(6*(a + b*x)) + 114
9*E^(8*(a + b*x)) - 240*E^(10*(a + b*x)) + E^(12*(a + b*x)))*ArcTanh[Sqrt[E^(2*(a + b*x))]])/Sqrt[E^(2*(a + b*
x))] + 256*E^(8*(a + b*x))*(1 + E^(2*(a + b*x)))^3*HypergeometricPFQ[{3/2, 2, 2, 2, 2}, {1, 1, 1, 11/2}, E^(2*
(a + b*x))])/(60480*b*E^(5*(a + b*x)))

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Maple [A]  time = 0.119, size = 79, normalized size = 1.1 \begin{align*}{\frac{{{\rm e}^{3\,bx+3\,a}}}{6\,b}}+{\frac{5\,{{\rm e}^{bx+a}}}{2\,b}}-2\,{\frac{{{\rm e}^{bx+a}}}{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }}-2\,{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{b}}+2\,{\frac{\ln \left ({{\rm e}^{bx+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)^2,x)

[Out]

1/6*exp(3*b*x+3*a)/b+5/2*exp(b*x+a)/b-2/b*exp(b*x+a)/(exp(2*b*x+2*a)-1)-2/b*ln(1+exp(b*x+a))+2/b*ln(exp(b*x+a)
-1)

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Maxima [A]  time = 1.04947, size = 117, normalized size = 1.6 \begin{align*} -\frac{2 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac{2 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{b} + \frac{14 \, e^{\left (-2 \, b x - 2 \, a\right )} - 27 \, e^{\left (-4 \, b x - 4 \, a\right )} + 1}{6 \, b{\left (e^{\left (-3 \, b x - 3 \, a\right )} - e^{\left (-5 \, b x - 5 \, a\right )}\right )}} \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)^2,x, algorithm="maxima")

[Out]

-2*log(e^(-b*x - a) + 1)/b + 2*log(e^(-b*x - a) - 1)/b + 1/6*(14*e^(-2*b*x - 2*a) - 27*e^(-4*b*x - 4*a) + 1)/(
b*(e^(-3*b*x - 3*a) - e^(-5*b*x - 5*a)))

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Fricas [B]  time = 1.86758, size = 797, normalized size = 10.92 \begin{align*} \frac{\cosh \left (b x + a\right )^{5} + 5 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + \sinh \left (b x + a\right )^{5} + 2 \,{\left (5 \, \cosh \left (b x + a\right )^{2} + 7\right )} \sinh \left (b x + a\right )^{3} + 14 \, \cosh \left (b x + a\right )^{3} + 2 \,{\left (5 \, \cosh \left (b x + a\right )^{3} + 21 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 12 \,{\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 ) + 12 \,{\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 (5 \, \cosh \left (b x + a\right )^{4} + 42 \, \cosh \left (b x + a\right )^{2} - 27\right )} \sinh \left (b x + a\right ) - 27 \, \cosh \left (b x + a\right )}{6 \,{\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(2*b*x+2*a)*cosh(b*x+a)^3*csch(b*x+a)^2,x, algorithm="fricas")

[Out]

1/6*(cosh(b*x + a)^5 + 5*cosh(b*x + a)*sinh(b*x + a)^4 + sinh(b*x + a)^5 + 2*(5*cosh(b*x + a)^2 + 7)*sinh(b*x
+ a)^3 + 14*cosh(b*x + a)^3 + 2*(5*cosh(b*x + a)^3 + 21*cosh(b*x + a))*sinh(b*x + a)^2 - 12*(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) + 12*(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) + (5*cosh(b
*x + a)^4 + 42*cosh(b*x + a)^2 - 27)*sinh(b*x + a) - 27*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)

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

[Out]

Timed out

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Giac [A]  time = 1.19178, size = 101, normalized size = 1.38 \begin{align*} \frac{{\left (e^{\left (3 \, b x + 15 \, a\right )} + 15 \, e^{\left (b x + 13 \, a\right )}\right )} e^{\left (-12 \, a\right )} - \frac{12 \, e^{\left (b x + a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} - 1} - 12 \, \log \left (e^{\left (b x + a\right )} + 1\right ) + 12 \, \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{6 \, 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)^2,x, algorithm="giac")

[Out]

1/6*((e^(3*b*x + 15*a) + 15*e^(b*x + 13*a))*e^(-12*a) - 12*e^(b*x + a)/(e^(2*b*x + 2*a) - 1) - 12*log(e^(b*x +
a) + 1) + 12*log(abs(e^(b*x + a) - 1)))/b