### 3.930 $$\int e^{2 (a+b x)} \coth ^2(a+b x) \text{csch}(a+b x) \, dx$$

Optimal. Leaf size=85 $\frac{2 e^{a+b x}}{b}+\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}-\frac{5 \tanh ^{-1}\left (e^{a+b x}\right )}{b}$

[Out]

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

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Rubi [A]  time = 0.0684383, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {2282, 12, 463, 455, 388, 207} $\frac{2 e^{a+b x}}{b}+\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}-\frac{5 \tanh ^{-1}\left (e^{a+b x}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*E^(a + b*x))/b - (2*E^(3*a + 3*b*x))/(b*(1 - E^(2*a + 2*b*x))^2) + (3*E^(a + b*x))/(b*(1 - E^(2*a + 2*b*x))
) - (5*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 463

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

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 388

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

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^{2 (a+b x)} \coth ^2(a+b x) \text{csch}(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{2 x^2 \left (1+x^2\right )^2}{\left (-1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{x^2 \left (1+x^2\right )^2}{\left (-1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=-\frac{2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (8+4 x^2\right )}{\left (-1+x^2\right )^2} \, dx,x,e^{a+b x}\right )}{2 b}\\ &=-\frac{2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{\operatorname{Subst}\left (\int \frac{-12-8 x^2}{-1+x^2} \, dx,x,e^{a+b x}\right )}{4 b}\\ &=\frac{2 e^{a+b x}}{b}-\frac{2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{2 e^{a+b x}}{b}-\frac{2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac{3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{5 \tanh ^{-1}\left (e^{a+b x}\right )}{b}\\ \end{align*}

Mathematica [C]  time = 4.02851, size = 247, normalized size = 2.91 $-\frac{e^{-3 (a+b x)} \left (128 e^{8 (a+b x)} \left (e^{2 (a+b x)}+1\right )^2 \, _5F_4\left (2,2,2,2,\frac{5}{2};1,1,1,\frac{11}{2};e^{2 (a+b x)}\right )+128 e^{8 (a+b x)} \left (16 e^{2 (a+b x)}+7 e^{4 (a+b x)}+9\right ) \, _4F_3\left (2,2,2,\frac{5}{2};1,1,\frac{11}{2};e^{2 (a+b x)}\right )-21 \left (62725 e^{2 (a+b x)}-12071 e^{4 (a+b x)}-19353 e^{6 (a+b x)}+768 e^{8 (a+b x)}+56595\right )+\frac{315 \left (2924 e^{2 (a+b x)}-2534 e^{4 (a+b x)}-1548 e^{6 (a+b x)}+297 e^{8 (a+b x)}+3773\right ) \tanh ^{-1}\left (\sqrt{e^{2 (a+b x)}}\right )}{\sqrt{e^{2 (a+b x)}}}\right )}{10080 b}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(-21*(56595 + 62725*E^(2*(a + b*x)) - 12071*E^(4*(a + b*x)) - 19353*E^(6*(a + b*x)) + 768*E^(8*(a + b*x))) +
(315*(3773 + 2924*E^(2*(a + b*x)) - 2534*E^(4*(a + b*x)) - 1548*E^(6*(a + b*x)) + 297*E^(8*(a + b*x)))*ArcTanh
[Sqrt[E^(2*(a + b*x))]])/Sqrt[E^(2*(a + b*x))] + 128*E^(8*(a + b*x))*(9 + 16*E^(2*(a + b*x)) + 7*E^(4*(a + b*x
)))*HypergeometricPFQ[{2, 2, 2, 5/2}, {1, 1, 11/2}, E^(2*(a + b*x))] + 128*E^(8*(a + b*x))*(1 + E^(2*(a + b*x)
))^2*HypergeometricPFQ[{2, 2, 2, 2, 5/2}, {1, 1, 1, 11/2}, E^(2*(a + b*x))])/(10080*b*E^(3*(a + b*x)))

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Maple [A]  time = 0.08, size = 78, normalized size = 0.9 \begin{align*} 2\,{\frac{{{\rm e}^{bx+a}}}{b}}-{\frac{{{\rm e}^{bx+a}} \left ( 5\,{{\rm e}^{2\,bx+2\,a}}-3 \right ) }{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) ^{2}}}-{\frac{5\,\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{2\,b}}+{\frac{5\,\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{2\,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)^2*csch(b*x+a)^3,x)

[Out]

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

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Maxima [A]  time = 1.23156, size = 130, normalized size = 1.53 \begin{align*} -\frac{5 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{2 \, b} + \frac{5 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{2 \, b} - \frac{9 \, e^{\left (-2 \, b x - 2 \, a\right )} - 5 \, e^{\left (-4 \, b x - 4 \, a\right )} - 2}{b{\left (e^{\left (-b x - a\right )} - 2 \, 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)^2*csch(b*x+a)^3,x, algorithm="maxima")

[Out]

-5/2*log(e^(-b*x - a) + 1)/b + 5/2*log(e^(-b*x - a) - 1)/b - (9*e^(-2*b*x - 2*a) - 5*e^(-4*b*x - 4*a) - 2)/(b*
(e^(-b*x - a) - 2*e^(-3*b*x - 3*a) + e^(-5*b*x - 5*a)))

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Fricas [B]  time = 1.84851, size = 1295, normalized size = 15.24 \begin{align*} \frac{4 \, \cosh \left (b x + a\right )^{5} + 20 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 4 \, \sinh \left (b x + a\right )^{5} + 2 \,{\left (20 \, \cosh \left (b x + a\right )^{2} - 9\right )} \sinh \left (b x + a\right )^{3} - 18 \, \cosh \left (b x + a\right )^{3} + 2 \,{\left (20 \, \cosh \left (b x + a\right )^{3} - 27 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 5 \,{\left (\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} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \,{\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 5 \,{\left (\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} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \,{\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 2 \,{\left (10 \, \cosh \left (b x + a\right )^{4} - 27 \, \cosh \left (b x + a\right )^{2} + 5\right )} \sinh \left (b x + a\right ) + 10 \, \cosh \left (b x + a\right )}{2 \,{\left (b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} - 2 \, b \cosh \left (b x + a\right )^{2} + 2 \,{\left (3 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{2} + 4 \,{\left (b \cosh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 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)^2*csch(b*x+a)^3,x, algorithm="fricas")

[Out]

1/2*(4*cosh(b*x + a)^5 + 20*cosh(b*x + a)*sinh(b*x + a)^4 + 4*sinh(b*x + a)^5 + 2*(20*cosh(b*x + a)^2 - 9)*sin
h(b*x + a)^3 - 18*cosh(b*x + a)^3 + 2*(20*cosh(b*x + a)^3 - 27*cosh(b*x + a))*sinh(b*x + a)^2 - 5*(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 - 1)*sinh(b*x + a)^2 - 2*cosh(
b*x + a)^2 + 4*(cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 5
*(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 - 1)*sinh(b*x + a
)^2 - 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + sinh(b*x
+ a) - 1) + 2*(10*cosh(b*x + a)^4 - 27*cosh(b*x + a)^2 + 5)*sinh(b*x + a) + 10*cosh(b*x + a))/(b*cosh(b*x + a)
^4 + 4*b*cosh(b*x + a)*sinh(b*x + a)^3 + b*sinh(b*x + a)^4 - 2*b*cosh(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^2 - b)
*sinh(b*x + a)^2 + 4*(b*cosh(b*x + a)^3 - b*cosh(b*x + a))*sinh(b*x + a) + 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)**2*csch(b*x+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.18629, size = 97, normalized size = 1.14 \begin{align*} -\frac{\frac{2 \,{\left (5 \, e^{\left (3 \, b x + 3 \, a\right )} - 3 \, e^{\left (b x + a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}} - 4 \, e^{\left (b x + a\right )} + 5 \, \log \left (e^{\left (b x + a\right )} + 1\right ) - 5 \, \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{2 \, 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)^2*csch(b*x+a)^3,x, algorithm="giac")

[Out]

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