### 3.936 $$\int e^{2 (a+b x)} \coth ^3(a+b x) \, dx$$

Optimal. Leaf size=80 $\frac{e^{2 a+2 b x}}{2 b}+\frac{6}{b \left (1-e^{2 a+2 b x}\right )}-\frac{2}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac{3 \log \left (1-e^{2 a+2 b x}\right )}{b}$

[Out]

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

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Rubi [A]  time = 0.06267, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {2282, 444, 43} $\frac{e^{2 a+2 b x}}{2 b}+\frac{6}{b \left (1-e^{2 a+2 b x}\right )}-\frac{2}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac{3 \log \left (1-e^{2 a+2 b x}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(2*(a + b*x))*Coth[a + b*x]^3,x]

[Out]

E^(2*a + 2*b*x)/(2*b) - 2/(b*(1 - E^(2*a + 2*b*x))^2) + 6/(b*(1 - E^(2*a + 2*b*x))) + (3*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 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int e^{2 (a+b x)} \coth ^3(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x \left (1+x^2\right )^3}{\left (-1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(1+x)^3}{(-1+x)^3} \, dx,x,e^{2 a+2 b x}\right )}{2 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (1+\frac{8}{(-1+x)^3}+\frac{12}{(-1+x)^2}+\frac{6}{-1+x}\right ) \, dx,x,e^{2 a+2 b x}\right )}{2 b}\\ &=\frac{e^{2 a+2 b x}}{2 b}-\frac{2}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac{6}{b \left (1-e^{2 a+2 b x}\right )}+\frac{3 \log \left (1-e^{2 a+2 b x}\right )}{b}\\ \end{align*}

Mathematica [A]  time = 0.109898, size = 60, normalized size = 0.75 $\frac{\frac{8-12 e^{2 (a+b x)}}{\left (e^{2 (a+b x)}-1\right )^2}+e^{2 (a+b x)}+6 \log \left (1-e^{2 (a+b x)}\right )}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^(2*(a + b*x))*Coth[a + b*x]^3,x]

[Out]

(E^(2*(a + b*x)) + (8 - 12*E^(2*(a + b*x)))/(-1 + E^(2*(a + b*x)))^2 + 6*Log[1 - E^(2*(a + b*x))])/(2*b)

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Maple [A]  time = 0.127, size = 70, normalized size = 0.9 \begin{align*}{\frac{{{\rm e}^{2\,bx+2\,a}}}{2\,b}}-6\,{\frac{a}{b}}-2\,{\frac{3\,{{\rm e}^{2\,bx+2\,a}}-2}{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) ^{2}}}+3\,{\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)^3,x)

[Out]

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

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Maxima [A]  time = 1.0164, size = 143, normalized size = 1.79 \begin{align*} \frac{6 \,{\left (b x + a\right )}}{b} + \frac{3 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac{3 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{b} - \frac{10 \, e^{\left (-2 \, b x - 2 \, a\right )} - 5 \, e^{\left (-4 \, b x - 4 \, a\right )} - 1}{2 \, b{\left (e^{\left (-2 \, b x - 2 \, a\right )} - 2 \, e^{\left (-4 \, b x - 4 \, a\right )} + e^{\left (-6 \, b x - 6 \, 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)^3,x, algorithm="maxima")

[Out]

6*(b*x + a)/b + 3*log(e^(-b*x - a) + 1)/b + 3*log(e^(-b*x - a) - 1)/b - 1/2*(10*e^(-2*b*x - 2*a) - 5*e^(-4*b*x
- 4*a) - 1)/(b*(e^(-2*b*x - 2*a) - 2*e^(-4*b*x - 4*a) + e^(-6*b*x - 6*a)))

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Fricas [B]  time = 1.867, size = 1099, normalized size = 13.74 \begin{align*} \frac{\cosh \left (b x + a\right )^{6} + 6 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + \sinh \left (b x + a\right )^{6} +{\left (15 \, \cosh \left (b x + a\right )^{2} - 2\right )} \sinh \left (b x + a\right )^{4} - 2 \, \cosh \left (b x + a\right )^{4} + 4 \,{\left (5 \, \cosh \left (b x + a\right )^{3} - 2 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} +{\left (15 \, \cosh \left (b x + a\right )^{4} - 12 \, \cosh \left (b x + a\right )^{2} - 11\right )} \sinh \left (b x + a\right )^{2} - 11 \, \cosh \left (b x + a\right )^{2} + 6 \,{\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 (\frac{2 \, \sinh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right ) + 2 \,{\left (3 \, \cosh \left (b x + a\right )^{5} - 4 \, \cosh \left (b x + a\right )^{3} - 11 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 8}{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)^3*csch(b*x+a)^3,x, algorithm="fricas")

[Out]

1/2*(cosh(b*x + a)^6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 + sinh(b*x + a)^6 + (15*cosh(b*x + a)^2 - 2)*sinh(b*x +
a)^4 - 2*cosh(b*x + a)^4 + 4*(5*cosh(b*x + a)^3 - 2*cosh(b*x + a))*sinh(b*x + a)^3 + (15*cosh(b*x + a)^4 - 12
*cosh(b*x + a)^2 - 11)*sinh(b*x + a)^2 - 11*cosh(b*x + a)^2 + 6*(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(2*sinh(b*x + a)/(cosh(b*x + a) - sinh(b*x + a))) + 2*(3*cosh(b*x + a)^5
- 4*cosh(b*x + a)^3 - 11*cosh(b*x + a))*sinh(b*x + a) + 8)/(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)**3*csch(b*x+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.23137, size = 95, normalized size = 1.19 \begin{align*} -\frac{\frac{9 \, e^{\left (4 \, b x + 4 \, a\right )} - 6 \, e^{\left (2 \, b x + 2 \, a\right )} + 1}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}} - e^{\left (2 \, b x + 2 \, a\right )} - 6 \, \log \left ({\left | e^{\left (2 \, b x + 2 \, 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)^3*csch(b*x+a)^3,x, algorithm="giac")

[Out]

-1/2*((9*e^(4*b*x + 4*a) - 6*e^(2*b*x + 2*a) + 1)/(e^(2*b*x + 2*a) - 1)^2 - e^(2*b*x + 2*a) - 6*log(abs(e^(2*b
*x + 2*a) - 1)))/b