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

Optimal. Leaf size=62 \[ \frac {4}{b \left (1-e^{2 a+2 b x}\right )}-\frac {2}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {\log \left (1-e^{2 a+2 b x}\right )}{b} \]

[Out]

-2/b/(1-exp(2*b*x+2*a))^2+4/b/(1-exp(2*b*x+2*a))+ln(1-exp(2*b*x+2*a))/b

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Rubi [A]  time = 0.07, antiderivative size = 62, normalized size of antiderivative = 1.00, 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, 444, 43} \[ \frac {4}{b \left (1-e^{2 a+2 b x}\right )}-\frac {2}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {\log \left (1-e^{2 a+2 b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2/(b*(1 - E^(2*a + 2*b*x))^2) + 4/(b*(1 - E^(2*a + 2*b*x))) + Log[1 - E^(2*a + 2*b*x)]/b

Rule 12

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

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])

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 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]]]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 46, normalized size = 0.74 \[ \frac {\frac {2-4 e^{2 (a+b x)}}{\left (e^{2 (a+b x)}-1\right )^2}+\log \left (1-e^{2 (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2 - 4*E^(2*(a + b*x)))/(-1 + E^(2*(a + b*x)))^2 + Log[1 - E^(2*(a + b*x))])/b

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fricas [B]  time = 0.47, size = 262, normalized size = 4.23 \[ -\frac {4 \, \cosh \left (b x + a\right )^{2} - {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\frac {2 \, \sinh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right ) + 8 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 4 \, \sinh \left (b x + a\right )^{2} - 2}{b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} - 2 \, b \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b \cosh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b} \]

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)^3,x, algorithm="fricas")

[Out]

-(4*cosh(b*x + a)^2 - (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*si
nh(b*x + a)/(cosh(b*x + a) - sinh(b*x + a))) + 8*cosh(b*x + a)*sinh(b*x + a) + 4*sinh(b*x + a)^2 - 2)/(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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giac [A]  time = 0.13, size = 59, normalized size = 0.95 \[ -\frac {\frac {3 \, e^{\left (4 \, b x + 4 \, a\right )} + 2 \, e^{\left (2 \, b x + 2 \, a\right )} - 1}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}} - 2 \, \log \left ({\left | e^{\left (2 \, b x + 2 \, a\right )} - 1 \right |}\right )}{2 \, b} \]

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)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.17, size = 43, normalized size = 0.69 \[ x +\frac {a}{b}-\frac {\coth \left (b x +a \right )}{b}+\frac {\ln \left (\sinh \left (b x +a \right )\right )}{b}-\frac {\coth ^{2}\left (b x +a \right )}{2 b} \]

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)^3,x)

[Out]

x+a/b-coth(b*x+a)/b+ln(sinh(b*x+a))/b-1/2*coth(b*x+a)^2/b

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maxima [A]  time = 0.34, size = 69, normalized size = 1.11 \[ \frac {\log \left (e^{\left (b x + a\right )} + 1\right )}{b} + \frac {\log \left (e^{\left (b x + a\right )} - 1\right )}{b} - \frac {2 \, {\left (2 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}}{b {\left (e^{\left (4 \, b x + 4 \, a\right )} - 2 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} \]

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)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.84, size = 65, normalized size = 1.05 \[ \frac {\ln \left ({\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1\right )}{b}-\frac {4}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )}-\frac {2}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(a + b*x)^2*exp(a + b*x))/sinh(a + b*x)^3,x)

[Out]

log(exp(2*a)*exp(2*b*x) - 1)/b - 4/(b*(exp(2*a + 2*b*x) - 1)) - 2/(b*(exp(4*a + 4*b*x) - 2*exp(2*a + 2*b*x) +
1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**3,x)

[Out]

Timed out

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