3.877 \(\int e^{2 (a+b x)} \text {csch}^3(a+b x) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2282, 12, 288, 207} \[ \frac {3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}-\frac {3 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-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))) - (3*ArcTanh[E^(a
+ 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 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])

Rule 288

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

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^{2 (a+b x)} \text {csch}^3(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {8 x^4}{\left (-1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {8 \operatorname {Subst}\left (\int \frac {x^4}{\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 {6 \operatorname {Subst}\left (\int \frac {x^2}{\left (-1+x^2\right )^2} \, 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 {3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, 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 {3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 61, normalized size = 0.84 \[ \frac {3 e^{a+b x}-5 e^{3 (a+b x)}-3 \left (e^{2 (a+b x)}-1\right )^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b \left (e^{2 (a+b x)}-1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.45, size = 388, normalized size = 5.32 \[ -\frac {10 \, \cosh \left (b x + a\right )^{3} + 30 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + 10 \, \sinh \left (b x + a\right )^{3} + 3 \, {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) - 3 \, {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 6 \, {\left (5 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) - 6 \, \cosh \left (b x + a\right )}{2 \, {\left (b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} - 2 \, b \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b \cosh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*csch(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/2*(10*cosh(b*x + a)^3 + 30*cosh(b*x + a)*sinh(b*x + a)^2 + 10*sinh(b*x + a)^3 + 3*(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) - 3*(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) + 6
*(5*cosh(b*x + a)^2 - 1)*sinh(b*x + a) - 6*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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giac [A]  time = 0.13, size = 78, normalized size = 1.07 \[ -\frac {{\left (3 \, e^{\left (-2 \, a\right )} \log \left (e^{\left (b x + a\right )} + 1\right ) - 3 \, e^{\left (-2 \, a\right )} \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right ) + \frac {2 \, {\left (5 \, e^{\left (3 \, b x + 2 \, a\right )} - 3 \, e^{\left (b x\right )}\right )} e^{\left (-a\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}}\right )} e^{\left (2 \, a\right )}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*csch(b*x+a)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.34, size = 67, normalized size = 0.92 \[ -\frac {{\mathrm e}^{b x +a} \left (5 \,{\mathrm e}^{2 b x +2 a}-3\right )}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )^{2}}-\frac {3 \ln \left (1+{\mathrm e}^{b x +a}\right )}{2 b}+\frac {3 \ln \left ({\mathrm e}^{b x +a}-1\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*b*x+2*a)*csch(b*x+a)^3,x)

[Out]

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

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maxima [A]  time = 0.31, size = 88, normalized size = 1.21 \[ -\frac {3 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{2 \, b} + \frac {3 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{2 \, b} + \frac {5 \, e^{\left (-b x - a\right )} - 3 \, e^{\left (-3 \, b x - 3 \, a\right )}}{b {\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*csch(b*x+a)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.81, size = 90, normalized size = 1.23 \[ -\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}-\frac {2\,{\mathrm {e}}^{3\,a+3\,b\,x}}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {3\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (3*atan((exp(b*x)*exp(a)*(-b^2)^(1/2))/b))/(-b^2)^(1/2) - (2*exp(3*a + 3*b*x))/(b*(exp(4*a + 4*b*x) - 2*exp(
2*a + 2*b*x) + 1)) - (3*exp(a + b*x))/(b*(exp(2*a + 2*b*x) - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{2 a} \int e^{2 b x} \operatorname {csch}^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*csch(b*x+a)**3,x)

[Out]

exp(2*a)*Integral(exp(2*b*x)*csch(a + b*x)**3, x)

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