3.600 \(\int e^{2 x} \text {csch}^4(x) \, dx\)

Optimal. Leaf size=20 \[ \frac {8 e^{6 x}}{3 \left (1-e^{2 x}\right )^3} \]

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Rubi [A]  time = 0.02, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2282, 12, 264} \[ \frac {8 e^{6 x}}{3 \left (1-e^{2 x}\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*x)*Csch[x]^4,x]

[Out]

(8*E^(6*x))/(3*(1 - E^(2*x))^3)

Rule 12

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

Rule 264

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

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 x} \text {csch}^4(x) \, dx &=\operatorname {Subst}\left (\int \frac {16 x^5}{\left (1-x^2\right )^4} \, dx,x,e^x\right )\\ &=16 \operatorname {Subst}\left (\int \frac {x^5}{\left (1-x^2\right )^4} \, dx,x,e^x\right )\\ &=\frac {8 e^{6 x}}{3 \left (1-e^{2 x}\right )^3}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 1.00 \[ \frac {8 e^{6 x}}{3 \left (1-e^{2 x}\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*x)*Csch[x]^4,x]

[Out]

(8*E^(6*x))/(3*(1 - E^(2*x))^3)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{2 x} \text {csch}^4(x) \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[E^(2*x)*Csch[x]^4,x]

[Out]

Could not integrate

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fricas [B]  time = 1.31, size = 75, normalized size = 3.75 \[ -\frac {8 \, {\left (4 \, \cosh \relax (x)^{2} + 4 \, \cosh \relax (x) \sinh \relax (x) + 4 \, \sinh \relax (x)^{2} - 3\right )}}{3 \, {\left (\cosh \relax (x)^{4} + 4 \, \cosh \relax (x) \sinh \relax (x)^{3} + \sinh \relax (x)^{4} + 2 \, {\left (3 \, \cosh \relax (x)^{2} - 2\right )} \sinh \relax (x)^{2} - 4 \, \cosh \relax (x)^{2} + 4 \, {\left (\cosh \relax (x)^{3} - \cosh \relax (x)\right )} \sinh \relax (x) + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)/sinh(x)^4,x, algorithm="fricas")

[Out]

-8/3*(4*cosh(x)^2 + 4*cosh(x)*sinh(x) + 4*sinh(x)^2 - 3)/(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 2*(3*c
osh(x)^2 - 2)*sinh(x)^2 - 4*cosh(x)^2 + 4*(cosh(x)^3 - cosh(x))*sinh(x) + 3)

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giac [A]  time = 0.62, size = 24, normalized size = 1.20 \[ -\frac {8 \, {\left (3 \, e^{\left (4 \, x\right )} - 3 \, e^{\left (2 \, x\right )} + 1\right )}}{3 \, {\left (e^{\left (2 \, x\right )} - 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)/sinh(x)^4,x, algorithm="giac")

[Out]

-8/3*(3*e^(4*x) - 3*e^(2*x) + 1)/(e^(2*x) - 1)^3

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maple [A]  time = 0.10, size = 25, normalized size = 1.25

method	result	size
risch	\(-\frac {8 \left (3 \,{\mathrm e}^{4 x}-3 \,{\mathrm e}^{2 x}+1\right )}{3 \left (-1+{\mathrm e}^{2 x}\right )^{3}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)/sinh(x)^4,x,method=_RETURNVERBOSE)

[Out]

-8/3*(3*exp(4*x)-3*exp(2*x)+1)/(-1+exp(2*x))^3

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maxima [A]  time = 0.43, size = 22, normalized size = 1.10 \[ \frac {8}{3 \, {\left (3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)/sinh(x)^4,x, algorithm="maxima")

[Out]

8/3/(3*e^(-2*x) - 3*e^(-4*x) + e^(-6*x) - 1)

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mupad [B]  time = 0.33, size = 24, normalized size = 1.20 \[ -\frac {8\,\left (3\,{\mathrm {e}}^{4\,x}-3\,{\mathrm {e}}^{2\,x}+1\right )}{3\,{\left ({\mathrm {e}}^{2\,x}-1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)/sinh(x)^4,x)

[Out]

-(8*(3*exp(4*x) - 3*exp(2*x) + 1))/(3*(exp(2*x) - 1)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{2 x}}{\sinh ^{4}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)/sinh(x)**4,x)

[Out]

Integral(exp(2*x)/sinh(x)**4, x)

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