3.309 \(\int e^{a+b x} \text{csch}^5(a+b x) \, dx\)

Optimal. Leaf size=66 \[ -\frac{8}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac{32}{3 b \left (1-e^{2 a+2 b x}\right )^3}-\frac{4}{b \left (1-e^{2 a+2 b x}\right )^4} \]

[Out]

-4/(b*(1 - E^(2*a + 2*b*x))^4) + 32/(3*b*(1 - E^(2*a + 2*b*x))^3) - 8/(b*(1 - E^(2*a + 2*b*x))^2)

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Rubi [A]  time = 0.056439, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2282, 12, 266, 43} \[ -\frac{8}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac{32}{3 b \left (1-e^{2 a+2 b x}\right )^3}-\frac{4}{b \left (1-e^{2 a+2 b x}\right )^4} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Csch[a + b*x]^5,x]

[Out]

-4/(b*(1 - E^(2*a + 2*b*x))^4) + 32/(3*b*(1 - E^(2*a + 2*b*x))^3) - 8/(b*(1 - E^(2*a + 2*b*x))^2)

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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^{a+b x} \text{csch}^5(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{32 x^5}{\left (-1+x^2\right )^5} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{32 \operatorname{Subst}\left (\int \frac{x^5}{\left (-1+x^2\right )^5} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{16 \operatorname{Subst}\left (\int \frac{x^2}{(-1+x)^5} \, dx,x,e^{2 a+2 b x}\right )}{b}\\ &=\frac{16 \operatorname{Subst}\left (\int \left (\frac{1}{(-1+x)^5}+\frac{2}{(-1+x)^4}+\frac{1}{(-1+x)^3}\right ) \, dx,x,e^{2 a+2 b x}\right )}{b}\\ &=-\frac{4}{b \left (1-e^{2 a+2 b x}\right )^4}+\frac{32}{3 b \left (1-e^{2 a+2 b x}\right )^3}-\frac{8}{b \left (1-e^{2 a+2 b x}\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.0350101, size = 44, normalized size = 0.67 \[ -\frac{4 \left (-4 e^{2 (a+b x)}+6 e^{4 (a+b x)}+1\right )}{3 b \left (e^{2 (a+b x)}-1\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(a + b*x)*Csch[a + b*x]^5,x]

[Out]

(-4*(1 - 4*E^(2*(a + b*x)) + 6*E^(4*(a + b*x))))/(3*b*(-1 + E^(2*(a + b*x)))^4)

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Maple [A]  time = 0.048, size = 61, normalized size = 0.9 \begin{align*}{\frac{1}{b} \left ( \left ({\frac{2}{3}}-{\frac{ \left ({\rm csch} \left (bx+a\right ) \right ) ^{2}}{3}} \right ){\rm coth} \left (bx+a\right )-{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{4\, \left ( \sinh \left ( bx+a \right ) \right ) ^{4}}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{4\, \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*csch(b*x+a)^5,x)

[Out]

1/b*((2/3-1/3*csch(b*x+a)^2)*coth(b*x+a)-1/4/sinh(b*x+a)^4*cosh(b*x+a)^2+1/4/sinh(b*x+a)^2*cosh(b*x+a)^2)

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Maxima [B]  time = 1.01289, size = 232, normalized size = 3.52 \begin{align*} -\frac{8 \, e^{\left (4 \, b x + 4 \, a\right )}}{b{\left (e^{\left (8 \, b x + 8 \, a\right )} - 4 \, e^{\left (6 \, b x + 6 \, a\right )} + 6 \, e^{\left (4 \, b x + 4 \, a\right )} - 4 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} + \frac{16 \, e^{\left (2 \, b x + 2 \, a\right )}}{3 \, b{\left (e^{\left (8 \, b x + 8 \, a\right )} - 4 \, e^{\left (6 \, b x + 6 \, a\right )} + 6 \, e^{\left (4 \, b x + 4 \, a\right )} - 4 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} - \frac{4}{3 \, b{\left (e^{\left (8 \, b x + 8 \, a\right )} - 4 \, e^{\left (6 \, b x + 6 \, a\right )} + 6 \, e^{\left (4 \, b x + 4 \, a\right )} - 4 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*csch(b*x+a)^5,x, algorithm="maxima")

[Out]

-8*e^(4*b*x + 4*a)/(b*(e^(8*b*x + 8*a) - 4*e^(6*b*x + 6*a) + 6*e^(4*b*x + 4*a) - 4*e^(2*b*x + 2*a) + 1)) + 16/
3*e^(2*b*x + 2*a)/(b*(e^(8*b*x + 8*a) - 4*e^(6*b*x + 6*a) + 6*e^(4*b*x + 4*a) - 4*e^(2*b*x + 2*a) + 1)) - 4/3/
(b*(e^(8*b*x + 8*a) - 4*e^(6*b*x + 6*a) + 6*e^(4*b*x + 4*a) - 4*e^(2*b*x + 2*a) + 1))

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Fricas [B]  time = 1.87212, size = 635, normalized size = 9.62 \begin{align*} -\frac{4 \,{\left (7 \, \cosh \left (b x + a\right )^{2} + 10 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 7 \, \sinh \left (b x + a\right )^{2} - 4\right )}}{3 \,{\left (b \cosh \left (b x + a\right )^{6} + 6 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + b \sinh \left (b x + a\right )^{6} - 4 \, b \cosh \left (b x + a\right )^{4} +{\left (15 \, b \cosh \left (b x + a\right )^{2} - 4 \, b\right )} \sinh \left (b x + a\right )^{4} + 4 \,{\left (5 \, b \cosh \left (b x + a\right )^{3} - 4 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 7 \, b \cosh \left (b x + a\right )^{2} +{\left (15 \, b \cosh \left (b x + a\right )^{4} - 24 \, b \cosh \left (b x + a\right )^{2} + 7 \, b\right )} \sinh \left (b x + a\right )^{2} + 2 \,{\left (3 \, b \cosh \left (b x + a\right )^{5} - 8 \, b \cosh \left (b x + a\right )^{3} + 5 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - 4 \, b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*csch(b*x+a)^5,x, algorithm="fricas")

[Out]

-4/3*(7*cosh(b*x + a)^2 + 10*cosh(b*x + a)*sinh(b*x + a) + 7*sinh(b*x + a)^2 - 4)/(b*cosh(b*x + a)^6 + 6*b*cos
h(b*x + a)*sinh(b*x + a)^5 + b*sinh(b*x + a)^6 - 4*b*cosh(b*x + a)^4 + (15*b*cosh(b*x + a)^2 - 4*b)*sinh(b*x +
a)^4 + 4*(5*b*cosh(b*x + a)^3 - 4*b*cosh(b*x + a))*sinh(b*x + a)^3 + 7*b*cosh(b*x + a)^2 + (15*b*cosh(b*x + a
)^4 - 24*b*cosh(b*x + a)^2 + 7*b)*sinh(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^5 - 8*b*cosh(b*x + a)^3 + 5*b*cosh(b*
x + a))*sinh(b*x + a) - 4*b)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{a} \int e^{b x} \operatorname{csch}^{5}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*csch(b*x+a)**5,x)

[Out]

exp(a)*Integral(exp(b*x)*csch(a + b*x)**5, x)

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Giac [A]  time = 1.11199, size = 57, normalized size = 0.86 \begin{align*} -\frac{4 \,{\left (6 \, e^{\left (4 \, b x + 4 \, a\right )} - 4 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}{3 \, b{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*csch(b*x+a)^5,x, algorithm="giac")

[Out]

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