3.308 \(\int e^{a+b x} \text{csch}^4(a+b x) \, dx\)

Optimal. Leaf size=101 \[ \frac{e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac{8 e^{3 a+3 b x}}{3 b \left (1-e^{2 a+2 b x}\right )^3}+\frac{\tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.0536303, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {2282, 12, 288, 199, 206} \[ \frac{e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac{8 e^{3 a+3 b x}}{3 b \left (1-e^{2 a+2 b x}\right )^3}+\frac{\tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Csch[a + b*x]^4,x]

[Out]

(8*E^(3*a + 3*b*x))/(3*b*(1 - E^(2*a + 2*b*x))^3) - (2*E^(a + b*x))/(b*(1 - E^(2*a + 2*b*x))^2) + E^(a + b*x)/
(b*(1 - E^(2*a + 2*b*x))) + ArcTanh[E^(a + 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 12

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

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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0726207, size = 75, normalized size = 0.74 \[ \frac{3 e^{a+b x}-8 e^{3 (a+b x)}-3 e^{5 (a+b x)}+3 \left (e^{2 (a+b x)}-1\right )^3 \tanh ^{-1}\left (e^{a+b x}\right )}{3 b \left (e^{2 (a+b x)}-1\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(a + b*x)*Csch[a + b*x]^4,x]

[Out]

(3*E^(a + b*x) - 8*E^(3*(a + b*x)) - 3*E^(5*(a + b*x)) + 3*(-1 + E^(2*(a + b*x)))^3*ArcTanh[E^(a + b*x)])/(3*b
*(-1 + E^(2*(a + b*x)))^3)

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Maple [A]  time = 0.045, size = 71, normalized size = 0.7 \begin{align*}{\frac{1}{b} \left ( -{\frac{{\rm csch} \left (bx+a\right ){\rm coth} \left (bx+a\right )}{2}}+{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) -{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{3\, \left ( \sinh \left ( bx+a \right ) \right ) ^{3}}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{3\,\sinh \left ( bx+a \right ) }}-{\frac{\sinh \left ( bx+a \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*csch(b*x+a)^4,x)

[Out]

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

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Maxima [A]  time = 1.16916, size = 135, normalized size = 1.34 \begin{align*} \frac{\log \left (e^{\left (b x + a\right )} + 1\right )}{2 \, b} - \frac{\log \left (e^{\left (b x + a\right )} - 1\right )}{2 \, b} - \frac{3 \, e^{\left (5 \, b x + 5 \, a\right )} + 8 \, e^{\left (3 \, b x + 3 \, a\right )} - 3 \, e^{\left (b x + a\right )}}{3 \, b{\left (e^{\left (6 \, b x + 6 \, a\right )} - 3 \, e^{\left (4 \, b x + 4 \, a\right )} + 3 \, 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)^4,x, algorithm="maxima")

[Out]

1/2*log(e^(b*x + a) + 1)/b - 1/2*log(e^(b*x + a) - 1)/b - 1/3*(3*e^(5*b*x + 5*a) + 8*e^(3*b*x + 3*a) - 3*e^(b*
x + a))/(b*(e^(6*b*x + 6*a) - 3*e^(4*b*x + 4*a) + 3*e^(2*b*x + 2*a) - 1))

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Fricas [B]  time = 2.0076, size = 1963, normalized size = 19.44 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*csch(b*x+a)^4,x, algorithm="fricas")

[Out]

-1/6*(6*cosh(b*x + a)^5 + 30*cosh(b*x + a)*sinh(b*x + a)^4 + 6*sinh(b*x + a)^5 + 4*(15*cosh(b*x + a)^2 + 4)*si
nh(b*x + a)^3 + 16*cosh(b*x + a)^3 + 12*(5*cosh(b*x + a)^3 + 4*cosh(b*x + a))*sinh(b*x + a)^2 - 3*(cosh(b*x +
a)^6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 + sinh(b*x + a)^6 + 3*(5*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^4 - 3*cosh(
b*x + a)^4 + 4*(5*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x + a)^3 + 3*(5*cosh(b*x + a)^4 - 6*cosh(b*x + a)^
2 + 1)*sinh(b*x + a)^2 + 3*cosh(b*x + a)^2 + 6*(cosh(b*x + a)^5 - 2*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)^6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 + sinh
(b*x + a)^6 + 3*(5*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^4 - 3*cosh(b*x + a)^4 + 4*(5*cosh(b*x + a)^3 - 3*cosh(b*
x + a))*sinh(b*x + a)^3 + 3*(5*cosh(b*x + a)^4 - 6*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 3*cosh(b*x + a)^2 +
6*(cosh(b*x + a)^5 - 2*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)^4 + 8*cosh(b*x + a)^2 - 1)*sinh(b*x + a) - 6*cosh(b*x + a))/(b*cosh(b*x + a)^6 + 6*b*
cosh(b*x + a)*sinh(b*x + a)^5 + b*sinh(b*x + a)^6 - 3*b*cosh(b*x + a)^4 + 3*(5*b*cosh(b*x + a)^2 - b)*sinh(b*x
+ a)^4 + 4*(5*b*cosh(b*x + a)^3 - 3*b*cosh(b*x + a))*sinh(b*x + a)^3 + 3*b*cosh(b*x + a)^2 + 3*(5*b*cosh(b*x
+ a)^4 - 6*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^2 + 6*(b*cosh(b*x + a)^5 - 2*b*cosh(b*x + a)^3 + b*cosh(b*x +
a))*sinh(b*x + a) - b)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{a} \int e^{b x} \operatorname{csch}^{4}{\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)**4,x)

[Out]

exp(a)*Integral(exp(b*x)*csch(a + b*x)**4, x)

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Giac [A]  time = 1.12661, size = 107, normalized size = 1.06 \begin{align*} \frac{\log \left (e^{\left (b x + a\right )} + 1\right )}{2 \, b} - \frac{\log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{2 \, b} - \frac{3 \, e^{\left (5 \, b x + 5 \, a\right )} + 8 \, e^{\left (3 \, b x + 3 \, a\right )} - 3 \, e^{\left (b x + a\right )}}{3 \, b{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*csch(b*x+a)^4,x, algorithm="giac")

[Out]

1/2*log(e^(b*x + a) + 1)/b - 1/2*log(abs(e^(b*x + a) - 1))/b - 1/3*(3*e^(5*b*x + 5*a) + 8*e^(3*b*x + 3*a) - 3*
e^(b*x + a))/(b*(e^(2*b*x + 2*a) - 1)^3)