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]

8/3*exp(3*b*x+3*a)/b/(1-exp(2*b*x+2*a))^3-2*exp(b*x+a)/b/(1-exp(2*b*x+2*a))^2+exp(b*x+a)/b/(1-exp(2*b*x+2*a))+
arctanh(exp(b*x+a))/b

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Rubi [A]  time = 0.05, antiderivative size = 101, normalized size of antiderivative = 1.00, 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 12

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

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^{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*}

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Mathematica [A]  time = 0.07, 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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fricas [B]  time = 2.05, size = 705, normalized size = 6.98 \[ -\frac {6 \, \cosh \left (b x + a\right )^{5} + 30 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 6 \, \sinh \left (b x + a\right )^{5} + 4 \, {\left (15 \, \cosh \left (b x + a\right )^{2} + 4\right )} \sinh \left (b x + a\right )^{3} + 16 \, \cosh \left (b x + a\right )^{3} + 12 \, {\left (5 \, \cosh \left (b x + a\right )^{3} + 4 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 3 \, {\left (\cosh \left (b x + a\right )^{6} + 6 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + \sinh \left (b x + a\right )^{6} + 3 \, {\left (5 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{4} - 3 \, \cosh \left (b x + a\right )^{4} + 4 \, {\left (5 \, \cosh \left (b x + a\right )^{3} - 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 3 \, {\left (5 \, \cosh \left (b x + a\right )^{4} - 6 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 3 \, \cosh \left (b x + a\right )^{2} + 6 \, {\left (\cosh \left (b x + a\right )^{5} - 2 \, \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 )^{6} + 6 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + \sinh \left (b x + a\right )^{6} + 3 \, {\left (5 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{4} - 3 \, \cosh \left (b x + a\right )^{4} + 4 \, {\left (5 \, \cosh \left (b x + a\right )^{3} - 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 3 \, {\left (5 \, \cosh \left (b x + a\right )^{4} - 6 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 3 \, \cosh \left (b x + a\right )^{2} + 6 \, {\left (\cosh \left (b x + a\right )^{5} - 2 \, \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 )^{4} + 8 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) - 6 \, \cosh \left (b x + a\right )}{6 \, {\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} - 3 \, b \cosh \left (b x + a\right )^{4} + 3 \, {\left (5 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{4} + 4 \, {\left (5 \, b \cosh \left (b x + a\right )^{3} - 3 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right )^{2} + 3 \, {\left (5 \, b \cosh \left (b x + a\right )^{4} - 6 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{2} + 6 \, {\left (b \cosh \left (b x + a\right )^{5} - 2 \, 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(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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giac [A]  time = 0.14, size = 75, normalized size = 0.74 \[ -\frac {\frac {2 \, {\left (3 \, e^{\left (5 \, b x + 5 \, a\right )} + 8 \, e^{\left (3 \, b x + 3 \, a\right )} - 3 \, e^{\left (b x + a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{3}} - 3 \, \log \left (e^{\left (b x + a\right )} + 1\right ) + 3 \, \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{6 \, b} \]

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]

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

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maple [A]  time = 0.03, size = 37, normalized size = 0.37 \[ \frac {-\frac {\mathrm {csch}\left (b x +a \right ) \coth \left (b x +a \right )}{2}+\arctanh \left ({\mathrm e}^{b x +a}\right )-\frac {1}{3 \sinh \left (b x +a \right )^{3}}}{b} \]

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)

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maxima [A]  time = 0.30, size = 100, normalized size = 0.99 \[ \frac {\log \left (e^{\left (b x + a\right )} + 1\right )}{2 \, b} - \frac {\log \left (e^{\left (b x + a\right )} - 1\right )}{2 \, b} - \frac {3 \, e^{\left (5 \, b x + 5 \, a\right )} + 8 \, e^{\left (3 \, b x + 3 \, a\right )} - 3 \, e^{\left (b x + a\right )}}{3 \, b {\left (e^{\left (6 \, b x + 6 \, a\right )} - 3 \, e^{\left (4 \, b x + 4 \, a\right )} + 3 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}} \]

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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mupad [B]  time = 0.60, size = 135, normalized size = 1.34 \[ \frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {8\,{\mathrm {e}}^{3\,a+3\,b\,x}}{3\,b\,\left (3\,{\mathrm {e}}^{2\,a+2\,b\,x}-3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}-1\right )}-\frac {{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(a + b*x)/sinh(a + b*x)^4,x)

[Out]

atan((exp(b*x)*exp(a)*(-b^2)^(1/2))/b)/(-b^2)^(1/2) - (2*exp(a + b*x))/(b*(exp(4*a + 4*b*x) - 2*exp(2*a + 2*b*
x) + 1)) - (8*exp(3*a + 3*b*x))/(3*b*(3*exp(2*a + 2*b*x) - 3*exp(4*a + 4*b*x) + exp(6*a + 6*b*x) - 1)) - 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 \[ e^{a} \int e^{b x} \operatorname {csch}^{4}{\left (a + b x \right )}\, dx \]

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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