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

Optimal. Leaf size=58 \[ -\frac {2 e^{4 c (a+b x)} \sinh (a c+b c x) \sqrt {\text {csch}^2(a c+b c x)}}{b c \left (1-e^{2 c (a+b x)}\right )^2} \]

[Out]

-2*exp(4*c*(b*x+a))*sinh(b*c*x+a*c)*(csch(b*c*x+a*c)^2)^(1/2)/b/c/(1-exp(2*c*(b*x+a)))^2

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Rubi [A]  time = 0.12, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6720, 2282, 12, 264} \[ -\frac {2 e^{4 c (a+b x)} \sinh (a c+b c x) \sqrt {\text {csch}^2(a c+b c x)}}{b c \left (1-e^{2 c (a+b x)}\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*E^(4*c*(a + b*x))*Sqrt[Csch[a*c + b*c*x]^2]*Sinh[a*c + b*c*x])/(b*c*(1 - E^(2*c*(a + b*x)))^2)

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]]]

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 56, normalized size = 0.97 \[ -\frac {2 e^{4 c (a+b x)} \sinh ^3(c (a+b x)) \text {csch}^2(c (a+b x))^{3/2}}{b c \left (e^{2 c (a+b x)}-1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*E^(4*c*(a + b*x))*(Csch[c*(a + b*x)]^2)^(3/2)*Sinh[c*(a + b*x)]^3)/(b*c*(-1 + E^(2*c*(a + b*x)))^2)

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fricas [B]  time = 1.06, size = 121, normalized size = 2.09 \[ -\frac {2 \, {\left (\cosh \left (b c x + a c\right ) + 3 \, \sinh \left (b c x + a c\right )\right )}}{b c \cosh \left (b c x + a c\right )^{3} + 3 \, b c \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{2} + b c \sinh \left (b c x + a c\right )^{3} - b c \cosh \left (b c x + a c\right ) + 3 \, {\left (b c \cosh \left (b c x + a c\right )^{2} - b c\right )} \sinh \left (b c x + a c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(cosh(b*c*x + a*c) + 3*sinh(b*c*x + a*c))/(b*c*cosh(b*c*x + a*c)^3 + 3*b*c*cosh(b*c*x + a*c)*sinh(b*c*x + a
*c)^2 + b*c*sinh(b*c*x + a*c)^3 - b*c*cosh(b*c*x + a*c) + 3*(b*c*cosh(b*c*x + a*c)^2 - b*c)*sinh(b*c*x + a*c))

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giac [A]  time = 0.13, size = 64, normalized size = 1.10 \[ -\frac {2 \, {\left (2 \, e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}}{b c {\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}^{2} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(2*e^(2*b*c*x + 2*a*c) - 1)/(b*c*(e^(2*b*c*x + 2*a*c) - 1)^2*sgn(e^(b*c*x + a*c) - e^(-b*c*x - a*c)))

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maple [A]  time = 0.60, size = 69, normalized size = 1.19 \[ -\frac {2 \left (2 \,{\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}\, {\mathrm e}^{-c \left (b x +a \right )}}{b c \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 84, normalized size = 1.45 \[ -\frac {4 \, e^{\left (2 \, b c x + 2 \, a c\right )}}{b c {\left (e^{\left (4 \, b c x + 4 \, a c\right )} - 2 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} + \frac {2}{b c {\left (e^{\left (4 \, b c x + 4 \, a c\right )} - 2 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.55, size = 78, normalized size = 1.34 \[ -\frac {{\mathrm {e}}^{-a\,c-b\,c\,x}\,\left (2\,{\mathrm {e}}^{2\,a\,c+2\,b\,c\,x}-1\right )\,\sqrt {\frac {1}{{\left (\frac {{\mathrm {e}}^{a\,c+b\,c\,x}}{2}-\frac {{\mathrm {e}}^{-a\,c-b\,c\,x}}{2}\right )}^2}}}{b\,c\,\left ({\mathrm {e}}^{2\,a\,c+2\,b\,c\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(a + b*x))*(1/sinh(a*c + b*c*x)^2)^(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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