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*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)

________________________________________________________________________________________

Rubi [A]  time = 0.117562, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, 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 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])

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 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]

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*}

Mathematica [A]  time = 0.0427245, 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)

________________________________________________________________________________________

Maple [A]  time = 0.16, size = 69, normalized size = 1.2 \begin{align*} -2\,{\frac{ \left ( 2\,{{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ){{\rm e}^{-c \left ( bx+a \right ) }}}{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) cb}\sqrt{{\frac{{{\rm e}^{2\,c \left ( bx+a \right ) }}}{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) ^{2}}}}} \end{align*}

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

________________________________________________________________________________________

Maxima [A]  time = 1.57095, size = 113, normalized size = 1.95 \begin{align*} -\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 )}} \end{align*}

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))

________________________________________________________________________________________

Fricas [B]  time = 1.57986, size = 300, normalized size = 5.17 \begin{align*} -\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 )} \end{align*}

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))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{a c} \int \left (\operatorname{csch}^{2}{\left (a c + b c x \right )}\right )^{\frac{3}{2}} e^{b c x}\, dx \end{align*}

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)

________________________________________________________________________________________

Giac [A]  time = 1.16995, size = 86, normalized size = 1.48 \begin{align*} -\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 )} \end{align*}

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)))