3.333 \(\int \frac{e^{c (a+b x)}}{\sinh ^2(a c+b c x)^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.143144, 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} \[ -\frac{2 e^{4 c (a+b x)} \sinh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt{\sinh ^2(a c+b c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0693504, size = 46, normalized size = 0.79 \[ -\frac{4 e^{5 c (a+b x)} \sqrt{\sinh ^2(c (a+b x))}}{b c \left (e^{2 c (a+b x)}-1\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*E^(5*c*(a + b*x))*Sqrt[Sinh[c*(a + b*x)]^2])/(b*c*(-1 + E^(2*c*(a + b*x)))^3)

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{{{\rm e}^{c \left ( bx+a \right ) }} \left ( \left ( \sinh \left ( bcx+ac \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.56679, 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))/(sinh(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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Fricas [B]  time = 1.75967, 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))/(sinh(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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.23097, size = 117, normalized size = 2.02 \begin{align*} -\frac{2 \,{\left (2 \, e^{\left (2 \, b c x + 2 \, a c\right )} \mathrm{sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) - \mathrm{sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right )\right )}}{b c{\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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