3.334 \(\int \frac{e^{c (a+b x)}}{\sinh ^2(a c+b c x)^{5/2}} \, dx\)

Optimal. Leaf size=147 \[ -\frac{8 \sinh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt{\sinh ^2(a c+b c x)}}+\frac{32 \sinh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt{\sinh ^2(a c+b c x)}}-\frac{4 \sinh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt{\sinh ^2(a c+b c x)}} \]

[Out]

(-4*Sinh[a*c + b*c*x])/(b*c*(1 - E^(2*c*(a + b*x)))^4*Sqrt[Sinh[a*c + b*c*x]^2]) + (32*Sinh[a*c + b*c*x])/(3*b
*c*(1 - E^(2*c*(a + b*x)))^3*Sqrt[Sinh[a*c + b*c*x]^2]) - (8*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 = 0.213861, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6720, 2282, 12, 266, 43} \[ -\frac{8 \sinh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt{\sinh ^2(a c+b c x)}}+\frac{32 \sinh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt{\sinh ^2(a c+b c x)}}-\frac{4 \sinh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt{\sinh ^2(a c+b c x)}} \]

Antiderivative was successfully verified.

[In]

Int[E^(c*(a + b*x))/(Sinh[a*c + b*c*x]^2)^(5/2),x]

[Out]

(-4*Sinh[a*c + b*c*x])/(b*c*(1 - E^(2*c*(a + b*x)))^4*Sqrt[Sinh[a*c + b*c*x]^2]) + (32*Sinh[a*c + b*c*x])/(3*b
*c*(1 - E^(2*c*(a + b*x)))^3*Sqrt[Sinh[a*c + b*c*x]^2]) - (8*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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{e^{c (a+b x)}}{\sinh ^2(a c+b c x)^{5/2}} \, dx &=\frac{\sinh (a c+b c x) \int e^{c (a+b x)} \text{csch}^5(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{32 x^5}{\left (-1+x^2\right )^5} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt{\sinh ^2(a c+b c x)}}\\ &=\frac{(32 \sinh (a c+b c x)) \operatorname{Subst}\left (\int \frac{x^5}{\left (-1+x^2\right )^5} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt{\sinh ^2(a c+b c x)}}\\ &=\frac{(16 \sinh (a c+b c x)) \operatorname{Subst}\left (\int \frac{x^2}{(-1+x)^5} \, dx,x,e^{2 c (a+b x)}\right )}{b c \sqrt{\sinh ^2(a c+b c x)}}\\ &=\frac{(16 \sinh (a c+b c x)) \operatorname{Subst}\left (\int \left (\frac{1}{(-1+x)^5}+\frac{2}{(-1+x)^4}+\frac{1}{(-1+x)^3}\right ) \, dx,x,e^{2 c (a+b x)}\right )}{b c \sqrt{\sinh ^2(a c+b c x)}}\\ &=-\frac{4 \sinh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt{\sinh ^2(a c+b c x)}}+\frac{32 \sinh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt{\sinh ^2(a c+b c x)}}-\frac{8 \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 = 0.0700153, size = 72, normalized size = 0.49 \[ -\frac{4 \left (-4 e^{2 c (a+b x)}+6 e^{4 c (a+b x)}+1\right ) \sinh (c (a+b x))}{3 b c \left (e^{2 c (a+b x)}-1\right )^4 \sqrt{\sinh ^2(c (a+b x))}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(c*(a + b*x))/(Sinh[a*c + b*c*x]^2)^(5/2),x]

[Out]

(-4*(1 - 4*E^(2*c*(a + b*x)) + 6*E^(4*c*(a + b*x)))*Sinh[c*(a + b*x)])/(3*b*c*(-1 + E^(2*c*(a + b*x)))^4*Sqrt[
Sinh[c*(a + b*x)]^2])

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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{5}{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)^(5/2),x)

[Out]

int(exp(c*(b*x+a))/(sinh(b*c*x+a*c)^2)^(5/2),x)

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Maxima [A]  time = 1.62597, size = 282, normalized size = 1.92 \begin{align*} -\frac{8 \, e^{\left (4 \, b c x + 4 \, a c\right )}}{b c{\left (e^{\left (8 \, b c x + 8 \, a c\right )} - 4 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 6 \, e^{\left (4 \, b c x + 4 \, a c\right )} - 4 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} + \frac{16 \, e^{\left (2 \, b c x + 2 \, a c\right )}}{3 \, b c{\left (e^{\left (8 \, b c x + 8 \, a c\right )} - 4 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 6 \, e^{\left (4 \, b c x + 4 \, a c\right )} - 4 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} - \frac{4}{3 \, b c{\left (e^{\left (8 \, b c x + 8 \, a c\right )} - 4 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 6 \, e^{\left (4 \, b c x + 4 \, a c\right )} - 4 \, 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)^(5/2),x, algorithm="maxima")

[Out]

-8*e^(4*b*c*x + 4*a*c)/(b*c*(e^(8*b*c*x + 8*a*c) - 4*e^(6*b*c*x + 6*a*c) + 6*e^(4*b*c*x + 4*a*c) - 4*e^(2*b*c*
x + 2*a*c) + 1)) + 16/3*e^(2*b*c*x + 2*a*c)/(b*c*(e^(8*b*c*x + 8*a*c) - 4*e^(6*b*c*x + 6*a*c) + 6*e^(4*b*c*x +
4*a*c) - 4*e^(2*b*c*x + 2*a*c) + 1)) - 4/3/(b*c*(e^(8*b*c*x + 8*a*c) - 4*e^(6*b*c*x + 6*a*c) + 6*e^(4*b*c*x +
4*a*c) - 4*e^(2*b*c*x + 2*a*c) + 1))

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Fricas [B]  time = 1.87394, size = 797, normalized size = 5.42 \begin{align*} -\frac{4 \,{\left (7 \, \cosh \left (b c x + a c\right )^{2} + 10 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right ) + 7 \, \sinh \left (b c x + a c\right )^{2} - 4\right )}}{3 \,{\left (b c \cosh \left (b c x + a c\right )^{6} + 6 \, b c \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{5} + b c \sinh \left (b c x + a c\right )^{6} - 4 \, b c \cosh \left (b c x + a c\right )^{4} +{\left (15 \, b c \cosh \left (b c x + a c\right )^{2} - 4 \, b c\right )} \sinh \left (b c x + a c\right )^{4} + 7 \, b c \cosh \left (b c x + a c\right )^{2} + 4 \,{\left (5 \, b c \cosh \left (b c x + a c\right )^{3} - 4 \, b c \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )^{3} +{\left (15 \, b c \cosh \left (b c x + a c\right )^{4} - 24 \, b c \cosh \left (b c x + a c\right )^{2} + 7 \, b c\right )} \sinh \left (b c x + a c\right )^{2} - 4 \, b c + 2 \,{\left (3 \, b c \cosh \left (b c x + a c\right )^{5} - 8 \, b c \cosh \left (b c x + a c\right )^{3} + 5 \, b c \cosh \left (b c x + a c\right )\right )} \sinh \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))/(sinh(b*c*x+a*c)^2)^(5/2),x, algorithm="fricas")

[Out]

-4/3*(7*cosh(b*c*x + a*c)^2 + 10*cosh(b*c*x + a*c)*sinh(b*c*x + a*c) + 7*sinh(b*c*x + a*c)^2 - 4)/(b*c*cosh(b*
c*x + a*c)^6 + 6*b*c*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^5 + b*c*sinh(b*c*x + a*c)^6 - 4*b*c*cosh(b*c*x + a*c)
^4 + (15*b*c*cosh(b*c*x + a*c)^2 - 4*b*c)*sinh(b*c*x + a*c)^4 + 7*b*c*cosh(b*c*x + a*c)^2 + 4*(5*b*c*cosh(b*c*
x + a*c)^3 - 4*b*c*cosh(b*c*x + a*c))*sinh(b*c*x + a*c)^3 + (15*b*c*cosh(b*c*x + a*c)^4 - 24*b*c*cosh(b*c*x +
a*c)^2 + 7*b*c)*sinh(b*c*x + a*c)^2 - 4*b*c + 2*(3*b*c*cosh(b*c*x + a*c)^5 - 8*b*c*cosh(b*c*x + a*c)^3 + 5*b*c
*cosh(b*c*x + a*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)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.29122, size = 165, normalized size = 1.12 \begin{align*} -\frac{4 \,{\left (6 \, e^{\left (4 \, b c x + 4 \, a c\right )} \mathrm{sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) - 4 \, 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 )}}{3 \, b c{\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}^{4}} \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)^(5/2),x, algorithm="giac")

[Out]

-4/3*(6*e^(4*b*c*x + 4*a*c)*sgn(e^(b*c*x + a*c) - e^(-b*c*x - a*c)) - 4*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)^4)