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(b*c*x+a*c)/b/c/(1-exp(2*c*(b*x+a)))^4/(sinh(b*c*x+a*c)^2)^(1/2)+32/3*sinh(b*c*x+a*c)/b/c/(1-exp(2*c*(b
*x+a)))^3/(sinh(b*c*x+a*c)^2)^(1/2)-8*sinh(b*c*x+a*c)/b/c/(1-exp(2*c*(b*x+a)))^2/(sinh(b*c*x+a*c)^2)^(1/2)

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Rubi [A]  time = 0.21, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 12

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

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])

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 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 \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*}

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Mathematica [A]  time = 0.07, 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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fricas [B]  time = 0.68, size = 315, normalized size = 2.14 \[ -\frac {4 \, {\left (7 \, \cosh \left (b c x + a c\right )^{2} + 10 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right ) + 7 \, \sinh \left (b c x + a c\right )^{2} - 4\right )}}{3 \, {\left (b c \cosh \left (b c x + a c\right )^{6} + 6 \, b c \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{5} + b c \sinh \left (b c x + a c\right )^{6} - 4 \, b c \cosh \left (b c x + a c\right )^{4} + {\left (15 \, b c \cosh \left (b c x + a c\right )^{2} - 4 \, b c\right )} \sinh \left (b c x + a c\right )^{4} + 7 \, b c \cosh \left (b c x + a c\right )^{2} + 4 \, {\left (5 \, b c \cosh \left (b c x + a c\right )^{3} - 4 \, b c \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )^{3} + {\left (15 \, b c \cosh \left (b c x + a c\right )^{4} - 24 \, b c \cosh \left (b c x + a c\right )^{2} + 7 \, b c\right )} \sinh \left (b c x + a c\right )^{2} - 4 \, b c + 2 \, {\left (3 \, b c \cosh \left (b c x + a c\right )^{5} - 8 \, b c \cosh \left (b c x + a c\right )^{3} + 5 \, b c \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )\right )}} \]

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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giac [A]  time = 0.20, size = 122, normalized size = 0.83 \[ -\frac {4 \, {\left (6 \, e^{\left (4 \, b c x + 4 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) - 4 \, e^{\left (2 \, b c x + 2 \, a c\right )} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) + \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right )\right )}}{3 \, b c {\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}^{4}} \]

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)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {32 \,{\mathrm e}^{c \left (b x +a \right )}}{\left (-2+2 \cosh \left (2 b c x +2 a c \right )\right )^{\frac {5}{2}}}\, dx \]

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 = 0.42, size = 209, normalized size = 1.42 \[ -\frac {8 \, e^{\left (4 \, b c x + 4 \, a c\right )}}{b c {\left (e^{\left (8 \, b c x + 8 \, a c\right )} - 4 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 6 \, e^{\left (4 \, b c x + 4 \, a c\right )} - 4 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} + \frac {16 \, e^{\left (2 \, b c x + 2 \, a c\right )}}{3 \, b c {\left (e^{\left (8 \, b c x + 8 \, a c\right )} - 4 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 6 \, e^{\left (4 \, b c x + 4 \, a c\right )} - 4 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} - \frac {4}{3 \, b c {\left (e^{\left (8 \, b c x + 8 \, a c\right )} - 4 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 6 \, e^{\left (4 \, b c x + 4 \, a c\right )} - 4 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} \]

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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mupad [B]  time = 0.62, size = 89, normalized size = 0.61 \[ -\frac {8\,{\mathrm {e}}^{a\,c+b\,c\,x}\,\sqrt {{\left (\frac {{\mathrm {e}}^{a\,c+b\,c\,x}}{2}-\frac {{\mathrm {e}}^{-a\,c-b\,c\,x}}{2}\right )}^2}\,\left (6\,{\mathrm {e}}^{4\,a\,c+4\,b\,c\,x}-4\,{\mathrm {e}}^{2\,a\,c+2\,b\,c\,x}+1\right )}{3\,b\,c\,{\left ({\mathrm {e}}^{2\,a\,c+2\,b\,c\,x}-1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(8*exp(a*c + b*c*x)*((exp(a*c + b*c*x)/2 - exp(- a*c - b*c*x)/2)^2)^(1/2)*(6*exp(4*a*c + 4*b*c*x) - 4*exp(2*a
*c + 2*b*c*x) + 1))/(3*b*c*(exp(2*a*c + 2*b*c*x) - 1)^5)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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