3.335 \(\int \frac{e^{c (a+b x)}}{\sinh ^2(a c+b c x)^{7/2}} \, dx\)

Optimal. Leaf size=199 \[ \frac{64 \sinh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt{\sinh ^2(a c+b c x)}}-\frac{48 \sinh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt{\sinh ^2(a c+b c x)}}+\frac{192 \sinh (a c+b c x)}{5 b c \left (1-e^{2 c (a+b x)}\right )^5 \sqrt{\sinh ^2(a c+b c x)}}-\frac{32 \sinh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^6 \sqrt{\sinh ^2(a c+b c x)}} \]

[Out]

(-32*Sinh[a*c + b*c*x])/(3*b*c*(1 - E^(2*c*(a + b*x)))^6*Sqrt[Sinh[a*c + b*c*x]^2]) + (192*Sinh[a*c + b*c*x])/
(5*b*c*(1 - E^(2*c*(a + b*x)))^5*Sqrt[Sinh[a*c + b*c*x]^2]) - (48*Sinh[a*c + b*c*x])/(b*c*(1 - E^(2*c*(a + b*x
)))^4*Sqrt[Sinh[a*c + b*c*x]^2]) + (64*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])

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Rubi [A]  time = 0.279457, antiderivative size = 199, 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} \[ \frac{64 \sinh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt{\sinh ^2(a c+b c x)}}-\frac{48 \sinh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt{\sinh ^2(a c+b c x)}}+\frac{192 \sinh (a c+b c x)}{5 b c \left (1-e^{2 c (a+b x)}\right )^5 \sqrt{\sinh ^2(a c+b c x)}}-\frac{32 \sinh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^6 \sqrt{\sinh ^2(a c+b c x)}} \]

Antiderivative was successfully verified.

[In]

Int[E^(c*(a + b*x))/(Sinh[a*c + b*c*x]^2)^(7/2),x]

[Out]

(-32*Sinh[a*c + b*c*x])/(3*b*c*(1 - E^(2*c*(a + b*x)))^6*Sqrt[Sinh[a*c + b*c*x]^2]) + (192*Sinh[a*c + b*c*x])/
(5*b*c*(1 - E^(2*c*(a + b*x)))^5*Sqrt[Sinh[a*c + b*c*x]^2]) - (48*Sinh[a*c + b*c*x])/(b*c*(1 - E^(2*c*(a + b*x
)))^4*Sqrt[Sinh[a*c + b*c*x]^2]) + (64*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])

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)^{7/2}} \, dx &=\frac{\sinh (a c+b c x) \int e^{c (a+b x)} \text{csch}^7(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{128 x^7}{\left (-1+x^2\right )^7} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt{\sinh ^2(a c+b c x)}}\\ &=\frac{(128 \sinh (a c+b c x)) \operatorname{Subst}\left (\int \frac{x^7}{\left (-1+x^2\right )^7} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt{\sinh ^2(a c+b c x)}}\\ &=\frac{(64 \sinh (a c+b c x)) \operatorname{Subst}\left (\int \frac{x^3}{(-1+x)^7} \, dx,x,e^{2 c (a+b x)}\right )}{b c \sqrt{\sinh ^2(a c+b c x)}}\\ &=\frac{(64 \sinh (a c+b c x)) \operatorname{Subst}\left (\int \left (\frac{1}{(-1+x)^7}+\frac{3}{(-1+x)^6}+\frac{3}{(-1+x)^5}+\frac{1}{(-1+x)^4}\right ) \, dx,x,e^{2 c (a+b x)}\right )}{b c \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 )^6 \sqrt{\sinh ^2(a c+b c x)}}+\frac{192 \sinh (a c+b c x)}{5 b c \left (1-e^{2 c (a+b x)}\right )^5 \sqrt{\sinh ^2(a c+b c x)}}-\frac{48 \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{64 \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)}}\\ \end{align*}

Mathematica [A]  time = 0.078297, size = 84, normalized size = 0.42 \[ -\frac{16 \left (6 e^{2 c (a+b x)}-15 e^{4 c (a+b x)}+20 e^{6 c (a+b x)}-1\right ) \sinh (c (a+b x))}{15 b c \left (e^{2 c (a+b x)}-1\right )^6 \sqrt{\sinh ^2(c (a+b x))}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(c*(a + b*x))/(Sinh[a*c + b*c*x]^2)^(7/2),x]

[Out]

(-16*(-1 + 6*E^(2*c*(a + b*x)) - 15*E^(4*c*(a + b*x)) + 20*E^(6*c*(a + b*x)))*Sinh[c*(a + b*x)])/(15*b*c*(-1 +
E^(2*c*(a + b*x)))^6*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{7}{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)^(7/2),x)

[Out]

int(exp(c*(b*x+a))/(sinh(b*c*x+a*c)^2)^(7/2),x)

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Maxima [B]  time = 1.7505, size = 521, normalized size = 2.62 \begin{align*} -\frac{64 \, e^{\left (6 \, b c x + 6 \, a c\right )}}{3 \, b c{\left (e^{\left (12 \, b c x + 12 \, a c\right )} - 6 \, e^{\left (10 \, b c x + 10 \, a c\right )} + 15 \, e^{\left (8 \, b c x + 8 \, a c\right )} - 20 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 15 \, e^{\left (4 \, b c x + 4 \, a c\right )} - 6 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} + \frac{16 \, e^{\left (4 \, b c x + 4 \, a c\right )}}{b c{\left (e^{\left (12 \, b c x + 12 \, a c\right )} - 6 \, e^{\left (10 \, b c x + 10 \, a c\right )} + 15 \, e^{\left (8 \, b c x + 8 \, a c\right )} - 20 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 15 \, e^{\left (4 \, b c x + 4 \, a c\right )} - 6 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} - \frac{32 \, e^{\left (2 \, b c x + 2 \, a c\right )}}{5 \, b c{\left (e^{\left (12 \, b c x + 12 \, a c\right )} - 6 \, e^{\left (10 \, b c x + 10 \, a c\right )} + 15 \, e^{\left (8 \, b c x + 8 \, a c\right )} - 20 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 15 \, e^{\left (4 \, b c x + 4 \, a c\right )} - 6 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} + \frac{16}{15 \, b c{\left (e^{\left (12 \, b c x + 12 \, a c\right )} - 6 \, e^{\left (10 \, b c x + 10 \, a c\right )} + 15 \, e^{\left (8 \, b c x + 8 \, a c\right )} - 20 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 15 \, e^{\left (4 \, b c x + 4 \, a c\right )} - 6 \, 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)^(7/2),x, algorithm="maxima")

[Out]

-64/3*e^(6*b*c*x + 6*a*c)/(b*c*(e^(12*b*c*x + 12*a*c) - 6*e^(10*b*c*x + 10*a*c) + 15*e^(8*b*c*x + 8*a*c) - 20*
e^(6*b*c*x + 6*a*c) + 15*e^(4*b*c*x + 4*a*c) - 6*e^(2*b*c*x + 2*a*c) + 1)) + 16*e^(4*b*c*x + 4*a*c)/(b*c*(e^(1
2*b*c*x + 12*a*c) - 6*e^(10*b*c*x + 10*a*c) + 15*e^(8*b*c*x + 8*a*c) - 20*e^(6*b*c*x + 6*a*c) + 15*e^(4*b*c*x
+ 4*a*c) - 6*e^(2*b*c*x + 2*a*c) + 1)) - 32/5*e^(2*b*c*x + 2*a*c)/(b*c*(e^(12*b*c*x + 12*a*c) - 6*e^(10*b*c*x
+ 10*a*c) + 15*e^(8*b*c*x + 8*a*c) - 20*e^(6*b*c*x + 6*a*c) + 15*e^(4*b*c*x + 4*a*c) - 6*e^(2*b*c*x + 2*a*c) +
1)) + 16/15/(b*c*(e^(12*b*c*x + 12*a*c) - 6*e^(10*b*c*x + 10*a*c) + 15*e^(8*b*c*x + 8*a*c) - 20*e^(6*b*c*x +
6*a*c) + 15*e^(4*b*c*x + 4*a*c) - 6*e^(2*b*c*x + 2*a*c) + 1))

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Fricas [B]  time = 1.98596, size = 1512, normalized size = 7.6 \begin{align*} -\frac{16 \,{\left (19 \, \cosh \left (b c x + a c\right )^{3} + 57 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{2} + 21 \, \sinh \left (b c x + a c\right )^{3} + 21 \,{\left (3 \, \cosh \left (b c x + a c\right )^{2} - 1\right )} \sinh \left (b c x + a c\right ) - 9 \, \cosh \left (b c x + a c\right )\right )}}{15 \,{\left (b c \cosh \left (b c x + a c\right )^{9} + 9 \, b c \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{8} + b c \sinh \left (b c x + a c\right )^{9} - 6 \, b c \cosh \left (b c x + a c\right )^{7} + 6 \,{\left (6 \, b c \cosh \left (b c x + a c\right )^{2} - b c\right )} \sinh \left (b c x + a c\right )^{7} + 15 \, b c \cosh \left (b c x + a c\right )^{5} + 42 \,{\left (2 \, b c \cosh \left (b c x + a c\right )^{3} - b c \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )^{6} + 3 \,{\left (42 \, b c \cosh \left (b c x + a c\right )^{4} - 42 \, b c \cosh \left (b c x + a c\right )^{2} + 5 \, b c\right )} \sinh \left (b c x + a c\right )^{5} - 19 \, b c \cosh \left (b c x + a c\right )^{3} + 3 \,{\left (42 \, b c \cosh \left (b c x + a c\right )^{5} - 70 \, b c \cosh \left (b c x + a c\right )^{3} + 25 \, b c \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )^{4} + 3 \,{\left (28 \, b c \cosh \left (b c x + a c\right )^{6} - 70 \, b c \cosh \left (b c x + a c\right )^{4} + 50 \, b c \cosh \left (b c x + a c\right )^{2} - 7 \, b c\right )} \sinh \left (b c x + a c\right )^{3} + 9 \, b c \cosh \left (b c x + a c\right ) + 3 \,{\left (12 \, b c \cosh \left (b c x + a c\right )^{7} - 42 \, b c \cosh \left (b c x + a c\right )^{5} + 50 \, b c \cosh \left (b c x + a c\right )^{3} - 19 \, b c \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )^{2} + 3 \,{\left (3 \, b c \cosh \left (b c x + a c\right )^{8} - 14 \, b c \cosh \left (b c x + a c\right )^{6} + 25 \, b c \cosh \left (b c x + a c\right )^{4} - 21 \, b c \cosh \left (b c x + a c\right )^{2} + 7 \, b c\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)^(7/2),x, algorithm="fricas")

[Out]

-16/15*(19*cosh(b*c*x + a*c)^3 + 57*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^2 + 21*sinh(b*c*x + a*c)^3 + 21*(3*cos
h(b*c*x + a*c)^2 - 1)*sinh(b*c*x + a*c) - 9*cosh(b*c*x + a*c))/(b*c*cosh(b*c*x + a*c)^9 + 9*b*c*cosh(b*c*x + a
*c)*sinh(b*c*x + a*c)^8 + b*c*sinh(b*c*x + a*c)^9 - 6*b*c*cosh(b*c*x + a*c)^7 + 6*(6*b*c*cosh(b*c*x + a*c)^2 -
b*c)*sinh(b*c*x + a*c)^7 + 15*b*c*cosh(b*c*x + a*c)^5 + 42*(2*b*c*cosh(b*c*x + a*c)^3 - b*c*cosh(b*c*x + a*c)
)*sinh(b*c*x + a*c)^6 + 3*(42*b*c*cosh(b*c*x + a*c)^4 - 42*b*c*cosh(b*c*x + a*c)^2 + 5*b*c)*sinh(b*c*x + a*c)^
5 - 19*b*c*cosh(b*c*x + a*c)^3 + 3*(42*b*c*cosh(b*c*x + a*c)^5 - 70*b*c*cosh(b*c*x + a*c)^3 + 25*b*c*cosh(b*c*
x + a*c))*sinh(b*c*x + a*c)^4 + 3*(28*b*c*cosh(b*c*x + a*c)^6 - 70*b*c*cosh(b*c*x + a*c)^4 + 50*b*c*cosh(b*c*x
+ a*c)^2 - 7*b*c)*sinh(b*c*x + a*c)^3 + 9*b*c*cosh(b*c*x + a*c) + 3*(12*b*c*cosh(b*c*x + a*c)^7 - 42*b*c*cosh
(b*c*x + a*c)^5 + 50*b*c*cosh(b*c*x + a*c)^3 - 19*b*c*cosh(b*c*x + a*c))*sinh(b*c*x + a*c)^2 + 3*(3*b*c*cosh(b
*c*x + a*c)^8 - 14*b*c*cosh(b*c*x + a*c)^6 + 25*b*c*cosh(b*c*x + a*c)^4 - 21*b*c*cosh(b*c*x + a*c)^2 + 7*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)**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.33243, size = 217, normalized size = 1.09 \begin{align*} -\frac{16 \,{\left (20 \, e^{\left (6 \, b c x + 6 \, a c\right )} \mathrm{sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) - 15 \, 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 ) + 6 \, 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 )}}{15 \, b c{\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}^{6}} \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)^(7/2),x, algorithm="giac")

[Out]

-16/15*(20*e^(6*b*c*x + 6*a*c)*sgn(e^(b*c*x + a*c) - e^(-b*c*x - a*c)) - 15*e^(4*b*c*x + 4*a*c)*sgn(e^(b*c*x +
a*c) - e^(-b*c*x - a*c)) + 6*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)^6)