3.126 \(\int e^{c (a+b x)} \text {csch}^2(a c+b c x)^{5/2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.17, 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} \[ -\frac {8 \sinh (a c+b c x) \sqrt {\text {csch}^2(a c+b c x)}}{b c \left (1-e^{2 c (a+b x)}\right )^2}+\frac {32 \sinh (a c+b c x) \sqrt {\text {csch}^2(a c+b c x)}}{3 b c \left (1-e^{2 c (a+b x)}\right )^3}-\frac {4 \sinh (a c+b c x) \sqrt {\text {csch}^2(a c+b c x)}}{b c \left (1-e^{2 c (a+b x)}\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[Csch[a*c + b*c*x]^2]*Sinh[a*c + b*c*x])/(b*c*(1 - E^(2*c*(a + b*x)))^4) + (32*Sqrt[Csch[a*c + b*c*x]^
2]*Sinh[a*c + b*c*x])/(3*b*c*(1 - E^(2*c*(a + b*x)))^3) - (8*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 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 e^{c (a+b x)} \text {csch}^2(a c+b c x)^{5/2} \, 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}^5(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 {32 x^5}{\left (-1+x^2\right )^5} \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac {\left (32 \sqrt {\text {csch}^2(a c+b c x)} \sinh (a c+b c x)\right ) \operatorname {Subst}\left (\int \frac {x^5}{\left (-1+x^2\right )^5} \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac {\left (16 \sqrt {\text {csch}^2(a c+b c x)} \sinh (a c+b c x)\right ) \operatorname {Subst}\left (\int \frac {x^2}{(-1+x)^5} \, dx,x,e^{2 c (a+b x)}\right )}{b c}\\ &=\frac {\left (16 \sqrt {\text {csch}^2(a c+b c x)} \sinh (a c+b c x)\right ) \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}\\ &=-\frac {4 \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 )^4}+\frac {32 \sqrt {\text {csch}^2(a c+b c x)} \sinh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3}-\frac {8 \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*}

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Mathematica [A]  time = 0.07, size = 72, normalized size = 0.49 \[ -\frac {4 \left (-4 e^{2 c (a+b x)}+6 e^{4 c (a+b x)}+1\right ) \sinh (c (a+b x)) \sqrt {\text {csch}^2(c (a+b x))}}{3 b c \left (e^{2 c (a+b x)}-1\right )^4} \]

Antiderivative was successfully verified.

[In]

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

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fricas [B]  time = 1.98, 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))*(csch(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 = 0.15, size = 77, normalized size = 0.52 \[ -\frac {4 \, {\left (6 \, e^{\left (4 \, b c x + 4 \, a c\right )} - 4 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}}{3 \, b c {\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}^{4} \mathrm {sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*(csch(b*c*x+a*c)^2)^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.61, size = 80, normalized size = 0.54 \[ -\frac {4 \left (6 \,{\mathrm e}^{4 c \left (b x +a \right )}-4 \,{\mathrm e}^{2 c \left (b x +a \right )}+1\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}\, {\mathrm e}^{-c \left (b x +a \right )}}{3 b c \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))*(csch(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 = 1.53, size = 91, normalized size = 0.62 \[ -\frac {2\,{\mathrm {e}}^{-a\,c-b\,c\,x}\,\sqrt {\frac {1}{{\left (\frac {{\mathrm {e}}^{a\,c+b\,c\,x}}{2}-\frac {{\mathrm {e}}^{-a\,c-b\,c\,x}}{2}\right )}^2}}\,\left (6\,{\mathrm {e}}^{4\,a\,c+4\,b\,c\,x}-4\,{\mathrm {e}}^{2\,a\,c+2\,b\,c\,x}+1\right )}{3\,b\,c\,{\left ({\mathrm {e}}^{2\,a\,c+2\,b\,c\,x}-1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(a + b*x))*(1/sinh(a*c + b*c*x)^2)^(5/2),x)

[Out]

-(2*exp(- a*c - b*c*x)*(1/(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)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{a c} \int \left (\operatorname {csch}^{2}{\left (a c + b c x \right )}\right )^{\frac {5}{2}} e^{b c x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*(csch(b*c*x+a*c)**2)**(5/2),x)

[Out]

exp(a*c)*Integral((csch(a*c + b*c*x)**2)**(5/2)*exp(b*c*x), x)

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