3.130 \(\int \frac{e^{c (a+b x)}}{\text{csch}^2(a c+b c x)^{3/2}} \, dx\)

Optimal. Leaf size=162 \[ \frac{e^{-2 c (a+b x)} \text{csch}(a c+b c x)}{16 b c \sqrt{\text{csch}^2(a c+b c x)}}-\frac{3 e^{2 c (a+b x)} \text{csch}(a c+b c x)}{16 b c \sqrt{\text{csch}^2(a c+b c x)}}+\frac{e^{4 c (a+b x)} \text{csch}(a c+b c x)}{32 b c \sqrt{\text{csch}^2(a c+b c x)}}+\frac{3 x \text{csch}(a c+b c x)}{8 \sqrt{\text{csch}^2(a c+b c x)}} \]

[Out]

Csch[a*c + b*c*x]/(16*b*c*E^(2*c*(a + b*x))*Sqrt[Csch[a*c + b*c*x]^2]) - (3*E^(2*c*(a + b*x))*Csch[a*c + b*c*x
])/(16*b*c*Sqrt[Csch[a*c + b*c*x]^2]) + (E^(4*c*(a + b*x))*Csch[a*c + b*c*x])/(32*b*c*Sqrt[Csch[a*c + b*c*x]^2
]) + (3*x*Csch[a*c + b*c*x])/(8*Sqrt[Csch[a*c + b*c*x]^2])

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Rubi [A]  time = 0.153345, antiderivative size = 162, 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} \[ \frac{e^{-2 c (a+b x)} \text{csch}(a c+b c x)}{16 b c \sqrt{\text{csch}^2(a c+b c x)}}-\frac{3 e^{2 c (a+b x)} \text{csch}(a c+b c x)}{16 b c \sqrt{\text{csch}^2(a c+b c x)}}+\frac{e^{4 c (a+b x)} \text{csch}(a c+b c x)}{32 b c \sqrt{\text{csch}^2(a c+b c x)}}+\frac{3 x \text{csch}(a c+b c x)}{8 \sqrt{\text{csch}^2(a c+b c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0633141, size = 76, normalized size = 0.47 \[ \frac{\left (e^{-2 c (a+b x)}-3 e^{2 c (a+b x)}+\frac{1}{2} e^{4 c (a+b x)}+6 b c x\right ) \text{csch}^3(c (a+b x))}{16 b c \text{csch}^2(c (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((E^(-2*c*(a + b*x)) - 3*E^(2*c*(a + b*x)) + E^(4*c*(a + b*x))/2 + 6*b*c*x)*Csch[c*(a + b*x)]^3)/(16*b*c*(Csch
[c*(a + b*x)]^2)^(3/2))

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Maple [A]  time = 0.171, size = 216, normalized size = 1.3 \begin{align*}{\frac{3\,x{{\rm e}^{c \left ( bx+a \right ) }}}{8\,{{\rm e}^{2\,c \left ( bx+a \right ) }}-8}{\frac{1}{\sqrt{{\frac{{{\rm e}^{2\,c \left ( bx+a \right ) }}}{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) ^{2}}}}}}}+{\frac{{{\rm e}^{5\,c \left ( bx+a \right ) }}}{ \left ( 32\,{{\rm e}^{2\,c \left ( bx+a \right ) }}-32 \right ) cb}{\frac{1}{\sqrt{{\frac{{{\rm e}^{2\,c \left ( bx+a \right ) }}}{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) ^{2}}}}}}}-{\frac{3\,{{\rm e}^{3\,c \left ( bx+a \right ) }}}{ \left ( 16\,{{\rm e}^{2\,c \left ( bx+a \right ) }}-16 \right ) cb}{\frac{1}{\sqrt{{\frac{{{\rm e}^{2\,c \left ( bx+a \right ) }}}{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) ^{2}}}}}}}+{\frac{{{\rm e}^{-c \left ( bx+a \right ) }}}{ \left ( 16\,{{\rm e}^{2\,c \left ( bx+a \right ) }}-16 \right ) cb}{\frac{1}{\sqrt{{\frac{{{\rm e}^{2\,c \left ( bx+a \right ) }}}{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8/(1/(exp(2*c*(b*x+a))-1)^2*exp(2*c*(b*x+a)))^(1/2)/(exp(2*c*(b*x+a))-1)*x*exp(c*(b*x+a))+1/32/(1/(exp(2*c*(
b*x+a))-1)^2*exp(2*c*(b*x+a)))^(1/2)/(exp(2*c*(b*x+a))-1)/c/b*exp(5*c*(b*x+a))-3/16/(1/(exp(2*c*(b*x+a))-1)^2*
exp(2*c*(b*x+a)))^(1/2)/(exp(2*c*(b*x+a))-1)/c/b*exp(3*c*(b*x+a))+1/16/(1/(exp(2*c*(b*x+a))-1)^2*exp(2*c*(b*x+
a)))^(1/2)/(exp(2*c*(b*x+a))-1)/c/b*exp(-c*(b*x+a))

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Maxima [A]  time = 1.55534, size = 84, normalized size = 0.52 \begin{align*} \frac{{\left (e^{\left (6 \, b c x + 6 \, a c\right )} - 6 \, e^{\left (4 \, b c x + 4 \, a c\right )} + 2\right )} e^{\left (-2 \, b c x - 2 \, a c\right )}}{32 \, b c} + \frac{3 \,{\left (b c x + a c\right )}}{8 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(e^(6*b*c*x + 6*a*c) - 6*e^(4*b*c*x + 4*a*c) + 2)*e^(-2*b*c*x - 2*a*c)/(b*c) + 3/8*(b*c*x + a*c)/(b*c)

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Fricas [A]  time = 1.54929, size = 319, normalized size = 1.97 \begin{align*} \frac{3 \, \cosh \left (b c x + a c\right )^{3} + 9 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{2} - \sinh \left (b c x + a c\right )^{3} + 6 \,{\left (2 \, b c x - 1\right )} \cosh \left (b c x + a c\right ) - 3 \,{\left (4 \, b c x + \cosh \left (b c x + a c\right )^{2} + 2\right )} \sinh \left (b c x + a c\right )}{32 \,{\left (b c \cosh \left (b c x + a c\right ) - b c \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))/(csch(b*c*x+a*c)^2)^(3/2),x, algorithm="fricas")

[Out]

1/32*(3*cosh(b*c*x + a*c)^3 + 9*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^2 - sinh(b*c*x + a*c)^3 + 6*(2*b*c*x - 1)*
cosh(b*c*x + a*c) - 3*(4*b*c*x + cosh(b*c*x + a*c)^2 + 2)*sinh(b*c*x + a*c))/(b*c*cosh(b*c*x + a*c) - b*c*sinh
(b*c*x + a*c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{a c} \int \frac{e^{b c x}}{\left (\operatorname{csch}^{2}{\left (a c + b c x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(a*c)*Integral(exp(b*c*x)/(csch(a*c + b*c*x)**2)**(3/2), x)

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Giac [A]  time = 1.19345, size = 275, normalized size = 1.7 \begin{align*} \frac{{\left (12 \, b c x e^{\left (-a c\right )} \mathrm{sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) - 2 \,{\left (3 \, 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 )} e^{\left (-2 \, b c x - 3 \, a c\right )} +{\left (e^{\left (4 \, b c x + 9 \, 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 + 7 \, a c\right )} \mathrm{sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right )\right )} e^{\left (-6 \, a c\right )}\right )} e^{\left (a c\right )}}{32 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(12*b*c*x*e^(-a*c)*sgn(e^(b*c*x + a*c) - e^(-b*c*x - a*c)) - 2*(3*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)))*e^(-2*b*c*x - 3*a*c) + (e^(4*b*c*x + 9*a*c)*sg
n(e^(b*c*x + a*c) - e^(-b*c*x - a*c)) - 6*e^(2*b*c*x + 7*a*c)*sgn(e^(b*c*x + a*c) - e^(-b*c*x - a*c)))*e^(-6*a
*c))*e^(a*c)/(b*c)