3.156 \(\int \frac{1}{(a \sinh ^4(x))^{3/2}} \, dx\)

Optimal. Leaf size=68 \[ -\frac{\sinh (x) \cosh (x)}{a \sqrt{a \sinh ^4(x)}}-\frac{\cosh ^2(x) \coth ^3(x)}{5 a \sqrt{a \sinh ^4(x)}}+\frac{2 \cosh ^2(x) \coth (x)}{3 a \sqrt{a \sinh ^4(x)}} \]

[Out]

(2*Cosh[x]^2*Coth[x])/(3*a*Sqrt[a*Sinh[x]^4]) - (Cosh[x]^2*Coth[x]^3)/(5*a*Sqrt[a*Sinh[x]^4]) - (Cosh[x]*Sinh[
x])/(a*Sqrt[a*Sinh[x]^4])

________________________________________________________________________________________

Rubi [A]  time = 0.0220898, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3207, 3767} \[ -\frac{\sinh (x) \cosh (x)}{a \sqrt{a \sinh ^4(x)}}-\frac{\cosh ^2(x) \coth ^3(x)}{5 a \sqrt{a \sinh ^4(x)}}+\frac{2 \cosh ^2(x) \coth (x)}{3 a \sqrt{a \sinh ^4(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sinh[x]^4)^(-3/2),x]

[Out]

(2*Cosh[x]^2*Coth[x])/(3*a*Sqrt[a*Sinh[x]^4]) - (Cosh[x]^2*Coth[x]^3)/(5*a*Sqrt[a*Sinh[x]^4]) - (Cosh[x]*Sinh[
x])/(a*Sqrt[a*Sinh[x]^4])

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin{align*} \int \frac{1}{\left (a \sinh ^4(x)\right )^{3/2}} \, dx &=\frac{\sinh ^2(x) \int \text{csch}^6(x) \, dx}{a \sqrt{a \sinh ^4(x)}}\\ &=-\frac{\left (i \sinh ^2(x)\right ) \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-i \coth (x)\right )}{a \sqrt{a \sinh ^4(x)}}\\ &=\frac{2 \cosh ^2(x) \coth (x)}{3 a \sqrt{a \sinh ^4(x)}}-\frac{\cosh ^2(x) \coth ^3(x)}{5 a \sqrt{a \sinh ^4(x)}}-\frac{\cosh (x) \sinh (x)}{a \sqrt{a \sinh ^4(x)}}\\ \end{align*}

Mathematica [A]  time = 0.038011, size = 34, normalized size = 0.5 \[ -\frac{\sinh ^5(x) \cosh (x) \left (3 \text{csch}^4(x)-4 \text{csch}^2(x)+8\right )}{15 \left (a \sinh ^4(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sinh[x]^4)^(-3/2),x]

[Out]

-(Cosh[x]*(8 - 4*Csch[x]^2 + 3*Csch[x]^4)*Sinh[x]^5)/(15*(a*Sinh[x]^4)^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.086, size = 80, normalized size = 1.2 \begin{align*} -{\frac{\sqrt{8}\sqrt{2} \left ( 2\, \left ( \cosh \left ( 2\,x \right ) \right ) ^{2}-6\,\cosh \left ( 2\,x \right ) +7 \right ) }{15\,{a}^{2} \left ( -1+\cosh \left ( 2\,x \right ) \right ) ^{2}\sinh \left ( 2\,x \right ) }\sqrt{a \left ( \sinh \left ( 2\,x \right ) \right ) ^{2}}\sqrt{a \left ( -1+\cosh \left ( 2\,x \right ) \right ) \left ( \cosh \left ( 2\,x \right ) +1 \right ) }{\frac{1}{\sqrt{a \left ( -1+\cosh \left ( 2\,x \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sinh(x)^4)^(3/2),x)

[Out]

-1/15*8^(1/2)/a^2*2^(1/2)*(2*cosh(2*x)^2-6*cosh(2*x)+7)*(a*sinh(2*x)^2)^(1/2)*(a*(-1+cosh(2*x))*(cosh(2*x)+1))
^(1/2)/(-1+cosh(2*x))^2/sinh(2*x)/(a*(-1+cosh(2*x))^2)^(1/2)

________________________________________________________________________________________

Maxima [B]  time = 1.93976, size = 231, normalized size = 3.4 \begin{align*} -\frac{16 \, e^{\left (-2 \, x\right )}}{3 \,{\left (5 \, a^{\frac{3}{2}} e^{\left (-2 \, x\right )} - 10 \, a^{\frac{3}{2}} e^{\left (-4 \, x\right )} + 10 \, a^{\frac{3}{2}} e^{\left (-6 \, x\right )} - 5 \, a^{\frac{3}{2}} e^{\left (-8 \, x\right )} + a^{\frac{3}{2}} e^{\left (-10 \, x\right )} - a^{\frac{3}{2}}\right )}} + \frac{32 \, e^{\left (-4 \, x\right )}}{3 \,{\left (5 \, a^{\frac{3}{2}} e^{\left (-2 \, x\right )} - 10 \, a^{\frac{3}{2}} e^{\left (-4 \, x\right )} + 10 \, a^{\frac{3}{2}} e^{\left (-6 \, x\right )} - 5 \, a^{\frac{3}{2}} e^{\left (-8 \, x\right )} + a^{\frac{3}{2}} e^{\left (-10 \, x\right )} - a^{\frac{3}{2}}\right )}} + \frac{16}{15 \,{\left (5 \, a^{\frac{3}{2}} e^{\left (-2 \, x\right )} - 10 \, a^{\frac{3}{2}} e^{\left (-4 \, x\right )} + 10 \, a^{\frac{3}{2}} e^{\left (-6 \, x\right )} - 5 \, a^{\frac{3}{2}} e^{\left (-8 \, x\right )} + a^{\frac{3}{2}} e^{\left (-10 \, x\right )} - a^{\frac{3}{2}}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sinh(x)^4)^(3/2),x, algorithm="maxima")

[Out]

-16/3*e^(-2*x)/(5*a^(3/2)*e^(-2*x) - 10*a^(3/2)*e^(-4*x) + 10*a^(3/2)*e^(-6*x) - 5*a^(3/2)*e^(-8*x) + a^(3/2)*
e^(-10*x) - a^(3/2)) + 32/3*e^(-4*x)/(5*a^(3/2)*e^(-2*x) - 10*a^(3/2)*e^(-4*x) + 10*a^(3/2)*e^(-6*x) - 5*a^(3/
2)*e^(-8*x) + a^(3/2)*e^(-10*x) - a^(3/2)) + 16/15/(5*a^(3/2)*e^(-2*x) - 10*a^(3/2)*e^(-4*x) + 10*a^(3/2)*e^(-
6*x) - 5*a^(3/2)*e^(-8*x) + a^(3/2)*e^(-10*x) - a^(3/2))

________________________________________________________________________________________

Fricas [B]  time = 2.04043, size = 3163, normalized size = 46.51 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sinh(x)^4)^(3/2),x, algorithm="fricas")

[Out]

-16/15*(40*cosh(x)*e^(2*x)*sinh(x)^3 + 10*e^(2*x)*sinh(x)^4 + 5*(12*cosh(x)^2 - 1)*e^(2*x)*sinh(x)^2 + 10*(4*c
osh(x)^3 - cosh(x))*e^(2*x)*sinh(x) + (10*cosh(x)^4 - 5*cosh(x)^2 + 1)*e^(2*x))*sqrt(a*e^(8*x) - 4*a*e^(6*x) +
6*a*e^(4*x) - 4*a*e^(2*x) + a)*e^(-2*x)/(a^2*cosh(x)^10 + (a^2*e^(4*x) - 2*a^2*e^(2*x) + a^2)*sinh(x)^10 - 5*
a^2*cosh(x)^8 + 10*(a^2*cosh(x)*e^(4*x) - 2*a^2*cosh(x)*e^(2*x) + a^2*cosh(x))*sinh(x)^9 + 5*(9*a^2*cosh(x)^2
- a^2 + (9*a^2*cosh(x)^2 - a^2)*e^(4*x) - 2*(9*a^2*cosh(x)^2 - a^2)*e^(2*x))*sinh(x)^8 + 10*a^2*cosh(x)^6 + 40
*(3*a^2*cosh(x)^3 - a^2*cosh(x) + (3*a^2*cosh(x)^3 - a^2*cosh(x))*e^(4*x) - 2*(3*a^2*cosh(x)^3 - a^2*cosh(x))*
e^(2*x))*sinh(x)^7 + 10*(21*a^2*cosh(x)^4 - 14*a^2*cosh(x)^2 + a^2 + (21*a^2*cosh(x)^4 - 14*a^2*cosh(x)^2 + a^
2)*e^(4*x) - 2*(21*a^2*cosh(x)^4 - 14*a^2*cosh(x)^2 + a^2)*e^(2*x))*sinh(x)^6 - 10*a^2*cosh(x)^4 + 4*(63*a^2*c
osh(x)^5 - 70*a^2*cosh(x)^3 + 15*a^2*cosh(x) + (63*a^2*cosh(x)^5 - 70*a^2*cosh(x)^3 + 15*a^2*cosh(x))*e^(4*x)
- 2*(63*a^2*cosh(x)^5 - 70*a^2*cosh(x)^3 + 15*a^2*cosh(x))*e^(2*x))*sinh(x)^5 + 10*(21*a^2*cosh(x)^6 - 35*a^2*
cosh(x)^4 + 15*a^2*cosh(x)^2 - a^2 + (21*a^2*cosh(x)^6 - 35*a^2*cosh(x)^4 + 15*a^2*cosh(x)^2 - a^2)*e^(4*x) -
2*(21*a^2*cosh(x)^6 - 35*a^2*cosh(x)^4 + 15*a^2*cosh(x)^2 - a^2)*e^(2*x))*sinh(x)^4 + 5*a^2*cosh(x)^2 + 40*(3*
a^2*cosh(x)^7 - 7*a^2*cosh(x)^5 + 5*a^2*cosh(x)^3 - a^2*cosh(x) + (3*a^2*cosh(x)^7 - 7*a^2*cosh(x)^5 + 5*a^2*c
osh(x)^3 - a^2*cosh(x))*e^(4*x) - 2*(3*a^2*cosh(x)^7 - 7*a^2*cosh(x)^5 + 5*a^2*cosh(x)^3 - a^2*cosh(x))*e^(2*x
))*sinh(x)^3 + 5*(9*a^2*cosh(x)^8 - 28*a^2*cosh(x)^6 + 30*a^2*cosh(x)^4 - 12*a^2*cosh(x)^2 + a^2 + (9*a^2*cosh
(x)^8 - 28*a^2*cosh(x)^6 + 30*a^2*cosh(x)^4 - 12*a^2*cosh(x)^2 + a^2)*e^(4*x) - 2*(9*a^2*cosh(x)^8 - 28*a^2*co
sh(x)^6 + 30*a^2*cosh(x)^4 - 12*a^2*cosh(x)^2 + a^2)*e^(2*x))*sinh(x)^2 - a^2 + (a^2*cosh(x)^10 - 5*a^2*cosh(x
)^8 + 10*a^2*cosh(x)^6 - 10*a^2*cosh(x)^4 + 5*a^2*cosh(x)^2 - a^2)*e^(4*x) - 2*(a^2*cosh(x)^10 - 5*a^2*cosh(x)
^8 + 10*a^2*cosh(x)^6 - 10*a^2*cosh(x)^4 + 5*a^2*cosh(x)^2 - a^2)*e^(2*x) + 10*(a^2*cosh(x)^9 - 4*a^2*cosh(x)^
7 + 6*a^2*cosh(x)^5 - 4*a^2*cosh(x)^3 + a^2*cosh(x) + (a^2*cosh(x)^9 - 4*a^2*cosh(x)^7 + 6*a^2*cosh(x)^5 - 4*a
^2*cosh(x)^3 + a^2*cosh(x))*e^(4*x) - 2*(a^2*cosh(x)^9 - 4*a^2*cosh(x)^7 + 6*a^2*cosh(x)^5 - 4*a^2*cosh(x)^3 +
a^2*cosh(x))*e^(2*x))*sinh(x))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \sinh ^{4}{\left (x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sinh(x)**4)**(3/2),x)

[Out]

Integral((a*sinh(x)**4)**(-3/2), x)

________________________________________________________________________________________

Giac [A]  time = 1.27163, size = 47, normalized size = 0.69 \begin{align*} -\frac{16 \,{\left (10 \, \sqrt{a} e^{\left (4 \, x\right )} - 5 \, \sqrt{a} e^{\left (2 \, x\right )} + \sqrt{a}\right )}}{15 \, a^{2}{\left (e^{\left (2 \, x\right )} - 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sinh(x)^4)^(3/2),x, algorithm="giac")

[Out]

-16/15*(10*sqrt(a)*e^(4*x) - 5*sqrt(a)*e^(2*x) + sqrt(a))/(a^2*(e^(2*x) - 1)^5)