3.151 \(\int \frac {1}{(a \sinh ^3(x))^{5/2}} \, dx\)

Optimal. Leaf size=135 \[ \frac {154 \sinh (x) \cosh (x)}{195 a^2 \sqrt {a \sinh ^3(x)}}-\frac {154 \coth (x)}{585 a^2 \sqrt {a \sinh ^3(x)}}-\frac {154 i \sinh ^2(x) E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right )}{195 a^2 \sqrt {i \sinh (x)} \sqrt {a \sinh ^3(x)}}-\frac {2 \coth (x) \text {csch}^4(x)}{13 a^2 \sqrt {a \sinh ^3(x)}}+\frac {22 \coth (x) \text {csch}^2(x)}{117 a^2 \sqrt {a \sinh ^3(x)}} \]

[Out]

-154/585*coth(x)/a^2/(a*sinh(x)^3)^(1/2)+22/117*coth(x)*csch(x)^2/a^2/(a*sinh(x)^3)^(1/2)-2/13*coth(x)*csch(x)
^4/a^2/(a*sinh(x)^3)^(1/2)+154/195*cosh(x)*sinh(x)/a^2/(a*sinh(x)^3)^(1/2)-154/195*I*(sin(1/4*Pi+1/2*I*x)^2)^(
1/2)/sin(1/4*Pi+1/2*I*x)*EllipticE(cos(1/4*Pi+1/2*I*x),2^(1/2))*sinh(x)^2/a^2/(I*sinh(x))^(1/2)/(a*sinh(x)^3)^
(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3207, 2636, 2640, 2639} \[ \frac {154 \sinh (x) \cosh (x)}{195 a^2 \sqrt {a \sinh ^3(x)}}-\frac {154 \coth (x)}{585 a^2 \sqrt {a \sinh ^3(x)}}-\frac {154 i \sinh ^2(x) E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right )}{195 a^2 \sqrt {i \sinh (x)} \sqrt {a \sinh ^3(x)}}-\frac {2 \coth (x) \text {csch}^4(x)}{13 a^2 \sqrt {a \sinh ^3(x)}}+\frac {22 \coth (x) \text {csch}^2(x)}{117 a^2 \sqrt {a \sinh ^3(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sinh[x]^3)^(-5/2),x]

[Out]

(-154*Coth[x])/(585*a^2*Sqrt[a*Sinh[x]^3]) + (22*Coth[x]*Csch[x]^2)/(117*a^2*Sqrt[a*Sinh[x]^3]) - (2*Coth[x]*C
sch[x]^4)/(13*a^2*Sqrt[a*Sinh[x]^3]) + (154*Cosh[x]*Sinh[x])/(195*a^2*Sqrt[a*Sinh[x]^3]) - (((154*I)/195)*Elli
pticE[Pi/4 - (I/2)*x, 2]*Sinh[x]^2)/(a^2*Sqrt[I*Sinh[x]]*Sqrt[a*Sinh[x]^3])

Rule 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

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
]])

Rubi steps

\begin {align*} \int \frac {1}{\left (a \sinh ^3(x)\right )^{5/2}} \, dx &=\frac {\sinh ^{\frac {3}{2}}(x) \int \frac {1}{\sinh ^{\frac {15}{2}}(x)} \, dx}{a^2 \sqrt {a \sinh ^3(x)}}\\ &=-\frac {2 \coth (x) \text {csch}^4(x)}{13 a^2 \sqrt {a \sinh ^3(x)}}-\frac {\left (11 \sinh ^{\frac {3}{2}}(x)\right ) \int \frac {1}{\sinh ^{\frac {11}{2}}(x)} \, dx}{13 a^2 \sqrt {a \sinh ^3(x)}}\\ &=\frac {22 \coth (x) \text {csch}^2(x)}{117 a^2 \sqrt {a \sinh ^3(x)}}-\frac {2 \coth (x) \text {csch}^4(x)}{13 a^2 \sqrt {a \sinh ^3(x)}}+\frac {\left (77 \sinh ^{\frac {3}{2}}(x)\right ) \int \frac {1}{\sinh ^{\frac {7}{2}}(x)} \, dx}{117 a^2 \sqrt {a \sinh ^3(x)}}\\ &=-\frac {154 \coth (x)}{585 a^2 \sqrt {a \sinh ^3(x)}}+\frac {22 \coth (x) \text {csch}^2(x)}{117 a^2 \sqrt {a \sinh ^3(x)}}-\frac {2 \coth (x) \text {csch}^4(x)}{13 a^2 \sqrt {a \sinh ^3(x)}}-\frac {\left (77 \sinh ^{\frac {3}{2}}(x)\right ) \int \frac {1}{\sinh ^{\frac {3}{2}}(x)} \, dx}{195 a^2 \sqrt {a \sinh ^3(x)}}\\ &=-\frac {154 \coth (x)}{585 a^2 \sqrt {a \sinh ^3(x)}}+\frac {22 \coth (x) \text {csch}^2(x)}{117 a^2 \sqrt {a \sinh ^3(x)}}-\frac {2 \coth (x) \text {csch}^4(x)}{13 a^2 \sqrt {a \sinh ^3(x)}}+\frac {154 \cosh (x) \sinh (x)}{195 a^2 \sqrt {a \sinh ^3(x)}}-\frac {\left (77 \sinh ^{\frac {3}{2}}(x)\right ) \int \sqrt {\sinh (x)} \, dx}{195 a^2 \sqrt {a \sinh ^3(x)}}\\ &=-\frac {154 \coth (x)}{585 a^2 \sqrt {a \sinh ^3(x)}}+\frac {22 \coth (x) \text {csch}^2(x)}{117 a^2 \sqrt {a \sinh ^3(x)}}-\frac {2 \coth (x) \text {csch}^4(x)}{13 a^2 \sqrt {a \sinh ^3(x)}}+\frac {154 \cosh (x) \sinh (x)}{195 a^2 \sqrt {a \sinh ^3(x)}}-\frac {\left (77 \sinh ^2(x)\right ) \int \sqrt {i \sinh (x)} \, dx}{195 a^2 \sqrt {i \sinh (x)} \sqrt {a \sinh ^3(x)}}\\ &=-\frac {154 \coth (x)}{585 a^2 \sqrt {a \sinh ^3(x)}}+\frac {22 \coth (x) \text {csch}^2(x)}{117 a^2 \sqrt {a \sinh ^3(x)}}-\frac {2 \coth (x) \text {csch}^4(x)}{13 a^2 \sqrt {a \sinh ^3(x)}}+\frac {154 \cosh (x) \sinh (x)}{195 a^2 \sqrt {a \sinh ^3(x)}}-\frac {154 i E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right ) \sinh ^2(x)}{195 a^2 \sqrt {i \sinh (x)} \sqrt {a \sinh ^3(x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.20, size = 69, normalized size = 0.51 \[ \frac {462 \sinh (x) \cosh (x)-2 \coth (x) \left (45 \text {csch}^4(x)-55 \text {csch}^2(x)+77\right )+462 i (i \sinh (x))^{3/2} E\left (\left .\frac {1}{4} (\pi -2 i x)\right |2\right )}{585 a^2 \sqrt {a \sinh ^3(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sinh[x]^3)^(-5/2),x]

[Out]

(-2*Coth[x]*(77 - 55*Csch[x]^2 + 45*Csch[x]^4) + (462*I)*EllipticE[(Pi - (2*I)*x)/4, 2]*(I*Sinh[x])^(3/2) + 46
2*Cosh[x]*Sinh[x])/(585*a^2*Sqrt[a*Sinh[x]^3])

________________________________________________________________________________________

fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \sinh \relax (x)^{3}}}{a^{3} \sinh \relax (x)^{9}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sinh(x)^3)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(a*sinh(x)^3)/(a^3*sinh(x)^9), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \sinh \relax (x)^{3}\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sinh(x)^3)^(5/2),x, algorithm="giac")

[Out]

integrate((a*sinh(x)^3)^(-5/2), x)

________________________________________________________________________________________

maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \left (\sinh ^{3}\relax (x )\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sinh(x)^3)^(5/2),x)

[Out]

int(1/(a*sinh(x)^3)^(5/2),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \sinh \relax (x)^{3}\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sinh(x)^3)^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sinh(x)^3)^(-5/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a\,{\mathrm {sinh}\relax (x)}^3\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sinh(x)^3)^(5/2),x)

[Out]

int(1/(a*sinh(x)^3)^(5/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \sinh ^{3}{\relax (x )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sinh(x)**3)**(5/2),x)

[Out]

Integral((a*sinh(x)**3)**(-5/2), x)

________________________________________________________________________________________