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

Optimal. Leaf size=61 \[ \frac {3 \coth (x)}{8 a^2 \sqrt {a \sinh ^2(x)}}-\frac {3 \sinh (x) \tanh ^{-1}(\cosh (x))}{8 a^2 \sqrt {a \sinh ^2(x)}}-\frac {\coth (x)}{4 a \left (a \sinh ^2(x)\right )^{3/2}} \]

[Out]

-1/4*coth(x)/a/(a*sinh(x)^2)^(3/2)+3/8*coth(x)/a^2/(a*sinh(x)^2)^(1/2)-3/8*arctanh(cosh(x))*sinh(x)/a^2/(a*sin
h(x)^2)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3204, 3207, 3770} \[ \frac {3 \coth (x)}{8 a^2 \sqrt {a \sinh ^2(x)}}-\frac {3 \sinh (x) \tanh ^{-1}(\cosh (x))}{8 a^2 \sqrt {a \sinh ^2(x)}}-\frac {\coth (x)}{4 a \left (a \sinh ^2(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Coth[x]/(4*a*(a*Sinh[x]^2)^(3/2)) + (3*Coth[x])/(8*a^2*Sqrt[a*Sinh[x]^2]) - (3*ArcTanh[Cosh[x]]*Sinh[x])/(8*a
^2*Sqrt[a*Sinh[x]^2])

Rule 3204

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(Cot[e + f*x]*(b*Sin[e + f*x]^2)^(p + 1))/(b*f*(
2*p + 1)), x] + Dist[(2*(p + 1))/(b*(2*p + 1)), Int[(b*Sin[e + f*x]^2)^(p + 1), x], x] /; FreeQ[{b, e, f}, x]
&&  !IntegerQ[p] && LtQ[p, -1]

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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {1}{\left (a \sinh ^2(x)\right )^{5/2}} \, dx &=-\frac {\coth (x)}{4 a \left (a \sinh ^2(x)\right )^{3/2}}-\frac {3 \int \frac {1}{\left (a \sinh ^2(x)\right )^{3/2}} \, dx}{4 a}\\ &=-\frac {\coth (x)}{4 a \left (a \sinh ^2(x)\right )^{3/2}}+\frac {3 \coth (x)}{8 a^2 \sqrt {a \sinh ^2(x)}}+\frac {3 \int \frac {1}{\sqrt {a \sinh ^2(x)}} \, dx}{8 a^2}\\ &=-\frac {\coth (x)}{4 a \left (a \sinh ^2(x)\right )^{3/2}}+\frac {3 \coth (x)}{8 a^2 \sqrt {a \sinh ^2(x)}}+\frac {(3 \sinh (x)) \int \text {csch}(x) \, dx}{8 a^2 \sqrt {a \sinh ^2(x)}}\\ &=-\frac {\coth (x)}{4 a \left (a \sinh ^2(x)\right )^{3/2}}+\frac {3 \coth (x)}{8 a^2 \sqrt {a \sinh ^2(x)}}-\frac {3 \tanh ^{-1}(\cosh (x)) \sinh (x)}{8 a^2 \sqrt {a \sinh ^2(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 67, normalized size = 1.10 \[ -\frac {\text {csch}(x) \sqrt {a \sinh ^2(x)} \left (\text {csch}^4\left (\frac {x}{2}\right )-6 \text {csch}^2\left (\frac {x}{2}\right )-\text {sech}^4\left (\frac {x}{2}\right )-6 \text {sech}^2\left (\frac {x}{2}\right )-24 \log \left (\tanh \left (\frac {x}{2}\right )\right )\right )}{64 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/64*(Csch[x]*(-6*Csch[x/2]^2 + Csch[x/2]^4 - 24*Log[Tanh[x/2]] - 6*Sech[x/2]^2 - Sech[x/2]^4)*Sqrt[a*Sinh[x]
^2])/a^3

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fricas [B]  time = 0.57, size = 875, normalized size = 14.34 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(42*cosh(x)*e^x*sinh(x)^6 + 6*e^x*sinh(x)^7 + 2*(63*cosh(x)^2 - 11)*e^x*sinh(x)^5 + 10*(21*cosh(x)^3 - 11
*cosh(x))*e^x*sinh(x)^4 + 2*(105*cosh(x)^4 - 110*cosh(x)^2 - 11)*e^x*sinh(x)^3 + 2*(63*cosh(x)^5 - 110*cosh(x)
^3 - 33*cosh(x))*e^x*sinh(x)^2 + 2*(21*cosh(x)^6 - 55*cosh(x)^4 - 33*cosh(x)^2 + 3)*e^x*sinh(x) + 2*(3*cosh(x)
^7 - 11*cosh(x)^5 - 11*cosh(x)^3 + 3*cosh(x))*e^x + 3*(8*cosh(x)*e^x*sinh(x)^7 + e^x*sinh(x)^8 + 4*(7*cosh(x)^
2 - 1)*e^x*sinh(x)^6 + 8*(7*cosh(x)^3 - 3*cosh(x))*e^x*sinh(x)^5 + 2*(35*cosh(x)^4 - 30*cosh(x)^2 + 3)*e^x*sin
h(x)^4 + 8*(7*cosh(x)^5 - 10*cosh(x)^3 + 3*cosh(x))*e^x*sinh(x)^3 + 4*(7*cosh(x)^6 - 15*cosh(x)^4 + 9*cosh(x)^
2 - 1)*e^x*sinh(x)^2 + 8*(cosh(x)^7 - 3*cosh(x)^5 + 3*cosh(x)^3 - cosh(x))*e^x*sinh(x) + (cosh(x)^8 - 4*cosh(x
)^6 + 6*cosh(x)^4 - 4*cosh(x)^2 + 1)*e^x)*log((cosh(x) + sinh(x) - 1)/(cosh(x) + sinh(x) + 1)))*sqrt(a*e^(4*x)
- 2*a*e^(2*x) + a)*e^(-x)/(a^3*cosh(x)^8 - 4*a^3*cosh(x)^6 - (a^3*e^(2*x) - a^3)*sinh(x)^8 - 8*(a^3*cosh(x)*e
^(2*x) - a^3*cosh(x))*sinh(x)^7 + 6*a^3*cosh(x)^4 + 4*(7*a^3*cosh(x)^2 - a^3 - (7*a^3*cosh(x)^2 - a^3)*e^(2*x)
)*sinh(x)^6 + 8*(7*a^3*cosh(x)^3 - 3*a^3*cosh(x) - (7*a^3*cosh(x)^3 - 3*a^3*cosh(x))*e^(2*x))*sinh(x)^5 - 4*a^
3*cosh(x)^2 + 2*(35*a^3*cosh(x)^4 - 30*a^3*cosh(x)^2 + 3*a^3 - (35*a^3*cosh(x)^4 - 30*a^3*cosh(x)^2 + 3*a^3)*e
^(2*x))*sinh(x)^4 + 8*(7*a^3*cosh(x)^5 - 10*a^3*cosh(x)^3 + 3*a^3*cosh(x) - (7*a^3*cosh(x)^5 - 10*a^3*cosh(x)^
3 + 3*a^3*cosh(x))*e^(2*x))*sinh(x)^3 + a^3 + 4*(7*a^3*cosh(x)^6 - 15*a^3*cosh(x)^4 + 9*a^3*cosh(x)^2 - a^3 -
(7*a^3*cosh(x)^6 - 15*a^3*cosh(x)^4 + 9*a^3*cosh(x)^2 - a^3)*e^(2*x))*sinh(x)^2 - (a^3*cosh(x)^8 - 4*a^3*cosh(
x)^6 + 6*a^3*cosh(x)^4 - 4*a^3*cosh(x)^2 + a^3)*e^(2*x) + 8*(a^3*cosh(x)^7 - 3*a^3*cosh(x)^5 + 3*a^3*cosh(x)^3
- a^3*cosh(x) - (a^3*cosh(x)^7 - 3*a^3*cosh(x)^5 + 3*a^3*cosh(x)^3 - a^3*cosh(x))*e^(2*x))*sinh(x))

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giac [A]  time = 0.22, size = 52, normalized size = 0.85 \[ \frac {3 \, {\left (e^{\left (-x\right )} + e^{x}\right )}^{3} - 20 \, e^{\left (-x\right )} - 20 \, e^{x}}{4 \, {\left ({\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4\right )}^{2} a^{\frac {5}{2}} \mathrm {sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(3*(e^(-x) + e^x)^3 - 20*e^(-x) - 20*e^x)/(((e^(-x) + e^x)^2 - 4)^2*a^(5/2)*sgn(e^(3*x) - e^x))

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maple [A]  time = 0.10, size = 89, normalized size = 1.46 \[ \frac {\sqrt {a \left (\cosh ^{2}\relax (x )\right )}\, \left (-3 \ln \left (\frac {2 \sqrt {a}\, \sqrt {a \left (\cosh ^{2}\relax (x )\right )}+2 a}{\sinh \relax (x )}\right ) a \left (\sinh ^{4}\relax (x )\right )+3 \sqrt {a \left (\cosh ^{2}\relax (x )\right )}\, \left (\sinh ^{2}\relax (x )\right ) \sqrt {a}-2 \sqrt {a}\, \sqrt {a \left (\cosh ^{2}\relax (x )\right )}\right )}{8 a^{\frac {7}{2}} \sinh \relax (x )^{3} \cosh \relax (x ) \sqrt {a \left (\sinh ^{2}\relax (x )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/a^(7/2)/sinh(x)^3*(a*cosh(x)^2)^(1/2)*(-3*ln(2*(a^(1/2)*(a*cosh(x)^2)^(1/2)+a)/sinh(x))*a*sinh(x)^4+3*(a*c
osh(x)^2)^(1/2)*sinh(x)^2*a^(1/2)-2*a^(1/2)*(a*cosh(x)^2)^(1/2))/cosh(x)/(a*sinh(x)^2)^(1/2)

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maxima [A]  time = 0.42, size = 96, normalized size = 1.57 \[ \frac {3 \, e^{\left (-x\right )} - 11 \, e^{\left (-3 \, x\right )} - 11 \, e^{\left (-5 \, x\right )} + 3 \, e^{\left (-7 \, x\right )}}{4 \, {\left (4 \, a^{\frac {5}{2}} e^{\left (-2 \, x\right )} - 6 \, a^{\frac {5}{2}} e^{\left (-4 \, x\right )} + 4 \, a^{\frac {5}{2}} e^{\left (-6 \, x\right )} - a^{\frac {5}{2}} e^{\left (-8 \, x\right )} - a^{\frac {5}{2}}\right )}} + \frac {3 \, \log \left (e^{\left (-x\right )} + 1\right )}{8 \, a^{\frac {5}{2}}} - \frac {3 \, \log \left (e^{\left (-x\right )} - 1\right )}{8 \, a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(3*e^(-x) - 11*e^(-3*x) - 11*e^(-5*x) + 3*e^(-7*x))/(4*a^(5/2)*e^(-2*x) - 6*a^(5/2)*e^(-4*x) + 4*a^(5/2)*e
^(-6*x) - a^(5/2)*e^(-8*x) - a^(5/2)) + 3/8*log(e^(-x) + 1)/a^(5/2) - 3/8*log(e^(-x) - 1)/a^(5/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\left (a\,{\mathrm {sinh}\relax (x)}^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(a*sinh(x)^2)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*sinh(x)**2)**(-5/2), x)

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