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

Optimal. Leaf size=87 \[ \frac {10 \cosh (x)}{21 a \sqrt {a \sinh ^3(x)}}+\frac {10 i \sqrt {i \sinh (x)} \sinh (x) F\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right )}{21 a \sqrt {a \sinh ^3(x)}}-\frac {2 \coth (x) \text {csch}(x)}{7 a \sqrt {a \sinh ^3(x)}} \]

[Out]

10/21*cosh(x)/a/(a*sinh(x)^3)^(1/2)-2/7*coth(x)*csch(x)/a/(a*sinh(x)^3)^(1/2)+10/21*I*(sin(1/4*Pi+1/2*I*x)^2)^
(1/2)/sin(1/4*Pi+1/2*I*x)*EllipticF(cos(1/4*Pi+1/2*I*x),2^(1/2))*sinh(x)*(I*sinh(x))^(1/2)/a/(a*sinh(x)^3)^(1/
2)

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Rubi [A]  time = 0.04, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3207, 2636, 2642, 2641} \[ \frac {10 \cosh (x)}{21 a \sqrt {a \sinh ^3(x)}}+\frac {10 i \sqrt {i \sinh (x)} \sinh (x) F\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right )}{21 a \sqrt {a \sinh ^3(x)}}-\frac {2 \coth (x) \text {csch}(x)}{7 a \sqrt {a \sinh ^3(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*Cosh[x])/(21*a*Sqrt[a*Sinh[x]^3]) - (2*Coth[x]*Csch[x])/(7*a*Sqrt[a*Sinh[x]^3]) + (((10*I)/21)*EllipticF[P
i/4 - (I/2)*x, 2]*Sqrt[I*Sinh[x]]*Sinh[x])/(a*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 2641

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

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[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 )^{3/2}} \, dx &=\frac {\sinh ^{\frac {3}{2}}(x) \int \frac {1}{\sinh ^{\frac {9}{2}}(x)} \, dx}{a \sqrt {a \sinh ^3(x)}}\\ &=-\frac {2 \coth (x) \text {csch}(x)}{7 a \sqrt {a \sinh ^3(x)}}-\frac {\left (5 \sinh ^{\frac {3}{2}}(x)\right ) \int \frac {1}{\sinh ^{\frac {5}{2}}(x)} \, dx}{7 a \sqrt {a \sinh ^3(x)}}\\ &=\frac {10 \cosh (x)}{21 a \sqrt {a \sinh ^3(x)}}-\frac {2 \coth (x) \text {csch}(x)}{7 a \sqrt {a \sinh ^3(x)}}+\frac {\left (5 \sinh ^{\frac {3}{2}}(x)\right ) \int \frac {1}{\sqrt {\sinh (x)}} \, dx}{21 a \sqrt {a \sinh ^3(x)}}\\ &=\frac {10 \cosh (x)}{21 a \sqrt {a \sinh ^3(x)}}-\frac {2 \coth (x) \text {csch}(x)}{7 a \sqrt {a \sinh ^3(x)}}+\frac {\left (5 \sqrt {i \sinh (x)} \sinh (x)\right ) \int \frac {1}{\sqrt {i \sinh (x)}} \, dx}{21 a \sqrt {a \sinh ^3(x)}}\\ &=\frac {10 \cosh (x)}{21 a \sqrt {a \sinh ^3(x)}}-\frac {2 \coth (x) \text {csch}(x)}{7 a \sqrt {a \sinh ^3(x)}}+\frac {10 i F\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right ) \sqrt {i \sinh (x)} \sinh (x)}{21 a \sqrt {a \sinh ^3(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 53, normalized size = 0.61 \[ \frac {2 \left (5 \cosh (x)-3 \coth (x) \text {csch}(x)+5 (i \sinh (x))^{3/2} F\left (\left .\frac {1}{4} (\pi -2 i x)\right |2\right )\right )}{21 a \sqrt {a \sinh ^3(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*Cosh[x] - 3*Coth[x]*Csch[x] + 5*EllipticF[(Pi - (2*I)*x)/4, 2]*(I*Sinh[x])^(3/2)))/(21*a*Sqrt[a*Sinh[x]^
3])

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fricas [F]  time = 0.94, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \sinh \relax (x)^{3}}}{a^{2} \sinh \relax (x)^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a*sinh(x)^3)/(a^2*sinh(x)^6), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sinh(x)^3)^(-3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sinh(x)^3)^(-3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(a*sinh(x)^3)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*sinh(x)**3)**(-3/2), x)

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