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

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

[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])

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Rubi [A]  time = 0.0419059, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, 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 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 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 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 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]

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*}

Mathematica [A]  time = 0.0820777, size = 53, normalized size = 0.61 \[ \frac{2 \left (5 (i \sinh (x))^{3/2} \text{EllipticF}\left (\frac{1}{4} (\pi -2 i x),2\right )+5 \cosh (x)-3 \coth (x) \text{csch}(x)\right )}{21 a \sqrt{a \sinh ^3(x)}} \]

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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Maple [F]  time = 0.062, size = 0, normalized size = 0. \begin{align*} \int \left ( a \left ( \sinh \left ( x \right ) \right ) ^{3} \right ) ^{-{\frac{3}{2}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \sinh \left (x\right )^{3}\right )^{\frac{3}{2}}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{a \sinh \left (x\right )^{3}}}{a^{2} \sinh \left (x\right )^{6}}, x\right ) \end{align*}

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

[Out]

Integral((a*sinh(x)**3)**(-3/2), x)

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

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)