3.40 \(\int \frac {1}{(a \text {csch}^3(x))^{3/2}} \, dx\)

Optimal. Leaf size=89 \[ -\frac {14 \cosh (x)}{45 a \sqrt {a \text {csch}^3(x)}}+\frac {2 \sinh ^2(x) \cosh (x)}{9 a \sqrt {a \text {csch}^3(x)}}+\frac {14 i \text {csch}(x) E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right )}{15 a \sqrt {i \sinh (x)} \sqrt {a \text {csch}^3(x)}} \]

[Out]

-14/45*cosh(x)/a/(a*csch(x)^3)^(1/2)+2/9*cosh(x)*sinh(x)^2/a/(a*csch(x)^3)^(1/2)+14/15*I*csch(x)*(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))/a/(a*csch(x)^3)^(1/2)/(I*sinh(x))^
(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 89, 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 = {4123, 3769, 3771, 2639} \[ -\frac {14 \cosh (x)}{45 a \sqrt {a \text {csch}^3(x)}}+\frac {2 \sinh ^2(x) \cosh (x)}{9 a \sqrt {a \text {csch}^3(x)}}+\frac {14 i \text {csch}(x) E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right )}{15 a \sqrt {i \sinh (x)} \sqrt {a \text {csch}^3(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-14*Cosh[x])/(45*a*Sqrt[a*Csch[x]^3]) + (((14*I)/15)*Csch[x]*EllipticE[Pi/4 - (I/2)*x, 2])/(a*Sqrt[a*Csch[x]^
3]*Sqrt[I*Sinh[x]]) + (2*Cosh[x]*Sinh[x]^2)/(9*a*Sqrt[a*Csch[x]^3])

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 3769

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

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 4123

Int[((b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[(b^IntPart[p]*(b*(c*Sec[e + f*x])^n)^
FracPart[p])/(c*Sec[e + f*x])^(n*FracPart[p]), Int[(c*Sec[e + f*x])^(n*p), x], x] /; FreeQ[{b, c, e, f, n, p},
x] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {1}{\left (a \text {csch}^3(x)\right )^{3/2}} \, dx &=-\frac {\left (i (i \text {csch}(x))^{3/2}\right ) \int \frac {1}{(i \text {csch}(x))^{9/2}} \, dx}{a \sqrt {a \text {csch}^3(x)}}\\ &=\frac {2 \cosh (x) \sinh ^2(x)}{9 a \sqrt {a \text {csch}^3(x)}}-\frac {\left (7 i (i \text {csch}(x))^{3/2}\right ) \int \frac {1}{(i \text {csch}(x))^{5/2}} \, dx}{9 a \sqrt {a \text {csch}^3(x)}}\\ &=-\frac {14 \cosh (x)}{45 a \sqrt {a \text {csch}^3(x)}}+\frac {2 \cosh (x) \sinh ^2(x)}{9 a \sqrt {a \text {csch}^3(x)}}-\frac {\left (7 i (i \text {csch}(x))^{3/2}\right ) \int \frac {1}{\sqrt {i \text {csch}(x)}} \, dx}{15 a \sqrt {a \text {csch}^3(x)}}\\ &=-\frac {14 \cosh (x)}{45 a \sqrt {a \text {csch}^3(x)}}+\frac {2 \cosh (x) \sinh ^2(x)}{9 a \sqrt {a \text {csch}^3(x)}}+\frac {(7 \text {csch}(x)) \int \sqrt {i \sinh (x)} \, dx}{15 a \sqrt {a \text {csch}^3(x)} \sqrt {i \sinh (x)}}\\ &=-\frac {14 \cosh (x)}{45 a \sqrt {a \text {csch}^3(x)}}+\frac {14 i \text {csch}(x) E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right )}{15 a \sqrt {a \text {csch}^3(x)} \sqrt {i \sinh (x)}}+\frac {2 \cosh (x) \sinh ^2(x)}{9 a \sqrt {a \text {csch}^3(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 57, normalized size = 0.64 \[ \frac {-33 \cosh (x)+5 \cosh (3 x)+84 \sqrt {i \sinh (x)} \text {csch}^2(x) E\left (\left .\frac {1}{4} (\pi -2 i x)\right |2\right )}{90 a \sqrt {a \text {csch}^3(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-33*Cosh[x] + 5*Cosh[3*x] + 84*Csch[x]^2*EllipticE[(Pi - (2*I)*x)/4, 2]*Sqrt[I*Sinh[x]])/(90*a*Sqrt[a*Csch[x]
^3])

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \operatorname {csch}\relax (x)^{3}}}{a^{2} \operatorname {csch}\relax (x)^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*csch(x)^3)^(-3/2), x)

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maple [F]  time = 0.20, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \mathrm {csch}\relax (x )^{3}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*csch(x)^3)^(-3/2), x)

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

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 \[ \int \frac {1}{\left (a \operatorname {csch}^{3}{\relax (x )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*csch(x)**3)**(-3/2), x)

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