3.19 \(\int \frac {1}{(b \text {csch}(c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=90 \[ \frac {2 \cosh (c+d x)}{5 b d (b \text {csch}(c+d x))^{3/2}}+\frac {6 i E\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right )}{5 b^2 d \sqrt {i \sinh (c+d x)} \sqrt {b \text {csch}(c+d x)}} \]

[Out]

2/5*cosh(d*x+c)/b/d/(b*csch(d*x+c))^(3/2)-6/5*I*(sin(1/2*I*c+1/4*Pi+1/2*I*d*x)^2)^(1/2)/sin(1/2*I*c+1/4*Pi+1/2
*I*d*x)*EllipticE(cos(1/2*I*c+1/4*Pi+1/2*I*d*x),2^(1/2))/b^2/d/(b*csch(d*x+c))^(1/2)/(I*sinh(d*x+c))^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3769, 3771, 2639} \[ \frac {2 \cosh (c+d x)}{5 b d (b \text {csch}(c+d x))^{3/2}}+\frac {6 i E\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right )}{5 b^2 d \sqrt {i \sinh (c+d x)} \sqrt {b \text {csch}(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Csch[c + d*x])^(-5/2),x]

[Out]

(2*Cosh[c + d*x])/(5*b*d*(b*Csch[c + d*x])^(3/2)) + (((6*I)/5)*EllipticE[(I*c - Pi/2 + I*d*x)/2, 2])/(b^2*d*Sq
rt[b*Csch[c + d*x]]*Sqrt[I*Sinh[c + d*x]])

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]

Rubi steps

\begin {align*} \int \frac {1}{(b \text {csch}(c+d x))^{5/2}} \, dx &=\frac {2 \cosh (c+d x)}{5 b d (b \text {csch}(c+d x))^{3/2}}-\frac {3 \int \frac {1}{\sqrt {b \text {csch}(c+d x)}} \, dx}{5 b^2}\\ &=\frac {2 \cosh (c+d x)}{5 b d (b \text {csch}(c+d x))^{3/2}}-\frac {3 \int \sqrt {i \sinh (c+d x)} \, dx}{5 b^2 \sqrt {b \text {csch}(c+d x)} \sqrt {i \sinh (c+d x)}}\\ &=\frac {2 \cosh (c+d x)}{5 b d (b \text {csch}(c+d x))^{3/2}}+\frac {6 i E\left (\left .\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right )\right |2\right )}{5 b^2 d \sqrt {b \text {csch}(c+d x)} \sqrt {i \sinh (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 68, normalized size = 0.76 \[ \frac {\sinh (2 (c+d x))-\frac {6 i E\left (\left .\frac {1}{4} (-2 i c-2 i d x+\pi )\right |2\right )}{\sqrt {i \sinh (c+d x)}}}{5 b^2 d \sqrt {b \text {csch}(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Csch[c + d*x])^(-5/2),x]

[Out]

(((-6*I)*EllipticE[((-2*I)*c + Pi - (2*I)*d*x)/4, 2])/Sqrt[I*Sinh[c + d*x]] + Sinh[2*(c + d*x)])/(5*b^2*d*Sqrt
[b*Csch[c + d*x]])

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \operatorname {csch}\left (d x + c\right )}}{b^{3} \operatorname {csch}\left (d x + c\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*csch(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*csch(d*x + c))/(b^3*csch(d*x + c)^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \operatorname {csch}\left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*csch(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*csch(d*x + c))^(-5/2), x)

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maple [F]  time = 0.21, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,\mathrm {csch}\left (d x +c \right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*csch(d*x+c))^(5/2),x)

[Out]

int(1/(b*csch(d*x+c))^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \operatorname {csch}\left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*csch(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*csch(d*x + c))^(-5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b/sinh(c + d*x))^(5/2),x)

[Out]

int(1/(b/sinh(c + d*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \operatorname {csch}{\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*csch(d*x+c))**(5/2),x)

[Out]

Integral((b*csch(c + d*x))**(-5/2), x)

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