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

Optimal. Leaf size=88 \[ -\frac {2 b \cosh (c+d x) (b \text {csch}(c+d x))^{3/2}}{3 d}+\frac {2 i b^2 \sqrt {i \sinh (c+d x)} F\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \text {csch}(c+d x)}}{3 d} \]

[Out]

-2/3*b*cosh(d*x+c)*(b*csch(d*x+c))^(3/2)/d-2/3*I*b^2*(sin(1/2*I*c+1/4*Pi+1/2*I*d*x)^2)^(1/2)/sin(1/2*I*c+1/4*P
i+1/2*I*d*x)*EllipticF(cos(1/2*I*c+1/4*Pi+1/2*I*d*x),2^(1/2))*(b*csch(d*x+c))^(1/2)*(I*sinh(d*x+c))^(1/2)/d

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 88, 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 = {3768, 3771, 2641} \[ -\frac {2 b \cosh (c+d x) (b \text {csch}(c+d x))^{3/2}}{3 d}+\frac {2 i b^2 \sqrt {i \sinh (c+d x)} F\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \text {csch}(c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Csch[c + d*x])^(5/2),x]

[Out]

(-2*b*Cosh[c + d*x]*(b*Csch[c + d*x])^(3/2))/(3*d) + (((2*I)/3)*b^2*Sqrt[b*Csch[c + d*x]]*EllipticF[(I*c - Pi/
2 + I*d*x)/2, 2]*Sqrt[I*Sinh[c + d*x]])/d

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
 1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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 (b \text {csch}(c+d x))^{5/2} \, dx &=-\frac {2 b \cosh (c+d x) (b \text {csch}(c+d x))^{3/2}}{3 d}-\frac {1}{3} b^2 \int \sqrt {b \text {csch}(c+d x)} \, dx\\ &=-\frac {2 b \cosh (c+d x) (b \text {csch}(c+d x))^{3/2}}{3 d}-\frac {1}{3} \left (b^2 \sqrt {b \text {csch}(c+d x)} \sqrt {i \sinh (c+d x)}\right ) \int \frac {1}{\sqrt {i \sinh (c+d x)}} \, dx\\ &=-\frac {2 b \cosh (c+d x) (b \text {csch}(c+d x))^{3/2}}{3 d}+\frac {2 i b^2 \sqrt {b \text {csch}(c+d x)} F\left (\left .\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right )\right |2\right ) \sqrt {i \sinh (c+d x)}}{3 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 66, normalized size = 0.75 \[ -\frac {2 b^2 \sqrt {b \text {csch}(c+d x)} \left (\coth (c+d x)+i \sqrt {i \sinh (c+d x)} F\left (\left .\frac {1}{4} (-2 i c-2 i d x+\pi )\right |2\right )\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Csch[c + d*x])^(5/2),x]

[Out]

(-2*b^2*Sqrt[b*Csch[c + d*x]]*(Coth[c + d*x] + I*EllipticF[((-2*I)*c + Pi - (2*I)*d*x)/4, 2]*Sqrt[I*Sinh[c + d
*x]]))/(3*d)

________________________________________________________________________________________

fricas [F]  time = 0.93, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \operatorname {csch}\left (d x + c\right )} b^{2} \operatorname {csch}\left (d x + c\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*csch(d*x + c))*b^2*csch(d*x + c)^2, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*csch(d*x + c))^(5/2), x)

________________________________________________________________________________________

maple [F]  time = 0.29, size = 0, normalized size = 0.00 \[ \int \left (b \,\mathrm {csch}\left (d x +c \right )\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*csch(d*x + c))^(5/2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/sinh(c + d*x))^(5/2),x)

[Out]

int((b/sinh(c + d*x))^(5/2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*csch(c + d*x))**(5/2), x)

________________________________________________________________________________________