\(\int \text {csch}^{\frac {5}{2}}(a+b x) \, dx\) [7]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 10, antiderivative size = 80 \[ \int \text {csch}^{\frac {5}{2}}(a+b x) \, dx=-\frac {2 \cosh (a+b x) \text {csch}^{\frac {3}{2}}(a+b x)}{3 b}+\frac {2 i \sqrt {\text {csch}(a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right ),2\right ) \sqrt {i \sinh (a+b x)}}{3 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3853, 3856, 2720} \[ \int \text {csch}^{\frac {5}{2}}(a+b x) \, dx=-\frac {2 \cosh (a+b x) \text {csch}^{\frac {3}{2}}(a+b x)}{3 b}+\frac {2 i \sqrt {i \sinh (a+b x)} \sqrt {\text {csch}(a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right ),2\right )}{3 b} \]

[In]

Int[Csch[a + b*x]^(5/2),x]

[Out]

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

Rule 2720

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

Rule 3853

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 3856

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*} \text {integral}& = -\frac {2 \cosh (a+b x) \text {csch}^{\frac {3}{2}}(a+b x)}{3 b}-\frac {1}{3} \int \sqrt {\text {csch}(a+b x)} \, dx \\ & = -\frac {2 \cosh (a+b x) \text {csch}^{\frac {3}{2}}(a+b x)}{3 b}-\frac {1}{3} \left (\sqrt {\text {csch}(a+b x)} \sqrt {i \sinh (a+b x)}\right ) \int \frac {1}{\sqrt {i \sinh (a+b x)}} \, dx \\ & = -\frac {2 \cosh (a+b x) \text {csch}^{\frac {3}{2}}(a+b x)}{3 b}+\frac {2 i \sqrt {\text {csch}(a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right ),2\right ) \sqrt {i \sinh (a+b x)}}{3 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.76 \[ \int \text {csch}^{\frac {5}{2}}(a+b x) \, dx=-\frac {2 \sqrt {\text {csch}(a+b x)} \left (\coth (a+b x)+i \operatorname {EllipticF}\left (\frac {1}{4} (-2 i a+\pi -2 i b x),2\right ) \sqrt {i \sinh (a+b x)}\right )}{3 b} \]

[In]

Integrate[Csch[a + b*x]^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.26

method result size
default \(-\frac {i \sqrt {1-i \sinh \left (b x +a \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (b x +a \right )}\, \sqrt {i \sinh \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {1-i \sinh \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right ) \sinh \left (b x +a \right )+2 \cosh \left (b x +a \right )^{2}}{3 \sinh \left (b x +a \right )^{\frac {3}{2}} \cosh \left (b x +a \right ) b}\) \(101\)

[In]

int(csch(b*x+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/sinh(b*x+a)^(3/2)*(I*(1-I*sinh(b*x+a))^(1/2)*2^(1/2)*(1+I*sinh(b*x+a))^(1/2)*(I*sinh(b*x+a))^(1/2)*Ellipt
icF((1-I*sinh(b*x+a))^(1/2),1/2*2^(1/2))*sinh(b*x+a)+2*cosh(b*x+a)^2)/cosh(b*x+a)/b

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.08 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.42 \[ \int \text {csch}^{\frac {5}{2}}(a+b x) \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )} \sqrt {\frac {\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1}} + {\left (\sqrt {2} \cosh \left (b x + a\right )^{2} + 2 \, \sqrt {2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sqrt {2} \sinh \left (b x + a\right )^{2} - \sqrt {2}\right )} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )\right )}}{3 \, {\left (b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2} - b\right )}} \]

[In]

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

[Out]

-2/3*(sqrt(2)*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)*sqrt((cosh(b*x + a) + si
nh(b*x + a))/(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 - 1)) + (sqrt(2)*cosh(b*x + a)
^2 + 2*sqrt(2)*cosh(b*x + a)*sinh(b*x + a) + sqrt(2)*sinh(b*x + a)^2 - sqrt(2))*weierstrassPInverse(4, 0, cosh
(b*x + a) + sinh(b*x + a)))/(b*cosh(b*x + a)^2 + 2*b*cosh(b*x + a)*sinh(b*x + a) + b*sinh(b*x + a)^2 - b)

Sympy [F]

\[ \int \text {csch}^{\frac {5}{2}}(a+b x) \, dx=\int \operatorname {csch}^{\frac {5}{2}}{\left (a + b x \right )}\, dx \]

[In]

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

[Out]

Integral(csch(a + b*x)**(5/2), x)

Maxima [F]

\[ \int \text {csch}^{\frac {5}{2}}(a+b x) \, dx=\int { \operatorname {csch}\left (b x + a\right )^{\frac {5}{2}} \,d x } \]

[In]

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

[Out]

integrate(csch(b*x + a)^(5/2), x)

Giac [F]

\[ \int \text {csch}^{\frac {5}{2}}(a+b x) \, dx=\int { \operatorname {csch}\left (b x + a\right )^{\frac {5}{2}} \,d x } \]

[In]

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

[Out]

integrate(csch(b*x + a)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \text {csch}^{\frac {5}{2}}(a+b x) \, dx=\int {\left (\frac {1}{\mathrm {sinh}\left (a+b\,x\right )}\right )}^{5/2} \,d x \]

[In]

int((1/sinh(a + b*x))^(5/2),x)

[Out]

int((1/sinh(a + b*x))^(5/2), x)