\(\int \frac {1}{(b \sinh (c+d x))^{5/2}} \, dx\) [21]

Optimal result

Integrand size = 12, antiderivative size = 90 \[ \int \frac {1}{(b \sinh (c+d x))^{5/2}} \, dx=-\frac {2 \cosh (c+d x)}{3 b d (b \sinh (c+d x))^{3/2}}+\frac {2 i \operatorname {EllipticF}\left (\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right ),2\right ) \sqrt {i \sinh (c+d x)}}{3 b^2 d \sqrt {b \sinh (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2716, 2721, 2720} \[ \int \frac {1}{(b \sinh (c+d x))^{5/2}} \, dx=-\frac {2 \cosh (c+d x)}{3 b d (b \sinh (c+d x))^{3/2}}+\frac {2 i \sqrt {i \sinh (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right ),2\right )}{3 b^2 d \sqrt {b \sinh (c+d x)}} \]

[In]

Int[(b*Sinh[c + d*x])^(-5/2),x]

[Out]

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

Rule 2716

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 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 2721

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

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.08 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(b \sinh (c+d x))^{5/2}} \, dx=-\frac {2 \left (\coth (c+d x)+\sqrt {2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cosh (2 (c+d x))+\sinh (2 (c+d x))\right ) \sqrt {-\left ((1+\coth (c+d x)) \sinh ^2(c+d x)\right )}\right )}{3 b^2 d \sqrt {b \sinh (c+d x)}} \]

[In]

Integrate[(b*Sinh[c + d*x])^(-5/2),x]

[Out]

(-2*(Coth[c + d*x] + Sqrt[2]*Hypergeometric2F1[1/4, 1/2, 5/4, Cosh[2*(c + d*x)] + Sinh[2*(c + d*x)]]*Sqrt[-((1
+ Coth[c + d*x])*Sinh[c + d*x]^2)]))/(3*b^2*d*Sqrt[b*Sinh[c + d*x]])

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.27

method	result	size
default	\(-\frac {i \sqrt {1-i \sinh \left (d x +c \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) \sinh \left (d x +c \right )+2 \cosh \left (d x +c \right )^{2}}{3 b^{2} \sinh \left (d x +c \right ) \cosh \left (d x +c \right ) \sqrt {b \sinh \left (d x +c \right )}\, d}\)	\(114\)

[In]

int(1/(b*sinh(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/b^2/sinh(d*x+c)*(I*(1-I*sinh(d*x+c))^(1/2)*2^(1/2)*(1+I*sinh(d*x+c))^(1/2)*(I*sinh(d*x+c))^(1/2)*Elliptic
F((1-I*sinh(d*x+c))^(1/2),1/2*2^(1/2))*sinh(d*x+c)+2*cosh(d*x+c)^2)/cosh(d*x+c)/(b*sinh(d*x+c))^(1/2)/d

Fricas [C] (verification not implemented)

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

Time = 0.09 (sec) , antiderivative size = 347, normalized size of antiderivative = 3.86 \[ \int \frac {1}{(b \sinh (c+d x))^{5/2}} \, dx=-\frac {2 \, {\left ({\left (\sqrt {2} \cosh \left (d x + c\right )^{4} + 4 \, \sqrt {2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + \sqrt {2} \sinh \left (d x + c\right )^{4} + 2 \, {\left (3 \, \sqrt {2} \cosh \left (d x + c\right )^{2} - \sqrt {2}\right )} \sinh \left (d x + c\right )^{2} - 2 \, \sqrt {2} \cosh \left (d x + c\right )^{2} + 4 \, {\left (\sqrt {2} \cosh \left (d x + c\right )^{3} - \sqrt {2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) + 2 \, {\left (\cosh \left (d x + c\right )^{3} + 3 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + \sinh \left (d x + c\right )^{3} + {\left (3 \, \cosh \left (d x + c\right )^{2} + 1\right )} \sinh \left (d x + c\right ) + \cosh \left (d x + c\right )\right )} \sqrt {b \sinh \left (d x + c\right )}\right )}}{3 \, {\left (b^{3} d \cosh \left (d x + c\right )^{4} + 4 \, b^{3} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b^{3} d \sinh \left (d x + c\right )^{4} - 2 \, b^{3} d \cosh \left (d x + c\right )^{2} + b^{3} d + 2 \, {\left (3 \, b^{3} d \cosh \left (d x + c\right )^{2} - b^{3} d\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (b^{3} d \cosh \left (d x + c\right )^{3} - b^{3} d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}} \]

[In]

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

[Out]

-2/3*((sqrt(2)*cosh(d*x + c)^4 + 4*sqrt(2)*cosh(d*x + c)*sinh(d*x + c)^3 + sqrt(2)*sinh(d*x + c)^4 + 2*(3*sqrt
(2)*cosh(d*x + c)^2 - sqrt(2))*sinh(d*x + c)^2 - 2*sqrt(2)*cosh(d*x + c)^2 + 4*(sqrt(2)*cosh(d*x + c)^3 - sqrt
(2)*cosh(d*x + c))*sinh(d*x + c) + sqrt(2))*sqrt(b)*weierstrassPInverse(4, 0, cosh(d*x + c) + sinh(d*x + c)) +
2*(cosh(d*x + c)^3 + 3*cosh(d*x + c)*sinh(d*x + c)^2 + sinh(d*x + c)^3 + (3*cosh(d*x + c)^2 + 1)*sinh(d*x + c
) + cosh(d*x + c))*sqrt(b*sinh(d*x + c)))/(b^3*d*cosh(d*x + c)^4 + 4*b^3*d*cosh(d*x + c)*sinh(d*x + c)^3 + b^3
*d*sinh(d*x + c)^4 - 2*b^3*d*cosh(d*x + c)^2 + b^3*d + 2*(3*b^3*d*cosh(d*x + c)^2 - b^3*d)*sinh(d*x + c)^2 + 4
*(b^3*d*cosh(d*x + c)^3 - b^3*d*cosh(d*x + c))*sinh(d*x + c))

Sympy [F]

\[ \int \frac {1}{(b \sinh (c+d x))^{5/2}} \, dx=\int \frac {1}{\left (b \sinh {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{(b \sinh (c+d x))^{5/2}} \, dx=\int { \frac {1}{\left (b \sinh \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {1}{(b \sinh (c+d x))^{5/2}} \, dx=\int { \frac {1}{\left (b \sinh \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(b*sinh(c + d*x))^(5/2),x)

[Out]

int(1/(b*sinh(c + d*x))^(5/2), x)