Integrand size = 12, antiderivative size = 90 \[ \int \frac {1}{(b \sinh (c+d x))^{5/2}} \, dx=-\frac {2 \cosh (c+d x)}{3 b d (b \sinh (c+d x))^{3/2}}+\frac {2 i \operatorname {EllipticF}\left (\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right ),2\right ) \sqrt {i \sinh (c+d x)}}{3 b^2 d \sqrt {b \sinh (c+d x)}} \] Output:
-2/3*cosh(d*x+c)/b/d/(b*sinh(d*x+c))^(3/2)+2/3*I*InverseJacobiAM(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)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.07 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(b \sinh (c+d x))^{5/2}} \, dx=-\frac {2 \left (\coth (c+d x)+\sqrt {2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cosh (2 (c+d x))+\sinh (2 (c+d x))\right ) \sqrt {-\left ((1+\coth (c+d x)) \sinh ^2(c+d x)\right )}\right )}{3 b^2 d \sqrt {b \sinh (c+d x)}} \] Input:
Integrate[(b*Sinh[c + d*x])^(-5/2),x]
Output:
(-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]])
Time = 0.35 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3116, 3042, 3121, 3042, 3120}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{(b \sinh (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(-i b \sin (i c+i d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle -\frac {\int \frac {1}{\sqrt {b \sinh (c+d x)}}dx}{3 b^2}-\frac {2 \cosh (c+d x)}{3 b d (b \sinh (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 \cosh (c+d x)}{3 b d (b \sinh (c+d x))^{3/2}}-\frac {\int \frac {1}{\sqrt {-i b \sin (i c+i d x)}}dx}{3 b^2}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle -\frac {2 \cosh (c+d x)}{3 b d (b \sinh (c+d x))^{3/2}}-\frac {\sqrt {i \sinh (c+d x)} \int \frac {1}{\sqrt {i \sinh (c+d x)}}dx}{3 b^2 \sqrt {b \sinh (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 \cosh (c+d x)}{3 b d (b \sinh (c+d x))^{3/2}}-\frac {\sqrt {i \sinh (c+d x)} \int \frac {1}{\sqrt {\sin (i c+i d x)}}dx}{3 b^2 \sqrt {b \sinh (c+d x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle -\frac {2 \cosh (c+d x)}{3 b d (b \sinh (c+d x))^{3/2}}+\frac {2 i \sqrt {i \sinh (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right ),2\right )}{3 b^2 d \sqrt {b \sinh (c+d x)}}\) |
Input:
Int[(b*Sinh[c + d*x])^(-5/2),x]
Output:
(-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*Sinh[c + d*x]])/(b^2*d*Sqrt[b*Sinh[c + d *x]])
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] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
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]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Time = 0.23 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.27
method | result | size |
default | \(-\frac {i \sqrt {-i \sinh \left (d x +c \right )+1}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-i \sinh \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \sinh \left (d x +c \right )+2 \cosh \left (d x +c \right )^{2}}{3 b^{2} \sinh \left (d x +c \right ) \cosh \left (d x +c \right ) \sqrt {b \sinh \left (d x +c \right )}\, d}\) | \(114\) |
Input:
int(1/(b*sinh(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/3/b^2/sinh(d*x+c)*(I*(-I*sinh(d*x+c)+1)^(1/2)*2^(1/2)*(1+I*sinh(d*x+c)) ^(1/2)*(I*sinh(d*x+c))^(1/2)*EllipticF((-I*sinh(d*x+c)+1)^(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
Leaf count of result is larger than twice the leaf count of optimal. 319 vs. \(2 (68) = 136\).
Time = 0.09 (sec) , antiderivative size = 319, normalized size of antiderivative = 3.54 \[ \int \frac {1}{(b \sinh (c+d x))^{5/2}} \, dx=-\frac {4 \, {\left (\sqrt {\frac {1}{2}} {\left (\cosh \left (d x + c\right )^{4} + 4 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + \sinh \left (d x + c\right )^{4} + 2 \, {\left (3 \, \cosh \left (d x + c\right )^{2} - 1\right )} \sinh \left (d x + c\right )^{2} - 2 \, \cosh \left (d x + c\right )^{2} + 4 \, {\left (\cosh \left (d x + c\right )^{3} - \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 1\right )} \sqrt {b} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) + {\left (\cosh \left (d x + c\right )^{3} + 3 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + \sinh \left (d x + c\right )^{3} + {\left (3 \, \cosh \left (d x + c\right )^{2} + 1\right )} \sinh \left (d x + c\right ) + \cosh \left (d x + c\right )\right )} \sqrt {b \sinh \left (d x + c\right )}\right )}}{3 \, {\left (b^{3} d \cosh \left (d x + c\right )^{4} + 4 \, b^{3} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b^{3} d \sinh \left (d x + c\right )^{4} - 2 \, b^{3} d \cosh \left (d x + c\right )^{2} + b^{3} d + 2 \, {\left (3 \, b^{3} d \cosh \left (d x + c\right )^{2} - b^{3} d\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (b^{3} d \cosh \left (d x + c\right )^{3} - b^{3} d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}} \] Input:
integrate(1/(b*sinh(d*x+c))^(5/2),x, algorithm="fricas")
Output:
-4/3*(sqrt(1/2)*(cosh(d*x + c)^4 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh( d*x + c)^4 + 2*(3*cosh(d*x + c)^2 - 1)*sinh(d*x + c)^2 - 2*cosh(d*x + c)^2 + 4*(cosh(d*x + c)^3 - cosh(d*x + c))*sinh(d*x + c) + 1)*sqrt(b)*weierstr assPInverse(4, 0, cosh(d*x + c) + sinh(d*x + c)) + (cosh(d*x + c)^3 + 3*co sh(d*x + c)*sinh(d*x + c)^2 + sinh(d*x + c)^3 + (3*cosh(d*x + c)^2 + 1)*si nh(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))
\[ \int \frac {1}{(b \sinh (c+d x))^{5/2}} \, dx=\int \frac {1}{\left (b \sinh {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/(b*sinh(d*x+c))**(5/2),x)
Output:
Integral((b*sinh(c + d*x))**(-5/2), x)
\[ \int \frac {1}{(b \sinh (c+d x))^{5/2}} \, dx=\int { \frac {1}{\left (b \sinh \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(b*sinh(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate((b*sinh(d*x + c))^(-5/2), x)
\[ \int \frac {1}{(b \sinh (c+d x))^{5/2}} \, dx=\int { \frac {1}{\left (b \sinh \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(b*sinh(d*x+c))^(5/2),x, algorithm="giac")
Output:
integrate((b*sinh(d*x + c))^(-5/2), x)
Timed out. \[ \int \frac {1}{(b \sinh (c+d x))^{5/2}} \, dx=\int \frac {1}{{\left (b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:
int(1/(b*sinh(c + d*x))^(5/2),x)
Output:
int(1/(b*sinh(c + d*x))^(5/2), x)
\[ \int \frac {1}{(b \sinh (c+d x))^{5/2}} \, dx=\frac {\sqrt {b}\, \left (\int \frac {\sqrt {\sinh \left (d x +c \right )}}{\sinh \left (d x +c \right )^{3}}d x \right )}{b^{3}} \] Input:
int(1/(b*sinh(d*x+c))^(5/2),x)
Output:
(sqrt(b)*int(sqrt(sinh(c + d*x))/sinh(c + d*x)**3,x))/b**3