Integrand size = 10, antiderivative size = 80 \[ \int \frac {1}{\sinh ^{\frac {5}{2}}(a+b x)} \, dx=-\frac {2 \cosh (a+b x)}{3 b \sinh ^{\frac {3}{2}}(a+b x)}+\frac {2 i \operatorname {EllipticF}\left (\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right ),2\right ) \sqrt {i \sinh (a+b x)}}{3 b \sqrt {\sinh (a+b x)}} \] Output:
-2/3*cosh(b*x+a)/b/sinh(b*x+a)^(3/2)+2/3*I*InverseJacobiAM(1/2*I*a-1/4*Pi+ 1/2*I*b*x,2^(1/2))*(I*sinh(b*x+a))^(1/2)/b/sinh(b*x+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.06 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\sinh ^{\frac {5}{2}}(a+b x)} \, dx=-\frac {2 \left (\cosh (a+b x)+\operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cosh (2 (a+b x))+\sinh (2 (a+b x))\right ) \sinh (a+b x) \sqrt {1-\cosh (2 a+2 b x)-\sinh (2 a+2 b x)}\right )}{3 b \sinh ^{\frac {3}{2}}(a+b x)} \] Input:
Integrate[Sinh[a + b*x]^(-5/2),x]
Output:
(-2*(Cosh[a + b*x] + Hypergeometric2F1[1/4, 1/2, 5/4, Cosh[2*(a + b*x)] + Sinh[2*(a + b*x)]]*Sinh[a + b*x]*Sqrt[1 - Cosh[2*a + 2*b*x] - Sinh[2*a + 2 *b*x]]))/(3*b*Sinh[a + b*x]^(3/2))
Time = 0.31 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, 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.
| \(\displaystyle \int \frac {1}{\sinh ^{\frac {5}{2}}(a+b x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(-i \sin (i a+i b x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle -\frac {1}{3} \int \frac {1}{\sqrt {\sinh (a+b x)}}dx-\frac {2 \cosh (a+b x)}{3 b \sinh ^{\frac {3}{2}}(a+b x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 \cosh (a+b x)}{3 b \sinh ^{\frac {3}{2}}(a+b x)}-\frac {1}{3} \int \frac {1}{\sqrt {-i \sin (i a+i b x)}}dx\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle -\frac {2 \cosh (a+b x)}{3 b \sinh ^{\frac {3}{2}}(a+b x)}-\frac {\sqrt {i \sinh (a+b x)} \int \frac {1}{\sqrt {i \sinh (a+b x)}}dx}{3 \sqrt {\sinh (a+b x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 \cosh (a+b x)}{3 b \sinh ^{\frac {3}{2}}(a+b x)}-\frac {\sqrt {i \sinh (a+b x)} \int \frac {1}{\sqrt {\sin (i a+i b x)}}dx}{3 \sqrt {\sinh (a+b x)}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {2 \cosh (a+b x)}{3 b \sinh ^{\frac {3}{2}}(a+b x)}+\frac {2 i \sqrt {i \sinh (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right ),2\right )}{3 b \sqrt {\sinh (a+b x)}}\) |
Input:
Int[Sinh[a + b*x]^(-5/2),x]
Output:
(-2*Cosh[a + b*x])/(3*b*Sinh[a + b*x]^(3/2)) + (((2*I)/3)*EllipticF[(I*a - Pi/2 + I*b*x)/2, 2]*Sqrt[I*Sinh[a + b*x]])/(b*Sqrt[Sinh[a + b*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 = 0.15 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.26
| method | result | size |
| default | \(-\frac {i \sqrt {1-i \sinh \left (b x +a \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (b x +a \right )}\, \sqrt {i \sinh \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {1-i \sinh \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right ) \sinh \left (b x +a \right )+2 \cosh \left (b x +a \right )^{2}}{3 \sinh \left (b x +a \right )^{\frac {3}{2}} \cosh \left (b x +a \right ) b}\) | \(101\) |
Input:
int(1/sinh(b*x+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
-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)*EllipticF((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
Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (58) = 116\).
Time = 0.09 (sec) , antiderivative size = 314, normalized size of antiderivative = 3.92 \[ \int \frac {1}{\sinh ^{\frac {5}{2}}(a+b x)} \, dx=-\frac {2 \, {\left ({\left (\sqrt {2} \cosh \left (b x + a\right )^{4} + 4 \, \sqrt {2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sqrt {2} \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \sqrt {2} \cosh \left (b x + a\right )^{2} - \sqrt {2}\right )} \sinh \left (b x + a\right )^{2} - 2 \, \sqrt {2} \cosh \left (b x + a\right )^{2} + 4 \, {\left (\sqrt {2} \cosh \left (b x + a\right )^{3} - \sqrt {2} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + \sqrt {2}\right )} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 2 \, {\left (\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3} + {\left (3 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right ) + \cosh \left (b x + a\right )\right )} \sqrt {\sinh \left (b x + a\right )}\right )}}{3 \, {\left (b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} - 2 \, b \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b \cosh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b\right )}} \] Input:
integrate(1/sinh(b*x+a)^(5/2),x, algorithm="fricas")
Output:
-2/3*((sqrt(2)*cosh(b*x + a)^4 + 4*sqrt(2)*cosh(b*x + a)*sinh(b*x + a)^3 + sqrt(2)*sinh(b*x + a)^4 + 2*(3*sqrt(2)*cosh(b*x + a)^2 - sqrt(2))*sinh(b* x + a)^2 - 2*sqrt(2)*cosh(b*x + a)^2 + 4*(sqrt(2)*cosh(b*x + a)^3 - sqrt(2 )*cosh(b*x + a))*sinh(b*x + a) + sqrt(2))*weierstrassPInverse(4, 0, cosh(b *x + a) + sinh(b*x + a)) + 2*(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 + sinh(b*x + a)^3 + (3*cosh(b*x + a)^2 + 1)*sinh(b*x + a) + cosh(b*x + a))*sqrt(sinh(b*x + a)))/(b*cosh(b*x + a)^4 + 4*b*cosh(b*x + a)*sinh(b* x + a)^3 + b*sinh(b*x + a)^4 - 2*b*cosh(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^ 2 - b)*sinh(b*x + a)^2 + 4*(b*cosh(b*x + a)^3 - b*cosh(b*x + a))*sinh(b*x + a) + b)
\[ \int \frac {1}{\sinh ^{\frac {5}{2}}(a+b x)} \, dx=\int \frac {1}{\sinh ^{\frac {5}{2}}{\left (a + b x \right )}}\, dx \] Input:
integrate(1/sinh(b*x+a)**(5/2),x)
Output:
Integral(sinh(a + b*x)**(-5/2), x)
\[ \int \frac {1}{\sinh ^{\frac {5}{2}}(a+b x)} \, dx=\int { \frac {1}{\sinh \left (b x + a\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/sinh(b*x+a)^(5/2),x, algorithm="maxima")
Output:
integrate(sinh(b*x + a)^(-5/2), x)
\[ \int \frac {1}{\sinh ^{\frac {5}{2}}(a+b x)} \, dx=\int { \frac {1}{\sinh \left (b x + a\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/sinh(b*x+a)^(5/2),x, algorithm="giac")
Output:
integrate(sinh(b*x + a)^(-5/2), x)
Timed out. \[ \int \frac {1}{\sinh ^{\frac {5}{2}}(a+b x)} \, dx=\int \frac {1}{{\mathrm {sinh}\left (a+b\,x\right )}^{5/2}} \,d x \] Input:
int(1/sinh(a + b*x)^(5/2),x)
Output:
int(1/sinh(a + b*x)^(5/2), x)
\[ \int \frac {1}{\sinh ^{\frac {5}{2}}(a+b x)} \, dx=\int \frac {\sqrt {\sinh \left (b x +a \right )}}{\sinh \left (b x +a \right )^{3}}d x \] Input:
int(1/sinh(b*x+a)^(5/2),x)
Output:
int(sqrt(sinh(a + b*x))/sinh(a + b*x)**3,x)