Integrand size = 10, antiderivative size = 80 \[ \int \text {csch}^{\frac {5}{2}}(a+b x) \, dx=-\frac {2 \cosh (a+b x) \text {csch}^{\frac {3}{2}}(a+b x)}{3 b}+\frac {2 i \sqrt {\text {csch}(a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right ),2\right ) \sqrt {i \sinh (a+b x)}}{3 b} \] Output:
-2/3*cosh(b*x+a)*csch(b*x+a)^(3/2)/b+2/3*I*csch(b*x+a)^(1/2)*InverseJacobi AM(1/2*I*a-1/4*Pi+1/2*I*b*x,2^(1/2))*(I*sinh(b*x+a))^(1/2)/b
Time = 0.09 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.76 \[ \int \text {csch}^{\frac {5}{2}}(a+b x) \, dx=-\frac {2 \sqrt {\text {csch}(a+b x)} \left (\coth (a+b x)+i \operatorname {EllipticF}\left (\frac {1}{4} (-2 i a+\pi -2 i b x),2\right ) \sqrt {i \sinh (a+b x)}\right )}{3 b} \] Input:
Integrate[Csch[a + b*x]^(5/2),x]
Output:
(-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)
Time = 0.34 (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, 4255, 3042, 4258, 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.
\(\displaystyle \int \text {csch}^{\frac {5}{2}}(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (i \csc (i a+i b x))^{5/2}dx\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle -\frac {1}{3} \int \sqrt {\text {csch}(a+b x)}dx-\frac {2 \cosh (a+b x) \text {csch}^{\frac {3}{2}}(a+b x)}{3 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 \cosh (a+b x) \text {csch}^{\frac {3}{2}}(a+b x)}{3 b}-\frac {1}{3} \int \sqrt {i \csc (i a+i b x)}dx\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle -\frac {2 \cosh (a+b x) \text {csch}^{\frac {3}{2}}(a+b x)}{3 b}-\frac {1}{3} \sqrt {i \sinh (a+b x)} \sqrt {\text {csch}(a+b x)} \int \frac {1}{\sqrt {i \sinh (a+b x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 \cosh (a+b x) \text {csch}^{\frac {3}{2}}(a+b x)}{3 b}-\frac {1}{3} \sqrt {i \sinh (a+b x)} \sqrt {\text {csch}(a+b x)} \int \frac {1}{\sqrt {\sin (i a+i b x)}}dx\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle -\frac {2 \cosh (a+b x) \text {csch}^{\frac {3}{2}}(a+b x)}{3 b}+\frac {2 i \sqrt {i \sinh (a+b x)} \sqrt {\text {csch}(a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right ),2\right )}{3 b}\) |
Input:
Int[Csch[a + b*x]^(5/2),x]
Output:
(-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
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[(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] + Simp[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]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(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]
Time = 0.22 (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(csch(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. 194 vs. \(2 (58) = 116\).
Time = 0.09 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.42 \[ \int \text {csch}^{\frac {5}{2}}(a+b x) \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )} \sqrt {\frac {\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1}} + {\left (\sqrt {2} \cosh \left (b x + a\right )^{2} + 2 \, \sqrt {2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sqrt {2} \sinh \left (b x + a\right )^{2} - \sqrt {2}\right )} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )\right )}}{3 \, {\left (b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2} - b\right )}} \] Input:
integrate(csch(b*x+a)^(5/2),x, algorithm="fricas")
Output:
-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) + sinh(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)
\[ \int \text {csch}^{\frac {5}{2}}(a+b x) \, dx=\int \operatorname {csch}^{\frac {5}{2}}{\left (a + b x \right )}\, dx \] Input:
integrate(csch(b*x+a)**(5/2),x)
Output:
Integral(csch(a + b*x)**(5/2), x)
\[ \int \text {csch}^{\frac {5}{2}}(a+b x) \, dx=\int { \operatorname {csch}\left (b x + a\right )^{\frac {5}{2}} \,d x } \] Input:
integrate(csch(b*x+a)^(5/2),x, algorithm="maxima")
Output:
integrate(csch(b*x + a)^(5/2), x)
\[ \int \text {csch}^{\frac {5}{2}}(a+b x) \, dx=\int { \operatorname {csch}\left (b x + a\right )^{\frac {5}{2}} \,d x } \] Input:
integrate(csch(b*x+a)^(5/2),x, algorithm="giac")
Output:
integrate(csch(b*x + a)^(5/2), x)
Timed out. \[ \int \text {csch}^{\frac {5}{2}}(a+b x) \, dx=\int {\left (\frac {1}{\mathrm {sinh}\left (a+b\,x\right )}\right )}^{5/2} \,d x \] Input:
int((1/sinh(a + b*x))^(5/2),x)
Output:
int((1/sinh(a + b*x))^(5/2), x)
\[ \int \text {csch}^{\frac {5}{2}}(a+b x) \, dx=\int \sqrt {\mathrm {csch}\left (b x +a \right )}\, \mathrm {csch}\left (b x +a \right )^{2}d x \] Input:
int(csch(b*x+a)^(5/2),x)
Output:
int(sqrt(csch(a + b*x))*csch(a + b*x)**2,x)