Integrand size = 10, antiderivative size = 46 \[ \int \frac {1}{\cosh ^{\frac {5}{2}}(a+b x)} \, dx=-\frac {2 i \operatorname {EllipticF}\left (\frac {1}{2} i (a+b x),2\right )}{3 b}+\frac {2 \sinh (a+b x)}{3 b \cosh ^{\frac {3}{2}}(a+b x)} \] Output:
-2/3*I*InverseJacobiAM(1/2*I*(b*x+a),2^(1/2))/b+2/3*sinh(b*x+a)/b/cosh(b*x +a)^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.05 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.83 \[ \int \frac {1}{\cosh ^{\frac {5}{2}}(a+b x)} \, dx=\frac {2 \left (\sinh (a+b x)+\cosh (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\cosh (2 (a+b x))-\sinh (2 (a+b x))\right ) \sqrt {1+\cosh (2 (a+b x))+\sinh (2 (a+b x))}\right )}{3 b \cosh ^{\frac {3}{2}}(a+b x)} \] Input:
Integrate[Cosh[a + b*x]^(-5/2),x]
Output:
(2*(Sinh[a + b*x] + Cosh[a + b*x]*Hypergeometric2F1[1/4, 1/2, 5/4, -Cosh[2 *(a + b*x)] - Sinh[2*(a + b*x)]]*Sqrt[1 + Cosh[2*(a + b*x)] + Sinh[2*(a + b*x)]]))/(3*b*Cosh[a + b*x]^(3/2))
Time = 0.25 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3116, 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.
\(\displaystyle \int \frac {1}{\cosh ^{\frac {5}{2}}(a+b x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin \left (i a+i b x+\frac {\pi }{2}\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \frac {1}{3} \int \frac {1}{\sqrt {\cosh (a+b x)}}dx+\frac {2 \sinh (a+b x)}{3 b \cosh ^{\frac {3}{2}}(a+b x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \sinh (a+b x)}{3 b \cosh ^{\frac {3}{2}}(a+b x)}+\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (i a+i b x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 \sinh (a+b x)}{3 b \cosh ^{\frac {3}{2}}(a+b x)}-\frac {2 i \operatorname {EllipticF}\left (\frac {1}{2} i (a+b x),2\right )}{3 b}\) |
Input:
Int[Cosh[a + b*x]^(-5/2),x]
Output:
(((-2*I)/3)*EllipticF[(I/2)*(a + b*x), 2])/b + (2*Sinh[a + b*x])/(3*b*Cosh [a + b*x]^(3/2))
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]
Leaf count of result is larger than twice the leaf count of optimal. \(189\) vs. \(2(38)=76\).
Time = 1.13 (sec) , antiderivative size = 190, normalized size of antiderivative = 4.13
method | result | size |
default | \(\frac {\sqrt {\left (2 \cosh \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right ) \sinh \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}\, \left (\frac {\cosh \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {2 \sinh \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}+\sinh \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}}{3 \left (\cosh \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-\frac {1}{2}\right )^{2}}+\frac {2 \sqrt {-\sinh \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}\, \sqrt {-2 \cosh \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cosh \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )}{3 \sqrt {2 \sinh \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}+\sinh \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}}\right )}{\sinh \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {2 \cosh \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1}\, b}\) | \(190\) |
Input:
int(1/cosh(b*x+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
((2*cosh(1/2*b*x+1/2*a)^2-1)*sinh(1/2*b*x+1/2*a)^2)^(1/2)*(1/3*cosh(1/2*b* x+1/2*a)*(2*sinh(1/2*b*x+1/2*a)^4+sinh(1/2*b*x+1/2*a)^2)^(1/2)/(cosh(1/2*b *x+1/2*a)^2-1/2)^2+2/3*(-sinh(1/2*b*x+1/2*a)^2)^(1/2)*(-2*cosh(1/2*b*x+1/2 *a)^2+1)^(1/2)/(2*sinh(1/2*b*x+1/2*a)^4+sinh(1/2*b*x+1/2*a)^2)^(1/2)*Ellip ticF(cosh(1/2*b*x+1/2*a),2^(1/2)))/sinh(1/2*b*x+1/2*a)/(2*cosh(1/2*b*x+1/2 *a)^2-1)^(1/2)/b
Leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (37) = 74\).
Time = 0.10 (sec) , antiderivative size = 310, normalized size of antiderivative = 6.74 \[ \int \frac {1}{\cosh ^{\frac {5}{2}}(a+b x)} \, dx=\frac {2 \, {\left ({\left (\sqrt {2} \cosh \left (b x + a\right )^{4} + 4 \, \sqrt {2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sqrt {2} \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \sqrt {2} \cosh \left (b x + a\right )^{2} + \sqrt {2}\right )} \sinh \left (b x + a\right )^{2} + 2 \, \sqrt {2} \cosh \left (b x + a\right )^{2} + 4 \, {\left (\sqrt {2} \cosh \left (b x + a\right )^{3} + \sqrt {2} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 2 \, {\left (\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3} + {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) - \cosh \left (b x + a\right )\right )} \sqrt {\cosh \left (b x + a\right )}\right )}}{3 \, {\left (b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} + 2 \, b \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b \cosh \left (b x + a\right )^{3} + b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b\right )}} \] Input:
integrate(1/cosh(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(cosh(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)
\[ \int \frac {1}{\cosh ^{\frac {5}{2}}(a+b x)} \, dx=\int \frac {1}{\cosh ^{\frac {5}{2}}{\left (a + b x \right )}}\, dx \] Input:
integrate(1/cosh(b*x+a)**(5/2),x)
Output:
Integral(cosh(a + b*x)**(-5/2), x)
\[ \int \frac {1}{\cosh ^{\frac {5}{2}}(a+b x)} \, dx=\int { \frac {1}{\cosh \left (b x + a\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/cosh(b*x+a)^(5/2),x, algorithm="maxima")
Output:
integrate(cosh(b*x + a)^(-5/2), x)
\[ \int \frac {1}{\cosh ^{\frac {5}{2}}(a+b x)} \, dx=\int { \frac {1}{\cosh \left (b x + a\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/cosh(b*x+a)^(5/2),x, algorithm="giac")
Output:
integrate(cosh(b*x + a)^(-5/2), x)
Timed out. \[ \int \frac {1}{\cosh ^{\frac {5}{2}}(a+b x)} \, dx=\int \frac {1}{{\mathrm {cosh}\left (a+b\,x\right )}^{5/2}} \,d x \] Input:
int(1/cosh(a + b*x)^(5/2),x)
Output:
int(1/cosh(a + b*x)^(5/2), x)
\[ \int \frac {1}{\cosh ^{\frac {5}{2}}(a+b x)} \, dx=\int \frac {\sqrt {\cosh \left (b x +a \right )}}{\cosh \left (b x +a \right )^{3}}d x \] Input:
int(1/cosh(b*x+a)^(5/2),x)
Output:
int(sqrt(cosh(a + b*x))/cosh(a + b*x)**3,x)