Integrand size = 10, antiderivative size = 66 \[ \int \text {sech}^{\frac {5}{2}}(a+b x) \, dx=-\frac {2 i \sqrt {\cosh (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} i (a+b x),2\right ) \sqrt {\text {sech}(a+b x)}}{3 b}+\frac {2 \text {sech}^{\frac {3}{2}}(a+b x) \sinh (a+b x)}{3 b} \]
2/3*sech(b*x+a)^(3/2)*sinh(b*x+a)/b-2/3*I*(cosh(1/2*a+1/2*b*x)^2)^(1/2)/co sh(1/2*a+1/2*b*x)*EllipticF(I*sinh(1/2*a+1/2*b*x),2^(1/2))*cosh(b*x+a)^(1/ 2)*sech(b*x+a)^(1/2)/b
Time = 0.10 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.77 \[ \int \text {sech}^{\frac {5}{2}}(a+b x) \, dx=\frac {2 \text {sech}^{\frac {3}{2}}(a+b x) \left (-i \cosh ^{\frac {3}{2}}(a+b x) \operatorname {EllipticF}\left (\frac {1}{2} i (a+b x),2\right )+\sinh (a+b x)\right )}{3 b} \]
(2*Sech[a + b*x]^(3/2)*((-I)*Cosh[a + b*x]^(3/2)*EllipticF[(I/2)*(a + b*x) , 2] + Sinh[a + b*x]))/(3*b)
Time = 0.33 (sec) , antiderivative size = 66, 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.
\(\displaystyle \int \text {sech}^{\frac {5}{2}}(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (i a+i b x+\frac {\pi }{2}\right )^{5/2}dx\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {1}{3} \int \sqrt {\text {sech}(a+b x)}dx+\frac {2 \sinh (a+b x) \text {sech}^{\frac {3}{2}}(a+b x)}{3 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \sinh (a+b x) \text {sech}^{\frac {3}{2}}(a+b x)}{3 b}+\frac {1}{3} \int \sqrt {\csc \left (i a+i b x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {1}{3} \sqrt {\cosh (a+b x)} \sqrt {\text {sech}(a+b x)} \int \frac {1}{\sqrt {\cosh (a+b x)}}dx+\frac {2 \sinh (a+b x) \text {sech}^{\frac {3}{2}}(a+b x)}{3 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \sinh (a+b x) \text {sech}^{\frac {3}{2}}(a+b x)}{3 b}+\frac {1}{3} \sqrt {\cosh (a+b x)} \sqrt {\text {sech}(a+b x)} \int \frac {1}{\sqrt {\sin \left (i a+i b x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 \sinh (a+b x) \text {sech}^{\frac {3}{2}}(a+b x)}{3 b}-\frac {2 i \sqrt {\cosh (a+b x)} \sqrt {\text {sech}(a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} i (a+b x),2\right )}{3 b}\) |
(((-2*I)/3)*Sqrt[Cosh[a + b*x]]*EllipticF[(I/2)*(a + b*x), 2]*Sqrt[Sech[a + b*x]])/b + (2*Sech[a + b*x]^(3/2)*Sinh[a + b*x])/(3*b)
3.1.9.3.1 Defintions of rubi rules used
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]
Leaf count of result is larger than twice the leaf count of optimal. \(216\) vs. \(2(82)=164\).
Time = 1.74 (sec) , antiderivative size = 217, normalized size of antiderivative = 3.29
method | result | size |
default | \(\frac {2 \left (2 \sqrt {-\sinh \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}\, \sqrt {-2 \sinh \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cosh \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \sinh \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+2 \cosh \left (\frac {b x}{2}+\frac {a}{2}\right ) \sinh \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+\sqrt {-\sinh \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}\, \sqrt {-2 \sinh \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cosh \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )\right ) \sqrt {\left (-1+2 \cosh \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}\right ) \sinh \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}}{3 \sqrt {2 \sinh \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}+\sinh \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}\, \left (-1+2 \cosh \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}\right )^{\frac {3}{2}} \sinh \left (\frac {b x}{2}+\frac {a}{2}\right ) b}\) | \(217\) |
2/3*(2*(-sinh(1/2*b*x+1/2*a)^2)^(1/2)*(-2*sinh(1/2*b*x+1/2*a)^2-1)^(1/2)*E llipticF(cosh(1/2*b*x+1/2*a),2^(1/2))*sinh(1/2*b*x+1/2*a)^2+2*cosh(1/2*b*x +1/2*a)*sinh(1/2*b*x+1/2*a)^2+(-sinh(1/2*b*x+1/2*a)^2)^(1/2)*(-2*sinh(1/2* b*x+1/2*a)^2-1)^(1/2)*EllipticF(cosh(1/2*b*x+1/2*a),2^(1/2)))*((-1+2*cosh( 1/2*b*x+1/2*a)^2)*sinh(1/2*b*x+1/2*a)^2)^(1/2)/(2*sinh(1/2*b*x+1/2*a)^4+si nh(1/2*b*x+1/2*a)^2)^(1/2)/(-1+2*cosh(1/2*b*x+1/2*a)^2)^(3/2)/sinh(1/2*b*x +1/2*a)/b
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.88 \[ \int \text {sech}^{\frac {5}{2}}(a+b x) \, dx=\frac {2 \, {\left (\sqrt {2} {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} \sqrt {\frac {\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1}} + {\left (\sqrt {2} \cosh \left (b x + a\right )^{2} + 2 \, \sqrt {2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sqrt {2} \sinh \left (b x + a\right )^{2} + \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )\right )}}{3 \, {\left (b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2} + b\right )}} \]
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)
\[ \int \text {sech}^{\frac {5}{2}}(a+b x) \, dx=\int \operatorname {sech}^{\frac {5}{2}}{\left (a + b x \right )}\, dx \]
\[ \int \text {sech}^{\frac {5}{2}}(a+b x) \, dx=\int { \operatorname {sech}\left (b x + a\right )^{\frac {5}{2}} \,d x } \]
\[ \int \text {sech}^{\frac {5}{2}}(a+b x) \, dx=\int { \operatorname {sech}\left (b x + a\right )^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int \text {sech}^{\frac {5}{2}}(a+b x) \, dx=\int {\left (\frac {1}{\mathrm {cosh}\left (a+b\,x\right )}\right )}^{5/2} \,d x \]