Integrand size = 12, antiderivative size = 74 \[ \int (b \text {sech}(c+d x))^{5/2} \, dx=-\frac {2 i b^2 \sqrt {\cosh (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} i (c+d x),2\right ) \sqrt {b \text {sech}(c+d x)}}{3 d}+\frac {2 b (b \text {sech}(c+d x))^{3/2} \sinh (c+d x)}{3 d} \] Output:
-2/3*I*b^2*cosh(d*x+c)^(1/2)*InverseJacobiAM(1/2*I*(d*x+c),2^(1/2))*(b*sec h(d*x+c))^(1/2)/d+2/3*b*(b*sech(d*x+c))^(3/2)*sinh(d*x+c)/d
Time = 0.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.76 \[ \int (b \text {sech}(c+d x))^{5/2} \, dx=\frac {2 b^2 \sqrt {b \text {sech}(c+d x)} \left (-i \sqrt {\cosh (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} i (c+d x),2\right )+\tanh (c+d x)\right )}{3 d} \] Input:
Integrate[(b*Sech[c + d*x])^(5/2),x]
Output:
(2*b^2*Sqrt[b*Sech[c + d*x]]*((-I)*Sqrt[Cosh[c + d*x]]*EllipticF[(I/2)*(c + d*x), 2] + Tanh[c + d*x]))/(3*d)
Time = 0.36 (sec) , antiderivative size = 74, 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, 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.
\(\displaystyle \int (b \text {sech}(c+d x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (b \csc \left (i c+i d x+\frac {\pi }{2}\right )\right )^{5/2}dx\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {1}{3} b^2 \int \sqrt {b \text {sech}(c+d x)}dx+\frac {2 b \sinh (c+d x) (b \text {sech}(c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b \sinh (c+d x) (b \text {sech}(c+d x))^{3/2}}{3 d}+\frac {1}{3} b^2 \int \sqrt {b \csc \left (i c+i d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {1}{3} b^2 \sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)} \int \frac {1}{\sqrt {\cosh (c+d x)}}dx+\frac {2 b \sinh (c+d x) (b \text {sech}(c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b \sinh (c+d x) (b \text {sech}(c+d x))^{3/2}}{3 d}+\frac {1}{3} b^2 \sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)} \int \frac {1}{\sqrt {\sin \left (i c+i d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 b \sinh (c+d x) (b \text {sech}(c+d x))^{3/2}}{3 d}-\frac {2 i b^2 \sqrt {\cosh (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} i (c+d x),2\right ) \sqrt {b \text {sech}(c+d x)}}{3 d}\) |
Input:
Int[(b*Sech[c + d*x])^(5/2),x]
Output:
(((-2*I)/3)*b^2*Sqrt[Cosh[c + d*x]]*EllipticF[(I/2)*(c + d*x), 2]*Sqrt[b*S ech[c + d*x]])/d + (2*b*(b*Sech[c + d*x])^(3/2)*Sinh[c + d*x])/(3*d)
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]
\[\int \left (b \,\operatorname {sech}\left (d x +c \right )\right )^{\frac {5}{2}}d x\]
Input:
int((b*sech(d*x+c))^(5/2),x)
Output:
int((b*sech(d*x+c))^(5/2),x)
Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (61) = 122\).
Time = 0.08 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.91 \[ \int (b \text {sech}(c+d x))^{5/2} \, dx=\frac {2 \, {\left (\sqrt {2} {\left (b^{2} \cosh \left (d x + c\right )^{2} + 2 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{2} \sinh \left (d x + c\right )^{2} + b^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) + \sqrt {2} {\left (b^{2} \cosh \left (d x + c\right )^{2} + 2 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{2} \sinh \left (d x + c\right )^{2} - b^{2}\right )} \sqrt {\frac {b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} + 1}}\right )}}{3 \, {\left (d \cosh \left (d x + c\right )^{2} + 2 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + d \sinh \left (d x + c\right )^{2} + d\right )}} \] Input:
integrate((b*sech(d*x+c))^(5/2),x, algorithm="fricas")
Output:
2/3*(sqrt(2)*(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh(d*x + c) + b^ 2*sinh(d*x + c)^2 + b^2)*sqrt(b)*weierstrassPInverse(-4, 0, cosh(d*x + c) + sinh(d*x + c)) + sqrt(2)*(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh (d*x + c) + b^2*sinh(d*x + c)^2 - b^2)*sqrt((b*cosh(d*x + c) + b*sinh(d*x + c))/(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 + 1)))/(d*cosh(d*x + c)^2 + 2*d*cosh(d*x + c)*sinh(d*x + c) + d*sinh(d*x + c)^2 + d)
\[ \int (b \text {sech}(c+d x))^{5/2} \, dx=\int \left (b \operatorname {sech}{\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \] Input:
integrate((b*sech(d*x+c))**(5/2),x)
Output:
Integral((b*sech(c + d*x))**(5/2), x)
\[ \int (b \text {sech}(c+d x))^{5/2} \, dx=\int { \left (b \operatorname {sech}\left (d x + c\right )\right )^{\frac {5}{2}} \,d x } \] Input:
integrate((b*sech(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate((b*sech(d*x + c))^(5/2), x)
\[ \int (b \text {sech}(c+d x))^{5/2} \, dx=\int { \left (b \operatorname {sech}\left (d x + c\right )\right )^{\frac {5}{2}} \,d x } \] Input:
integrate((b*sech(d*x+c))^(5/2),x, algorithm="giac")
Output:
integrate((b*sech(d*x + c))^(5/2), x)
Timed out. \[ \int (b \text {sech}(c+d x))^{5/2} \, dx=\int {\left (\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}\right )}^{5/2} \,d x \] Input:
int((b/cosh(c + d*x))^(5/2),x)
Output:
int((b/cosh(c + d*x))^(5/2), x)
\[ \int (b \text {sech}(c+d x))^{5/2} \, dx=\sqrt {b}\, \left (\int \sqrt {\mathrm {sech}\left (d x +c \right )}\, \mathrm {sech}\left (d x +c \right )^{2}d x \right ) b^{2} \] Input:
int((b*sech(d*x+c))^(5/2),x)
Output:
sqrt(b)*int(sqrt(sech(c + d*x))*sech(c + d*x)**2,x)*b**2