Integrand size = 12, antiderivative size = 70 \[ \int (b \text {sech}(c+d x))^{3/2} \, dx=\frac {2 i b^2 E\left (\left .\frac {1}{2} i (c+d x)\right |2\right )}{d \sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}}+\frac {2 b \sqrt {b \text {sech}(c+d x)} \sinh (c+d x)}{d} \] Output:
2*I*b^2*EllipticE(I*sinh(1/2*d*x+1/2*c),2^(1/2))/d/cosh(d*x+c)^(1/2)/(b*se ch(d*x+c))^(1/2)+2*b*(b*sech(d*x+c))^(1/2)*sinh(d*x+c)/d
Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.74 \[ \int (b \text {sech}(c+d x))^{3/2} \, dx=\frac {2 b \sqrt {b \text {sech}(c+d x)} \left (i \sqrt {\cosh (c+d x)} E\left (\left .\frac {1}{2} i (c+d x)\right |2\right )+\sinh (c+d x)\right )}{d} \] Input:
Integrate[(b*Sech[c + d*x])^(3/2),x]
Output:
(2*b*Sqrt[b*Sech[c + d*x]]*(I*Sqrt[Cosh[c + d*x]]*EllipticE[(I/2)*(c + d*x ), 2] + Sinh[c + d*x]))/d
Time = 0.39 (sec) , antiderivative size = 70, 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, 3119}
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.
\(\displaystyle \int (b \text {sech}(c+d x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (b \csc \left (i c+i d x+\frac {\pi }{2}\right )\right )^{3/2}dx\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {2 b \sinh (c+d x) \sqrt {b \text {sech}(c+d x)}}{d}-b^2 \int \frac {1}{\sqrt {b \text {sech}(c+d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b \sinh (c+d x) \sqrt {b \text {sech}(c+d x)}}{d}-b^2 \int \frac {1}{\sqrt {b \csc \left (i c+i d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {2 b \sinh (c+d x) \sqrt {b \text {sech}(c+d x)}}{d}-\frac {b^2 \int \sqrt {\cosh (c+d x)}dx}{\sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b \sinh (c+d x) \sqrt {b \text {sech}(c+d x)}}{d}-\frac {b^2 \int \sqrt {\sin \left (i c+i d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {2 b \sinh (c+d x) \sqrt {b \text {sech}(c+d x)}}{d}+\frac {2 i b^2 E\left (\left .\frac {1}{2} i (c+d x)\right |2\right )}{d \sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}}\) |
Input:
Int[(b*Sech[c + d*x])^(3/2),x]
Output:
((2*I)*b^2*EllipticE[(I/2)*(c + d*x), 2])/(d*Sqrt[Cosh[c + d*x]]*Sqrt[b*Se ch[c + d*x]]) + (2*b*Sqrt[b*Sech[c + d*x]]*Sinh[c + d*x])/d
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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]
\[\int \left (b \,\operatorname {sech}\left (d x +c \right )\right )^{\frac {3}{2}}d x\]
Input:
int((b*sech(d*x+c))^(3/2),x)
Output:
int((b*sech(d*x+c))^(3/2),x)
Time = 0.09 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.53 \[ \int (b \text {sech}(c+d x))^{3/2} \, dx=\frac {2 \, {\left (\sqrt {2} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right )\right ) + \sqrt {2} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} + 1}}\right )}}{d} \] Input:
integrate((b*sech(d*x+c))^(3/2),x, algorithm="fricas")
Output:
2*(sqrt(2)*b^(3/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cosh( d*x + c) + sinh(d*x + c))) + sqrt(2)*(b*cosh(d*x + c) + b*sinh(d*x + c))*s qrt((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
\[ \int (b \text {sech}(c+d x))^{3/2} \, dx=\int \left (b \operatorname {sech}{\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \] Input:
integrate((b*sech(d*x+c))**(3/2),x)
Output:
Integral((b*sech(c + d*x))**(3/2), x)
\[ \int (b \text {sech}(c+d x))^{3/2} \, dx=\int { \left (b \operatorname {sech}\left (d x + c\right )\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((b*sech(d*x+c))^(3/2),x, algorithm="maxima")
Output:
integrate((b*sech(d*x + c))^(3/2), x)
\[ \int (b \text {sech}(c+d x))^{3/2} \, dx=\int { \left (b \operatorname {sech}\left (d x + c\right )\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((b*sech(d*x+c))^(3/2),x, algorithm="giac")
Output:
integrate((b*sech(d*x + c))^(3/2), x)
Timed out. \[ \int (b \text {sech}(c+d x))^{3/2} \, dx=\int {\left (\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}\right )}^{3/2} \,d x \] Input:
int((b/cosh(c + d*x))^(3/2),x)
Output:
int((b/cosh(c + d*x))^(3/2), x)
\[ \int (b \text {sech}(c+d x))^{3/2} \, dx=\sqrt {b}\, \left (\int \sqrt {\mathrm {sech}\left (d x +c \right )}\, \mathrm {sech}\left (d x +c \right )d x \right ) b \] Input:
int((b*sech(d*x+c))^(3/2),x)
Output:
sqrt(b)*int(sqrt(sech(c + d*x))*sech(c + d*x),x)*b