\(\int (b \text {sech}(c+d x))^{3/2} \, dx\) [17]

Optimal result

Integrand size = 12, antiderivative size = 70 \[ \int (b \text {sech}(c+d x))^{3/2} \, dx=\frac {2 i b^2 E\left (\left .\frac {1}{2} i (c+d x)\right |2\right )}{d \sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}}+\frac {2 b \sqrt {b \text {sech}(c+d x)} \sinh (c+d x)}{d} \]

[Out]

2*I*b^2*(cosh(1/2*d*x+1/2*c)^2)^(1/2)/cosh(1/2*d*x+1/2*c)*EllipticE(I*sinh(1/2*d*x+1/2*c),2^(1/2))/d/cosh(d*x+
c)^(1/2)/(b*sech(d*x+c))^(1/2)+2*b*sinh(d*x+c)*(b*sech(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3853, 3856, 2719} \[ \int (b \text {sech}(c+d x))^{3/2} \, dx=\frac {2 b \sinh (c+d x) \sqrt {b \text {sech}(c+d x)}}{d}+\frac {2 i b^2 E\left (\left .\frac {1}{2} i (c+d x)\right |2\right )}{d \sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}} \]

[In]

Int[(b*Sech[c + d*x])^(3/2),x]

[Out]

((2*I)*b^2*EllipticE[(I/2)*(c + d*x), 2])/(d*Sqrt[Cosh[c + d*x]]*Sqrt[b*Sech[c + d*x]]) + (2*b*Sqrt[b*Sech[c +
d*x]]*Sinh[c + d*x])/d

Rule 2719

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]

Rule 3853

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] + Dist[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]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 b \sqrt {b \text {sech}(c+d x)} \sinh (c+d x)}{d}-b^2 \int \frac {1}{\sqrt {b \text {sech}(c+d x)}} \, dx \\ & = \frac {2 b \sqrt {b \text {sech}(c+d x)} \sinh (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)}} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.74 \[ \int (b \text {sech}(c+d x))^{3/2} \, dx=\frac {2 b \sqrt {b \text {sech}(c+d x)} \left (i \sqrt {\cosh (c+d x)} E\left (\left .\frac {1}{2} i (c+d x)\right |2\right )+\sinh (c+d x)\right )}{d} \]

[In]

Integrate[(b*Sech[c + d*x])^(3/2),x]

[Out]

(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

Maple [F]

\[\int \left (b \,\operatorname {sech}\left (d x +c \right )\right )^{\frac {3}{2}}d x\]

[In]

int((b*sech(d*x+c))^(3/2),x)

[Out]

int((b*sech(d*x+c))^(3/2),x)

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.08 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.53 \[ \int (b \text {sech}(c+d x))^{3/2} \, dx=\frac {2 \, {\left (\sqrt {2} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right )\right ) + \sqrt {2} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} + 1}}\right )}}{d} \]

[In]

integrate((b*sech(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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))*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

Sympy [F]

\[ \int (b \text {sech}(c+d x))^{3/2} \, dx=\int \left (b \operatorname {sech}{\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \]

[In]

integrate((b*sech(d*x+c))**(3/2),x)

[Out]

Integral((b*sech(c + d*x))**(3/2), x)

Maxima [F]

\[ \int (b \text {sech}(c+d x))^{3/2} \, dx=\int { \left (b \operatorname {sech}\left (d x + c\right )\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate((b*sech(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sech(d*x + c))^(3/2), x)

Giac [F]

\[ \int (b \text {sech}(c+d x))^{3/2} \, dx=\int { \left (b \operatorname {sech}\left (d x + c\right )\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate((b*sech(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((b*sech(d*x + c))^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int (b \text {sech}(c+d x))^{3/2} \, dx=\int {\left (\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}\right )}^{3/2} \,d x \]

[In]

int((b/cosh(c + d*x))^(3/2),x)

[Out]

int((b/cosh(c + d*x))^(3/2), x)