3.1.20 \(\int \frac {1}{(b \sinh (c+d x))^{3/2}} \, dx\) [20]

Optimal. Leaf size=86 \[ -\frac {2 \cosh (c+d x)}{b d \sqrt {b \sinh (c+d x)}}-\frac {2 i E\left (\left .\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right )\right |2\right ) \sqrt {b \sinh (c+d x)}}{b^2 d \sqrt {i \sinh (c+d x)}} \]

[Out]

-2*cosh(d*x+c)/b/d/(b*sinh(d*x+c))^(1/2)+2*I*(sin(1/2*I*c+1/4*Pi+1/2*I*d*x)^2)^(1/2)/sin(1/2*I*c+1/4*Pi+1/2*I*
d*x)*EllipticE(cos(1/2*I*c+1/4*Pi+1/2*I*d*x),2^(1/2))*(b*sinh(d*x+c))^(1/2)/b^2/d/(I*sinh(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2716, 2721, 2719} \begin {gather*} -\frac {2 \cosh (c+d x)}{b d \sqrt {b \sinh (c+d x)}}-\frac {2 i E\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \sinh (c+d x)}}{b^2 d \sqrt {i \sinh (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Sinh[c + d*x])^(-3/2),x]

[Out]

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

Rule 2716

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] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

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 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rubi steps

\begin {align*} \int \frac {1}{(b \sinh (c+d x))^{3/2}} \, dx &=-\frac {2 \cosh (c+d x)}{b d \sqrt {b \sinh (c+d x)}}+\frac {\int \sqrt {b \sinh (c+d x)} \, dx}{b^2}\\ &=-\frac {2 \cosh (c+d x)}{b d \sqrt {b \sinh (c+d x)}}+\frac {\sqrt {b \sinh (c+d x)} \int \sqrt {i \sinh (c+d x)} \, dx}{b^2 \sqrt {i \sinh (c+d x)}}\\ &=-\frac {2 \cosh (c+d x)}{b d \sqrt {b \sinh (c+d x)}}-\frac {2 i E\left (\left .\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right )\right |2\right ) \sqrt {b \sinh (c+d x)}}{b^2 d \sqrt {i \sinh (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.05, size = 62, normalized size = 0.72 \begin {gather*} -\frac {2 \left (\cosh (c+d x)-E\left (\left .\frac {1}{4} (-2 i c+\pi -2 i d x)\right |2\right ) \sqrt {i \sinh (c+d x)}\right )}{b d \sqrt {b \sinh (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Sinh[c + d*x])^(-3/2),x]

[Out]

(-2*(Cosh[c + d*x] - EllipticE[((-2*I)*c + Pi - (2*I)*d*x)/4, 2]*Sqrt[I*Sinh[c + d*x]]))/(b*d*Sqrt[b*Sinh[c +
d*x]])

________________________________________________________________________________________

Maple [A]
time = 0.73, size = 159, normalized size = 1.85

method	result	size
default	\(\frac {2 \sqrt {1-i \sinh \left (d x +c \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \EllipticE \left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {1-i \sinh \left (d x +c \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \EllipticF \left (\sqrt {1-i \sinh \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \left (\cosh ^{2}\left (d x +c \right )\right )}{b \cosh \left (d x +c \right ) \sqrt {b \sinh \left (d x +c \right )}\, d}\)	\(159\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*sinh(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sinh(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sinh(d*x + c))^(-3/2), x)

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.08, size = 169, normalized size = 1.97 \begin {gather*} -\frac {2 \, {\left ({\left (\sqrt {2} \cosh \left (d x + c\right )^{2} + 2 \, \sqrt {2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sqrt {2} \sinh \left (d x + c\right )^{2} - \sqrt {2}\right )} \sqrt {b} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right )\right ) + 2 \, {\left (\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}\right )} \sqrt {b \sinh \left (d x + c\right )}\right )}}{b^{2} d \cosh \left (d x + c\right )^{2} + 2 \, b^{2} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{2} d \sinh \left (d x + c\right )^{2} - b^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sinh(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-2*((sqrt(2)*cosh(d*x + c)^2 + 2*sqrt(2)*cosh(d*x + c)*sinh(d*x + c) + sqrt(2)*sinh(d*x + c)^2 - sqrt(2))*sqrt
(b)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cosh(d*x + c) + sinh(d*x + c))) + 2*(cosh(d*x + c)^2 + 2*c
osh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)*sqrt(b*sinh(d*x + c)))/(b^2*d*cosh(d*x + c)^2 + 2*b^2*d*cosh(d*x
+ c)*sinh(d*x + c) + b^2*d*sinh(d*x + c)^2 - b^2*d)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b \sinh {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sinh(d*x+c))**(3/2),x)

[Out]

Integral((b*sinh(c + d*x))**(-3/2), x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sinh(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((b*sinh(d*x + c))^(-3/2), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*sinh(c + d*x))^(3/2),x)

[Out]

int(1/(b*sinh(c + d*x))^(3/2), x)

________________________________________________________________________________________