\(\int \frac {1}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx\) [12]

Optimal result

Integrand size = 10, antiderivative size = 76 \[ \int \frac {1}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx=-\frac {2 \cosh (a+b x)}{b \sqrt {\sinh (a+b x)}}-\frac {2 i E\left (\left .\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right )\right |2\right ) \sqrt {\sinh (a+b x)}}{b \sqrt {i \sinh (a+b x)}} \]

[Out]

-2*cosh(b*x+a)/b/sinh(b*x+a)^(1/2)+2*I*(sin(1/2*I*a+1/4*Pi+1/2*I*b*x)^2)^(1/2)/sin(1/2*I*a+1/4*Pi+1/2*I*b*x)*E
llipticE(cos(1/2*I*a+1/4*Pi+1/2*I*b*x),2^(1/2))*sinh(b*x+a)^(1/2)/b/(I*sinh(b*x+a))^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2716, 2721, 2719} \[ \int \frac {1}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx=-\frac {2 \cosh (a+b x)}{b \sqrt {\sinh (a+b x)}}-\frac {2 i \sqrt {\sinh (a+b x)} E\left (\left .\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right )\right |2\right )}{b \sqrt {i \sinh (a+b x)}} \]

[In]

Int[Sinh[a + b*x]^(-3/2),x]

[Out]

(-2*Cosh[a + b*x])/(b*Sqrt[Sinh[a + b*x]]) - ((2*I)*EllipticE[(I*a - Pi/2 + I*b*x)/2, 2]*Sqrt[Sinh[a + b*x]])/
(b*Sqrt[I*Sinh[a + b*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*} \text {integral}& = -\frac {2 \cosh (a+b x)}{b \sqrt {\sinh (a+b x)}}+\int \sqrt {\sinh (a+b x)} \, dx \\ & = -\frac {2 \cosh (a+b x)}{b \sqrt {\sinh (a+b x)}}+\frac {\sqrt {\sinh (a+b x)} \int \sqrt {i \sinh (a+b x)} \, dx}{\sqrt {i \sinh (a+b x)}} \\ & = -\frac {2 \cosh (a+b x)}{b \sqrt {\sinh (a+b x)}}-\frac {2 i E\left (\left .\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right )\right |2\right ) \sqrt {\sinh (a+b x)}}{b \sqrt {i \sinh (a+b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx=-\frac {2 \left (\cosh (a+b x)-E\left (\left .\frac {1}{4} (-2 i a+\pi -2 i b x)\right |2\right ) \sqrt {i \sinh (a+b x)}\right )}{b \sqrt {\sinh (a+b x)}} \]

[In]

Integrate[Sinh[a + b*x]^(-3/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.03

method	result	size
default	\(\frac {2 \sqrt {1-i \sinh \left (b x +a \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (b x +a \right )}\, \sqrt {i \sinh \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {1-i \sinh \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {1-i \sinh \left (b x +a \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (b x +a \right )}\, \sqrt {i \sinh \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {1-i \sinh \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )-2 \cosh \left (b x +a \right )^{2}}{\cosh \left (b x +a \right ) \sqrt {\sinh \left (b x +a \right )}\, b}\)	\(154\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

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

Time = 0.07 (sec) , antiderivative size = 152, normalized size of antiderivative = 2.00 \[ \int \frac {1}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx=-\frac {2 \, {\left ({\left (\sqrt {2} \cosh \left (b x + a\right )^{2} + 2 \, \sqrt {2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sqrt {2} \sinh \left (b x + a\right )^{2} - \sqrt {2}\right )} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )\right ) + 2 \, {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2}\right )} \sqrt {\sinh \left (b x + a\right )}\right )}}{b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2} - b} \]

[In]

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

[Out]

-2*((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))*weie
rstrassZeta(4, 0, weierstrassPInverse(4, 0, cosh(b*x + a) + sinh(b*x + a))) + 2*(cosh(b*x + a)^2 + 2*cosh(b*x
+ a)*sinh(b*x + a) + sinh(b*x + a)^2)*sqrt(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)

Sympy [F]

\[ \int \frac {1}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx=\int \frac {1}{\sinh ^{\frac {3}{2}}{\left (a + b x \right )}}\, dx \]

[In]

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

[Out]

Integral(sinh(a + b*x)**(-3/2), x)

Maxima [F]

\[ \int \frac {1}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx=\int { \frac {1}{\sinh \left (b x + a\right )^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

integrate(sinh(b*x + a)^(-3/2), x)

Giac [F]

\[ \int \frac {1}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx=\int { \frac {1}{\sinh \left (b x + a\right )^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

integrate(sinh(b*x + a)^(-3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx=\int \frac {1}{{\mathrm {sinh}\left (a+b\,x\right )}^{3/2}} \,d x \]

[In]

int(1/sinh(a + b*x)^(3/2),x)

[Out]

int(1/sinh(a + b*x)^(3/2), x)