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

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

[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]])

________________________________________________________________________________________

Rubi [A]  time = 0.0379691, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2636, 2640, 2639} \[ -\frac{2 \cosh (c+d x)}{b d \sqrt{b \sinh (c+d x)}}-\frac{2 i E\left (\left .\frac{1}{2} \left (i c+i d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{b \sinh (c+d x)}}{b^2 d \sqrt{i \sinh (c+d x)}} \]

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 2636

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 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

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 = 0.0615495, size = 62, normalized size = 0.72 \[ -\frac{2 \left (\cosh (c+d x)-\sqrt{i \sinh (c+d x)} E\left (\left .\frac{1}{4} (-2 i c-2 i d x+\pi )\right |2\right )\right )}{b d \sqrt{b \sinh (c+d x)}} \]

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 = 0.046, size = 159, normalized size = 1.9 \begin{align*}{\frac{1}{bd\cosh \left ( dx+c \right ) } \left ( 2\,\sqrt{1-i\sinh \left ( dx+c \right ) }\sqrt{2}\sqrt{1+i\sinh \left ( dx+c \right ) }\sqrt{i\sinh \left ( dx+c \right ) }{\it EllipticE} \left ( \sqrt{1-i\sinh \left ( dx+c \right ) },1/2\,\sqrt{2} \right ) -\sqrt{1-i\sinh \left ( dx+c \right ) }\sqrt{2}\sqrt{1+i\sinh \left ( dx+c \right ) }\sqrt{i\sinh \left ( dx+c \right ) }{\it EllipticF} \left ( \sqrt{1-i\sinh \left ( dx+c \right ) },{\frac{\sqrt{2}}{2}} \right ) -2\, \left ( \cosh \left ( dx+c \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{b\sinh \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \sinh \left (d x + c\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \sinh \left (d x + c\right )}}{b^{2} \sinh \left (d x + c\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sinh(d*x + c))/(b^2*sinh(d*x + c)^2), x)

________________________________________________________________________________________

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

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \sinh \left (d x + c\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

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)