3.19 \(\int \frac{1}{\sqrt{b \sinh (c+d x)}} \, dx\)

Optimal. Leaf size=56 \[ -\frac{2 i \sqrt{i \sinh (c+d x)} \text{EllipticF}\left (\frac{1}{2} \left (i c+i d x-\frac{\pi }{2}\right ),2\right )}{d \sqrt{b \sinh (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0218033, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2642, 2641} \[ -\frac{2 i \sqrt{i \sinh (c+d x)} F\left (\left .\frac{1}{2} \left (i c+i d x-\frac{\pi }{2}\right )\right |2\right )}{d \sqrt{b \sinh (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[b*Sinh[c + d*x]],x]

[Out]

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

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

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

Rubi steps

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

Mathematica [A]  time = 0.0340537, size = 54, normalized size = 0.96 \[ \frac{2 i \sqrt{i \sinh (c+d x)} \text{EllipticF}\left (\frac{1}{2} \left (\frac{\pi }{2}-i (c+d x)\right ),2\right )}{d \sqrt{b \sinh (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[b*Sinh[c + d*x]],x]

[Out]

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

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Maple [A]  time = 0.034, size = 89, normalized size = 1.6 \begin{align*}{\frac{i\sqrt{2}}{d\cosh \left ( dx+c \right ) }\sqrt{-i \left ( \sinh \left ( dx+c \right ) +i \right ) }\sqrt{-i \left ( i-\sinh \left ( dx+c \right ) \right ) }\sqrt{i\sinh \left ( dx+c \right ) }{\it EllipticF} \left ( \sqrt{-i \left ( \sinh \left ( dx+c \right ) +i \right ) },{\frac{\sqrt{2}}{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))^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(b*sinh(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(b*sinh(c + d*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(b*sinh(d*x + c)), x)