3.129 \(\int \frac{1}{\sqrt{a+b \sinh ^2(x)}} \, dx\)

Optimal. Leaf size=42 \[ -\frac{i \sqrt{\frac{b \sinh ^2(x)}{a}+1} \text{EllipticF}\left (i x,\frac{b}{a}\right )}{\sqrt{a+b \sinh ^2(x)}} \]

[Out]

((-I)*EllipticF[I*x, b/a]*Sqrt[1 + (b*Sinh[x]^2)/a])/Sqrt[a + b*Sinh[x]^2]

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Rubi [A]  time = 0.0329238, antiderivative size = 42, 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 = {3183, 3182} \[ -\frac{i \sqrt{\frac{b \sinh ^2(x)}{a}+1} F\left (i x\left |\frac{b}{a}\right .\right )}{\sqrt{a+b \sinh ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + b*Sinh[x]^2],x]

[Out]

((-I)*EllipticF[I*x, b/a]*Sqrt[1 + (b*Sinh[x]^2)/a])/Sqrt[a + b*Sinh[x]^2]

Rule 3183

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[Sqrt[1 + (b*Sin[e + f*x]^2)/a]/Sqrt[a +
b*Sin[e + f*x]^2], Int[1/Sqrt[1 + (b*Sin[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] &&  !GtQ[a, 0]

Rule 3182

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(1*EllipticF[e + f*x, -(b/a)])/(Sqrt[a]*
f), x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rubi steps

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

Mathematica [A]  time = 0.0507961, size = 53, normalized size = 1.26 \[ -\frac{i \sqrt{\frac{2 a+b \cosh (2 x)-b}{a}} \text{EllipticF}\left (i x,\frac{b}{a}\right )}{\sqrt{2 a+b \cosh (2 x)-b}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a + b*Sinh[x]^2],x]

[Out]

((-I)*Sqrt[(2*a - b + b*Cosh[2*x])/a]*EllipticF[I*x, b/a])/Sqrt[2*a - b + b*Cosh[2*x]]

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Maple [A]  time = 0.059, size = 63, normalized size = 1.5 \begin{align*}{\frac{1}{\cosh \left ( x \right ) }\sqrt{{\frac{a+b \left ( \sinh \left ( x \right ) \right ) ^{2}}{a}}}\sqrt{ \left ( \cosh \left ( x \right ) \right ) ^{2}}{\it EllipticF} \left ( \sinh \left ( x \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ){\frac{1}{\sqrt{-{\frac{b}{a}}}}}{\frac{1}{\sqrt{a+b \left ( \sinh \left ( x \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sinh(x)^2)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(b*sinh(x)^2 + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{\sqrt{b \sinh \left (x\right )^{2} + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/sqrt(b*sinh(x)^2 + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(a + b*sinh(x)**2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(b*sinh(x)^2 + a), x)