3.585 \(\int \frac{1}{\sqrt{a-b x^2}} \, dx\)

Optimal. Leaf size=26 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a-b x^2}}\right )}{\sqrt{b}} \]

[Out]

ArcTan[(Sqrt[b]*x)/Sqrt[a - b*x^2]]/Sqrt[b]

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Rubi [A]  time = 0.0048264, antiderivative size = 26, 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 = {217, 203} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a-b x^2}}\right )}{\sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a - b*x^2],x]

[Out]

ArcTan[(Sqrt[b]*x)/Sqrt[a - b*x^2]]/Sqrt[b]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a-b x^2}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{1+b x^2} \, dx,x,\frac{x}{\sqrt{a-b x^2}}\right )\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a-b x^2}}\right )}{\sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.0055323, size = 26, normalized size = 1. \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a-b x^2}}\right )}{\sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a - b*x^2],x]

[Out]

ArcTan[(Sqrt[b]*x)/Sqrt[a - b*x^2]]/Sqrt[b]

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Maple [A]  time = 0.003, size = 21, normalized size = 0.8 \begin{align*}{\arctan \left ({x\sqrt{b}{\frac{1}{\sqrt{-b{x}^{2}+a}}}} \right ){\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(x*b^(1/2)/(-b*x^2+a)^(1/2))/b^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.20117, size = 171, normalized size = 6.58 \begin{align*} \left [-\frac{\sqrt{-b} \log \left (2 \, b x^{2} - 2 \, \sqrt{-b x^{2} + a} \sqrt{-b} x - a\right )}{2 \, b}, -\frac{\arctan \left (\frac{\sqrt{-b x^{2} + a} \sqrt{b} x}{b x^{2} - a}\right )}{\sqrt{b}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*sqrt(-b)*log(2*b*x^2 - 2*sqrt(-b*x^2 + a)*sqrt(-b)*x - a)/b, -arctan(sqrt(-b*x^2 + a)*sqrt(b)*x/(b*x^2 -
a))/sqrt(b)]

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Sympy [A]  time = 1.05413, size = 48, normalized size = 1.85 \begin{align*} \begin{cases} - \frac{i \operatorname{acosh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{\sqrt{b}} & \text{for}\: \frac{\left |{b x^{2}}\right |}{\left |{a}\right |} > 1 \\\frac{\operatorname{asin}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{\sqrt{b}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-I*acosh(sqrt(b)*x/sqrt(a))/sqrt(b), Abs(b*x**2)/Abs(a) > 1), (asin(sqrt(b)*x/sqrt(a))/sqrt(b), Tru
e))

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Giac [A]  time = 2.08344, size = 38, normalized size = 1.46 \begin{align*} -\frac{\log \left ({\left | -\sqrt{-b} x + \sqrt{-b x^{2} + a} \right |}\right )}{\sqrt{-b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(abs(-sqrt(-b)*x + sqrt(-b*x^2 + a)))/sqrt(-b)