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

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Rubi [A]  time = 0.00, antiderivative size = 26, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 1.00 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a-b x^2}}\right )}{\sqrt {b}} \]

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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fricas [A]  time = 0.77, size = 72, normalized size = 2.77 \[ \left [-\frac {\sqrt {-b} \log \left (2 \, b x^{2} - 2 \, \sqrt {-b x^{2} + a} \sqrt {-b} x - a\right )}{2 \, b}, -\frac {\arctan \left (\frac {\sqrt {-b x^{2} + a} \sqrt {b} x}{b x^{2} - a}\right )}{\sqrt {b}}\right ] \]

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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giac [A]  time = 1.15, size = 28, normalized size = 1.08 \[ -\frac {\log \left ({\left | -\sqrt {-b} x + \sqrt {-b x^{2} + a} \right |}\right )}{\sqrt {-b}} \]

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)

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maple [A]  time = 0.00, size = 21, normalized size = 0.81 \[ \frac {\arctan \left (\frac {\sqrt {b}\, x}{\sqrt {-b \,x^{2}+a}}\right )}{\sqrt {b}} \]

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 [A]  time = 2.96, size = 13, normalized size = 0.50 \[ \frac {\arcsin \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsin(b*x/sqrt(a*b))/sqrt(b)

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mupad [B]  time = 0.11, size = 25, normalized size = 0.96 \[ \frac {\ln \left (\sqrt {a-b\,x^2}+\sqrt {-b}\,x\right )}{\sqrt {-b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a - b*x^2)^(1/2),x)

[Out]

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

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sympy [A]  time = 1.07, size = 46, normalized size = 1.77 \[ \begin {cases} - \frac {i \operatorname {acosh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{\sqrt {b}} & \text {for}\: \left |{\frac {b x^{2}}{a}}\right | > 1 \\\frac {\operatorname {asin}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{\sqrt {b}} & \text {otherwise} \end {cases} \]

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/a) > 1), (asin(sqrt(b)*x/sqrt(a))/sqrt(b), True))

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