\(\int \frac {1}{\sqrt {a^2-x^2}} \, dx\) [87]

Optimal result

Integrand size = 13, antiderivative size = 16 \[ \int \frac {1}{\sqrt {a^2-x^2}} \, dx=\arctan \left (\frac {x}{\sqrt {a^2-x^2}}\right ) \]

[Out]

arctan(x/(a^2-x^2)^(1/2))

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {223, 209} \[ \int \frac {1}{\sqrt {a^2-x^2}} \, dx=\arctan \left (\frac {x}{\sqrt {a^2-x^2}}\right ) \]

[In]

Int[1/Sqrt[a^2 - x^2],x]

[Out]

ArcTan[x/Sqrt[a^2 - x^2]]

Rule 209

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

Rule 223

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*} \text {integral}& = \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt {a^2-x^2}}\right ) \\ & = \arctan \left (\frac {x}{\sqrt {a^2-x^2}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {a^2-x^2}} \, dx=\arctan \left (\frac {x}{\sqrt {a^2-x^2}}\right ) \]

[In]

Integrate[1/Sqrt[a^2 - x^2],x]

[Out]

ArcTan[x/Sqrt[a^2 - x^2]]

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
default	\(\arctan \left (\frac {x}{\sqrt {a^{2}-x^{2}}}\right )\)	\(15\)
pseudoelliptic	\(-\arctan \left (\frac {\sqrt {a^{2}-x^{2}}}{x}\right )\)	\(19\)

[In]

int(1/(a^2-x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

arctan(x/(a^2-x^2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.44 \[ \int \frac {1}{\sqrt {a^2-x^2}} \, dx=-2 \, \arctan \left (-\frac {a - \sqrt {a^{2} - x^{2}}}{x}\right ) \]

[In]

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

[Out]

-2*arctan(-(a - sqrt(a^2 - x^2))/x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.52 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19 \[ \int \frac {1}{\sqrt {a^2-x^2}} \, dx=\begin {cases} - i \operatorname {acosh}{\left (\frac {x}{a} \right )} & \text {for}\: \left |{\frac {x^{2}}{a^{2}}}\right | > 1 \\\operatorname {asin}{\left (\frac {x}{a} \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-I*acosh(x/a), Abs(x**2/a**2) > 1), (asin(x/a), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.38 \[ \int \frac {1}{\sqrt {a^2-x^2}} \, dx=\arcsin \left (\frac {x}{a}\right ) \]

[In]

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

[Out]

arcsin(x/a)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.75 \[ \int \frac {1}{\sqrt {a^2-x^2}} \, dx=\frac {1}{2} \, a^{2} \arcsin \left (\frac {x}{a}\right ) \mathrm {sgn}\left (a\right ) + \frac {1}{2} \, \sqrt {a^{2} - x^{2}} x \]

[In]

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

[Out]

1/2*a^2*arcsin(x/a)*sgn(a) + 1/2*sqrt(a^2 - x^2)*x

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\sqrt {a^2-x^2}} \, dx=\mathrm {atan}\left (\frac {x}{\sqrt {a^2-x^2}}\right ) \]

[In]

int(1/(a^2 - x^2)^(1/2),x)

[Out]

atan(x/(a^2 - x^2)^(1/2))