3.9.42 \(\int \frac {1}{\sqrt {15-2 x-x^2}} \, dx\) [842]

Optimal. Leaf size=12 \[ -\sin ^{-1}\left (\frac {1}{4} (-1-x)\right ) \]

[Out]

arcsin(1/4+1/4*x)

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Rubi [A]
time = 0.00, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {633, 222} \begin {gather*} -\text {ArcSin}\left (\frac {1}{4} (-x-1)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[15 - 2*x - x^2],x]

[Out]

-ArcSin[(-1 - x)/4]

Rule 222

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

Rule 633

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*(-4*(c/(b^2 - 4*a*c)))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {15-2 x-x^2}} \, dx &=-\left (\frac {1}{8} \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{64}}} \, dx,x,-2-2 x\right )\right )\\ &=-\sin ^{-1}\left (\frac {1}{4} (-1-x)\right )\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 23, normalized size = 1.92 \begin {gather*} -2 \tan ^{-1}\left (\frac {\sqrt {15-2 x-x^2}}{5+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[15 - 2*x - x^2],x]

[Out]

-2*ArcTan[Sqrt[15 - 2*x - x^2]/(5 + x)]

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Maple [A]
time = 0.60, size = 7, normalized size = 0.58

method	result	size
default	\(\arcsin \left (\frac {1}{4}+\frac {x}{4}\right )\)	\(7\)
trager	\(\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-x \RootOf \left (\textit {\_Z}^{2}+1\right )-\RootOf \left (\textit {\_Z}^{2}+1\right )+\sqrt {-x^{2}-2 x +15}\right )\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-x^2-2*x+15)^(1/2),x,method=_RETURNVERBOSE)

[Out]

arcsin(1/4+1/4*x)

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Maxima [A]
time = 0.49, size = 8, normalized size = 0.67 \begin {gather*} -\arcsin \left (-\frac {1}{4} \, x - \frac {1}{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-x^2-2*x+15)^(1/2),x, algorithm="maxima")

[Out]

-arcsin(-1/4*x - 1/4)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 29 vs. \(2 (6) = 12\).
time = 0.35, size = 29, normalized size = 2.42 \begin {gather*} -\arctan \left (\frac {\sqrt {-x^{2} - 2 \, x + 15} {\left (x + 1\right )}}{x^{2} + 2 \, x - 15}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-x^2-2*x+15)^(1/2),x, algorithm="fricas")

[Out]

-arctan(sqrt(-x^2 - 2*x + 15)*(x + 1)/(x^2 + 2*x - 15))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {- x^{2} - 2 x + 15}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-x**2-2*x+15)**(1/2),x)

[Out]

Integral(1/sqrt(-x**2 - 2*x + 15), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (6) = 12\).
time = 3.12, size = 26, normalized size = 2.17 \begin {gather*} \frac {1}{2} \, \sqrt {-x^{2} - 2 \, x + 15} {\left (x + 1\right )} + 8 \, \arcsin \left (\frac {1}{4} \, x + \frac {1}{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-x^2-2*x+15)^(1/2),x, algorithm="giac")

[Out]

1/2*sqrt(-x^2 - 2*x + 15)*(x + 1) + 8*arcsin(1/4*x + 1/4)

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Mupad [B]
time = 3.12, size = 6, normalized size = 0.50 \begin {gather*} \mathrm {asin}\left (\frac {x}{4}+\frac {1}{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(15 - x^2 - 2*x)^(1/2),x)

[Out]

asin(x/4 + 1/4)

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