3.843 \(\int \frac {1}{\sqrt {3-x} \sqrt {5+x}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {53, 619, 216} \[ -\sin ^{-1}\left (\frac {1}{4} (-x-1)\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[3 - x]*Sqrt[5 + x]),x]

[Out]

-ArcSin[(-1 - x)/4]

Rule 53

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

Rule 216

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 619

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 {3-x} \sqrt {5+x}} \, dx &=\int \frac {1}{\sqrt {15-2 x-x^2}} \, dx\\ &=-\left (\frac {1}{8} \operatorname {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 = 0.02, size = 21, normalized size = 1.75 \[ -2 \sin ^{-1}\left (\frac {\sqrt {3-x}}{2 \sqrt {2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[3 - x]*Sqrt[5 + x]),x]

[Out]

-2*ArcSin[Sqrt[3 - x]/(2*Sqrt[2])]

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fricas [B]  time = 0.41, size = 29, normalized size = 2.42 \[ -\arctan \left (\frac {\sqrt {x + 5} {\left (x + 1\right )} \sqrt {-x + 3}}{x^{2} + 2 \, x - 15}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3-x)^(1/2)/(5+x)^(1/2),x, algorithm="fricas")

[Out]

-arctan(sqrt(x + 5)*(x + 1)*sqrt(-x + 3)/(x^2 + 2*x - 15))

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giac [B]  time = 0.33, size = 13, normalized size = 1.08 \[ 2 \, \arcsin \left (\frac {1}{4} \, \sqrt {2} \sqrt {x + 5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3-x)^(1/2)/(5+x)^(1/2),x, algorithm="giac")

[Out]

2*arcsin(1/4*sqrt(2)*sqrt(x + 5))

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maple [B]  time = 0.01, size = 31, normalized size = 2.58 \[ \frac {\sqrt {\left (-x +3\right ) \left (x +5\right )}\, \arcsin \left (\frac {x}{4}+\frac {1}{4}\right )}{\sqrt {-x +3}\, \sqrt {x +5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-x+3)^(1/2)/(x+5)^(1/2),x)

[Out]

((-x+3)*(x+5))^(1/2)/(-x+3)^(1/2)/(x+5)^(1/2)*arcsin(1/4*x+1/4)

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maxima [A]  time = 1.96, size = 8, normalized size = 0.67 \[ -\arcsin \left (-\frac {1}{4} \, x - \frac {1}{4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3-x)^(1/2)/(5+x)^(1/2),x, algorithm="maxima")

[Out]

-arcsin(-1/4*x - 1/4)

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mupad [B]  time = 3.43, size = 30, normalized size = 2.50 \[ 4\,\mathrm {atan}\left (\frac {\sqrt {3}-\sqrt {3-x}}{\sqrt {x+5}-\sqrt {5}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((3 - x)^(1/2)*(x + 5)^(1/2)),x)

[Out]

4*atan((3^(1/2) - (3 - x)^(1/2))/((x + 5)^(1/2) - 5^(1/2)))

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sympy [B]  time = 1.02, size = 41, normalized size = 3.42 \[ \begin {cases} - 2 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {x + 5}}{4} \right )} & \text {for}\: \frac {\left |{x + 5}\right |}{8} > 1 \\2 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {x + 5}}{4} \right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3-x)**(1/2)/(5+x)**(1/2),x)

[Out]

Piecewise((-2*I*acosh(sqrt(2)*sqrt(x + 5)/4), Abs(x + 5)/8 > 1), (2*asin(sqrt(2)*sqrt(x + 5)/4), True))

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