3.2887 \(\int \frac {1}{\sqrt {4-x} \sqrt {-15+8 x-x^2}} \, dx\)

Optimal. Leaf size=14 \[ -2 \operatorname {EllipticF}\left (\sin ^{-1}\left (\sqrt {4-x}\right ),-1\right ) \]

[Out]

-2*EllipticF((4-x)^(1/2),I)

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Rubi [A]  time = 0.01, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {689, 221} \[ -2 F\left (\left .\sin ^{-1}\left (\sqrt {4-x}\right )\right |-1\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[4 - x]*Sqrt[-15 + 8*x - x^2]),x]

[Out]

-2*EllipticF[ArcSin[Sqrt[4 - x]], -1]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 689

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[(4*Sqrt[-(c/(b^2 -
4*a*c))])/e, Subst[Int[1/Sqrt[Simp[1 - (b^2*x^4)/(d^2*(b^2 - 4*a*c)), x]], x], x, Sqrt[d + e*x]], x] /; FreeQ[
{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] && LtQ[c/(b^2 - 4*a*c), 0]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 28, normalized size = 2.00 \[ -2 \sqrt {4-x} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};(4-x)^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[4 - x]*Sqrt[-15 + 8*x - x^2]),x]

[Out]

-2*Sqrt[4 - x]*Hypergeometric2F1[1/4, 1/2, 5/4, (4 - x)^2]

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-x^{2} + 8 \, x - 15} \sqrt {-x + 4}}{x^{3} - 12 \, x^{2} + 47 \, x - 60}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4-x)^(1/2)/(-x^2+8*x-15)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(-x^2 + 8*x - 15)*sqrt(-x + 4)/(x^3 - 12*x^2 + 47*x - 60), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-x^{2} + 8 \, x - 15} \sqrt {-x + 4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4-x)^(1/2)/(-x^2+8*x-15)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.01, size = 47, normalized size = 3.36 \[ \frac {2 \sqrt {x -3}\, \sqrt {-x +5}\, \sqrt {-x^{2}+8 x -15}\, \EllipticF \left (\sqrt {-x +4}, i\right )}{x^{2}-8 x +15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-x+4)^(1/2)/(-x^2+8*x-15)^(1/2),x)

[Out]

2*EllipticF((-x+4)^(1/2),I)*(x-3)^(1/2)*(-x+5)^(1/2)*(-x^2+8*x-15)^(1/2)/(x^2-8*x+15)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-x^{2} + 8 \, x - 15} \sqrt {-x + 4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4-x)^(1/2)/(-x^2+8*x-15)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.07 \[ \int \frac {1}{\sqrt {4-x}\,\sqrt {-x^2+8\,x-15}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((4 - x)^(1/2)*(8*x - x^2 - 15)^(1/2)),x)

[Out]

int(1/((4 - x)^(1/2)*(8*x - x^2 - 15)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {- \left (x - 5\right ) \left (x - 3\right )} \sqrt {4 - x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4-x)**(1/2)/(-x**2+8*x-15)**(1/2),x)

[Out]

Integral(1/(sqrt(-(x - 5)*(x - 3))*sqrt(4 - x)), x)

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