3.338 \(\int \frac {x^2}{\sqrt {5-4 x^2}} \, dx\)

Optimal. Leaf size=30 \[ \frac {5}{16} \sin ^{-1}\left (\frac {2 x}{\sqrt {5}}\right )-\frac {1}{8} x \sqrt {5-4 x^2} \]

[Out]

5/16*arcsin(2/5*x*5^(1/2))-1/8*x*(-4*x^2+5)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {321, 216} \[ \frac {5}{16} \sin ^{-1}\left (\frac {2 x}{\sqrt {5}}\right )-\frac {1}{8} x \sqrt {5-4 x^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[5 - 4*x^2],x]

[Out]

-(x*Sqrt[5 - 4*x^2])/8 + (5*ArcSin[(2*x)/Sqrt[5]])/16

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {x^2}{\sqrt {5-4 x^2}} \, dx &=-\frac {1}{8} x \sqrt {5-4 x^2}+\frac {5}{8} \int \frac {1}{\sqrt {5-4 x^2}} \, dx\\ &=-\frac {1}{8} x \sqrt {5-4 x^2}+\frac {5}{16} \sin ^{-1}\left (\frac {2 x}{\sqrt {5}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 30, normalized size = 1.00 \[ \frac {5}{16} \sin ^{-1}\left (\frac {2 x}{\sqrt {5}}\right )-\frac {1}{8} x \sqrt {5-4 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[5 - 4*x^2],x]

[Out]

-1/8*(x*Sqrt[5 - 4*x^2]) + (5*ArcSin[(2*x)/Sqrt[5]])/16

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fricas [A]  time = 0.41, size = 30, normalized size = 1.00 \[ -\frac {1}{8} \, \sqrt {-4 \, x^{2} + 5} x - \frac {5}{16} \, \arctan \left (\frac {\sqrt {-4 \, x^{2} + 5}}{2 \, x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-4*x^2+5)^(1/2),x, algorithm="fricas")

[Out]

-1/8*sqrt(-4*x^2 + 5)*x - 5/16*arctan(1/2*sqrt(-4*x^2 + 5)/x)

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giac [A]  time = 1.00, size = 22, normalized size = 0.73 \[ -\frac {1}{8} \, \sqrt {-4 \, x^{2} + 5} x + \frac {5}{16} \, \arcsin \left (\frac {2}{5} \, \sqrt {5} x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-4*x^2+5)^(1/2),x, algorithm="giac")

[Out]

-1/8*sqrt(-4*x^2 + 5)*x + 5/16*arcsin(2/5*sqrt(5)*x)

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maple [A]  time = 0.00, size = 23, normalized size = 0.77 \[ -\frac {\sqrt {-4 x^{2}+5}\, x}{8}+\frac {5 \arcsin \left (\frac {2 \sqrt {5}\, x}{5}\right )}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(-4*x^2+5)^(1/2),x)

[Out]

5/16*arcsin(2/5*5^(1/2)*x)-1/8*x*(-4*x^2+5)^(1/2)

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maxima [A]  time = 1.30, size = 22, normalized size = 0.73 \[ -\frac {1}{8} \, \sqrt {-4 \, x^{2} + 5} x + \frac {5}{16} \, \arcsin \left (\frac {2}{5} \, \sqrt {5} x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-4*x^2+5)^(1/2),x, algorithm="maxima")

[Out]

-1/8*sqrt(-4*x^2 + 5)*x + 5/16*arcsin(2/5*sqrt(5)*x)

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mupad [B]  time = 0.17, size = 22, normalized size = 0.73 \[ \frac {5\,\mathrm {asin}\left (\frac {2\,\sqrt {5}\,x}{5}\right )}{16}-\frac {x\,\sqrt {\frac {5}{4}-x^2}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(5 - 4*x^2)^(1/2),x)

[Out]

(5*asin((2*5^(1/2)*x)/5))/16 - (x*(5/4 - x^2)^(1/2))/4

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sympy [A]  time = 0.26, size = 27, normalized size = 0.90 \[ - \frac {x \sqrt {5 - 4 x^{2}}}{8} + \frac {5 \operatorname {asin}{\left (\frac {2 \sqrt {5} x}{5} \right )}}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(-4*x**2+5)**(1/2),x)

[Out]

-x*sqrt(5 - 4*x**2)/8 + 5*asin(2*sqrt(5)*x/5)/16

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