3.369 \(\int x^2 \sqrt{5-x^2} \, dx\)

Optimal. Leaf size=47 \[ \frac{1}{4} \sqrt{5-x^2} x^3-\frac{5}{8} \sqrt{5-x^2} x+\frac{25}{8} \sin ^{-1}\left (\frac{x}{\sqrt{5}}\right ) \]

[Out]

(-5*x*Sqrt[5 - x^2])/8 + (x^3*Sqrt[5 - x^2])/4 + (25*ArcSin[x/Sqrt[5]])/8

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Rubi [A]  time = 0.0098578, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {279, 321, 216} \[ \frac{1}{4} \sqrt{5-x^2} x^3-\frac{5}{8} \sqrt{5-x^2} x+\frac{25}{8} \sin ^{-1}\left (\frac{x}{\sqrt{5}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[5 - x^2],x]

[Out]

(-5*x*Sqrt[5 - x^2])/8 + (x^3*Sqrt[5 - x^2])/4 + (25*ArcSin[x/Sqrt[5]])/8

Rule 279

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

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]

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]

Rubi steps

\begin{align*} \int x^2 \sqrt{5-x^2} \, dx &=\frac{1}{4} x^3 \sqrt{5-x^2}+\frac{5}{4} \int \frac{x^2}{\sqrt{5-x^2}} \, dx\\ &=-\frac{5}{8} x \sqrt{5-x^2}+\frac{1}{4} x^3 \sqrt{5-x^2}+\frac{25}{8} \int \frac{1}{\sqrt{5-x^2}} \, dx\\ &=-\frac{5}{8} x \sqrt{5-x^2}+\frac{1}{4} x^3 \sqrt{5-x^2}+\frac{25}{8} \sin ^{-1}\left (\frac{x}{\sqrt{5}}\right )\\ \end{align*}

Mathematica [A]  time = 0.017908, size = 35, normalized size = 0.74 \[ \frac{1}{8} \left (x \sqrt{5-x^2} \left (2 x^2-5\right )+25 \sin ^{-1}\left (\frac{x}{\sqrt{5}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[5 - x^2],x]

[Out]

(x*Sqrt[5 - x^2]*(-5 + 2*x^2) + 25*ArcSin[x/Sqrt[5]])/8

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Maple [A]  time = 0.005, size = 35, normalized size = 0.7 \begin{align*} -{\frac{x}{4} \left ( -{x}^{2}+5 \right ) ^{{\frac{3}{2}}}}+{\frac{5\,x}{8}\sqrt{-{x}^{2}+5}}+{\frac{25}{8}\arcsin \left ({\frac{x\sqrt{5}}{5}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x*(-x^2+5)^(3/2)+5/8*x*(-x^2+5)^(1/2)+25/8*arcsin(1/5*x*5^(1/2))

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Maxima [A]  time = 1.40881, size = 46, normalized size = 0.98 \begin{align*} -\frac{1}{4} \,{\left (-x^{2} + 5\right )}^{\frac{3}{2}} x + \frac{5}{8} \, \sqrt{-x^{2} + 5} x + \frac{25}{8} \, \arcsin \left (\frac{1}{5} \, \sqrt{5} x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(-x^2 + 5)^(3/2)*x + 5/8*sqrt(-x^2 + 5)*x + 25/8*arcsin(1/5*sqrt(5)*x)

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Fricas [A]  time = 2.02214, size = 89, normalized size = 1.89 \begin{align*} \frac{1}{8} \,{\left (2 \, x^{3} - 5 \, x\right )} \sqrt{-x^{2} + 5} - \frac{25}{8} \, \arctan \left (\frac{\sqrt{-x^{2} + 5}}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*x^3 - 5*x)*sqrt(-x^2 + 5) - 25/8*arctan(sqrt(-x^2 + 5)/x)

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Sympy [A]  time = 2.63583, size = 122, normalized size = 2.6 \begin{align*} \begin{cases} \frac{i x^{5}}{4 \sqrt{x^{2} - 5}} - \frac{15 i x^{3}}{8 \sqrt{x^{2} - 5}} + \frac{25 i x}{8 \sqrt{x^{2} - 5}} - \frac{25 i \operatorname{acosh}{\left (\frac{\sqrt{5} x}{5} \right )}}{8} & \text{for}\: \frac{\left |{x^{2}}\right |}{5} > 1 \\- \frac{x^{5}}{4 \sqrt{5 - x^{2}}} + \frac{15 x^{3}}{8 \sqrt{5 - x^{2}}} - \frac{25 x}{8 \sqrt{5 - x^{2}}} + \frac{25 \operatorname{asin}{\left (\frac{\sqrt{5} x}{5} \right )}}{8} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((I*x**5/(4*sqrt(x**2 - 5)) - 15*I*x**3/(8*sqrt(x**2 - 5)) + 25*I*x/(8*sqrt(x**2 - 5)) - 25*I*acosh(s
qrt(5)*x/5)/8, Abs(x**2)/5 > 1), (-x**5/(4*sqrt(5 - x**2)) + 15*x**3/(8*sqrt(5 - x**2)) - 25*x/(8*sqrt(5 - x**
2)) + 25*asin(sqrt(5)*x/5)/8, True))

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Giac [A]  time = 1.06833, size = 39, normalized size = 0.83 \begin{align*} \frac{1}{8} \,{\left (2 \, x^{2} - 5\right )} \sqrt{-x^{2} + 5} x + \frac{25}{8} \, \arcsin \left (\frac{1}{5} \, \sqrt{5} x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*x^2 - 5)*sqrt(-x^2 + 5)*x + 25/8*arcsin(1/5*sqrt(5)*x)