3.4.69 \(\int x^2 \sqrt {5-x^2} \, dx\) [369]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {285, 327, 222} \begin {gather*} -\frac {5}{8} \sqrt {5-x^2} x+\frac {1}{4} \sqrt {5-x^2} x^3+\frac {25}{8} \sin ^{-1}\left (\frac {x}{\sqrt {5}}\right ) \end {gather*}

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 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 285

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 327

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 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*}

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Mathematica [A]
time = 0.03, size = 42, normalized size = 0.89 \begin {gather*} \frac {1}{8} x \sqrt {5-x^2} \left (-5+2 x^2\right )+\frac {25}{8} \tan ^{-1}\left (\frac {x}{\sqrt {5-x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[5 - x^2],x]

[Out]

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

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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 3.78, size = 98, normalized size = 2.09 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\frac {I \left (25 x-15 x^3+2 x^5-25 \text {ArcCosh}\left [\frac {\sqrt {5} x}{5}\right ] \sqrt {-5+x^2}\right )}{8 \sqrt {-5+x^2}},\text {Abs}\left [x^2\right ]>5\right \}\right \},\frac {-25 x}{8 \sqrt {5-x^2}}+\frac {15 x^3}{8 \sqrt {5-x^2}}-\frac {x^5}{4 \sqrt {5-x^2}}+\frac {25 \text {ArcSin}\left [\frac {\sqrt {5} x}{5}\right ]}{8}\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[x^2*Sqrt[5 - x^2],x]')

[Out]

Piecewise[{{I / 8 (25 x - 15 x ^ 3 + 2 x ^ 5 - 25 ArcCosh[Sqrt[5] x / 5] Sqrt[-5 + x ^ 2]) / Sqrt[-5 + x ^ 2],
Abs[x ^ 2] > 5}}, -25 x / (8 Sqrt[5 - x ^ 2]) + 15 x ^ 3 / (8 Sqrt[5 - x ^ 2]) - x ^ 5 / (4 Sqrt[5 - x ^ 2])
+ 25 ArcSin[Sqrt[5] x / 5] / 8]

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Maple [A]
time = 0.08, size = 35, normalized size = 0.74

method	result	size
default	\(-\frac {x \left (-x^{2}+5\right )^{\frac {3}{2}}}{4}+\frac {5 x \sqrt {-x^{2}+5}}{8}+\frac {25 \arcsin \left (\frac {x \sqrt {5}}{5}\right )}{8}\)	\(35\)
risch	\(-\frac {x \left (2 x^{2}-5\right ) \left (x^{2}-5\right )}{8 \sqrt {-x^{2}+5}}+\frac {25 \arcsin \left (\frac {x \sqrt {5}}{5}\right )}{8}\)	\(35\)
meijerg	\(-\frac {25 i \left (-\frac {i \sqrt {\pi }\, x \sqrt {5}\, \left (-\frac {6 x^{2}}{5}+3\right ) \sqrt {-\frac {x^{2}}{5}+1}}{30}+\frac {i \sqrt {\pi }\, \arcsin \left (\frac {x \sqrt {5}}{5}\right )}{2}\right )}{4 \sqrt {\pi }}\)	\(47\)
trager	\(\frac {x \left (2 x^{2}-5\right ) \sqrt {-x^{2}+5}}{8}+\frac {25 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{2}+5}+x \right )}{8}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.35, size = 34, normalized size = 0.72 \begin {gather*} -\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 {gather*}

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 = 0.35, size = 37, normalized size = 0.79 \begin {gather*} \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 {gather*}

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 = 1.65, size = 121, normalized size = 2.57 \begin {gather*} \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}\: \left |{x^{2}}\right | > 5 \\- \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 {gather*}

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), (-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 = 0.00, size = 37, normalized size = 0.79 \begin {gather*} 2 \left (\frac {1}{8} x x-\frac {5}{16}\right ) x \sqrt {-x^{2}+5}+\frac {25}{8} \arcsin \left (\frac {x}{\sqrt {5}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.04, size = 30, normalized size = 0.64 \begin {gather*} \frac {25\,\mathrm {asin}\left (\frac {\sqrt {5}\,x}{5}\right )}{8}-\sqrt {5-x^2}\,\left (\frac {5\,x}{8}-\frac {x^3}{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(5 - x^2)^(1/2),x)

[Out]

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

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