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

Optimal. Leaf size=65 \[ -\frac {25}{16} \sqrt {5-x^2} x+\frac {1}{6} \sqrt {5-x^2} x^5-\frac {5}{24} \sqrt {5-x^2} x^3+\frac {125}{16} \sin ^{-1}\left (\frac {x}{\sqrt {5}}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {279, 321, 216} \[ \frac {1}{6} \sqrt {5-x^2} x^5-\frac {5}{24} \sqrt {5-x^2} x^3-\frac {25}{16} \sqrt {5-x^2} x+\frac {125}{16} \sin ^{-1}\left (\frac {x}{\sqrt {5}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^4*Sqrt[5 - x^2],x]

[Out]

(-25*x*Sqrt[5 - x^2])/16 - (5*x^3*Sqrt[5 - x^2])/24 + (x^5*Sqrt[5 - x^2])/6 + (125*ArcSin[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 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]

Rubi steps

\begin {align*} \int x^4 \sqrt {5-x^2} \, dx &=\frac {1}{6} x^5 \sqrt {5-x^2}+\frac {5}{6} \int \frac {x^4}{\sqrt {5-x^2}} \, dx\\ &=-\frac {5}{24} x^3 \sqrt {5-x^2}+\frac {1}{6} x^5 \sqrt {5-x^2}+\frac {25}{8} \int \frac {x^2}{\sqrt {5-x^2}} \, dx\\ &=-\frac {25}{16} x \sqrt {5-x^2}-\frac {5}{24} x^3 \sqrt {5-x^2}+\frac {1}{6} x^5 \sqrt {5-x^2}+\frac {125}{16} \int \frac {1}{\sqrt {5-x^2}} \, dx\\ &=-\frac {25}{16} x \sqrt {5-x^2}-\frac {5}{24} x^3 \sqrt {5-x^2}+\frac {1}{6} x^5 \sqrt {5-x^2}+\frac {125}{16} \sin ^{-1}\left (\frac {x}{\sqrt {5}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 40, normalized size = 0.62 \[ \frac {1}{48} \left (x \sqrt {5-x^2} \left (8 x^4-10 x^2-75\right )+375 \sin ^{-1}\left (\frac {x}{\sqrt {5}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*Sqrt[5 - x^2],x]

[Out]

(x*Sqrt[5 - x^2]*(-75 - 10*x^2 + 8*x^4) + 375*ArcSin[x/Sqrt[5]])/48

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IntegrateAlgebraic [C]  time = 0.05, size = 54, normalized size = 0.83 \[ \frac {1}{48} \sqrt {5-x^2} \left (8 x^5-10 x^3-75 x\right )+\frac {125}{16} i \log \left (\sqrt {5-x^2}-i x\right ) \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^4*Sqrt[5 - x^2],x]

[Out]

(Sqrt[5 - x^2]*(-75*x - 10*x^3 + 8*x^5))/48 + ((125*I)/16)*Log[(-I)*x + Sqrt[5 - x^2]]

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fricas [A]  time = 0.53, size = 42, normalized size = 0.65 \[ \frac {1}{48} \, {\left (8 \, x^{5} - 10 \, x^{3} - 75 \, x\right )} \sqrt {-x^{2} + 5} - \frac {125}{16} \, \arctan \left (\frac {\sqrt {-x^{2} + 5}}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(8*x^5 - 10*x^3 - 75*x)*sqrt(-x^2 + 5) - 125/16*arctan(sqrt(-x^2 + 5)/x)

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giac [A]  time = 0.60, size = 36, normalized size = 0.55 \[ \frac {1}{48} \, {\left (2 \, {\left (4 \, x^{2} - 5\right )} x^{2} - 75\right )} \sqrt {-x^{2} + 5} x + \frac {125}{16} \, \arcsin \left (\frac {1}{5} \, \sqrt {5} x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(2*(4*x^2 - 5)*x^2 - 75)*sqrt(-x^2 + 5)*x + 125/16*arcsin(1/5*sqrt(5)*x)

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maple [A]  time = 0.31, size = 40, normalized size = 0.62

method	result	size
risch	\(-\frac {x \left (8 x^{4}-10 x^{2}-75\right ) \left (x^{2}-5\right )}{48 \sqrt {-x^{2}+5}}+\frac {125 \arcsin \left (\frac {x \sqrt {5}}{5}\right )}{16}\)	\(40\)
default	\(-\frac {x^{3} \left (-x^{2}+5\right )^{\frac {3}{2}}}{6}-\frac {5 x \left (-x^{2}+5\right )^{\frac {3}{2}}}{8}+\frac {25 x \sqrt {-x^{2}+5}}{16}+\frac {125 \arcsin \left (\frac {x \sqrt {5}}{5}\right )}{16}\)	\(49\)
meijerg	\(\frac {125 i \left (\frac {i \sqrt {\pi }\, x \sqrt {5}\, \left (-\frac {8}{5} x^{4}+2 x^{2}+15\right ) \sqrt {-\frac {x^{2}}{5}+1}}{300}-\frac {i \sqrt {\pi }\, \arcsin \left (\frac {x \sqrt {5}}{5}\right )}{4}\right )}{4 \sqrt {\pi }}\)	\(52\)
trager	\(\frac {x \left (8 x^{4}-10 x^{2}-75\right ) \sqrt {-x^{2}+5}}{48}+\frac {125 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{2}+5}+x \right )}{16}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*x*(8*x^4-10*x^2-75)*(x^2-5)/(-x^2+5)^(1/2)+125/16*arcsin(1/5*x*5^(1/2))

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maxima [A]  time = 1.22, size = 48, normalized size = 0.74 \[ -\frac {1}{6} \, {\left (-x^{2} + 5\right )}^{\frac {3}{2}} x^{3} - \frac {5}{8} \, {\left (-x^{2} + 5\right )}^{\frac {3}{2}} x + \frac {25}{16} \, \sqrt {-x^{2} + 5} x + \frac {125}{16} \, \arcsin \left (\frac {1}{5} \, \sqrt {5} x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(-x^2 + 5)^(3/2)*x^3 - 5/8*(-x^2 + 5)^(3/2)*x + 25/16*sqrt(-x^2 + 5)*x + 125/16*arcsin(1/5*sqrt(5)*x)

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mupad [B]  time = 0.03, size = 35, normalized size = 0.54 \[ \frac {125\,\mathrm {asin}\left (\frac {\sqrt {5}\,x}{5}\right )}{16}-\sqrt {5-x^2}\,\left (-\frac {x^5}{6}+\frac {5\,x^3}{24}+\frac {25\,x}{16}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(5 - x^2)^(1/2),x)

[Out]

(125*asin((5^(1/2)*x)/5))/16 - (5 - x^2)^(1/2)*((25*x)/16 + (5*x^3)/24 - x^5/6)

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sympy [A]  time = 4.51, size = 155, normalized size = 2.38 \[ \begin {cases} \frac {i x^{7}}{6 \sqrt {x^{2} - 5}} - \frac {25 i x^{5}}{24 \sqrt {x^{2} - 5}} - \frac {25 i x^{3}}{48 \sqrt {x^{2} - 5}} + \frac {125 i x}{16 \sqrt {x^{2} - 5}} - \frac {125 i \operatorname {acosh}{\left (\frac {\sqrt {5} x}{5} \right )}}{16} & \text {for}\: \frac {\left |{x^{2}}\right |}{5} > 1 \\- \frac {x^{7}}{6 \sqrt {5 - x^{2}}} + \frac {25 x^{5}}{24 \sqrt {5 - x^{2}}} + \frac {25 x^{3}}{48 \sqrt {5 - x^{2}}} - \frac {125 x}{16 \sqrt {5 - x^{2}}} + \frac {125 \operatorname {asin}{\left (\frac {\sqrt {5} x}{5} \right )}}{16} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((I*x**7/(6*sqrt(x**2 - 5)) - 25*I*x**5/(24*sqrt(x**2 - 5)) - 25*I*x**3/(48*sqrt(x**2 - 5)) + 125*I*x
/(16*sqrt(x**2 - 5)) - 125*I*acosh(sqrt(5)*x/5)/16, Abs(x**2)/5 > 1), (-x**7/(6*sqrt(5 - x**2)) + 25*x**5/(24*
sqrt(5 - x**2)) + 25*x**3/(48*sqrt(5 - x**2)) - 125*x/(16*sqrt(5 - x**2)) + 125*asin(sqrt(5)*x/5)/16, True))

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