3.3.50 \(\int x^4 \sqrt {5-x^2} \, dx\) [250]

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

[Out]

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

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Rubi [A]
time = 0.01, 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 = {285, 327, 222} \begin {gather*} \frac {125}{16} \text {ArcSin}\left (\frac {x}{\sqrt {5}}\right )-\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 \end {gather*}

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 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^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 = 0.04, size = 47, normalized size = 0.72 \begin {gather*} \frac {1}{48} x \sqrt {5-x^2} \left (-75-10 x^2+8 x^4\right )+\frac {125}{16} \tan ^{-1}\left (\frac {x}{\sqrt {5-x^2}}\right ) \end {gather*}

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))/48 + (125*ArcTan[x/Sqrt[5 - x^2]])/16

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Maple [A]
time = 0.08, size = 49, normalized size = 0.75

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]

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

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Maxima [A]
time = 1.39, size = 48, normalized size = 0.74 \begin {gather*} -\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 ) \end {gather*}

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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Fricas [A]
time = 0.53, size = 42, normalized size = 0.65 \begin {gather*} \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 ) \end {gather*}

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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Sympy [C] Result contains complex when optimal does not.
time = 4.51, size = 153, normalized size = 2.35 \begin {gather*} \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}\: \left |{x^{2}}\right | > 5 \\- \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} \end {gather*}

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), (-x**7/(6*sqrt(5 - x**2)) + 25*x**5/(24*sq
rt(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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Giac [A]
time = 0.80, size = 36, normalized size = 0.55 \begin {gather*} \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 ) \end {gather*}

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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Mupad [B]
time = 0.03, size = 35, normalized size = 0.54 \begin {gather*} \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 ) \end {gather*}

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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