3.1.12 \(\int (x-x^2)^{3/2} \, dx\) [12]

Optimal. Leaf size=51 \[ -\frac {3}{64} (1-2 x) \sqrt {x-x^2}-\frac {1}{8} (1-2 x) \left (x-x^2\right )^{3/2}-\frac {3}{128} \sin ^{-1}(1-2 x) \]

[Out]

-1/8*(1-2*x)*(-x^2+x)^(3/2)+3/128*arcsin(-1+2*x)-3/64*(1-2*x)*(-x^2+x)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {626, 633, 222} \begin {gather*} -\frac {3}{128} \text {ArcSin}(1-2 x)-\frac {1}{8} (1-2 x) \left (x-x^2\right )^{3/2}-\frac {3}{64} (1-2 x) \sqrt {x-x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x - x^2)^(3/2),x]

[Out]

(-3*(1 - 2*x)*Sqrt[x - x^2])/64 - ((1 - 2*x)*(x - x^2)^(3/2))/8 - (3*ArcSin[1 - 2*x])/128

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 626

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1
))), x] - Dist[p*((b^2 - 4*a*c)/(2*c*(2*p + 1))), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 633

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*(-4*(c/(b^2 - 4*a*c)))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rubi steps

\begin {align*} \int \left (x-x^2\right )^{3/2} \, dx &=-\frac {1}{8} (1-2 x) \left (x-x^2\right )^{3/2}+\frac {3}{16} \int \sqrt {x-x^2} \, dx\\ &=-\frac {3}{64} (1-2 x) \sqrt {x-x^2}-\frac {1}{8} (1-2 x) \left (x-x^2\right )^{3/2}+\frac {3}{128} \int \frac {1}{\sqrt {x-x^2}} \, dx\\ &=-\frac {3}{64} (1-2 x) \sqrt {x-x^2}-\frac {1}{8} (1-2 x) \left (x-x^2\right )^{3/2}-\frac {3}{128} \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,1-2 x\right )\\ &=-\frac {3}{64} (1-2 x) \sqrt {x-x^2}-\frac {1}{8} (1-2 x) \left (x-x^2\right )^{3/2}-\frac {3}{128} \sin ^{-1}(1-2 x)\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 67, normalized size = 1.31 \begin {gather*} \frac {x \left (-3+x+26 x^2-40 x^3+16 x^4\right )+6 \sqrt {-1+x} \sqrt {x} \tanh ^{-1}\left (\frac {\sqrt {-1+x}}{-1+\sqrt {x}}\right )}{64 \sqrt {-((-1+x) x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x - x^2)^(3/2),x]

[Out]

(x*(-3 + x + 26*x^2 - 40*x^3 + 16*x^4) + 6*Sqrt[-1 + x]*Sqrt[x]*ArcTanh[Sqrt[-1 + x]/(-1 + Sqrt[x])])/(64*Sqrt
[-((-1 + x)*x)])

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Maple [A]
time = 0.41, size = 42, normalized size = 0.82

method	result	size
risch	\(\frac {\left (16 x^{3}-24 x^{2}+2 x +3\right ) x \left (x -1\right )}{64 \sqrt {-x \left (x -1\right )}}+\frac {3 \arcsin \left (2 x -1\right )}{128}\)	\(39\)
default	\(-\frac {\left (1-2 x \right ) \left (-x^{2}+x \right )^{\frac {3}{2}}}{8}+\frac {3 \arcsin \left (2 x -1\right )}{128}-\frac {3 \left (1-2 x \right ) \sqrt {-x^{2}+x}}{64}\)	\(42\)
meijerg	\(-\frac {3 i \left (-\frac {i \sqrt {\pi }\, \sqrt {x}\, \left (80 x^{3}-120 x^{2}+10 x +15\right ) \sqrt {1-x}}{240}+\frac {i \sqrt {\pi }\, \arcsin \left (\sqrt {x}\right )}{16}\right )}{4 \sqrt {\pi }}\)	\(49\)
trager	\(\left (-\frac {1}{4} x^{3}+\frac {3}{8} x^{2}-\frac {1}{32} x -\frac {3}{64}\right ) \sqrt {-x^{2}+x}-\frac {3 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (2 x \RootOf \left (\textit {\_Z}^{2}+1\right )+2 \sqrt {-x^{2}+x}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right )}{128}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^2+x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/8*(1-2*x)*(-x^2+x)^(3/2)+3/128*arcsin(2*x-1)-3/64*(1-2*x)*(-x^2+x)^(1/2)

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Maxima [A]
time = 0.53, size = 55, normalized size = 1.08 \begin {gather*} \frac {1}{4} \, {\left (-x^{2} + x\right )}^{\frac {3}{2}} x - \frac {1}{8} \, {\left (-x^{2} + x\right )}^{\frac {3}{2}} + \frac {3}{32} \, \sqrt {-x^{2} + x} x - \frac {3}{64} \, \sqrt {-x^{2} + x} + \frac {3}{128} \, \arcsin \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+x)^(3/2),x, algorithm="maxima")

[Out]

1/4*(-x^2 + x)^(3/2)*x - 1/8*(-x^2 + x)^(3/2) + 3/32*sqrt(-x^2 + x)*x - 3/64*sqrt(-x^2 + x) + 3/128*arcsin(2*x
- 1)

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Fricas [A]
time = 1.42, size = 43, normalized size = 0.84 \begin {gather*} -\frac {1}{64} \, {\left (16 \, x^{3} - 24 \, x^{2} + 2 \, x + 3\right )} \sqrt {-x^{2} + x} - \frac {3}{64} \, \arctan \left (\frac {\sqrt {-x^{2} + x}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+x)^(3/2),x, algorithm="fricas")

[Out]

-1/64*(16*x^3 - 24*x^2 + 2*x + 3)*sqrt(-x^2 + x) - 3/64*arctan(sqrt(-x^2 + x)/x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (- x^{2} + x\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**2+x)**(3/2),x)

[Out]

Integral((-x**2 + x)**(3/2), x)

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Giac [A]
time = 1.91, size = 35, normalized size = 0.69 \begin {gather*} -\frac {1}{64} \, {\left (2 \, {\left (4 \, {\left (2 \, x - 3\right )} x + 1\right )} x + 3\right )} \sqrt {-x^{2} + x} + \frac {3}{128} \, \arcsin \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+x)^(3/2),x, algorithm="giac")

[Out]

-1/64*(2*(4*(2*x - 3)*x + 1)*x + 3)*sqrt(-x^2 + x) + 3/128*arcsin(2*x - 1)

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Mupad [B]
time = 0.19, size = 39, normalized size = 0.76 \begin {gather*} \frac {3\,\mathrm {asin}\left (2\,x-1\right )}{128}+\frac {3\,\sqrt {x-x^2}\,\left (\frac {x}{2}-\frac {1}{4}\right )}{16}+\frac {{\left (x-x^2\right )}^{3/2}\,\left (x-\frac {1}{2}\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - x^2)^(3/2),x)

[Out]

(3*asin(2*x - 1))/128 + (3*(x - x^2)^(1/2)*(x/2 - 1/4))/16 + ((x - x^2)^(3/2)*(x - 1/2))/4

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