3.128 \(\int \sqrt {1-(1-x^2)^3} \, dx\)

Optimal. Leaf size=86 \[ -\frac {\sqrt {x^6-3 x^4+3 x^2} \left (3-2 x^2\right )}{8 x}-\frac {3 \sqrt {x^6-3 x^4+3 x^2} \sinh ^{-1}\left (\frac {3-2 x^2}{\sqrt {3}}\right )}{16 x \sqrt {x^4-3 x^2+3}} \]

[Out]

-1/8*(-2*x^2+3)*(x^6-3*x^4+3*x^2)^(1/2)/x-3/16*arcsinh(1/3*(-2*x^2+3)*3^(1/2))*(x^6-3*x^4+3*x^2)^(1/2)/x/(x^4-
3*x^2+3)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {1996, 1903, 1107, 612, 619, 215} \[ -\frac {\sqrt {x^6-3 x^4+3 x^2} \left (3-2 x^2\right )}{8 x}-\frac {3 \sqrt {x^6-3 x^4+3 x^2} \sinh ^{-1}\left (\frac {3-2 x^2}{\sqrt {3}}\right )}{16 x \sqrt {x^4-3 x^2+3}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - (1 - x^2)^3],x]

[Out]

-((3 - 2*x^2)*Sqrt[3*x^2 - 3*x^4 + x^6])/(8*x) - (3*Sqrt[3*x^2 - 3*x^4 + x^6]*ArcSinh[(3 - 2*x^2)/Sqrt[3]])/(1
6*x*Sqrt[3 - 3*x^2 + x^4])

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 612

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 619

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]

Rule 1107

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

Rule 1903

Int[Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[Sqrt[a*x^q + b*x^n + c*x^(
2*n - q)]/(x^(q/2)*Sqrt[a + b*x^(n - q) + c*x^(2*(n - q))]), Int[x^(q/2)*Sqrt[a + b*x^(n - q) + c*x^(2*(n - q)
)], x], x] /; FreeQ[{a, b, c, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q]

Rule 1996

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && GeneralizedTrinomialQ[u, x] &&  !Gen
eralizedTrinomialMatchQ[u, x]

Rubi steps

\begin {align*} \int \sqrt {1-\left (1-x^2\right )^3} \, dx &=\int \sqrt {3 x^2-3 x^4+x^6} \, dx\\ &=\frac {\sqrt {3 x^2-3 x^4+x^6} \int x \sqrt {3-3 x^2+x^4} \, dx}{x \sqrt {3-3 x^2+x^4}}\\ &=\frac {\sqrt {3 x^2-3 x^4+x^6} \operatorname {Subst}\left (\int \sqrt {3-3 x+x^2} \, dx,x,x^2\right )}{2 x \sqrt {3-3 x^2+x^4}}\\ &=-\frac {\left (3-2 x^2\right ) \sqrt {3 x^2-3 x^4+x^6}}{8 x}+\frac {\left (3 \sqrt {3 x^2-3 x^4+x^6}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {3-3 x+x^2}} \, dx,x,x^2\right )}{16 x \sqrt {3-3 x^2+x^4}}\\ &=-\frac {\left (3-2 x^2\right ) \sqrt {3 x^2-3 x^4+x^6}}{8 x}+\frac {\left (\sqrt {3} \sqrt {3 x^2-3 x^4+x^6}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{3}}} \, dx,x,-3+2 x^2\right )}{16 x \sqrt {3-3 x^2+x^4}}\\ &=-\frac {\left (3-2 x^2\right ) \sqrt {3 x^2-3 x^4+x^6}}{8 x}-\frac {3 \sqrt {3 x^2-3 x^4+x^6} \sinh ^{-1}\left (\frac {3-2 x^2}{\sqrt {3}}\right )}{16 x \sqrt {3-3 x^2+x^4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.00, size = 70, normalized size = 0.81 \[ \frac {x \left (4 x^6-18 x^4+30 x^2+3 \sqrt {x^4-3 x^2+3} \sinh ^{-1}\left (\frac {2 x^2-3}{\sqrt {3}}\right )-18\right )}{16 \sqrt {x^2 \left (x^4-3 x^2+3\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - (1 - x^2)^3],x]

[Out]

(x*(-18 + 30*x^2 - 18*x^4 + 4*x^6 + 3*Sqrt[3 - 3*x^2 + x^4]*ArcSinh[(-3 + 2*x^2)/Sqrt[3]]))/(16*Sqrt[x^2*(3 -
3*x^2 + x^4)])

________________________________________________________________________________________

fricas [A]  time = 0.59, size = 70, normalized size = 0.81 \[ -\frac {12 \, x \log \left (-\frac {2 \, x^{3} - 3 \, x - 2 \, \sqrt {x^{6} - 3 \, x^{4} + 3 \, x^{2}}}{x}\right ) - 8 \, \sqrt {x^{6} - 3 \, x^{4} + 3 \, x^{2}} {\left (2 \, x^{2} - 3\right )} - 9 \, x}{64 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-(-x^2+1)^3)^(1/2),x, algorithm="fricas")

[Out]

-1/64*(12*x*log(-(2*x^3 - 3*x - 2*sqrt(x^6 - 3*x^4 + 3*x^2))/x) - 8*sqrt(x^6 - 3*x^4 + 3*x^2)*(2*x^2 - 3) - 9*
x)/x

________________________________________________________________________________________

giac [A]  time = 0.38, size = 69, normalized size = 0.80 \[ \frac {1}{16} \, {\left (2 \, \sqrt {x^{4} - 3 \, x^{2} + 3} {\left (2 \, x^{2} - 3\right )} - 3 \, \log \left (-2 \, x^{2} + 2 \, \sqrt {x^{4} - 3 \, x^{2} + 3} + 3\right )\right )} \mathrm {sgn}\relax (x) + \frac {3}{16} \, {\left (2 \, \sqrt {3} + \log \left (2 \, \sqrt {3} + 3\right )\right )} \mathrm {sgn}\relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-(-x^2+1)^3)^(1/2),x, algorithm="giac")

[Out]

1/16*(2*sqrt(x^4 - 3*x^2 + 3)*(2*x^2 - 3) - 3*log(-2*x^2 + 2*sqrt(x^4 - 3*x^2 + 3) + 3))*sgn(x) + 3/16*(2*sqrt
(3) + log(2*sqrt(3) + 3))*sgn(x)

________________________________________________________________________________________

maple [A]  time = 0.00, size = 81, normalized size = 0.94 \[ \frac {\sqrt {x^{6}-3 x^{4}+3 x^{2}}\, \left (4 \sqrt {x^{4}-3 x^{2}+3}\, x^{2}+3 \arcsinh \left (\frac {\sqrt {3}\, \left (2 x^{2}-3\right )}{3}\right )-6 \sqrt {x^{4}-3 x^{2}+3}\right )}{16 \sqrt {x^{4}-3 x^{2}+3}\, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-(-x^2+1)^3)^(1/2),x)

[Out]

1/16*(x^6-3*x^4+3*x^2)^(1/2)*(4*(x^4-3*x^2+3)^(1/2)*x^2+3*arcsinh(1/3*3^(1/2)*(2*x^2-3))-6*(x^4-3*x^2+3)^(1/2)
)/x/(x^4-3*x^2+3)^(1/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {{\left (x^{2} - 1\right )}^{3} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-(-x^2+1)^3)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt((x^2 - 1)^3 + 1), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {{\left (x^2-1\right )}^3+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2 - 1)^3 + 1)^(1/2),x)

[Out]

int(((x^2 - 1)^3 + 1)^(1/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {1 - \left (1 - x^{2}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-(-x**2+1)**3)**(1/2),x)

[Out]

Integral(sqrt(1 - (1 - x**2)**3), x)

________________________________________________________________________________________