3.445 \(\int x^2 (-\sqrt{1-x}-\sqrt{1+x}) (\sqrt{1-x}+\sqrt{1+x}) \, dx\)

Optimal. Leaf size=48 \[ -\frac{1}{2} \sqrt{1-x^2} x^3-\frac{2 x^3}{3}+\frac{1}{4} \sqrt{1-x^2} x-\frac{1}{4} \sin ^{-1}(x) \]

[Out]

(-2*x^3)/3 + (x*Sqrt[1 - x^2])/4 - (x^3*Sqrt[1 - x^2])/2 - ArcSin[x]/4

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Rubi [A]  time = 0.241919, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.119, Rules used = {6688, 6742, 279, 321, 216} \[ -\frac{1}{2} \sqrt{1-x^2} x^3-\frac{2 x^3}{3}+\frac{1}{4} \sqrt{1-x^2} x-\frac{1}{4} \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*(-Sqrt[1 - x] - Sqrt[1 + x])*(Sqrt[1 - x] + Sqrt[1 + x]),x]

[Out]

(-2*x^3)/3 + (x*Sqrt[1 - x^2])/4 - (x^3*Sqrt[1 - x^2])/2 - ArcSin[x]/4

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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]

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]

Rubi steps

\begin{align*} \int x^2 \left (-\sqrt{1-x}-\sqrt{1+x}\right ) \left (\sqrt{1-x}+\sqrt{1+x}\right ) \, dx &=-\int x^2 \left (\sqrt{1-x}+\sqrt{1+x}\right )^2 \, dx\\ &=-\int \left (2 x^2+2 x^2 \sqrt{1-x^2}\right ) \, dx\\ &=-\frac{2 x^3}{3}-2 \int x^2 \sqrt{1-x^2} \, dx\\ &=-\frac{2 x^3}{3}-\frac{1}{2} x^3 \sqrt{1-x^2}-\frac{1}{2} \int \frac{x^2}{\sqrt{1-x^2}} \, dx\\ &=-\frac{2 x^3}{3}+\frac{1}{4} x \sqrt{1-x^2}-\frac{1}{2} x^3 \sqrt{1-x^2}-\frac{1}{4} \int \frac{1}{\sqrt{1-x^2}} \, dx\\ &=-\frac{2 x^3}{3}+\frac{1}{4} x \sqrt{1-x^2}-\frac{1}{2} x^3 \sqrt{1-x^2}-\frac{1}{4} \sin ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.0403239, size = 43, normalized size = 0.9 \[ \frac{1}{12} \left (-\left (6 \sqrt{1-x^2}+8\right ) x^3+3 \sqrt{1-x^2} x-3 \sin ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(-Sqrt[1 - x] - Sqrt[1 + x])*(Sqrt[1 - x] + Sqrt[1 + x]),x]

[Out]

(3*x*Sqrt[1 - x^2] - x^3*(8 + 6*Sqrt[1 - x^2]) - 3*ArcSin[x])/12

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Maple [A]  time = 0.001, size = 59, normalized size = 1.2 \begin{align*} -{\frac{2\,{x}^{3}}{3}}-{\frac{1}{4}\sqrt{1-x}\sqrt{1+x} \left ( 2\,{x}^{3}\sqrt{-{x}^{2}+1}-x\sqrt{-{x}^{2}+1}+\arcsin \left ( x \right ) \right ){\frac{1}{\sqrt{-{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.50632, size = 46, normalized size = 0.96 \begin{align*} -\frac{2}{3} \, x^{3} + \frac{1}{2} \,{\left (-x^{2} + 1\right )}^{\frac{3}{2}} x - \frac{1}{4} \, \sqrt{-x^{2} + 1} x - \frac{1}{4} \, \arcsin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.05418, size = 135, normalized size = 2.81 \begin{align*} -\frac{2}{3} \, x^{3} - \frac{1}{4} \,{\left (2 \, x^{3} - x\right )} \sqrt{x + 1} \sqrt{-x + 1} + \frac{1}{2} \, \arctan \left (\frac{\sqrt{x + 1} \sqrt{-x + 1} - 1}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*x^3 - 1/4*(2*x^3 - x)*sqrt(x + 1)*sqrt(-x + 1) + 1/2*arctan((sqrt(x + 1)*sqrt(-x + 1) - 1)/x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.21582, size = 84, normalized size = 1.75 \begin{align*} -\frac{2}{3} \,{\left (x + 1\right )}^{3} + 2 \,{\left (x + 1\right )}^{2} - \frac{1}{4} \,{\left ({\left (2 \,{\left (x + 1\right )}{\left (x - 2\right )} + 5\right )}{\left (x + 1\right )} - 1\right )} \sqrt{x + 1} \sqrt{-x + 1} - 2 \, x - \frac{1}{2} \, \arcsin \left (\frac{1}{2} \, \sqrt{2} \sqrt{x + 1}\right ) - 2 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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