3.5.44 \(\int x^3 (-\sqrt {1-x}-\sqrt {1+x}) (\sqrt {1-x}+\sqrt {1+x}) \, dx\) [444]

Optimal. Leaf size=38 \[ -\frac {x^4}{2}+\frac {2}{3} \left (1-x^2\right )^{3/2}-\frac {2}{5} \left (1-x^2\right )^{5/2} \]

[Out]

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

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Rubi [A]
time = 0.24, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {6820, 6874, 272, 45} \begin {gather*} -\frac {x^4}{2}-\frac {2}{5} \left (1-x^2\right )^{5/2}+\frac {2}{3} \left (1-x^2\right )^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(-Sqrt[1 - x] - Sqrt[1 + x])*(Sqrt[1 - x] + Sqrt[1 + x]),x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 6820

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

Rule 6874

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

Rubi steps

\begin {align*} \int x^3 \left (-\sqrt {1-x}-\sqrt {1+x}\right ) \left (\sqrt {1-x}+\sqrt {1+x}\right ) \, dx &=-\int x^3 \left (\sqrt {1-x}+\sqrt {1+x}\right )^2 \, dx\\ &=-\int \left (2 x^3+2 x^3 \sqrt {1-x^2}\right ) \, dx\\ &=-\frac {x^4}{2}-2 \int x^3 \sqrt {1-x^2} \, dx\\ &=-\frac {x^4}{2}-\text {Subst}\left (\int \sqrt {1-x} x \, dx,x,x^2\right )\\ &=-\frac {x^4}{2}-\text {Subst}\left (\int \left (\sqrt {1-x}-(1-x)^{3/2}\right ) \, dx,x,x^2\right )\\ &=-\frac {x^4}{2}+\frac {2}{3} \left (1-x^2\right )^{3/2}-\frac {2}{5} \left (1-x^2\right )^{5/2}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 44, normalized size = 1.16 \begin {gather*} -\frac {1}{30} \left (-1+x^2\right ) \left (15+8 \sqrt {1-x^2}+3 x^2 \left (5+4 \sqrt {1-x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(-Sqrt[1 - x] - Sqrt[1 + x])*(Sqrt[1 - x] + Sqrt[1 + x]),x]

[Out]

-1/30*((-1 + x^2)*(15 + 8*Sqrt[1 - x^2] + 3*x^2*(5 + 4*Sqrt[1 - x^2])))

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Maple [A]
time = 0.26, size = 33, normalized size = 0.87

method	result	size
default	\(-\frac {x^{4}}{2}-\frac {2 \sqrt {1-x}\, \sqrt {1+x}\, \left (x^{2}-1\right ) \left (3 x^{2}+2\right )}{15}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(-(1-x)^(1/2)-(1+x)^(1/2))*((1-x)^(1/2)+(1+x)^(1/2)),x,method=_RETURNVERBOSE)

[Out]

-1/2*x^4-2/15*(1-x)^(1/2)*(1+x)^(1/2)*(x^2-1)*(3*x^2+2)

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Maxima [A]
time = 0.48, size = 31, normalized size = 0.82 \begin {gather*} -\frac {1}{2} \, x^{4} + \frac {2}{5} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} x^{2} + \frac {4}{15} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^4 + 2/5*(-x^2 + 1)^(3/2)*x^2 + 4/15*(-x^2 + 1)^(3/2)

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Fricas [A]
time = 0.33, size = 32, normalized size = 0.84 \begin {gather*} -\frac {1}{2} \, x^{4} - \frac {2}{15} \, {\left (3 \, x^{4} - x^{2} - 2\right )} \sqrt {x + 1} \sqrt {-x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^4 - 2/15*(3*x^4 - x^2 - 2)*sqrt(x + 1)*sqrt(-x + 1)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(-(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 [B] Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (28) = 56\).
time = 3.18, size = 77, normalized size = 2.03 \begin {gather*} -\frac {1}{2} \, x^{4} - \frac {1}{60} \, {\left ({\left (2 \, {\left (3 \, {\left (4 \, x - 17\right )} {\left (x + 1\right )} + 133\right )} {\left (x + 1\right )} - 295\right )} {\left (x + 1\right )} + 195\right )} \sqrt {x + 1} \sqrt {-x + 1} - \frac {1}{12} \, {\left ({\left (2 \, {\left (3 \, x - 10\right )} {\left (x + 1\right )} + 43\right )} {\left (x + 1\right )} - 39\right )} \sqrt {x + 1} \sqrt {-x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^4 - 1/60*((2*(3*(4*x - 17)*(x + 1) + 133)*(x + 1) - 295)*(x + 1) + 195)*sqrt(x + 1)*sqrt(-x + 1) - 1/12
*((2*(3*x - 10)*(x + 1) + 43)*(x + 1) - 39)*sqrt(x + 1)*sqrt(-x + 1)

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Mupad [B]
time = 3.06, size = 42, normalized size = 1.11 \begin {gather*} \sqrt {1-x}\,\left (\frac {4\,\sqrt {x+1}}{15}+\frac {2\,x^2\,\sqrt {x+1}}{15}-\frac {2\,x^4\,\sqrt {x+1}}{5}\right )-\frac {x^4}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1 - x)^(1/2)*((4*(x + 1)^(1/2))/15 + (2*x^2*(x + 1)^(1/2))/15 - (2*x^4*(x + 1)^(1/2))/5) - x^4/2

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