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

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

[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 = 0.11, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6742, 266, 43} \[ \frac {x^4}{2}+\frac {2}{5} \left (1-x^2\right )^{5/2}-\frac {2}{3} \left (1-x^2\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(Sqrt[1 - x] + Sqrt[1 + x])^2,x]

[Out]

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

Rule 43

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 266

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 6742

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 )^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}+\operatorname {Subst}\left (\int \sqrt {1-x} x \, dx,x,x^2\right )\\ &=\frac {x^4}{2}+\operatorname {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 = 0.05, size = 38, normalized size = 1.00 \[ \frac {x^4}{2}+\frac {2}{5} \left (1-x^2\right )^{5/2}-\frac {2}{3} \left (1-x^2\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(Sqrt[1 - x] + Sqrt[1 + x])^2,x]

[Out]

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

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fricas [A]  time = 0.48, size = 32, normalized size = 0.84 \[ \frac {1}{2} \, x^{4} + \frac {2}{15} \, {\left (3 \, x^{4} - x^{2} - 2\right )} \sqrt {x + 1} \sqrt {-x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*((1-x)^(1/2)+(1+x)^(1/2))^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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giac [B]  time = 0.22, size = 77, normalized size = 2.03 \[ \frac {1}{2} \, x^{4} + \frac {1}{60} \, {\left ({\left (2 \, {\left (3 \, {\left (4 \, x - 17\right )} {\left (x + 1\right )} + 133\right )} {\left (x + 1\right )} - 295\right )} {\left (x + 1\right )} + 195\right )} \sqrt {x + 1} \sqrt {-x + 1} + \frac {1}{12} \, {\left ({\left (2 \, {\left (3 \, x - 10\right )} {\left (x + 1\right )} + 43\right )} {\left (x + 1\right )} - 39\right )} \sqrt {x + 1} \sqrt {-x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*((1-x)^(1/2)+(1+x)^(1/2))^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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maple [A]  time = 0.01, size = 33, normalized size = 0.87 \[ \frac {x^{4}}{2}+\frac {2 \sqrt {x +1}\, \sqrt {-x +1}\, \left (x^{2}-1\right ) \left (3 x^{2}+2\right )}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.31, size = 31, normalized size = 0.82 \[ \frac {1}{2} \, x^{4} - \frac {2}{5} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} x^{2} - \frac {4}{15} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*((1-x)^(1/2)+(1+x)^(1/2))^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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mupad [B]  time = 3.02, size = 45, normalized size = 1.18 \[ \frac {x^4}{2}-\frac {\sqrt {1-x}\,\left (-\frac {2\,x^5}{5}-\frac {2\,x^4}{5}+\frac {2\,x^3}{15}+\frac {2\,x^2}{15}+\frac {4\,x}{15}+\frac {4}{15}\right )}{\sqrt {x+1}} \]

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]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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