3.3.92 \(\int x^8 \sqrt {\frac {1-x^3}{1+x^3}} \, dx\) [292]

Optimal. Leaf size=113 \[ \frac {1}{2} \sqrt {\frac {1-x^3}{1+x^3}} \left (1+x^3\right )-\frac {1}{6} \sqrt {\frac {1-x^3}{1+x^3}} \left (1+x^3\right )^2-\frac {1}{9} \left (\frac {1-x^3}{1+x^3}\right )^{3/2} \left (1+x^3\right )^3-\frac {1}{3} \tan ^{-1}\left (\sqrt {\frac {1-x^3}{1+x^3}}\right ) \]

[Out]

-1/9*((-x^3+1)/(x^3+1))^(3/2)*(x^3+1)^3-1/3*arctan(((-x^3+1)/(x^3+1))^(1/2))+1/2*(x^3+1)*((-x^3+1)/(x^3+1))^(1
/2)-1/6*(x^3+1)^2*((-x^3+1)/(x^3+1))^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {1981, 1980, 474, 466, 393, 210} \begin {gather*} -\frac {1}{3} \text {ArcTan}\left (\sqrt {\frac {1-x^3}{x^3+1}}\right )-\frac {1}{9} \left (\frac {1-x^3}{x^3+1}\right )^{3/2} \left (x^3+1\right )^3-\frac {1}{6} \sqrt {\frac {1-x^3}{x^3+1}} \left (x^3+1\right )^2+\frac {1}{2} \sqrt {\frac {1-x^3}{x^3+1}} \left (x^3+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8*Sqrt[(1 - x^3)/(1 + x^3)],x]

[Out]

(Sqrt[(1 - x^3)/(1 + x^3)]*(1 + x^3))/2 - (Sqrt[(1 - x^3)/(1 + x^3)]*(1 + x^3)^2)/6 - (((1 - x^3)/(1 + x^3))^(
3/2)*(1 + x^3)^3)/9 - ArcTan[Sqrt[(1 - x^3)/(1 + x^3)]]/3

Rule 210

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

Rule 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 466

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x
*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 474

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(-(b*c - a*
d)^2)*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*b^2*e*n*(p + 1))), x] + Dist[1/(a*b^2*n*(p + 1)), Int[(e*x)^m*(a +
b*x^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a
, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]

Rule 1980

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_Symbol] :> With[{q = Denominator[p]}
, Dist[q*e*(b*c - a*d), Subst[Int[x^(q*(p + 1) - 1)*(((-a)*e + c*x^q)^m/(b*e - d*x^q)^(m + 2)), x], x, (e*((a
+ b*x)/(c + d*x)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e, m}, x] && FractionQ[p] && IntegerQ[m]

Rule 1981

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

Rubi steps

\begin {align*} \int x^8 \sqrt {\frac {1-x^3}{1+x^3}} \, dx &=-\left (\frac {4}{3} \text {Subst}\left (\int \frac {x^2 \left (-1+x^2\right )^2}{\left (-1-x^2\right )^4} \, dx,x,\sqrt {\frac {1-x^3}{1+x^3}}\right )\right )\\ &=-\frac {1}{9} \left (\frac {1-x^3}{1+x^3}\right )^{3/2} \left (1+x^3\right )^3-\frac {2}{9} \text {Subst}\left (\int \frac {x^2 \left (6-6 x^2\right )}{\left (-1-x^2\right )^3} \, dx,x,\sqrt {\frac {1-x^3}{1+x^3}}\right )\\ &=-\frac {1}{6} \sqrt {\frac {1-x^3}{1+x^3}} \left (1+x^3\right )^2-\frac {1}{9} \left (\frac {1-x^3}{1+x^3}\right )^{3/2} \left (1+x^3\right )^3+\frac {1}{18} \text {Subst}\left (\int \frac {12-24 x^2}{\left (-1-x^2\right )^2} \, dx,x,\sqrt {\frac {1-x^3}{1+x^3}}\right )\\ &=\frac {1}{2} \sqrt {\frac {1-x^3}{1+x^3}} \left (1+x^3\right )-\frac {1}{6} \sqrt {\frac {1-x^3}{1+x^3}} \left (1+x^3\right )^2-\frac {1}{9} \left (\frac {1-x^3}{1+x^3}\right )^{3/2} \left (1+x^3\right )^3+\frac {1}{3} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {\frac {1-x^3}{1+x^3}}\right )\\ &=\frac {1}{2} \sqrt {\frac {1-x^3}{1+x^3}} \left (1+x^3\right )-\frac {1}{6} \sqrt {\frac {1-x^3}{1+x^3}} \left (1+x^3\right )^2-\frac {1}{9} \left (\frac {1-x^3}{1+x^3}\right )^{3/2} \left (1+x^3\right )^3-\frac {1}{3} \tan ^{-1}\left (\sqrt {\frac {1-x^3}{1+x^3}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.39, size = 95, normalized size = 0.84 \begin {gather*} \frac {\sqrt {\frac {1-x^3}{1+x^3}} \left (\sqrt {1-x^3} \left (4+x^3-x^6+2 x^9\right )-6 \sqrt {1+x^3} \tan ^{-1}\left (\frac {\sqrt {1-x^3}}{\sqrt {1+x^3}}\right )\right )}{18 \sqrt {1-x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8*Sqrt[(1 - x^3)/(1 + x^3)],x]

[Out]

(Sqrt[(1 - x^3)/(1 + x^3)]*(Sqrt[1 - x^3]*(4 + x^3 - x^6 + 2*x^9) - 6*Sqrt[1 + x^3]*ArcTan[Sqrt[1 - x^3]/Sqrt[
1 + x^3]]))/(18*Sqrt[1 - x^3])

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Maple [A]
time = 0.20, size = 80, normalized size = 0.71

method	result	size
risch	\(\frac {\left (x^{3}+1\right ) \left (2 x^{6}-3 x^{3}+4\right ) \sqrt {-\frac {x^{3}-1}{x^{3}+1}}}{18}-\frac {\arcsin \left (x^{3}\right ) \sqrt {-\frac {x^{3}-1}{x^{3}+1}}\, \sqrt {-\left (x^{3}-1\right ) \left (x^{3}+1\right )}}{6 \left (x^{3}-1\right )}\)	\(80\)
trager	\(\frac {\left (x^{3}+1\right ) \left (2 x^{6}-3 x^{3}+4\right ) \sqrt {-\frac {x^{3}-1}{x^{3}+1}}}{18}+\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {-\frac {x^{3}-1}{x^{3}+1}}\, x^{3}+x^{3}+\RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {-\frac {x^{3}-1}{x^{3}+1}}\right )}{6}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*((-x^3+1)/(x^3+1))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/18*(x^3+1)*(2*x^6-3*x^3+4)*(-(x^3-1)/(x^3+1))^(1/2)-1/6*arcsin(x^3)*(-(x^3-1)/(x^3+1))^(1/2)*(-(x^3-1)*(x^3+
1))^(1/2)/(x^3-1)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*((-x^3+1)/(x^3+1))^(1/2),x, algorithm="maxima")

[Out]

integrate(x^8*sqrt(-(x^3 - 1)/(x^3 + 1)), x)

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Fricas [A]
time = 0.34, size = 65, normalized size = 0.58 \begin {gather*} \frac {1}{18} \, {\left (2 \, x^{9} - x^{6} + x^{3} + 4\right )} \sqrt {-\frac {x^{3} - 1}{x^{3} + 1}} - \frac {1}{3} \, \arctan \left (\frac {{\left (x^{3} + 1\right )} \sqrt {-\frac {x^{3} - 1}{x^{3} + 1}} - 1}{x^{3}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*((-x^3+1)/(x^3+1))^(1/2),x, algorithm="fricas")

[Out]

1/18*(2*x^9 - x^6 + x^3 + 4)*sqrt(-(x^3 - 1)/(x^3 + 1)) - 1/3*arctan(((x^3 + 1)*sqrt(-(x^3 - 1)/(x^3 + 1)) - 1
)/x^3)

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*((-x^3+1)/(x^3+1))^(1/2),x, algorithm="giac")

[Out]

integrate(x^8*sqrt(-(x^3 - 1)/(x^3 + 1)), x)

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Mupad [B]
time = 2.66, size = 101, normalized size = 0.89 \begin {gather*} \frac {2\,\sqrt {-\frac {x^3-1}{x^3+1}}}{9}-\frac {\mathrm {atan}\left (\sqrt {-\frac {x^3-1}{x^3+1}}\right )}{3}+\frac {x^3\,\sqrt {-\frac {x^3-1}{x^3+1}}}{18}-\frac {x^6\,\sqrt {-\frac {x^3-1}{x^3+1}}}{18}+\frac {x^9\,\sqrt {-\frac {x^3-1}{x^3+1}}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(-(x^3 - 1)/(x^3 + 1))^(1/2),x)

[Out]

(2*(-(x^3 - 1)/(x^3 + 1))^(1/2))/9 - atan((-(x^3 - 1)/(x^3 + 1))^(1/2))/3 + (x^3*(-(x^3 - 1)/(x^3 + 1))^(1/2))
/18 - (x^6*(-(x^3 - 1)/(x^3 + 1))^(1/2))/18 + (x^9*(-(x^3 - 1)/(x^3 + 1))^(1/2))/9

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