3.11.8 \(\int \frac {x^5}{\sqrt [3]{1-x^2} (3+x^2)} \, dx\) [1008]

Optimal. Leaf size=109 \[ \frac {3}{2} \left (1-x^2\right )^{2/3}+\frac {3}{10} \left (1-x^2\right )^{5/3}+\frac {9 \sqrt {3} \tan ^{-1}\left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )}{2\ 2^{2/3}}-\frac {9 \log \left (3+x^2\right )}{4\ 2^{2/3}}+\frac {27 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}} \]

[Out]

3/2*(-x^2+1)^(2/3)+3/10*(-x^2+1)^(5/3)-9/8*ln(x^2+3)*2^(1/3)+27/8*ln(2^(2/3)-(-x^2+1)^(1/3))*2^(1/3)+9/4*arcta
n(1/3*(1+(-2*x^2+2)^(1/3))*3^(1/2))*3^(1/2)*2^(1/3)

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Rubi [A]
time = 0.07, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {457, 90, 57, 631, 210, 31} \begin {gather*} \frac {9 \sqrt {3} \text {ArcTan}\left (\frac {\sqrt [3]{2-2 x^2}+1}{\sqrt {3}}\right )}{2\ 2^{2/3}}+\frac {3}{10} \left (1-x^2\right )^{5/3}+\frac {3}{2} \left (1-x^2\right )^{2/3}-\frac {9 \log \left (x^2+3\right )}{4\ 2^{2/3}}+\frac {27 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/((1 - x^2)^(1/3)*(3 + x^2)),x]

[Out]

(3*(1 - x^2)^(2/3))/2 + (3*(1 - x^2)^(5/3))/10 + (9*Sqrt[3]*ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3]])/(2*2^(2/3
)) - (9*Log[3 + x^2])/(4*2^(2/3)) + (27*Log[2^(2/3) - (1 - x^2)^(1/3)])/(4*2^(2/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {x^5}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {x^2}{\sqrt [3]{1-x} (3+x)} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \left (-\frac {2}{\sqrt [3]{1-x}}-(1-x)^{2/3}+\frac {9}{\sqrt [3]{1-x} (3+x)}\right ) \, dx,x,x^2\right )\\ &=\frac {3}{2} \left (1-x^2\right )^{2/3}+\frac {3}{10} \left (1-x^2\right )^{5/3}+\frac {9}{2} \text {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} (3+x)} \, dx,x,x^2\right )\\ &=\frac {3}{2} \left (1-x^2\right )^{2/3}+\frac {3}{10} \left (1-x^2\right )^{5/3}-\frac {9 \log \left (3+x^2\right )}{4\ 2^{2/3}}+\frac {27}{4} \text {Subst}\left (\int \frac {1}{2 \sqrt [3]{2}+2^{2/3} x+x^2} \, dx,x,\sqrt [3]{1-x^2}\right )-\frac {27 \text {Subst}\left (\int \frac {1}{2^{2/3}-x} \, dx,x,\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}}\\ &=\frac {3}{2} \left (1-x^2\right )^{2/3}+\frac {3}{10} \left (1-x^2\right )^{5/3}-\frac {9 \log \left (3+x^2\right )}{4\ 2^{2/3}}+\frac {27 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}}-\frac {27 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\sqrt [3]{2-2 x^2}\right )}{2\ 2^{2/3}}\\ &=\frac {3}{2} \left (1-x^2\right )^{2/3}+\frac {3}{10} \left (1-x^2\right )^{5/3}+\frac {9 \sqrt {3} \tan ^{-1}\left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )}{2\ 2^{2/3}}-\frac {9 \log \left (3+x^2\right )}{4\ 2^{2/3}}+\frac {27 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 121, normalized size = 1.11 \begin {gather*} \frac {1}{40} \left (72 \left (1-x^2\right )^{2/3}-12 x^2 \left (1-x^2\right )^{2/3}+90 \sqrt [3]{2} \sqrt {3} \tan ^{-1}\left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )+90 \sqrt [3]{2} \log \left (-2+\sqrt [3]{2-2 x^2}\right )-45 \sqrt [3]{2} \log \left (4+2 \sqrt [3]{2-2 x^2}+\left (2-2 x^2\right )^{2/3}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/((1 - x^2)^(1/3)*(3 + x^2)),x]

[Out]

(72*(1 - x^2)^(2/3) - 12*x^2*(1 - x^2)^(2/3) + 90*2^(1/3)*Sqrt[3]*ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3]] + 90
*2^(1/3)*Log[-2 + (2 - 2*x^2)^(1/3)] - 45*2^(1/3)*Log[4 + 2*(2 - 2*x^2)^(1/3) + (2 - 2*x^2)^(2/3)])/40

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 7.51, size = 482, normalized size = 4.42

method	result	size
risch	\(\frac {3 \left (x^{2}-6\right ) \left (x^{2}-1\right )}{10 \left (-x^{2}+1\right )^{\frac {1}{3}}}+\frac {9 \RootOf \left (\textit {\_Z}^{3}-2\right ) \ln \left (-\frac {8 \RootOf \left (\textit {\_Z}^{3}-2\right )^{3} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+16 \textit {\_Z}^{2}\right ) x^{2}+48 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+16 \textit {\_Z}^{2}\right )^{2} \RootOf \left (\textit {\_Z}^{3}-2\right )^{2} x^{2}-84 \left (-x^{2}+1\right )^{\frac {1}{3}} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+16 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}-2\right )-\RootOf \left (\textit {\_Z}^{3}-2\right ) x^{2}-6 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+16 \textit {\_Z}^{2}\right ) x^{2}-21 \left (-x^{2}+1\right )^{\frac {2}{3}}+21 \RootOf \left (\textit {\_Z}^{3}-2\right )+126 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+16 \textit {\_Z}^{2}\right )}{x^{2}+3}\right )}{4}+9 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+16 \textit {\_Z}^{2}\right ) \ln \left (\frac {24 \RootOf \left (\textit {\_Z}^{3}-2\right )^{3} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+16 \textit {\_Z}^{2}\right ) x^{2}+64 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+16 \textit {\_Z}^{2}\right )^{2} \RootOf \left (\textit {\_Z}^{3}-2\right )^{2} x^{2}+168 \left (-x^{2}+1\right )^{\frac {1}{3}} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+16 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}-2\right )+15 \RootOf \left (\textit {\_Z}^{3}-2\right ) x^{2}+40 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+16 \textit {\_Z}^{2}\right ) x^{2}+42 \left (-x^{2}+1\right )^{\frac {2}{3}}-63 \RootOf \left (\textit {\_Z}^{3}-2\right )-168 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-2\right )^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-2\right )+16 \textit {\_Z}^{2}\right )}{x^{2}+3}\right )\)	\(482\)
trager	\(\text {Expression too large to display}\)	\(649\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(-x^2+1)^(1/3)/(x^2+3),x,method=_RETURNVERBOSE)

[Out]

3/10*(x^2-6)*(x^2-1)/(-x^2+1)^(1/3)+9/4*RootOf(_Z^3-2)*ln(-(8*RootOf(_Z^3-2)^3*RootOf(RootOf(_Z^3-2)^2+4*_Z*Ro
otOf(_Z^3-2)+16*_Z^2)*x^2+48*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)^2*RootOf(_Z^3-2)^2*x^2-84*(-
x^2+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*RootOf(_Z^3-2)-RootOf(_Z^3-2)*x^2-6*RootOf(R
ootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*x^2-21*(-x^2+1)^(2/3)+21*RootOf(_Z^3-2)+126*RootOf(RootOf(_Z^3-2)
^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2))/(x^2+3))+9*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*ln((24*RootOf
(_Z^3-2)^3*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*x^2+64*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^
3-2)+16*_Z^2)^2*RootOf(_Z^3-2)^2*x^2+168*(-x^2+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*R
ootOf(_Z^3-2)+15*RootOf(_Z^3-2)*x^2+40*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*x^2+42*(-x^2+1)^(2
/3)-63*RootOf(_Z^3-2)-168*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2))/(x^2+3))

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Maxima [A]
time = 0.48, size = 108, normalized size = 0.99 \begin {gather*} \frac {9}{8} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) + \frac {3}{10} \, {\left (-x^{2} + 1\right )}^{\frac {5}{3}} - \frac {9}{16} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + \frac {9}{8} \cdot 4^{\frac {2}{3}} \log \left (-4^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) + \frac {3}{2} \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(-x^2+1)^(1/3)/(x^2+3),x, algorithm="maxima")

[Out]

9/8*4^(2/3)*sqrt(3)*arctan(1/12*4^(2/3)*sqrt(3)*(4^(1/3) + 2*(-x^2 + 1)^(1/3))) + 3/10*(-x^2 + 1)^(5/3) - 9/16
*4^(2/3)*log(4^(2/3) + 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^(2/3)) + 9/8*4^(2/3)*log(-4^(1/3) + (-x^2 + 1)^(1
/3)) + 3/2*(-x^2 + 1)^(2/3)

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Fricas [A]
time = 0.89, size = 102, normalized size = 0.94 \begin {gather*} -\frac {3}{10} \, {\left (x^{2} - 6\right )} {\left (-x^{2} + 1\right )}^{\frac {2}{3}} + \frac {9}{4} \cdot 4^{\frac {1}{6}} \sqrt {3} \arctan \left (\frac {1}{6} \cdot 4^{\frac {1}{6}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {9}{16} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + \frac {9}{8} \cdot 4^{\frac {2}{3}} \log \left (-4^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(-x^2+1)^(1/3)/(x^2+3),x, algorithm="fricas")

[Out]

-3/10*(x^2 - 6)*(-x^2 + 1)^(2/3) + 9/4*4^(1/6)*sqrt(3)*arctan(1/6*4^(1/6)*sqrt(3)*(4^(1/3) + 2*(-x^2 + 1)^(1/3
))) - 9/16*4^(2/3)*log(4^(2/3) + 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^(2/3)) + 9/8*4^(2/3)*log(-4^(1/3) + (-x
^2 + 1)^(1/3))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5}}{\sqrt [3]{- \left (x - 1\right ) \left (x + 1\right )} \left (x^{2} + 3\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(-x**2+1)**(1/3)/(x**2+3),x)

[Out]

Integral(x**5/((-(x - 1)*(x + 1))**(1/3)*(x**2 + 3)), x)

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Giac [A]
time = 0.62, size = 108, normalized size = 0.99 \begin {gather*} \frac {9}{8} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) + \frac {3}{10} \, {\left (-x^{2} + 1\right )}^{\frac {5}{3}} - \frac {9}{16} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + \frac {9}{8} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {1}{3}} - {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) + \frac {3}{2} \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(-x^2+1)^(1/3)/(x^2+3),x, algorithm="giac")

[Out]

9/8*4^(2/3)*sqrt(3)*arctan(1/12*4^(2/3)*sqrt(3)*(4^(1/3) + 2*(-x^2 + 1)^(1/3))) + 3/10*(-x^2 + 1)^(5/3) - 9/16
*4^(2/3)*log(4^(2/3) + 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^(2/3)) + 9/8*4^(2/3)*log(4^(1/3) - (-x^2 + 1)^(1/
3)) + 3/2*(-x^2 + 1)^(2/3)

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Mupad [B]
time = 0.50, size = 128, normalized size = 1.17 \begin {gather*} \frac {9\,2^{1/3}\,\ln \left (\frac {729\,{\left (1-x^2\right )}^{1/3}}{4}-\frac {729\,2^{2/3}}{4}\right )}{4}+\frac {3\,{\left (1-x^2\right )}^{2/3}}{2}+\frac {3\,{\left (1-x^2\right )}^{5/3}}{10}+\frac {9\,2^{1/3}\,\ln \left (\frac {729\,{\left (1-x^2\right )}^{1/3}}{4}-\frac {729\,2^{2/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{8}-\frac {9\,2^{1/3}\,\ln \left (\frac {729\,{\left (1-x^2\right )}^{1/3}}{4}-\frac {729\,2^{2/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(9*2^(1/3)*log((729*(1 - x^2)^(1/3))/4 - (729*2^(2/3))/4))/4 + (3*(1 - x^2)^(2/3))/2 + (3*(1 - x^2)^(5/3))/10
+ (9*2^(1/3)*log((729*(1 - x^2)^(1/3))/4 - (729*2^(2/3)*(3^(1/2)*1i - 1)^2)/16)*(3^(1/2)*1i - 1))/8 - (9*2^(1/
3)*log((729*(1 - x^2)^(1/3))/4 - (729*2^(2/3)*(3^(1/2)*1i + 1)^2)/16)*(3^(1/2)*1i + 1))/8

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