3.1020 \(\int \frac {x^5}{\sqrt [3]{1-x^2} (3+x^2)^2} \, dx\)

Optimal. Leaf size=116 \[ -\frac {3}{4} \left (1-x^2\right )^{2/3}-\frac {9 \left (1-x^2\right )^{2/3}}{8 \left (x^2+3\right )}+\frac {21 \log \left (x^2+3\right )}{16\ 2^{2/3}}-\frac {63 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}-\frac {21 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{2-2 x^2}+1}{\sqrt {3}}\right )}{8\ 2^{2/3}} \]

[Out]

-3/4*(-x^2+1)^(2/3)-9/8*(-x^2+1)^(2/3)/(x^2+3)+21/32*ln(x^2+3)*2^(1/3)-63/32*ln(2^(2/3)-(-x^2+1)^(1/3))*2^(1/3
)-21/16*arctan(1/3*(1+(-2*x^2+2)^(1/3))*3^(1/2))*3^(1/2)*2^(1/3)

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {446, 89, 80, 55, 617, 204, 31} \[ -\frac {3}{4} \left (1-x^2\right )^{2/3}-\frac {9 \left (1-x^2\right )^{2/3}}{8 \left (x^2+3\right )}+\frac {21 \log \left (x^2+3\right )}{16\ 2^{2/3}}-\frac {63 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}-\frac {21 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{2-2 x^2}+1}{\sqrt {3}}\right )}{8\ 2^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(1 - x^2)^(2/3))/4 - (9*(1 - x^2)^(2/3))/(8*(3 + x^2)) - (21*Sqrt[3]*ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3
]])/(8*2^(2/3)) + (21*Log[3 + x^2])/(16*2^(2/3)) - (63*Log[2^(2/3) - (1 - x^2)^(1/3)])/(16*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 55

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 80

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

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 204

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

Rule 446

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 617

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 )^2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [3]{1-x} (3+x)^2} \, dx,x,x^2\right )\\ &=-\frac {9 \left (1-x^2\right )^{2/3}}{8 \left (3+x^2\right )}+\frac {1}{8} \operatorname {Subst}\left (\int \frac {-9+4 x}{\sqrt [3]{1-x} (3+x)} \, dx,x,x^2\right )\\ &=-\frac {3}{4} \left (1-x^2\right )^{2/3}-\frac {9 \left (1-x^2\right )^{2/3}}{8 \left (3+x^2\right )}-\frac {21}{8} \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} (3+x)} \, dx,x,x^2\right )\\ &=-\frac {3}{4} \left (1-x^2\right )^{2/3}-\frac {9 \left (1-x^2\right )^{2/3}}{8 \left (3+x^2\right )}+\frac {21 \log \left (3+x^2\right )}{16\ 2^{2/3}}-\frac {63}{16} \operatorname {Subst}\left (\int \frac {1}{2 \sqrt [3]{2}+2^{2/3} x+x^2} \, dx,x,\sqrt [3]{1-x^2}\right )+\frac {63 \operatorname {Subst}\left (\int \frac {1}{2^{2/3}-x} \, dx,x,\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}\\ &=-\frac {3}{4} \left (1-x^2\right )^{2/3}-\frac {9 \left (1-x^2\right )^{2/3}}{8 \left (3+x^2\right )}+\frac {21 \log \left (3+x^2\right )}{16\ 2^{2/3}}-\frac {63 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}+\frac {63 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\sqrt [3]{2-2 x^2}\right )}{8\ 2^{2/3}}\\ &=-\frac {3}{4} \left (1-x^2\right )^{2/3}-\frac {9 \left (1-x^2\right )^{2/3}}{8 \left (3+x^2\right )}-\frac {21 \sqrt {3} \tan ^{-1}\left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )}{8\ 2^{2/3}}+\frac {21 \log \left (3+x^2\right )}{16\ 2^{2/3}}-\frac {63 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.17, size = 110, normalized size = 0.95 \[ \frac {3}{32} \left (-8 \left (1-x^2\right )^{2/3}-\frac {12 \left (1-x^2\right )^{2/3}}{x^2+3}+7 \sqrt [3]{2} \log \left (x^2+3\right )-21 \sqrt [3]{2} \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )-14 \sqrt [3]{2} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{2-2 x^2}+1}{\sqrt {3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(-8*(1 - x^2)^(2/3) - (12*(1 - x^2)^(2/3))/(3 + x^2) - 14*2^(1/3)*Sqrt[3]*ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sq
rt[3]] + 7*2^(1/3)*Log[3 + x^2] - 21*2^(1/3)*Log[2^(2/3) - (1 - x^2)^(1/3)]))/32

________________________________________________________________________________________

fricas [A]  time = 0.75, size = 153, normalized size = 1.32 \[ -\frac {3 \, {\left (28 \cdot 4^{\frac {1}{6}} \sqrt {3} \left (-1\right )^{\frac {1}{3}} {\left (x^{2} + 3\right )} \arctan \left (\frac {1}{6} \cdot 4^{\frac {1}{6}} \sqrt {3} {\left (2 \, \left (-1\right )^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} - 4^{\frac {1}{3}}\right )}\right ) + 7 \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{2} + 3\right )} \log \left (4^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} - 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) - 14 \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{2} + 3\right )} \log \left (-4^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) + 8 \, {\left (2 \, x^{2} + 9\right )} {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right )}}{64 \, {\left (x^{2} + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/64*(28*4^(1/6)*sqrt(3)*(-1)^(1/3)*(x^2 + 3)*arctan(1/6*4^(1/6)*sqrt(3)*(2*(-1)^(1/3)*(-x^2 + 1)^(1/3) - 4^(
1/3))) + 7*4^(2/3)*(-1)^(1/3)*(x^2 + 3)*log(4^(1/3)*(-1)^(2/3)*(-x^2 + 1)^(1/3) - 4^(2/3)*(-1)^(1/3) + (-x^2 +
1)^(2/3)) - 14*4^(2/3)*(-1)^(1/3)*(x^2 + 3)*log(-4^(1/3)*(-1)^(2/3) + (-x^2 + 1)^(1/3)) + 8*(2*x^2 + 9)*(-x^2
+ 1)^(2/3))/(x^2 + 3)

________________________________________________________________________________________

giac [A]  time = 0.42, size = 115, normalized size = 0.99 \[ -\frac {21}{32} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) + \frac {21}{64} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) - \frac {21}{32} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {1}{3}} - {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) - \frac {3}{4} \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}} - \frac {9 \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{8 \, {\left (x^{2} + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-21/32*4^(2/3)*sqrt(3)*arctan(1/12*4^(2/3)*sqrt(3)*(4^(1/3) + 2*(-x^2 + 1)^(1/3))) + 21/64*4^(2/3)*log(4^(2/3)
+ 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^(2/3)) - 21/32*4^(2/3)*log(4^(1/3) - (-x^2 + 1)^(1/3)) - 3/4*(-x^2 +
1)^(2/3) - 9/8*(-x^2 + 1)^(2/3)/(x^2 + 3)

________________________________________________________________________________________

maple [C]  time = 6.77, size = 490, normalized size = 4.22 \[ \frac {21 \RootOf \left (16 \textit {\_Z}^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}\right ) \ln \left (\frac {64 x^{2} \RootOf \left (16 \textit {\_Z}^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}\right )^{2} \RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+24 x^{2} \RootOf \left (16 \textit {\_Z}^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{3}+2\right )^{3}-40 x^{2} \RootOf \left (16 \textit {\_Z}^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}\right )-15 x^{2} \RootOf \left (\textit {\_Z}^{3}+2\right )+168 \left (-x^{2}+1\right )^{\frac {1}{3}} \RootOf \left (16 \textit {\_Z}^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{3}+2\right )+168 \RootOf \left (16 \textit {\_Z}^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}\right )+63 \RootOf \left (\textit {\_Z}^{3}+2\right )+42 \left (-x^{2}+1\right )^{\frac {2}{3}}}{x^{2}+3}\right )}{4}+\frac {21 \RootOf \left (\textit {\_Z}^{3}+2\right ) \ln \left (-\frac {48 x^{2} \RootOf \left (16 \textit {\_Z}^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}\right )^{2} \RootOf \left (\textit {\_Z}^{3}+2\right )^{2}+8 x^{2} \RootOf \left (16 \textit {\_Z}^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{3}+2\right )^{3}+6 x^{2} \RootOf \left (16 \textit {\_Z}^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}\right )+x^{2} \RootOf \left (\textit {\_Z}^{3}+2\right )-84 \left (-x^{2}+1\right )^{\frac {1}{3}} \RootOf \left (16 \textit {\_Z}^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{3}+2\right )-126 \RootOf \left (16 \textit {\_Z}^{2}+4 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+2\right )+\RootOf \left (\textit {\_Z}^{3}+2\right )^{2}\right )-21 \RootOf \left (\textit {\_Z}^{3}+2\right )-21 \left (-x^{2}+1\right )^{\frac {2}{3}}}{x^{2}+3}\right )}{16}+\frac {\frac {3}{4} x^{4}+\frac {21}{8} x^{2}-\frac {27}{8}}{\left (x^{2}+3\right ) \left (-x^{2}+1\right )^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*(2*x^4+7*x^2-9)/(x^2+3)/(-x^2+1)^(1/3)+21/16*RootOf(_Z^3+2)*ln(-(48*RootOf(RootOf(_Z^3+2)^2+4*_Z*RootOf(_Z
^3+2)+16*_Z^2)^2*RootOf(_Z^3+2)^2*x^2+8*RootOf(RootOf(_Z^3+2)^2+4*_Z*RootOf(_Z^3+2)+16*_Z^2)*RootOf(_Z^3+2)^3*
x^2-84*RootOf(_Z^3+2)*(-x^2+1)^(1/3)*RootOf(RootOf(_Z^3+2)^2+4*_Z*RootOf(_Z^3+2)+16*_Z^2)+6*RootOf(RootOf(_Z^3
+2)^2+4*_Z*RootOf(_Z^3+2)+16*_Z^2)*x^2+RootOf(_Z^3+2)*x^2-21*(-x^2+1)^(2/3)-126*RootOf(RootOf(_Z^3+2)^2+4*_Z*R
ootOf(_Z^3+2)+16*_Z^2)-21*RootOf(_Z^3+2))/(x^2+3))+21/4*RootOf(RootOf(_Z^3+2)^2+4*_Z*RootOf(_Z^3+2)+16*_Z^2)*l
n((64*RootOf(RootOf(_Z^3+2)^2+4*_Z*RootOf(_Z^3+2)+16*_Z^2)^2*RootOf(_Z^3+2)^2*x^2+24*RootOf(RootOf(_Z^3+2)^2+4
*_Z*RootOf(_Z^3+2)+16*_Z^2)*RootOf(_Z^3+2)^3*x^2+168*RootOf(_Z^3+2)*(-x^2+1)^(1/3)*RootOf(RootOf(_Z^3+2)^2+4*_
Z*RootOf(_Z^3+2)+16*_Z^2)-40*RootOf(RootOf(_Z^3+2)^2+4*_Z*RootOf(_Z^3+2)+16*_Z^2)*x^2-15*RootOf(_Z^3+2)*x^2+42
*(-x^2+1)^(2/3)+168*RootOf(RootOf(_Z^3+2)^2+4*_Z*RootOf(_Z^3+2)+16*_Z^2)+63*RootOf(_Z^3+2))/(x^2+3))

________________________________________________________________________________________

maxima [A]  time = 1.96, size = 115, normalized size = 0.99 \[ -\frac {21}{32} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) + \frac {21}{64} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) - \frac {21}{32} \cdot 4^{\frac {2}{3}} \log \left (-4^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) - \frac {3}{4} \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}} - \frac {9 \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{8 \, {\left (x^{2} + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-21/32*4^(2/3)*sqrt(3)*arctan(1/12*4^(2/3)*sqrt(3)*(4^(1/3) + 2*(-x^2 + 1)^(1/3))) + 21/64*4^(2/3)*log(4^(2/3)
+ 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^(2/3)) - 21/32*4^(2/3)*log(-4^(1/3) + (-x^2 + 1)^(1/3)) - 3/4*(-x^2 +
1)^(2/3) - 9/8*(-x^2 + 1)^(2/3)/(x^2 + 3)

________________________________________________________________________________________

mupad [B]  time = 0.89, size = 137, normalized size = 1.18 \[ -\frac {21\,2^{1/3}\,\ln \left (\frac {3969\,{\left (1-x^2\right )}^{1/3}}{64}-\frac {3969\,2^{2/3}}{64}\right )}{16}-\frac {9\,{\left (1-x^2\right )}^{2/3}}{8\,\left (x^2+3\right )}-\frac {3\,{\left (1-x^2\right )}^{2/3}}{4}-\frac {21\,2^{1/3}\,\ln \left (\frac {3969\,{\left (1-x^2\right )}^{1/3}}{64}-\frac {3969\,2^{2/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{256}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{32}+\frac {21\,2^{1/3}\,\ln \left (\frac {3969\,{\left (1-x^2\right )}^{1/3}}{64}-\frac {3969\,2^{2/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{256}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{32} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(21*2^(1/3)*log((3969*(1 - x^2)^(1/3))/64 - (3969*2^(2/3)*(3^(1/2)*1i + 1)^2)/256)*(3^(1/2)*1i + 1))/32 - (9*(
1 - x^2)^(2/3))/(8*(x^2 + 3)) - (3*(1 - x^2)^(2/3))/4 - (21*2^(1/3)*log((3969*(1 - x^2)^(1/3))/64 - (3969*2^(2
/3)*(3^(1/2)*1i - 1)^2)/256)*(3^(1/2)*1i - 1))/32 - (21*2^(1/3)*log((3969*(1 - x^2)^(1/3))/64 - (3969*2^(2/3))
/64))/16

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________