\(\int \frac {x^5}{\sqrt [3]{1-x^2} (3+x^2)^2} \, dx\) [1020]

Optimal result

Integrand size = 22, antiderivative size = 116 \[ \int \frac {x^5}{\sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=-\frac {3}{4} \left (1-x^2\right )^{2/3}-\frac {9 \left (1-x^2\right )^{2/3}}{8 \left (3+x^2\right )}-\frac {21 \sqrt {3} \arctan \left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )}{8\ 2^{2/3}}+\frac {21 \log \left (3+x^2\right )}{16\ 2^{2/3}}-\frac {63 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}} \]

[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] (verified)

Time = 0.06 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {457, 91, 81, 57, 631, 210, 31} \[ \int \frac {x^5}{\sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=-\frac {21 \sqrt {3} \arctan \left (\frac {\sqrt [3]{2-2 x^2}+1}{\sqrt {3}}\right )}{8\ 2^{2/3}}-\frac {3}{4} \left (1-x^2\right )^{2/3}-\frac {9 \left (1-x^2\right )^{2/3}}{8 \left (x^2+3\right )}+\frac {21 \log \left (x^2+3\right )}{16\ 2^{2/3}}-\frac {63 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}} \]

[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 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 81

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 91

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 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*} \text {integral}& = \frac {1}{2} \text {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} \text {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} \text {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} \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 {63 \text {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 \text {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] (verified)

Time = 0.25 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.03 \[ \int \frac {x^5}{\sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\frac {3}{32} \left (-\frac {4 \left (1-x^2\right )^{2/3} \left (9+2 x^2\right )}{3+x^2}-14 \sqrt [3]{2} \sqrt {3} \arctan \left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )-14 \sqrt [3]{2} \log \left (-2+\sqrt [3]{2-2 x^2}\right )+7 \sqrt [3]{2} \log \left (4+2 \sqrt [3]{2-2 x^2}+\left (2-2 x^2\right )^{2/3}\right )\right ) \]

[In]

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

[Out]

(3*((-4*(1 - x^2)^(2/3)*(9 + 2*x^2))/(3 + x^2) - 14*2^(1/3)*Sqrt[3]*ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3]] -
14*2^(1/3)*Log[-2 + (2 - 2*x^2)^(1/3)] + 7*2^(1/3)*Log[4 + 2*(2 - 2*x^2)^(1/3) + (2 - 2*x^2)^(2/3)]))/32

Maple [A] (verified)

Time = 8.04 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.98

method	result	size
pseudoelliptic	\(\frac {21 \left (x^{2}+3\right ) \left (-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (1+2^{\frac {1}{3}} \left (-x^{2}+1\right )^{\frac {1}{3}}\right )}{3}\right )+\ln \left (\left (-x^{2}+1\right )^{\frac {2}{3}}+2^{\frac {2}{3}} \left (-x^{2}+1\right )^{\frac {1}{3}}+2 \,2^{\frac {1}{3}}\right )-2 \ln \left (\left (-x^{2}+1\right )^{\frac {1}{3}}-2^{\frac {2}{3}}\right )\right ) 2^{\frac {1}{3}}-12 \left (-x^{2}+1\right )^{\frac {2}{3}} \left (2 x^{2}+9\right )}{32 x^{2}+96}\)	\(114\)
trager	\(\text {Expression too large to display}\)	\(656\)
risch	\(\text {Expression too large to display}\)	\(662\)

[In]

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

[Out]

(21*(x^2+3)*(-2*3^(1/2)*arctan(1/3*3^(1/2)*(1+2^(1/3)*(-x^2+1)^(1/3)))+ln((-x^2+1)^(2/3)+2^(2/3)*(-x^2+1)^(1/3
)+2*2^(1/3))-2*ln((-x^2+1)^(1/3)-2^(2/3)))*2^(1/3)-12*(-x^2+1)^(2/3)*(2*x^2+9))/(32*x^2+96)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.32 \[ \int \frac {x^5}{\sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=-\frac {3 \, {\left (28 \cdot 4^{\frac {1}{6}} \sqrt {3} \left (-1\right )^{\frac {1}{3}} {\left (x^{2} + 3\right )} \arctan \left (\frac {1}{6} \cdot 4^{\frac {1}{6}} \sqrt {3} {\left (2 \, \left (-1\right )^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} - 4^{\frac {1}{3}}\right )}\right ) + 7 \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{2} + 3\right )} \log \left (4^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} - 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) - 14 \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{2} + 3\right )} \log \left (-4^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) + 8 \, {\left (2 \, x^{2} + 9\right )} {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right )}}{64 \, {\left (x^{2} + 3\right )}} \]

[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)

Sympy [F]

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

[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)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.99 \[ \int \frac {x^5}{\sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=-\frac {21}{32} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) + \frac {21}{64} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) - \frac {21}{32} \cdot 4^{\frac {2}{3}} \log \left (-4^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) - \frac {3}{4} \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}} - \frac {9 \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{8 \, {\left (x^{2} + 3\right )}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.99 \[ \int \frac {x^5}{\sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=-\frac {21}{32} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) + \frac {21}{64} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) - \frac {21}{32} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {1}{3}} - {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) - \frac {3}{4} \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}} - \frac {9 \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{8 \, {\left (x^{2} + 3\right )}} \]

[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)

Mupad [B] (verification not implemented)

Time = 5.30 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.18 \[ \int \frac {x^5}{\sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=-\frac {21\,2^{1/3}\,\ln \left (\frac {3969\,{\left (1-x^2\right )}^{1/3}}{64}-\frac {3969\,2^{2/3}}{64}\right )}{16}-\frac {9\,{\left (1-x^2\right )}^{2/3}}{8\,\left (x^2+3\right )}-\frac {3\,{\left (1-x^2\right )}^{2/3}}{4}-\frac {21\,2^{1/3}\,\ln \left (\frac {3969\,{\left (1-x^2\right )}^{1/3}}{64}-\frac {3969\,2^{2/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{256}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{32}+\frac {21\,2^{1/3}\,\ln \left (\frac {3969\,{\left (1-x^2\right )}^{1/3}}{64}-\frac {3969\,2^{2/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{256}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{32} \]

[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