\(\int \frac {x}{\sqrt [3]{1-x^2} (3+x^2)} \, dx\) [1010]

Optimal result

Integrand size = 20, antiderivative size = 79 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )}{2\ 2^{2/3}}-\frac {\log \left (3+x^2\right )}{4\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}} \]

[Out]

-1/8*ln(x^2+3)*2^(1/3)+3/8*ln(2^(2/3)-(-x^2+1)^(1/3))*2^(1/3)+1/4*arctan(1/3*(1+(-2*x^2+2)^(1/3))*3^(1/2))*3^(
1/2)*2^(1/3)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {455, 57, 631, 210, 31} \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{2-2 x^2}+1}{\sqrt {3}}\right )}{2\ 2^{2/3}}-\frac {\log \left (x^2+3\right )}{4\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}} \]

[In]

Int[x/((1 - x^2)^(1/3)*(3 + x^2)),x]

[Out]

(Sqrt[3]*ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3]])/(2*2^(2/3)) - Log[3 + x^2]/(4*2^(2/3)) + (3*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 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 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.04 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )+2 \log \left (-2+\sqrt [3]{2-2 x^2}\right )-\log \left (4+2 \sqrt [3]{2-2 x^2}+\left (2-2 x^2\right )^{2/3}\right )}{4\ 2^{2/3}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.33 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.04

method	result	size
pseudoelliptic	\(\frac {2^{\frac {1}{3}} \left (2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (1+2^{\frac {1}{3}} \left (-x^{2}+1\right )^{\frac {1}{3}}\right )}{3}\right )+2 \ln \left (\left (-x^{2}+1\right )^{\frac {1}{3}}-2^{\frac {2}{3}}\right )-\ln \left (\left (-x^{2}+1\right )^{\frac {2}{3}}+2^{\frac {2}{3}} \left (-x^{2}+1\right )^{\frac {1}{3}}+2 \,2^{\frac {1}{3}}\right )\right )}{8}\)	\(82\)
trager	\(\text {Expression too large to display}\)	\(740\)

[In]

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

[Out]

1/8*2^(1/3)*(2*3^(1/2)*arctan(1/3*3^(1/2)*(1+2^(1/3)*(-x^2+1)^(1/3)))+2*ln((-x^2+1)^(1/3)-2^(2/3))-ln((-x^2+1)
^(2/3)+2^(2/3)*(-x^2+1)^(1/3)+2*2^(1/3)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.09 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {1}{4} \cdot 4^{\frac {1}{6}} \sqrt {3} \arctan \left (\frac {1}{6} \cdot 4^{\frac {1}{6}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{16} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{8} \cdot 4^{\frac {2}{3}} \log \left (-4^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 54.01 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.99 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\begin {cases} \sqrt [3]{2} \left (\frac {\log {\left (\sqrt [3]{2 - 2 x^{2}} - 2 \right )}}{4} - \frac {\log {\left (\left (2 - 2 x^{2}\right )^{\frac {2}{3}} + 2 \sqrt [3]{2 - 2 x^{2}} + 4 \right )}}{8} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {\sqrt {3} \left (\sqrt [3]{2 - 2 x^{2}} + 1\right )}{3} \right )}}{4}\right ) & \text {for}\: x > -1 \wedge x < 1 \end {cases} \]

[In]

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

[Out]

Piecewise((2**(1/3)*(log((2 - 2*x**2)**(1/3) - 2)/4 - log((2 - 2*x**2)**(2/3) + 2*(2 - 2*x**2)**(1/3) + 4)/8 +
sqrt(3)*atan(sqrt(3)*((2 - 2*x**2)**(1/3) + 1)/3)/4), (x > -1) & (x < 1)))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.09 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {1}{8} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{16} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{8} \cdot 4^{\frac {2}{3}} \log \left (-4^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.09 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {1}{8} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{16} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{8} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {1}{3}} - {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.35 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.34 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {2^{1/3}\,\ln \left (\frac {9\,{\left (1-x^2\right )}^{1/3}}{4}-\frac {9\,2^{2/3}}{4}\right )}{4}+\frac {2^{1/3}\,\ln \left (\frac {9\,{\left (1-x^2\right )}^{1/3}}{4}-\frac {9\,2^{2/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{8}-\frac {2^{1/3}\,\ln \left (\frac {9\,{\left (1-x^2\right )}^{1/3}}{4}-\frac {9\,2^{2/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{8} \]

[In]

int(x/((1 - x^2)^(1/3)*(x^2 + 3)),x)

[Out]

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