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

Optimal. Leaf size=79 \[ \frac {\sqrt {3} \tan ^{-1}\left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )}{2\ 2^{2/3}}-\frac {\log \left (3+x^2\right )}{4\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}} \]

[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)

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Rubi [A]
time = 0.04, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {455, 57, 631, 210, 31} \begin {gather*} \frac {\sqrt {3} \text {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}} \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]
time = 0.08, size = 82, normalized size = 1.04 \begin {gather*} \frac {2 \sqrt {3} \tan ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[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))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 3.58, size = 736, normalized size = 9.32

method	result	size
trager	\(\text {Expression too large to display}\)	\(736\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*RootOf(_Z^3-2)*ln(-(96*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)^2*RootOf(_Z^3-2)^2*x^2+8*RootO
f(_Z^3-2)^3*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*x^2+168*(-x^2+1)^(1/3)*RootOf(RootOf(_Z^3-2)^
2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*RootOf(_Z^3-2)+60*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*x^2+42*R
ootOf(_Z^3-2)^2*(-x^2+1)^(1/3)+5*RootOf(_Z^3-2)*x^2-42*(-x^2+1)^(2/3)-252*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(
_Z^3-2)+16*_Z^2)-21*RootOf(_Z^3-2))/(x^2+3))-1/4*ln((64*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(_Z^3-2)^3*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*x^2-168*(-x^2+1)
^(1/3)*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*RootOf(_Z^3-2)-8*RootOf(RootOf(_Z^3-2)^2+4*_Z*Root
Of(_Z^3-2)+16*_Z^2)*x^2-42*RootOf(_Z^3-2)^2*(-x^2+1)^(1/3)+RootOf(_Z^3-2)*x^2+42*(-x^2+1)^(2/3)+168*RootOf(Roo
tOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)-21*RootOf(_Z^3-2))/(x^2+3))*RootOf(_Z^3-2)-ln((64*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(_Z^3-2)^3*RootOf(RootOf(_Z^3-2)^2+4*_Z*Roo
tOf(_Z^3-2)+16*_Z^2)*x^2-168*(-x^2+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*RootOf(_Z^3-2
)-8*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*x^2-42*RootOf(_Z^3-2)^2*(-x^2+1)^(1/3)+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)-21*RootOf(_Z^3-2))/(x^2+3))*Ro
otOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)

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Maxima [A]
time = 0.49, size = 86, normalized size = 1.09 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Fricas [A]
time = 0.75, size = 86, normalized size = 1.09 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [A]
time = 41.52, size = 78, normalized size = 0.99 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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Giac [A]
time = 0.55, size = 86, normalized size = 1.09 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Mupad [B]
time = 0.59, size = 106, normalized size = 1.34 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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