3.1021 \(\int \frac {x^3}{\sqrt [3]{1-x^2} (3+x^2)^2} \, dx\)

Optimal. Leaf size=101 \[ \frac {3 \left (1-x^2\right )^{2/3}}{8 \left (x^2+3\right )}-\frac {3 \log \left (x^2+3\right )}{16\ 2^{2/3}}+\frac {9 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}+\frac {3 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{2-2 x^2}+1}{\sqrt {3}}\right )}{8\ 2^{2/3}} \]

[Out]

3/8*(-x^2+1)^(2/3)/(x^2+3)-3/32*ln(x^2+3)*2^(1/3)+9/32*ln(2^(2/3)-(-x^2+1)^(1/3))*2^(1/3)+3/16*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 = 0.07, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {446, 78, 55, 617, 204, 31} \[ \frac {3 \left (1-x^2\right )^{2/3}}{8 \left (x^2+3\right )}-\frac {3 \log \left (x^2+3\right )}{16\ 2^{2/3}}+\frac {9 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}+\frac {3 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{2-2 x^2}+1}{\sqrt {3}}\right )}{8\ 2^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(1 - x^2)^(2/3))/(8*(3 + x^2)) + (3*Sqrt[3]*ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3]])/(8*2^(2/3)) - (3*Log[3
+ x^2])/(16*2^(2/3)) + (9*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 78

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

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

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Mathematica [A]  time = 0.11, size = 97, normalized size = 0.96 \[ \frac {3}{32} \left (\frac {4 \left (1-x^2\right )^{2/3}}{x^2+3}-\sqrt [3]{2} \log \left (x^2+3\right )+3 \sqrt [3]{2} \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )+2 \sqrt [3]{2} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{2-2 x^2}+1}{\sqrt {3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.23, size = 121, normalized size = 1.20 \[ \frac {3 \, {\left (4 \cdot 4^{\frac {1}{6}} \sqrt {3} {\left (x^{2} + 3\right )} \arctan \left (\frac {1}{6} \cdot 4^{\frac {1}{6}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - 4^{\frac {2}{3}} {\left (x^{2} + 3\right )} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + 2 \cdot 4^{\frac {2}{3}} {\left (x^{2} + 3\right )} \log \left (-4^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) + 8 \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right )}}{64 \, {\left (x^{2} + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.41, size = 104, normalized size = 1.03 \[ \frac {3}{32} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {3}{64} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + \frac {3}{32} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {1}{3}} - {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) + \frac {3 \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{8 \, {\left (x^{2} + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 5.05, size = 758, normalized size = 7.50 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*(x^2-1)/(x^2+3)/(-x^2+1)^(1/3)+3/16*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*x^2*RootOf(_Z^3-2)^3*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^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)-42*RootOf(_Z^3-2)^2*(-x^2+1
)^(1/3)-60*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*x^2-5*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))-3/16*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*x^2*RootOf(_Z^3-2)^3*RootOf(RootOf(_Z^3-2)^2+4*_
Z*RootOf(_Z^3-2)+16*_Z^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)-42*RootOf(_Z^3-2)^2*(-x^2+1)^(1/3)-8*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*x^2+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))*R
ootOf(_Z^3-2)-3/4*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*x^2*Roo
tOf(_Z^3-2)^3*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^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)-42*RootOf(_Z^3-2)^2*(-x^2+1)^(1/3)-8*RootOf(RootOf(_Z^3-2)^2+4*_Z*
RootOf(_Z^3-2)+16*_Z^2)*x^2+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))*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)

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maxima [A]  time = 1.95, size = 104, normalized size = 1.03 \[ \frac {3}{32} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {3}{64} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + \frac {3}{32} \cdot 4^{\frac {2}{3}} \log \left (-4^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) + \frac {3 \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{8 \, {\left (x^{2} + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.86, size = 126, normalized size = 1.25 \[ \frac {3\,2^{1/3}\,\ln \left (\frac {81\,{\left (1-x^2\right )}^{1/3}}{64}-\frac {81\,2^{2/3}}{64}\right )}{16}+\frac {3\,{\left (1-x^2\right )}^{2/3}}{8\,\left (x^2+3\right )}+\frac {3\,2^{1/3}\,\ln \left (\frac {81\,{\left (1-x^2\right )}^{1/3}}{64}-\frac {81\,2^{2/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{256}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{32}-\frac {3\,2^{1/3}\,\ln \left (\frac {81\,{\left (1-x^2\right )}^{1/3}}{64}-\frac {81\,2^{2/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{256}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{32} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*2^(1/3)*log((81*(1 - x^2)^(1/3))/64 - (81*2^(2/3))/64))/16 + (3*(1 - x^2)^(2/3))/(8*(x^2 + 3)) + (3*2^(1/3)
*log((81*(1 - x^2)^(1/3))/64 - (81*2^(2/3)*(3^(1/2)*1i - 1)^2)/256)*(3^(1/2)*1i - 1))/32 - (3*2^(1/3)*log((81*
(1 - x^2)^(1/3))/64 - (81*2^(2/3)*(3^(1/2)*1i + 1)^2)/256)*(3^(1/2)*1i + 1))/32

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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