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

Optimal. Leaf size=158 \[ \frac {\left (1-x^2\right )^{2/3}}{24 \left (x^2+3\right )}+\frac {5 \log \left (x^2+3\right )}{144\ 2^{2/3}}+\frac {1}{12} \log \left (1-\sqrt [3]{1-x^2}\right )-\frac {5 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{48\ 2^{2/3}}-\frac {5 \tan ^{-1}\left (\frac {\sqrt [3]{2-2 x^2}+1}{\sqrt {3}}\right )}{24\ 2^{2/3} \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{1-x^2}+1}{\sqrt {3}}\right )}{6 \sqrt {3}}-\frac {\log (x)}{18} \]

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Rubi [A]  time = 0.11, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {446, 103, 156, 55, 618, 204, 31, 617} \begin {gather*} \frac {\left (1-x^2\right )^{2/3}}{24 \left (x^2+3\right )}+\frac {5 \log \left (x^2+3\right )}{144\ 2^{2/3}}+\frac {1}{12} \log \left (1-\sqrt [3]{1-x^2}\right )-\frac {5 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{48\ 2^{2/3}}-\frac {5 \tan ^{-1}\left (\frac {\sqrt [3]{2-2 x^2}+1}{\sqrt {3}}\right )}{24\ 2^{2/3} \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{1-x^2}+1}{\sqrt {3}}\right )}{6 \sqrt {3}}-\frac {\log (x)}{18} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(1 - x^2)^(2/3)/(24*(3 + x^2)) - (5*ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3]])/(24*2^(2/3)*Sqrt[3]) + ArcTan[(1
+ 2*(1 - x^2)^(1/3))/Sqrt[3]]/(6*Sqrt[3]) - Log[x]/18 + (5*Log[3 + x^2])/(144*2^(2/3)) + Log[1 - (1 - x^2)^(1/
3)]/12 - (5*Log[2^(2/3) - (1 - x^2)^(1/3)])/(48*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 103

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

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

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]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {1}{x \sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} x (3+x)^2} \, dx,x,x^2\right )\\ &=\frac {\left (1-x^2\right )^{2/3}}{24 \left (3+x^2\right )}+\frac {1}{24} \operatorname {Subst}\left (\int \frac {4-\frac {x}{3}}{\sqrt [3]{1-x} x (3+x)} \, dx,x,x^2\right )\\ &=\frac {\left (1-x^2\right )^{2/3}}{24 \left (3+x^2\right )}+\frac {1}{18} \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} x} \, dx,x,x^2\right )-\frac {5}{72} \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} (3+x)} \, dx,x,x^2\right )\\ &=\frac {\left (1-x^2\right )^{2/3}}{24 \left (3+x^2\right )}-\frac {\log (x)}{18}+\frac {5 \log \left (3+x^2\right )}{144\ 2^{2/3}}-\frac {1}{12} \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,\sqrt [3]{1-x^2}\right )+\frac {1}{12} \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\sqrt [3]{1-x^2}\right )-\frac {5}{48} \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 {5 \operatorname {Subst}\left (\int \frac {1}{2^{2/3}-x} \, dx,x,\sqrt [3]{1-x^2}\right )}{48\ 2^{2/3}}\\ &=\frac {\left (1-x^2\right )^{2/3}}{24 \left (3+x^2\right )}-\frac {\log (x)}{18}+\frac {5 \log \left (3+x^2\right )}{144\ 2^{2/3}}+\frac {1}{12} \log \left (1-\sqrt [3]{1-x^2}\right )-\frac {5 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{48\ 2^{2/3}}-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{1-x^2}\right )+\frac {5 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\sqrt [3]{2-2 x^2}\right )}{24\ 2^{2/3}}\\ &=\frac {\left (1-x^2\right )^{2/3}}{24 \left (3+x^2\right )}-\frac {5 \tan ^{-1}\left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )}{24\ 2^{2/3} \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {1+2 \sqrt [3]{1-x^2}}{\sqrt {3}}\right )}{6 \sqrt {3}}-\frac {\log (x)}{18}+\frac {5 \log \left (3+x^2\right )}{144\ 2^{2/3}}+\frac {1}{12} \log \left (1-\sqrt [3]{1-x^2}\right )-\frac {5 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{48\ 2^{2/3}}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 148, normalized size = 0.94 \begin {gather*} \frac {1}{288} \left (\frac {12 \left (1-x^2\right )^{2/3}}{x^2+3}+5 \sqrt [3]{2} \log \left (x^2+3\right )+24 \log \left (1-\sqrt [3]{1-x^2}\right )-15 \sqrt [3]{2} \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )-10 \sqrt [3]{2} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{2-2 x^2}+1}{\sqrt {3}}\right )+16 \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{1-x^2}+1}{\sqrt {3}}\right )-16 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((12*(1 - x^2)^(2/3))/(3 + x^2) - 10*2^(1/3)*Sqrt[3]*ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3]] + 16*Sqrt[3]*ArcT
an[(1 + 2*(1 - x^2)^(1/3))/Sqrt[3]] - 16*Log[x] + 5*2^(1/3)*Log[3 + x^2] + 24*Log[1 - (1 - x^2)^(1/3)] - 15*2^
(1/3)*Log[2^(2/3) - (1 - x^2)^(1/3)])/288

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IntegrateAlgebraic [A]  time = 0.26, size = 223, normalized size = 1.41 \begin {gather*} \frac {\left (1-x^2\right )^{2/3}}{24 \left (x^2+3\right )}+\frac {1}{18} \log \left (\sqrt [3]{1-x^2}-1\right )-\frac {5 \log \left (\sqrt [3]{2} \sqrt [3]{1-x^2}-2\right )}{72\ 2^{2/3}}-\frac {1}{36} \log \left (\left (1-x^2\right )^{2/3}+\sqrt [3]{1-x^2}+1\right )+\frac {5 \log \left (2^{2/3} \left (1-x^2\right )^{2/3}+2 \sqrt [3]{2} \sqrt [3]{1-x^2}+4\right )}{144\ 2^{2/3}}+\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{1-x^2}}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{6 \sqrt {3}}-\frac {5 \tan ^{-1}\left (\frac {\sqrt [3]{2} \sqrt [3]{1-x^2}}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{24\ 2^{2/3} \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(1 - x^2)^(2/3)/(24*(3 + x^2)) + ArcTan[1/Sqrt[3] + (2*(1 - x^2)^(1/3))/Sqrt[3]]/(6*Sqrt[3]) - (5*ArcTan[1/Sqr
t[3] + (2^(1/3)*(1 - x^2)^(1/3))/Sqrt[3]])/(24*2^(2/3)*Sqrt[3]) + Log[-1 + (1 - x^2)^(1/3)]/18 - (5*Log[-2 + 2
^(1/3)*(1 - x^2)^(1/3)])/(72*2^(2/3)) - Log[1 + (1 - x^2)^(1/3) + (1 - x^2)^(2/3)]/36 + (5*Log[4 + 2*2^(1/3)*(
1 - x^2)^(1/3) + 2^(2/3)*(1 - x^2)^(2/3)])/(144*2^(2/3))

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fricas [A]  time = 1.29, size = 227, normalized size = 1.44 \begin {gather*} -\frac {20 \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}} {\left (2 \, \sqrt {3} \left (-1\right )^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} - 4^{\frac {1}{3}} \sqrt {3}\right )}\right ) + 5 \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 ) - 10 \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 ) - 32 \, \sqrt {3} {\left (x^{2} + 3\right )} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) + 16 \, {\left (x^{2} + 3\right )} \log \left ({\left (-x^{2} + 1\right )}^{\frac {2}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) - 32 \, {\left (x^{2} + 3\right )} \log \left ({\left (-x^{2} + 1\right )}^{\frac {1}{3}} - 1\right ) - 24 \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{576 \, {\left (x^{2} + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/576*(20*4^(1/6)*sqrt(3)*(-1)^(1/3)*(x^2 + 3)*arctan(1/6*4^(1/6)*(2*sqrt(3)*(-1)^(1/3)*(-x^2 + 1)^(1/3) - 4^
(1/3)*sqrt(3))) + 5*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)) - 10*4^(2/3)*(-1)^(1/3)*(x^2 + 3)*log(-4^(1/3)*(-1)^(2/3) + (-x^2 + 1)^(1/3)) - 32*sqrt(3)
*(x^2 + 3)*arctan(2/3*sqrt(3)*(-x^2 + 1)^(1/3) + 1/3*sqrt(3)) + 16*(x^2 + 3)*log((-x^2 + 1)^(2/3) + (-x^2 + 1)
^(1/3) + 1) - 32*(x^2 + 3)*log((-x^2 + 1)^(1/3) - 1) - 24*(-x^2 + 1)^(2/3))/(x^2 + 3)

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giac [A]  time = 0.55, size = 167, normalized size = 1.06 \begin {gather*} -\frac {5}{288} \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 {5}{576} \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 {5}{288} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {1}{3}} - {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) + \frac {1}{18} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {{\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{24 \, {\left (x^{2} + 3\right )}} - \frac {1}{36} \, \log \left ({\left (-x^{2} + 1\right )}^{\frac {2}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{18} \, \log \left (-{\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/288*4^(2/3)*sqrt(3)*arctan(1/12*4^(2/3)*sqrt(3)*(4^(1/3) + 2*(-x^2 + 1)^(1/3))) + 5/576*4^(2/3)*log(4^(2/3)
+ 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^(2/3)) - 5/288*4^(2/3)*log(4^(1/3) - (-x^2 + 1)^(1/3)) + 1/18*sqrt(3)
*arctan(1/3*sqrt(3)*(2*(-x^2 + 1)^(1/3) + 1)) + 1/24*(-x^2 + 1)^(2/3)/(x^2 + 3) - 1/36*log((-x^2 + 1)^(2/3) +
(-x^2 + 1)^(1/3) + 1) + 1/18*log(-(-x^2 + 1)^(1/3) + 1)

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maple [F]  time = 1.63, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (-x^{2}+1\right )^{\frac {1}{3}} \left (x^{2}+3\right )^{2} x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{2} + 3\right )}^{2} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.93, size = 375, normalized size = 2.37 \begin {gather*} \frac {\ln \left (\frac {127}{512}-\frac {127\,{\left (1-x^2\right )}^{1/3}}{512}\right )}{18}-\frac {5\,2^{1/3}\,\ln \left (-\frac {25\,2^{2/3}\,\left (\frac {5\,2^{1/3}\,\left (\frac {30375\,2^{2/3}}{64}-\frac {68283\,{\left (1-x^2\right )}^{1/3}}{64}\right )}{144}-\frac {1647}{128}\right )}{20736}-\frac {25\,{\left (1-x^2\right )}^{1/3}}{384}\right )}{144}+\ln \left ({\left (-\frac {1}{36}+\frac {\sqrt {3}\,1{}\mathrm {i}}{36}\right )}^2\,\left (\left (-\frac {1}{36}+\frac {\sqrt {3}\,1{}\mathrm {i}}{36}\right )\,\left (393660\,{\left (-\frac {1}{36}+\frac {\sqrt {3}\,1{}\mathrm {i}}{36}\right )}^2-\frac {68283\,{\left (1-x^2\right )}^{1/3}}{64}\right )+\frac {1647}{128}\right )-\frac {25\,{\left (1-x^2\right )}^{1/3}}{384}\right )\,\left (-\frac {1}{36}+\frac {\sqrt {3}\,1{}\mathrm {i}}{36}\right )-\ln \left (-{\left (\frac {1}{36}+\frac {\sqrt {3}\,1{}\mathrm {i}}{36}\right )}^2\,\left (\left (\frac {1}{36}+\frac {\sqrt {3}\,1{}\mathrm {i}}{36}\right )\,\left (393660\,{\left (\frac {1}{36}+\frac {\sqrt {3}\,1{}\mathrm {i}}{36}\right )}^2-\frac {68283\,{\left (1-x^2\right )}^{1/3}}{64}\right )-\frac {1647}{128}\right )-\frac {25\,{\left (1-x^2\right )}^{1/3}}{384}\right )\,\left (\frac {1}{36}+\frac {\sqrt {3}\,1{}\mathrm {i}}{36}\right )+\frac {{\left (1-x^2\right )}^{2/3}}{24\,\left (x^2+3\right )}+\frac {5\,{\left (-1\right )}^{1/3}\,2^{1/3}\,\ln \left (\frac {25\,{\left (-1\right )}^{2/3}\,2^{2/3}\,\left (\frac {5\,{\left (-1\right )}^{1/3}\,2^{1/3}\,\left (\frac {30375\,{\left (-1\right )}^{2/3}\,2^{2/3}}{64}-\frac {68283\,{\left (1-x^2\right )}^{1/3}}{64}\right )}{144}+\frac {1647}{128}\right )}{20736}-\frac {25\,{\left (1-x^2\right )}^{1/3}}{384}\right )}{144}-\frac {5\,{\left (-1\right )}^{1/3}\,2^{1/3}\,\ln \left (-\frac {25\,{\left (1-x^2\right )}^{1/3}}{384}+\frac {25\,{\left (-1\right )}^{2/3}\,2^{2/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (\frac {5\,{\left (-1\right )}^{1/3}\,2^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (\frac {68283\,{\left (1-x^2\right )}^{1/3}}{64}-\frac {30375\,{\left (-1\right )}^{2/3}\,2^{2/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{256}\right )}{288}+\frac {1647}{128}\right )}{82944}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{288} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(127/512 - (127*(1 - x^2)^(1/3))/512)/18 - (5*2^(1/3)*log(- (25*2^(2/3)*((5*2^(1/3)*((30375*2^(2/3))/64 - (
68283*(1 - x^2)^(1/3))/64))/144 - 1647/128))/20736 - (25*(1 - x^2)^(1/3))/384))/144 + log(((3^(1/2)*1i)/36 - 1
/36)^2*(((3^(1/2)*1i)/36 - 1/36)*(393660*((3^(1/2)*1i)/36 - 1/36)^2 - (68283*(1 - x^2)^(1/3))/64) + 1647/128)
- (25*(1 - x^2)^(1/3))/384)*((3^(1/2)*1i)/36 - 1/36) - log(- ((3^(1/2)*1i)/36 + 1/36)^2*(((3^(1/2)*1i)/36 + 1/
36)*(393660*((3^(1/2)*1i)/36 + 1/36)^2 - (68283*(1 - x^2)^(1/3))/64) - 1647/128) - (25*(1 - x^2)^(1/3))/384)*(
(3^(1/2)*1i)/36 + 1/36) + (1 - x^2)^(2/3)/(24*(x^2 + 3)) + (5*(-1)^(1/3)*2^(1/3)*log((25*(-1)^(2/3)*2^(2/3)*((
5*(-1)^(1/3)*2^(1/3)*((30375*(-1)^(2/3)*2^(2/3))/64 - (68283*(1 - x^2)^(1/3))/64))/144 + 1647/128))/20736 - (2
5*(1 - x^2)^(1/3))/384))/144 - (5*(-1)^(1/3)*2^(1/3)*log((25*(-1)^(2/3)*2^(2/3)*(3^(1/2)*1i + 1)^2*((5*(-1)^(1
/3)*2^(1/3)*(3^(1/2)*1i + 1)*((68283*(1 - x^2)^(1/3))/64 - (30375*(-1)^(2/3)*2^(2/3)*(3^(1/2)*1i + 1)^2)/256))
/288 + 1647/128))/82944 - (25*(1 - x^2)^(1/3))/384)*(3^(1/2)*1i + 1))/288

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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