Integrand size = 22, antiderivative size = 183 \[ \int \frac {1}{x^3 \sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=-\frac {5 \left (1-x^2\right )^{2/3}}{72 \left (3+x^2\right )}-\frac {\left (1-x^2\right )^{2/3}}{6 x^2 \left (3+x^2\right )}+\frac {\arctan \left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )}{8\ 2^{2/3} \sqrt {3}}-\frac {\arctan \left (\frac {1+2 \sqrt [3]{1-x^2}}{\sqrt {3}}\right )}{18 \sqrt {3}}+\frac {\log (x)}{54}-\frac {\log \left (3+x^2\right )}{48\ 2^{2/3}}-\frac {1}{36} \log \left (1-\sqrt [3]{1-x^2}\right )+\frac {\log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}} \]
-5/72*(-x^2+1)^(2/3)/(x^2+3)-1/6*(-x^2+1)^(2/3)/x^2/(x^2+3)+1/54*ln(x)-1/9 6*ln(x^2+3)*2^(1/3)-1/36*ln(1-(-x^2+1)^(1/3))+1/32*ln(2^(2/3)-(-x^2+1)^(1/ 3))*2^(1/3)+1/48*arctan(1/3*(1+(-2*x^2+2)^(1/3))*3^(1/2))*3^(1/2)*2^(1/3)- 1/54*arctan(1/3*(1+2*(-x^2+1)^(1/3))*3^(1/2))*3^(1/2)
Time = 0.33 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^3 \sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\frac {1}{864} \left (-\frac {12 \left (1-x^2\right )^{2/3} \left (12+5 x^2\right )}{x^2 \left (3+x^2\right )}+18 \sqrt [3]{2} \sqrt {3} \arctan \left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )-16 \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{1-x^2}}{\sqrt {3}}\right )+18 \sqrt [3]{2} \log \left (-2+\sqrt [3]{2-2 x^2}\right )-9 \sqrt [3]{2} \log \left (4+2 \sqrt [3]{2-2 x^2}+\left (2-2 x^2\right )^{2/3}\right )-16 \log \left (-1+\sqrt [3]{1-x^2}\right )+8 \log \left (1+\sqrt [3]{1-x^2}+\left (1-x^2\right )^{2/3}\right )\right ) \]
((-12*(1 - x^2)^(2/3)*(12 + 5*x^2))/(x^2*(3 + x^2)) + 18*2^(1/3)*Sqrt[3]*A rcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3]] - 16*Sqrt[3]*ArcTan[(1 + 2*(1 - x^2 )^(1/3))/Sqrt[3]] + 18*2^(1/3)*Log[-2 + (2 - 2*x^2)^(1/3)] - 9*2^(1/3)*Log [4 + 2*(2 - 2*x^2)^(1/3) + (2 - 2*x^2)^(2/3)] - 16*Log[-1 + (1 - x^2)^(1/3 )] + 8*Log[1 + (1 - x^2)^(1/3) + (1 - x^2)^(2/3)])/864
Time = 0.32 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.12, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {354, 114, 27, 168, 174, 67, 16, 1082, 217, 1083, 217}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{x^3 \sqrt [3]{1-x^2} \left (x^2+3\right )^2} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \sqrt [3]{1-x^2} \left (x^2+3\right )^2}dx^2\) |
\(\Big \downarrow \) 114 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{3} \int \frac {3-4 x^2}{3 x^2 \sqrt [3]{1-x^2} \left (x^2+3\right )^2}dx^2-\frac {\left (1-x^2\right )^{2/3}}{3 x^2 \left (x^2+3\right )}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{9} \int \frac {3-4 x^2}{x^2 \sqrt [3]{1-x^2} \left (x^2+3\right )^2}dx^2-\frac {\left (1-x^2\right )^{2/3}}{3 x^2 \left (x^2+3\right )}\right )\) |
\(\Big \downarrow \) 168 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{9} \left (-\frac {1}{12} \int \frac {12-5 x^2}{x^2 \sqrt [3]{1-x^2} \left (x^2+3\right )}dx^2-\frac {5 \left (1-x^2\right )^{2/3}}{4 \left (x^2+3\right )}\right )-\frac {\left (1-x^2\right )^{2/3}}{3 x^2 \left (x^2+3\right )}\right )\) |
\(\Big \downarrow \) 174 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{9} \left (\frac {1}{12} \left (9 \int \frac {1}{\sqrt [3]{1-x^2} \left (x^2+3\right )}dx^2-4 \int \frac {1}{x^2 \sqrt [3]{1-x^2}}dx^2\right )-\frac {5 \left (1-x^2\right )^{2/3}}{4 \left (x^2+3\right )}\right )-\frac {\left (1-x^2\right )^{2/3}}{3 x^2 \left (x^2+3\right )}\right )\) |
\(\Big \downarrow \) 67 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{9} \left (\frac {1}{12} \left (9 \left (-\frac {3 \int \frac {1}{2^{2/3}-\sqrt [3]{1-x^2}}d\sqrt [3]{1-x^2}}{2\ 2^{2/3}}+\frac {3}{2} \int \frac {1}{x^4+2^{2/3} \sqrt [3]{1-x^2}+2 \sqrt [3]{2}}d\sqrt [3]{1-x^2}-\frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}\right )-4 \left (-\frac {3}{2} \int \frac {1}{1-\sqrt [3]{1-x^2}}d\sqrt [3]{1-x^2}+\frac {3}{2} \int \frac {1}{x^4+\sqrt [3]{1-x^2}+1}d\sqrt [3]{1-x^2}-\frac {1}{2} \log \left (x^2\right )\right )\right )-\frac {5 \left (1-x^2\right )^{2/3}}{4 \left (x^2+3\right )}\right )-\frac {\left (1-x^2\right )^{2/3}}{3 x^2 \left (x^2+3\right )}\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{9} \left (\frac {1}{12} \left (9 \left (\frac {3}{2} \int \frac {1}{x^4+2^{2/3} \sqrt [3]{1-x^2}+2 \sqrt [3]{2}}d\sqrt [3]{1-x^2}-\frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{2\ 2^{2/3}}\right )-4 \left (\frac {3}{2} \int \frac {1}{x^4+\sqrt [3]{1-x^2}+1}d\sqrt [3]{1-x^2}-\frac {1}{2} \log \left (x^2\right )+\frac {3}{2} \log \left (1-\sqrt [3]{1-x^2}\right )\right )\right )-\frac {5 \left (1-x^2\right )^{2/3}}{4 \left (x^2+3\right )}\right )-\frac {\left (1-x^2\right )^{2/3}}{3 x^2 \left (x^2+3\right )}\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{9} \left (\frac {1}{12} \left (9 \left (-\frac {3 \int \frac {1}{-x^4-3}d\left (\sqrt [3]{2} \sqrt [3]{1-x^2}+1\right )}{2^{2/3}}-\frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{2\ 2^{2/3}}\right )-4 \left (\frac {3}{2} \int \frac {1}{x^4+\sqrt [3]{1-x^2}+1}d\sqrt [3]{1-x^2}-\frac {1}{2} \log \left (x^2\right )+\frac {3}{2} \log \left (1-\sqrt [3]{1-x^2}\right )\right )\right )-\frac {5 \left (1-x^2\right )^{2/3}}{4 \left (x^2+3\right )}\right )-\frac {\left (1-x^2\right )^{2/3}}{3 x^2 \left (x^2+3\right )}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{9} \left (\frac {1}{12} \left (9 \left (\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{2} \sqrt [3]{1-x^2}+1}{\sqrt {3}}\right )}{2^{2/3}}-\frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{2\ 2^{2/3}}\right )-4 \left (\frac {3}{2} \int \frac {1}{x^4+\sqrt [3]{1-x^2}+1}d\sqrt [3]{1-x^2}-\frac {1}{2} \log \left (x^2\right )+\frac {3}{2} \log \left (1-\sqrt [3]{1-x^2}\right )\right )\right )-\frac {5 \left (1-x^2\right )^{2/3}}{4 \left (x^2+3\right )}\right )-\frac {\left (1-x^2\right )^{2/3}}{3 x^2 \left (x^2+3\right )}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{9} \left (\frac {1}{12} \left (9 \left (\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{2} \sqrt [3]{1-x^2}+1}{\sqrt {3}}\right )}{2^{2/3}}-\frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{2\ 2^{2/3}}\right )-4 \left (-3 \int \frac {1}{-x^4-3}d\left (2 \sqrt [3]{1-x^2}+1\right )-\frac {1}{2} \log \left (x^2\right )+\frac {3}{2} \log \left (1-\sqrt [3]{1-x^2}\right )\right )\right )-\frac {5 \left (1-x^2\right )^{2/3}}{4 \left (x^2+3\right )}\right )-\frac {\left (1-x^2\right )^{2/3}}{3 x^2 \left (x^2+3\right )}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{9} \left (\frac {1}{12} \left (9 \left (\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{2} \sqrt [3]{1-x^2}+1}{\sqrt {3}}\right )}{2^{2/3}}-\frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{2\ 2^{2/3}}\right )-4 \left (\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{1-x^2}+1}{\sqrt {3}}\right )-\frac {\log \left (x^2\right )}{2}+\frac {3}{2} \log \left (1-\sqrt [3]{1-x^2}\right )\right )\right )-\frac {5 \left (1-x^2\right )^{2/3}}{4 \left (x^2+3\right )}\right )-\frac {\left (1-x^2\right )^{2/3}}{3 x^2 \left (x^2+3\right )}\right )\) |
(-1/3*(1 - x^2)^(2/3)/(x^2*(3 + x^2)) + ((-5*(1 - x^2)^(2/3))/(4*(3 + x^2) ) + (-4*(Sqrt[3]*ArcTan[(1 + 2*(1 - x^2)^(1/3))/Sqrt[3]] - Log[x^2]/2 + (3 *Log[1 - (1 - x^2)^(1/3)])/2) + 9*((Sqrt[3]*ArcTan[(1 + 2^(1/3)*(1 - x^2)^ (1/3))/Sqrt[3]])/2^(2/3) - Log[3 + x^2]/(2*2^(2/3)) + (3*Log[2^(2/3) - (1 - x^2)^(1/3)])/(2*2^(2/3))))/12)/9)/2
3.11.24.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ {q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q), x ] + (Simp[3/(2*b) Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/3)], x] - Simp[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]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> 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] + Simp[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] && ILtQ[m, -1] && (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(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] + S imp[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*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(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]
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])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[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])] /; Fre eQ[{a, b, c}, x]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Time = 4.79 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.48
method | result | size |
pseudoelliptic | \(\frac {9 x^{2} \left (x^{2}+3\right ) \left (-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (1+2^{\frac {1}{3}} \left (-x^{2}+1\right )^{\frac {1}{3}}\right )}{3}\right )+\ln \left (\left (-x^{2}+1\right )^{\frac {2}{3}}+2^{\frac {2}{3}} \left (-x^{2}+1\right )^{\frac {1}{3}}+2 \,2^{\frac {1}{3}}\right )-2 \ln \left (\left (-x^{2}+1\right )^{\frac {1}{3}}-2^{\frac {2}{3}}\right )\right ) 2^{\frac {1}{3}}+16 x^{2} \sqrt {3}\, \left (x^{2}+3\right ) \arctan \left (\frac {\left (1+2 \left (-x^{2}+1\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )+\left (16 \ln \left (-1+\left (-x^{2}+1\right )^{\frac {1}{3}}\right )-8 \ln \left (1+\left (-x^{2}+1\right )^{\frac {1}{3}}+\left (-x^{2}+1\right )^{\frac {2}{3}}\right )\right ) x^{4}+\left (60 \left (-x^{2}+1\right )^{\frac {2}{3}}-24 \ln \left (1+\left (-x^{2}+1\right )^{\frac {1}{3}}+\left (-x^{2}+1\right )^{\frac {2}{3}}\right )+48 \ln \left (-1+\left (-x^{2}+1\right )^{\frac {1}{3}}\right )\right ) x^{2}+144 \left (-x^{2}+1\right )^{\frac {2}{3}}}{864 \left (-1+\left (-x^{2}+1\right )^{\frac {1}{3}}\right ) \left (x^{2}+3\right ) \left (1+\left (-x^{2}+1\right )^{\frac {1}{3}}+\left (-x^{2}+1\right )^{\frac {2}{3}}\right )}\) | \(271\) |
1/864*(9*x^2*(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)+16*x^2*3^(1/2)*(x^2+3)*arctan(1/3*(1+2*(-x^2+1)^(1/3 ))*3^(1/2))+(16*ln(-1+(-x^2+1)^(1/3))-8*ln(1+(-x^2+1)^(1/3)+(-x^2+1)^(2/3) ))*x^4+(60*(-x^2+1)^(2/3)-24*ln(1+(-x^2+1)^(1/3)+(-x^2+1)^(2/3))+48*ln(-1+ (-x^2+1)^(1/3)))*x^2+144*(-x^2+1)^(2/3))/(-1+(-x^2+1)^(1/3))/(x^2+3)/(1+(- x^2+1)^(1/3)+(-x^2+1)^(2/3))
Time = 0.29 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.30 \[ \int \frac {1}{x^3 \sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\frac {36 \cdot 4^{\frac {1}{6}} \sqrt {3} {\left (x^{4} + 3 \, x^{2}\right )} \arctan \left (\frac {1}{6} \cdot 4^{\frac {1}{6}} {\left (4^{\frac {1}{3}} \sqrt {3} + 2 \, \sqrt {3} {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - 9 \cdot 4^{\frac {2}{3}} {\left (x^{4} + 3 \, x^{2}\right )} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + 18 \cdot 4^{\frac {2}{3}} {\left (x^{4} + 3 \, x^{2}\right )} \log \left (-4^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) - 32 \, \sqrt {3} {\left (x^{4} + 3 \, x^{2}\right )} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) + 16 \, {\left (x^{4} + 3 \, x^{2}\right )} \log \left ({\left (-x^{2} + 1\right )}^{\frac {2}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) - 32 \, {\left (x^{4} + 3 \, x^{2}\right )} \log \left ({\left (-x^{2} + 1\right )}^{\frac {1}{3}} - 1\right ) - 24 \, {\left (5 \, x^{2} + 12\right )} {\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{1728 \, {\left (x^{4} + 3 \, x^{2}\right )}} \]
1/1728*(36*4^(1/6)*sqrt(3)*(x^4 + 3*x^2)*arctan(1/6*4^(1/6)*(4^(1/3)*sqrt( 3) + 2*sqrt(3)*(-x^2 + 1)^(1/3))) - 9*4^(2/3)*(x^4 + 3*x^2)*log(4^(2/3) + 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^(2/3)) + 18*4^(2/3)*(x^4 + 3*x^2)*lo g(-4^(1/3) + (-x^2 + 1)^(1/3)) - 32*sqrt(3)*(x^4 + 3*x^2)*arctan(2/3*sqrt( 3)*(-x^2 + 1)^(1/3) + 1/3*sqrt(3)) + 16*(x^4 + 3*x^2)*log((-x^2 + 1)^(2/3) + (-x^2 + 1)^(1/3) + 1) - 32*(x^4 + 3*x^2)*log((-x^2 + 1)^(1/3) - 1) - 24 *(5*x^2 + 12)*(-x^2 + 1)^(2/3))/(x^4 + 3*x^2)
\[ \int \frac {1}{x^3 \sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\int \frac {1}{x^{3} \sqrt [3]{- \left (x - 1\right ) \left (x + 1\right )} \left (x^{2} + 3\right )^{2}}\, dx \]
\[ \int \frac {1}{x^3 \sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\int { \frac {1}{{\left (x^{2} + 3\right )}^{2} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} x^{3}} \,d x } \]
Time = 0.32 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^3 \sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\frac {1}{96} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{192} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{96} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {1}{3}} - {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) - \frac {1}{54} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {5 \, {\left (-x^{2} + 1\right )}^{\frac {5}{3}} - 17 \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{72 \, {\left ({\left (x^{2} - 1\right )}^{2} + 5 \, x^{2} - 1\right )}} + \frac {1}{108} \, \log \left ({\left (-x^{2} + 1\right )}^{\frac {2}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {1}{54} \, \log \left (-{\left (-x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) \]
1/96*4^(2/3)*sqrt(3)*arctan(1/12*4^(2/3)*sqrt(3)*(4^(1/3) + 2*(-x^2 + 1)^( 1/3))) - 1/192*4^(2/3)*log(4^(2/3) + 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1) ^(2/3)) + 1/96*4^(2/3)*log(4^(1/3) - (-x^2 + 1)^(1/3)) - 1/54*sqrt(3)*arct an(1/3*sqrt(3)*(2*(-x^2 + 1)^(1/3) + 1)) + 1/72*(5*(-x^2 + 1)^(5/3) - 17*( -x^2 + 1)^(2/3))/((x^2 - 1)^2 + 5*x^2 - 1) + 1/108*log((-x^2 + 1)^(2/3) + (-x^2 + 1)^(1/3) + 1) - 1/54*log(-(-x^2 + 1)^(1/3) + 1)
Time = 5.40 (sec) , antiderivative size = 409, normalized size of antiderivative = 2.23 \[ \int \frac {1}{x^3 \sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx=\frac {2^{1/3}\,\ln \left (\frac {2^{2/3}\,\left (\frac {2^{1/3}\,\left (\frac {10935\,2^{2/3}}{64}-\frac {9099\,{\left (1-x^2\right )}^{1/3}}{64}\right )}{48}-\frac {665}{128}\right )}{2304}+\frac {{\left (1-x^2\right )}^{1/3}}{576}\right )}{48}-\frac {\ln \left (\frac {985\,{\left (1-x^2\right )}^{1/3}}{373248}-\frac {985}{373248}\right )}{54}+\ln \left ({\left (\frac {1}{108}+\frac {\sqrt {3}\,1{}\mathrm {i}}{108}\right )}^2\,\left (\left (\frac {1}{108}+\frac {\sqrt {3}\,1{}\mathrm {i}}{108}\right )\,\left (393660\,{\left (\frac {1}{108}+\frac {\sqrt {3}\,1{}\mathrm {i}}{108}\right )}^2-\frac {9099\,{\left (1-x^2\right )}^{1/3}}{64}\right )-\frac {665}{128}\right )+\frac {{\left (1-x^2\right )}^{1/3}}{576}\right )\,\left (\frac {1}{108}+\frac {\sqrt {3}\,1{}\mathrm {i}}{108}\right )-\ln \left (\frac {{\left (1-x^2\right )}^{1/3}}{576}-{\left (-\frac {1}{108}+\frac {\sqrt {3}\,1{}\mathrm {i}}{108}\right )}^2\,\left (\left (-\frac {1}{108}+\frac {\sqrt {3}\,1{}\mathrm {i}}{108}\right )\,\left (393660\,{\left (-\frac {1}{108}+\frac {\sqrt {3}\,1{}\mathrm {i}}{108}\right )}^2-\frac {9099\,{\left (1-x^2\right )}^{1/3}}{64}\right )+\frac {665}{128}\right )\right )\,\left (-\frac {1}{108}+\frac {\sqrt {3}\,1{}\mathrm {i}}{108}\right )-\frac {\frac {17\,{\left (1-x^2\right )}^{2/3}}{72}-\frac {5\,{\left (1-x^2\right )}^{5/3}}{72}}{{\left (x^2-1\right )}^2+5\,x^2-1}+\frac {2^{1/3}\,\ln \left (\frac {{\left (1-x^2\right )}^{1/3}}{576}+\frac {2^{2/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (\frac {2^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (\frac {10935\,2^{2/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{256}-\frac {9099\,{\left (1-x^2\right )}^{1/3}}{64}\right )}{96}-\frac {665}{128}\right )}{9216}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{96}-\frac {2^{1/3}\,\ln \left (\frac {{\left (1-x^2\right )}^{1/3}}{576}-\frac {2^{2/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (\frac {2^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (\frac {10935\,2^{2/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{256}-\frac {9099\,{\left (1-x^2\right )}^{1/3}}{64}\right )}{96}+\frac {665}{128}\right )}{9216}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{96} \]
(2^(1/3)*log((2^(2/3)*((2^(1/3)*((10935*2^(2/3))/64 - (9099*(1 - x^2)^(1/3 ))/64))/48 - 665/128))/2304 + (1 - x^2)^(1/3)/576))/48 - log((985*(1 - x^2 )^(1/3))/373248 - 985/373248)/54 + log(((3^(1/2)*1i)/108 + 1/108)^2*(((3^( 1/2)*1i)/108 + 1/108)*(393660*((3^(1/2)*1i)/108 + 1/108)^2 - (9099*(1 - x^ 2)^(1/3))/64) - 665/128) + (1 - x^2)^(1/3)/576)*((3^(1/2)*1i)/108 + 1/108) - log((1 - x^2)^(1/3)/576 - ((3^(1/2)*1i)/108 - 1/108)^2*(((3^(1/2)*1i)/1 08 - 1/108)*(393660*((3^(1/2)*1i)/108 - 1/108)^2 - (9099*(1 - x^2)^(1/3))/ 64) + 665/128))*((3^(1/2)*1i)/108 - 1/108) - ((17*(1 - x^2)^(2/3))/72 - (5 *(1 - x^2)^(5/3))/72)/((x^2 - 1)^2 + 5*x^2 - 1) + (2^(1/3)*log((1 - x^2)^( 1/3)/576 + (2^(2/3)*(3^(1/2)*1i - 1)^2*((2^(1/3)*(3^(1/2)*1i - 1)*((10935* 2^(2/3)*(3^(1/2)*1i - 1)^2)/256 - (9099*(1 - x^2)^(1/3))/64))/96 - 665/128 ))/9216)*(3^(1/2)*1i - 1))/96 - (2^(1/3)*log((1 - x^2)^(1/3)/576 - (2^(2/3 )*(3^(1/2)*1i + 1)^2*((2^(1/3)*(3^(1/2)*1i + 1)*((10935*2^(2/3)*(3^(1/2)*1 i + 1)^2)/256 - (9099*(1 - x^2)^(1/3))/64))/96 + 665/128))/9216)*(3^(1/2)* 1i + 1))/96