Integrand size = 22, antiderivative size = 82 \[ \int \frac {x^2}{\sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\frac {\arctan \left (\frac {1+2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {\log \left (1+x^3\right )}{6 \sqrt [3]{2}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}} \]
-1/12*ln(x^3+1)*2^(2/3)+1/4*ln(2^(1/3)-(-x^3+1)^(1/3))*2^(2/3)+1/6*arctan( 1/3*(1+2^(2/3)*(-x^3+1)^(1/3))*3^(1/2))*2^(2/3)*3^(1/2)
Time = 0.10 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.27 \[ \int \frac {x^2}{\sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {1+2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}\right )+2 \log \left (-2+2^{2/3} \sqrt [3]{1-x^3}\right )-\log \left (2+2^{2/3} \sqrt [3]{1-x^3}+\sqrt [3]{2} \left (1-x^3\right )^{2/3}\right )}{6 \sqrt [3]{2}} \]
(2*Sqrt[3]*ArcTan[(1 + 2^(2/3)*(1 - x^3)^(1/3))/Sqrt[3]] + 2*Log[-2 + 2^(2 /3)*(1 - x^3)^(1/3)] - Log[2 + 2^(2/3)*(1 - x^3)^(1/3) + 2^(1/3)*(1 - x^3) ^(2/3)])/(6*2^(1/3))
Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {946, 67, 16, 1082, 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.
\(\displaystyle \int \frac {x^2}{\sqrt [3]{1-x^3} \left (x^3+1\right )} \, dx\) |
\(\Big \downarrow \) 946 |
\(\displaystyle \frac {1}{3} \int \frac {1}{\sqrt [3]{1-x^3} \left (x^3+1\right )}dx^3\) |
\(\Big \downarrow \) 67 |
\(\displaystyle \frac {1}{3} \left (-\frac {3 \int \frac {1}{\sqrt [3]{2}-\sqrt [3]{1-x^3}}d\sqrt [3]{1-x^3}}{2 \sqrt [3]{2}}+\frac {3}{2} \int \frac {1}{x^6+\sqrt [3]{2} \sqrt [3]{1-x^3}+2^{2/3}}d\sqrt [3]{1-x^3}-\frac {\log \left (x^3+1\right )}{2 \sqrt [3]{2}}\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{3} \left (\frac {3}{2} \int \frac {1}{x^6+\sqrt [3]{2} \sqrt [3]{1-x^3}+2^{2/3}}d\sqrt [3]{1-x^3}-\frac {\log \left (x^3+1\right )}{2 \sqrt [3]{2}}+\frac {3 \log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{3} \left (-\frac {3 \int \frac {1}{-x^6-3}d\left (2^{2/3} \sqrt [3]{1-x^3}+1\right )}{\sqrt [3]{2}}-\frac {\log \left (x^3+1\right )}{2 \sqrt [3]{2}}+\frac {3 \log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{3} \left (\frac {\sqrt {3} \arctan \left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{\sqrt [3]{2}}-\frac {\log \left (x^3+1\right )}{2 \sqrt [3]{2}}+\frac {3 \log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}\right )\) |
((Sqrt[3]*ArcTan[(1 + 2^(2/3)*(1 - x^3)^(1/3))/Sqrt[3]])/2^(1/3) - Log[1 + x^3]/(2*2^(1/3)) + (3*Log[2^(1/3) - (1 - x^3)^(1/3)])/(2*2^(1/3)))/3
3.1.99.3.1 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[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_)^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_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[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]
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]
Time = 3.60 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.98
method | result | size |
pseudoelliptic | \(\frac {2^{\frac {2}{3}} \left (2 \arctan \left (\frac {\left (1+2^{\frac {2}{3}} \left (-x^{3}+1\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right ) \sqrt {3}+2 \ln \left (\left (-x^{3}+1\right )^{\frac {1}{3}}-2^{\frac {1}{3}}\right )-\ln \left (\left (-x^{3}+1\right )^{\frac {2}{3}}+2^{\frac {1}{3}} \left (-x^{3}+1\right )^{\frac {1}{3}}+2^{\frac {2}{3}}\right )\right )}{12}\) | \(80\) |
trager | \(\text {Expression too large to display}\) | \(655\) |
1/12*2^(2/3)*(2*arctan(1/3*(1+2^(2/3)*(-x^3+1)^(1/3))*3^(1/2))*3^(1/2)+2*l n((-x^3+1)^(1/3)-2^(1/3))-ln((-x^3+1)^(2/3)+2^(1/3)*(-x^3+1)^(1/3)+2^(2/3) ))
Time = 0.25 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.10 \[ \int \frac {x^2}{\sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\frac {1}{6} \, \sqrt {6} 2^{\frac {1}{6}} \arctan \left (\frac {1}{6} \cdot 2^{\frac {1}{6}} {\left (\sqrt {6} 2^{\frac {1}{3}} + 2 \, \sqrt {6} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{12} \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{6} \cdot 2^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right ) \]
1/6*sqrt(6)*2^(1/6)*arctan(1/6*2^(1/6)*(sqrt(6)*2^(1/3) + 2*sqrt(6)*(-x^3 + 1)^(1/3))) - 1/12*2^(2/3)*log(2^(2/3) + 2^(1/3)*(-x^3 + 1)^(1/3) + (-x^3 + 1)^(2/3)) + 1/6*2^(2/3)*log(-2^(1/3) + (-x^3 + 1)^(1/3))
\[ \int \frac {x^2}{\sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\int \frac {x^{2}}{\sqrt [3]{- \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x + 1\right ) \left (x^{2} - x + 1\right )}\, dx \]
Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.05 \[ \int \frac {x^2}{\sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} + 2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{12} \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{6} \cdot 2^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right ) \]
1/6*sqrt(3)*2^(2/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3) + 2*(-x^3 + 1)^(1/ 3))) - 1/12*2^(2/3)*log(2^(2/3) + 2^(1/3)*(-x^3 + 1)^(1/3) + (-x^3 + 1)^(2 /3)) + 1/6*2^(2/3)*log(-2^(1/3) + (-x^3 + 1)^(1/3))
Time = 0.32 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.06 \[ \int \frac {x^2}{\sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} + 2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{12} \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{6} \cdot 2^{\frac {2}{3}} \log \left ({\left | -2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} \right |}\right ) \]
1/6*sqrt(3)*2^(2/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3) + 2*(-x^3 + 1)^(1/ 3))) - 1/12*2^(2/3)*log(2^(2/3) + 2^(1/3)*(-x^3 + 1)^(1/3) + (-x^3 + 1)^(2 /3)) + 1/6*2^(2/3)*log(abs(-2^(1/3) + (-x^3 + 1)^(1/3)))
Time = 0.56 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.22 \[ \int \frac {x^2}{\sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx=\frac {2^{2/3}\,\ln \left ({\left (1-x^3\right )}^{1/3}-2^{1/3}\right )}{6}+\frac {2^{2/3}\,\ln \left ({\left (1-x^3\right )}^{1/3}-\frac {2^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{12}-\frac {2^{2/3}\,\ln \left ({\left (1-x^3\right )}^{1/3}-\frac {2^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{12} \]