Integrand size = 22, antiderivative size = 115 \[ \int \frac {x^8}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\frac {1}{2 \sqrt [3]{1-x^3}}+\frac {1}{2} \left (1-x^3\right )^{2/3}+\frac {\arctan \left (\frac {1+2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}-\frac {\log \left (1+x^3\right )}{12 \sqrt [3]{2}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}} \]
1/2/(-x^3+1)^(1/3)+1/2*(-x^3+1)^(2/3)-1/24*ln(x^3+1)*2^(2/3)+1/8*ln(2^(1/3 )-(-x^3+1)^(1/3))*2^(2/3)+1/12*arctan(1/3*(1+2^(2/3)*(-x^3+1)^(1/3))*3^(1/ 2))*2^(2/3)*3^(1/2)
Time = 0.26 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.15 \[ \int \frac {x^8}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\frac {1}{24} \left (-\frac {12 \left (-2+x^3\right )}{\sqrt [3]{1-x^3}}+2\ 2^{2/3} \sqrt {3} \arctan \left (\frac {1+2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}\right )+2\ 2^{2/3} \log \left (-2+2^{2/3} \sqrt [3]{1-x^3}\right )-2^{2/3} \log \left (2+2^{2/3} \sqrt [3]{1-x^3}+\sqrt [3]{2} \left (1-x^3\right )^{2/3}\right )\right ) \]
((-12*(-2 + x^3))/(1 - x^3)^(1/3) + 2*2^(2/3)*Sqrt[3]*ArcTan[(1 + 2^(2/3)* (1 - x^3)^(1/3))/Sqrt[3]] + 2*2^(2/3)*Log[-2 + 2^(2/3)*(1 - x^3)^(1/3)] - 2^(2/3)*Log[2 + 2^(2/3)*(1 - x^3)^(1/3) + 2^(1/3)*(1 - x^3)^(2/3)])/24
Time = 0.26 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {948, 98, 2009}
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.
\(\displaystyle \int \frac {x^8}{\left (1-x^3\right )^{4/3} \left (x^3+1\right )} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {x^6}{\left (1-x^3\right )^{4/3} \left (x^3+1\right )}dx^3\) |
\(\Big \downarrow \) 98 |
\(\displaystyle \frac {1}{3} \int \left (\frac {1}{\sqrt [3]{1-x^3} \left (1-x^6\right )}-\frac {1}{\sqrt [3]{1-x^3}}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {\sqrt {3} \arctan \left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{2}}+\frac {3}{2} \left (1-x^3\right )^{2/3}+\frac {3}{2 \sqrt [3]{1-x^3}}-\frac {\log \left (x^3+1\right )}{4 \sqrt [3]{2}}+\frac {3 \log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}\right )\) |
(3/(2*(1 - x^3)^(1/3)) + (3*(1 - x^3)^(2/3))/2 + (Sqrt[3]*ArcTan[(1 + 2^(2 /3)*(1 - x^3)^(1/3))/Sqrt[3]])/(2*2^(1/3)) - Log[1 + x^3]/(4*2^(1/3)) + (3 *Log[2^(1/3) - (1 - x^3)^(1/3)])/(4*2^(1/3)))/3
3.7.40.3.1 Defintions of rubi rules used
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x _)), x_] :> Int[ExpandIntegrand[(e + f*x)^FractionalPart[p], (c + d*x)^n*(( e + f*x)^IntegerPart[p]/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[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]]
Time = 8.37 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.10
method | result | size |
pseudoelliptic | \(-\frac {-2 \arctan \left (\frac {\left (1+2^{\frac {2}{3}} \left (-x^{3}+1\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right ) 2^{\frac {2}{3}} \sqrt {3}\, \left (-x^{3}+1\right )^{\frac {1}{3}}+12 x^{3}+2^{\frac {2}{3}} \ln \left (\left (-x^{3}+1\right )^{\frac {2}{3}}+2^{\frac {1}{3}} \left (-x^{3}+1\right )^{\frac {1}{3}}+2^{\frac {2}{3}}\right ) \left (-x^{3}+1\right )^{\frac {1}{3}}-2 \,2^{\frac {2}{3}} \ln \left (\left (-x^{3}+1\right )^{\frac {1}{3}}-2^{\frac {1}{3}}\right ) \left (-x^{3}+1\right )^{\frac {1}{3}}-24}{24 \left (-x^{3}+1\right )^{\frac {1}{3}}}\) | \(127\) |
risch | \(\text {Expression too large to display}\) | \(775\) |
trager | \(\text {Expression too large to display}\) | \(782\) |
-1/24*(-2*arctan(1/3*(1+2^(2/3)*(-x^3+1)^(1/3))*3^(1/2))*2^(2/3)*3^(1/2)*( -x^3+1)^(1/3)+12*x^3+2^(2/3)*ln((-x^3+1)^(2/3)+2^(1/3)*(-x^3+1)^(1/3)+2^(2 /3))*(-x^3+1)^(1/3)-2*2^(2/3)*ln((-x^3+1)^(1/3)-2^(1/3))*(-x^3+1)^(1/3)-24 )/(-x^3+1)^(1/3)
Time = 0.33 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.13 \[ \int \frac {x^8}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\frac {2 \, \sqrt {6} 2^{\frac {1}{6}} {\left (x^{3} - 1\right )} \arctan \left (\frac {1}{6} \cdot 2^{\frac {1}{6}} {\left (\sqrt {6} 2^{\frac {1}{3}} + 2 \, \sqrt {6} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}\right ) - 2^{\frac {2}{3}} {\left (x^{3} - 1\right )} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} {\left (x^{3} - 1\right )} \log \left (-2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right ) + 12 \, {\left (x^{3} - 2\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{24 \, {\left (x^{3} - 1\right )}} \]
1/24*(2*sqrt(6)*2^(1/6)*(x^3 - 1)*arctan(1/6*2^(1/6)*(sqrt(6)*2^(1/3) + 2* sqrt(6)*(-x^3 + 1)^(1/3))) - 2^(2/3)*(x^3 - 1)*log(2^(2/3) + 2^(1/3)*(-x^3 + 1)^(1/3) + (-x^3 + 1)^(2/3)) + 2*2^(2/3)*(x^3 - 1)*log(-2^(1/3) + (-x^3 + 1)^(1/3)) + 12*(x^3 - 2)*(-x^3 + 1)^(2/3))/(x^3 - 1)
\[ \int \frac {x^8}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\int \frac {x^{8}}{\left (- \left (x - 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac {4}{3}} \left (x + 1\right ) \left (x^{2} - x + 1\right )}\, dx \]
Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.94 \[ \int \frac {x^8}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\frac {1}{12} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} + 2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{24} \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{12} \cdot 2^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right ) + \frac {1}{2} \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} + \frac {1}{2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}} \]
1/12*sqrt(3)*2^(2/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3) + 2*(-x^3 + 1)^(1 /3))) - 1/24*2^(2/3)*log(2^(2/3) + 2^(1/3)*(-x^3 + 1)^(1/3) + (-x^3 + 1)^( 2/3)) + 1/12*2^(2/3)*log(-2^(1/3) + (-x^3 + 1)^(1/3)) + 1/2*(-x^3 + 1)^(2/ 3) + 1/2/(-x^3 + 1)^(1/3)
Time = 0.30 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.95 \[ \int \frac {x^8}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\frac {1}{12} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} + 2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{24} \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{12} \cdot 2^{\frac {2}{3}} \log \left ({\left | -2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} \right |}\right ) + \frac {1}{2} \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} + \frac {1}{2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}} \]
1/12*sqrt(3)*2^(2/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3) + 2*(-x^3 + 1)^(1 /3))) - 1/24*2^(2/3)*log(2^(2/3) + 2^(1/3)*(-x^3 + 1)^(1/3) + (-x^3 + 1)^( 2/3)) + 1/12*2^(2/3)*log(abs(-2^(1/3) + (-x^3 + 1)^(1/3))) + 1/2*(-x^3 + 1 )^(2/3) + 1/2/(-x^3 + 1)^(1/3)
Time = 8.48 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.11 \[ \int \frac {x^8}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx=\frac {2^{2/3}\,\ln \left (\frac {{\left (1-x^3\right )}^{1/3}}{4}-\frac {2^{1/3}}{4}\right )}{12}+\frac {1}{2\,{\left (1-x^3\right )}^{1/3}}+\frac {{\left (1-x^3\right )}^{2/3}}{2}+\frac {2^{2/3}\,\ln \left (\frac {{\left (1-x^3\right )}^{1/3}}{4}-\frac {2^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{24}-\frac {2^{2/3}\,\ln \left (\frac {{\left (1-x^3\right )}^{1/3}}{4}-\frac {2^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{24} \]