Integrand size = 32, antiderivative size = 493 \[ \int \frac {a+b x+c x^2}{\left (1-x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\frac {(a+b) \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {(a+b) \arctan \left (\frac {1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}-\frac {c \arctan \left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {(a-c) \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {(b+c) \arctan \left (\frac {1+2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {(a+b) \log \left ((1-x) (1+x)^2\right )}{12 \sqrt [3]{2}}-\frac {(a-c) \log \left (1+x^3\right )}{6 \sqrt [3]{2}}-\frac {(b+c) \log \left (1+x^3\right )}{6 \sqrt [3]{2}}+\frac {(a+b) \log \left (1+\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{6 \sqrt [3]{2}}-\frac {(a+b) \log \left (1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{3 \sqrt [3]{2}}+\frac {(b+c) \log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}+\frac {(a-c) \log \left (-\sqrt [3]{2} x-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}+\frac {1}{2} c \log \left (x+\sqrt [3]{1-x^3}\right )-\frac {(a+b) \log \left (-1+x+2^{2/3} \sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}} \]
1/24*(a+b)*ln((1-x)*(1+x)^2)*2^(2/3)-1/12*(a-c)*ln(x^3+1)*2^(2/3)-1/12*(b+ c)*ln(x^3+1)*2^(2/3)+1/12*(a+b)*ln(1+2^(2/3)*(1-x)^2/(-x^3+1)^(2/3)-2^(1/3 )*(1-x)/(-x^3+1)^(1/3))*2^(2/3)-1/6*(a+b)*ln(1+2^(1/3)*(1-x)/(-x^3+1)^(1/3 ))*2^(2/3)+1/4*(b+c)*ln(2^(1/3)-(-x^3+1)^(1/3))*2^(2/3)+1/4*(a-c)*ln(-2^(1 /3)*x-(-x^3+1)^(1/3))*2^(2/3)+1/2*c*ln(x+(-x^3+1)^(1/3))-1/8*(a+b)*ln(-1+x +2^(2/3)*(-x^3+1)^(1/3))*2^(2/3)+1/6*(a+b)*arctan(1/3*(1-2*2^(1/3)*(1-x)/( -x^3+1)^(1/3))*3^(1/2))*2^(2/3)*3^(1/2)+1/12*(a+b)*arctan(1/3*(1+2^(1/3)*( 1-x)/(-x^3+1)^(1/3))*3^(1/2))*2^(2/3)*3^(1/2)-1/3*c*arctan(1/3*(1-2*x/(-x^ 3+1)^(1/3))*3^(1/2))*3^(1/2)-1/6*(a-c)*arctan(1/3*(1-2*2^(1/3)*x/(-x^3+1)^ (1/3))*3^(1/2))*2^(2/3)*3^(1/2)+1/6*(b+c)*arctan(1/3*(1+2^(2/3)*(-x^3+1)^( 1/3))*3^(1/2))*2^(2/3)*3^(1/2)
\[ \int \frac {a+b x+c x^2}{\left (1-x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\int \frac {a+b x+c x^2}{\left (1-x+x^2\right ) \sqrt [3]{1-x^3}} \, dx \]
Time = 0.78 (sec) , antiderivative size = 570, normalized size of antiderivative = 1.16, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2583, 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.
\(\displaystyle \int \frac {a+b x+c x^2}{\left (x^2-x+1\right ) \sqrt [3]{1-x^3}} \, dx\) |
\(\Big \downarrow \) 2583 |
\(\displaystyle \int \left (\frac {x (a+b)}{\sqrt [3]{1-x^3} \left (x^3+1\right )}+\frac {a}{\sqrt [3]{1-x^3} \left (x^3+1\right )}+\frac {x^2 (b+c)}{\sqrt [3]{1-x^3} \left (x^3+1\right )}+\frac {c x^3}{\sqrt [3]{1-x^3} \left (x^3+1\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b) \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {(a+b) \arctan \left (\frac {\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}-\frac {a \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {(a+b) \log \left (\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )}{6 \sqrt [3]{2}}-\frac {(a+b) \log \left (\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )}{3 \sqrt [3]{2}}-\frac {(a+b) \log \left (2^{2/3} \sqrt [3]{1-x^3}+x-1\right )}{4 \sqrt [3]{2}}+\frac {(a+b) \log \left ((1-x) (x+1)^2\right )}{12 \sqrt [3]{2}}-\frac {a \log \left (x^3+1\right )}{6 \sqrt [3]{2}}+\frac {a \log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{2} x\right )}{2 \sqrt [3]{2}}+\frac {\arctan \left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right ) (b+c)}{\sqrt [3]{2} \sqrt {3}}-\frac {c \arctan \left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {c \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {(b+c) \log \left (x^3+1\right )}{6 \sqrt [3]{2}}+\frac {(b+c) \log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}+\frac {c \log \left (x^3+1\right )}{6 \sqrt [3]{2}}-\frac {c \log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{2} x\right )}{2 \sqrt [3]{2}}+\frac {1}{2} c \log \left (\sqrt [3]{1-x^3}+x\right )\) |
((a + b)*ArcTan[(1 - (2*2^(1/3)*(1 - x))/(1 - x^3)^(1/3))/Sqrt[3]])/(2^(1/ 3)*Sqrt[3]) + ((a + b)*ArcTan[(1 + (2^(1/3)*(1 - x))/(1 - x^3)^(1/3))/Sqrt [3]])/(2*2^(1/3)*Sqrt[3]) - (c*ArcTan[(1 - (2*x)/(1 - x^3)^(1/3))/Sqrt[3]] )/Sqrt[3] - (a*ArcTan[(1 - (2*2^(1/3)*x)/(1 - x^3)^(1/3))/Sqrt[3]])/(2^(1/ 3)*Sqrt[3]) + (c*ArcTan[(1 - (2*2^(1/3)*x)/(1 - x^3)^(1/3))/Sqrt[3]])/(2^( 1/3)*Sqrt[3]) + ((b + c)*ArcTan[(1 + 2^(2/3)*(1 - x^3)^(1/3))/Sqrt[3]])/(2 ^(1/3)*Sqrt[3]) + ((a + b)*Log[(1 - x)*(1 + x)^2])/(12*2^(1/3)) - (a*Log[1 + x^3])/(6*2^(1/3)) + (c*Log[1 + x^3])/(6*2^(1/3)) - ((b + c)*Log[1 + x^3 ])/(6*2^(1/3)) + ((a + b)*Log[1 + (2^(2/3)*(1 - x)^2)/(1 - x^3)^(2/3) - (2 ^(1/3)*(1 - x))/(1 - x^3)^(1/3)])/(6*2^(1/3)) - ((a + b)*Log[1 + (2^(1/3)* (1 - x))/(1 - x^3)^(1/3)])/(3*2^(1/3)) + ((b + c)*Log[2^(1/3) - (1 - x^3)^ (1/3)])/(2*2^(1/3)) + (a*Log[-(2^(1/3)*x) - (1 - x^3)^(1/3)])/(2*2^(1/3)) - (c*Log[-(2^(1/3)*x) - (1 - x^3)^(1/3)])/(2*2^(1/3)) + (c*Log[x + (1 - x^ 3)^(1/3)])/2 - ((a + b)*Log[-1 + x + 2^(2/3)*(1 - x^3)^(1/3)])/(4*2^(1/3))
3.1.46.3.1 Defintions of rubi rules used
Int[(Px_.)*((c_) + (d_.)*(x_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^3)^(p _.), x_Symbol] :> Simp[1/c^q Int[ExpandIntegrand[(c^3 - d^3*x^3)^q*(a + b *x^3)^p, Px/(c - d*x)^q, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Poly Q[Px, x] && EqQ[d^2 - c*e, 0] && ILtQ[q, 0] && RationalQ[p] && EqQ[Denomina tor[p], 3]
\[\int \frac {c \,x^{2}+b x +a}{\left (x^{2}-x +1\right ) \left (-x^{3}+1\right )^{\frac {1}{3}}}d x\]
Exception generated. \[ \int \frac {a+b x+c x^2}{\left (1-x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (residue poly has multiple non-linear fac tors)
\[ \int \frac {a+b x+c x^2}{\left (1-x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\int \frac {a + b x + c x^{2}}{\sqrt [3]{- \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x^{2} - x + 1\right )}\, dx \]
\[ \int \frac {a+b x+c x^2}{\left (1-x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\int { \frac {c x^{2} + b x + a}{{\left (-x^{3} + 1\right )}^{\frac {1}{3}} {\left (x^{2} - x + 1\right )}} \,d x } \]
\[ \int \frac {a+b x+c x^2}{\left (1-x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\int { \frac {c x^{2} + b x + a}{{\left (-x^{3} + 1\right )}^{\frac {1}{3}} {\left (x^{2} - x + 1\right )}} \,d x } \]
Timed out. \[ \int \frac {a+b x+c x^2}{\left (1-x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\int \frac {c\,x^2+b\,x+a}{{\left (1-x^3\right )}^{1/3}\,\left (x^2-x+1\right )} \,d x \]