\(\int \frac {(1+x^3)^{2/3} (4+x^3)}{x^6 (8-4 x^3+x^6)} \, dx\) [1666]

Optimal result

Integrand size = 30, antiderivative size = 112 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (4+x^3\right )}{x^6 \left (8-4 x^3+x^6\right )} \, dx=\frac {\left (-8-23 x^3\right ) \left (1+x^3\right )^{2/3}}{80 x^5}+\frac {1}{96} \text {RootSum}\left [13-20 \text {$\#$1}^3+8 \text {$\#$1}^6\&,\frac {-39 \log (x)+39 \log \left (\sqrt [3]{1+x^3}-x \text {$\#$1}\right )+32 \log (x) \text {$\#$1}^3-32 \log \left (\sqrt [3]{1+x^3}-x \text {$\#$1}\right ) \text {$\#$1}^3}{-5 \text {$\#$1}+4 \text {$\#$1}^4}\&\right ] \]

[Out]

Unintegrable

Rubi [F]

\[ \int \frac {\left (1+x^3\right )^{2/3} \left (4+x^3\right )}{x^6 \left (8-4 x^3+x^6\right )} \, dx=\int \frac {\left (1+x^3\right )^{2/3} \left (4+x^3\right )}{x^6 \left (8-4 x^3+x^6\right )} \, dx \]

[In]

Int[((1 + x^3)^(2/3)*(4 + x^3))/(x^6*(8 - 4*x^3 + x^6)),x]

[Out]

(1 + x^3)^(2/3)/96 - (3*(1 + x^3)^(2/3))/(16*x^2) - (1 + x^3)^(5/3)/(10*x^5) - (7/384 + (3*I)/128)*x^2*AppellF
1[2/3, 1/3, 1, 5/3, -x^3, (1/4 - I/4)*x^3] - (7/384 - (3*I)/128)*x^2*AppellF1[2/3, 1/3, 1, 5/3, -x^3, (1/4 + I
/4)*x^3] - ArcTan[(1 + (2*x)/(1 + x^3)^(1/3))/Sqrt[3]]/(8*Sqrt[3]) + (Sqrt[3]*ArcTan[(1 + (2*x)/(1 + x^3)^(1/3
))/Sqrt[3]])/8 + ((1/32 + I/96)*(5 - I)^(2/3)*ArcTan[(1 + ((10 - 2*I)^(1/3)*x)/(1 + x^3)^(1/3))/Sqrt[3]])/(2^(
1/3)*Sqrt[3]) + ((1/32 - I/96)*(5 + I)^(2/3)*ArcTan[(1 + ((10 + 2*I)^(1/3)*x)/(1 + x^3)^(1/3))/Sqrt[3]])/(2^(1
/3)*Sqrt[3]) + ((1/96 + I/32)*(3 - 2*I)^(2/3)*ArcTan[((3 - 2*I)^(1/3) + 2*(1 + x^3)^(1/3))/(Sqrt[3]*(3 - 2*I)^
(1/3))])/Sqrt[3] + ((1/96 - I/32)*(3 + 2*I)^(2/3)*ArcTan[((3 + 2*I)^(1/3) + 2*(1 + x^3)^(1/3))/(Sqrt[3]*(3 + 2
*I)^(1/3))])/Sqrt[3] + (x^2*Hypergeometric2F1[1/3, 2/3, 5/3, -x^3])/48 - (1/576 + I/192)*(3 - 2*I)^(2/3)*Log[(
2 - 2*I) - x^3] - (1/576 - I/192)*(3 + 2*I)^(2/3)*Log[(2 + 2*I) - x^3] + ((1/36 + I/288)*Log[(-16 - 16*I) + 8*
x^3])/(10 - 2*I)^(1/3) + ((1/36 - I/288)*Log[(-16 + 16*I) + 8*x^3])/(10 + 2*I)^(1/3) + (1/192 + I/64)*(3 - 2*I
)^(2/3)*Log[(3 - 2*I)^(1/3) - (1 + x^3)^(1/3)] + (1/192 - I/64)*(3 + 2*I)^(2/3)*Log[(3 + 2*I)^(1/3) - (1 + x^3
)^(1/3)] - ((1/12 + I/96)*Log[((5 - I)^(1/3)*x)/2^(2/3) - (1 + x^3)^(1/3)])/(10 - 2*I)^(1/3) - ((1/12 - I/96)*
Log[((5 + I)^(1/3)*x)/2^(2/3) - (1 + x^3)^(1/3)])/(10 + 2*I)^(1/3) - Log[-x + (1 + x^3)^(1/3)]/8 + Defer[Int][
(1 + x^3)^(2/3)/(4 - 4*x + 2*x^2 - 2*x^3 + x^4), x]/3 - (5*Defer[Int][(x*(1 + x^3)^(2/3))/(4 - 4*x + 2*x^2 - 2
*x^3 + x^4), x])/24 - Defer[Int][(x^3*(1 + x^3)^(2/3))/(4 - 4*x + 2*x^2 - 2*x^3 + x^4), x]/48

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\left (1+x^3\right )^{2/3}}{2 x^6}+\frac {3 \left (1+x^3\right )^{2/3}}{8 x^3}+\frac {(4+x) \left (1+x^3\right )^{2/3}}{48 \left (2+2 x+x^2\right )}+\frac {\left (16-10 x-x^3\right ) \left (1+x^3\right )^{2/3}}{48 \left (4-4 x+2 x^2-2 x^3+x^4\right )}\right ) \, dx \\ & = \frac {1}{48} \int \frac {(4+x) \left (1+x^3\right )^{2/3}}{2+2 x+x^2} \, dx+\frac {1}{48} \int \frac {\left (16-10 x-x^3\right ) \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\frac {3}{8} \int \frac {\left (1+x^3\right )^{2/3}}{x^3} \, dx+\frac {1}{2} \int \frac {\left (1+x^3\right )^{2/3}}{x^6} \, dx \\ & = -\frac {3 \left (1+x^3\right )^{2/3}}{16 x^2}-\frac {\left (1+x^3\right )^{5/3}}{10 x^5}+\frac {1}{48} \int \left (\frac {(1-3 i) \left (1+x^3\right )^{2/3}}{(2-2 i)+2 x}+\frac {(1+3 i) \left (1+x^3\right )^{2/3}}{(2+2 i)+2 x}\right ) \, dx+\frac {1}{48} \int \left (\frac {16 \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4}-\frac {10 x \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4}-\frac {x^3 \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4}\right ) \, dx+\frac {3}{8} \int \frac {1}{\sqrt [3]{1+x^3}} \, dx \\ & = -\frac {3 \left (1+x^3\right )^{2/3}}{16 x^2}-\frac {\left (1+x^3\right )^{5/3}}{10 x^5}+\frac {1}{8} \sqrt {3} \arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )-\frac {3}{16} \log \left (-x+\sqrt [3]{1+x^3}\right )-\frac {1}{48} \int \frac {x^3 \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\left (\frac {1}{48}-\frac {i}{16}\right ) \int \frac {\left (1+x^3\right )^{2/3}}{(2-2 i)+2 x} \, dx+\left (\frac {1}{48}+\frac {i}{16}\right ) \int \frac {\left (1+x^3\right )^{2/3}}{(2+2 i)+2 x} \, dx-\frac {5}{24} \int \frac {x \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\frac {1}{3} \int \frac {\left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx \\ & = \frac {1}{96} \left (1+x^3\right )^{2/3}-\frac {3 \left (1+x^3\right )^{2/3}}{16 x^2}-\frac {\left (1+x^3\right )^{5/3}}{10 x^5}+\frac {1}{8} \sqrt {3} \arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )-\frac {3}{16} \log \left (-x+\sqrt [3]{1+x^3}\right )+\left (\frac {1}{192}-\frac {i}{64}\right ) \int \frac {4-8 i x}{((2-2 i)+2 x) \sqrt [3]{1+x^3}} \, dx+\left (\frac {1}{192}+\frac {i}{64}\right ) \int \frac {4+8 i x}{((2+2 i)+2 x) \sqrt [3]{1+x^3}} \, dx-\frac {1}{48} \int \frac {x^3 \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\left (\frac {1}{48}-\frac {i}{24}\right ) \int \frac {x}{\sqrt [3]{1+x^3}} \, dx+\left (\frac {1}{48}+\frac {i}{24}\right ) \int \frac {x}{\sqrt [3]{1+x^3}} \, dx-\frac {5}{24} \int \frac {x \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\frac {1}{3} \int \frac {\left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx \\ & = \frac {1}{96} \left (1+x^3\right )^{2/3}-\frac {3 \left (1+x^3\right )^{2/3}}{16 x^2}-\frac {\left (1+x^3\right )^{5/3}}{10 x^5}+\frac {1}{8} \sqrt {3} \arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )+\frac {1}{48} x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-x^3\right )-\frac {3}{16} \log \left (-x+\sqrt [3]{1+x^3}\right )+\left (-\frac {1}{16}-\frac {i}{48}\right ) \int \frac {1}{\sqrt [3]{1+x^3}} \, dx+\left (-\frac {1}{16}+\frac {i}{48}\right ) \int \frac {1}{\sqrt [3]{1+x^3}} \, dx-\frac {1}{48} \int \frac {x^3 \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\left (\frac {3}{16}-\frac {7 i}{48}\right ) \int \frac {1}{((2-2 i)+2 x) \sqrt [3]{1+x^3}} \, dx+\left (\frac {3}{16}+\frac {7 i}{48}\right ) \int \frac {1}{((2+2 i)+2 x) \sqrt [3]{1+x^3}} \, dx-\frac {5}{24} \int \frac {x \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\frac {1}{3} \int \frac {\left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx \\ & = \frac {1}{96} \left (1+x^3\right )^{2/3}-\frac {3 \left (1+x^3\right )^{2/3}}{16 x^2}-\frac {\left (1+x^3\right )^{5/3}}{10 x^5}-\frac {\arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{8 \sqrt {3}}+\frac {1}{8} \sqrt {3} \arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )+\frac {1}{48} x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-x^3\right )-\frac {1}{8} \log \left (-x+\sqrt [3]{1+x^3}\right )-\frac {1}{48} \int \frac {x^3 \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\left (\frac {3}{16}-\frac {7 i}{48}\right ) \int \left (-\frac {8 i}{\sqrt [3]{1+x^3} \left ((-16-16 i)+8 x^3\right )}-\frac {(4-4 i) x}{\sqrt [3]{1+x^3} \left ((-16-16 i)+8 x^3\right )}+\frac {4 x^2}{\sqrt [3]{1+x^3} \left ((-16-16 i)+8 x^3\right )}\right ) \, dx+\left (\frac {3}{16}+\frac {7 i}{48}\right ) \int \left (\frac {8 i}{\sqrt [3]{1+x^3} \left ((-16+16 i)+8 x^3\right )}-\frac {(4+4 i) x}{\sqrt [3]{1+x^3} \left ((-16+16 i)+8 x^3\right )}+\frac {4 x^2}{\sqrt [3]{1+x^3} \left ((-16+16 i)+8 x^3\right )}\right ) \, dx-\frac {5}{24} \int \frac {x \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\frac {1}{3} \int \frac {\left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx \\ & = \frac {1}{96} \left (1+x^3\right )^{2/3}-\frac {3 \left (1+x^3\right )^{2/3}}{16 x^2}-\frac {\left (1+x^3\right )^{5/3}}{10 x^5}-\frac {\arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{8 \sqrt {3}}+\frac {1}{8} \sqrt {3} \arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )+\frac {1}{48} x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-x^3\right )-\frac {1}{8} \log \left (-x+\sqrt [3]{1+x^3}\right )+\left (-\frac {7}{6}-\frac {3 i}{2}\right ) \int \frac {1}{\sqrt [3]{1+x^3} \left ((-16-16 i)+8 x^3\right )} \, dx+\left (-\frac {7}{6}+\frac {3 i}{2}\right ) \int \frac {1}{\sqrt [3]{1+x^3} \left ((-16+16 i)+8 x^3\right )} \, dx+\left (-\frac {1}{6}-\frac {4 i}{3}\right ) \int \frac {x}{\sqrt [3]{1+x^3} \left ((-16+16 i)+8 x^3\right )} \, dx+\left (-\frac {1}{6}+\frac {4 i}{3}\right ) \int \frac {x}{\sqrt [3]{1+x^3} \left ((-16-16 i)+8 x^3\right )} \, dx-\frac {1}{48} \int \frac {x^3 \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx-\frac {5}{24} \int \frac {x \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\frac {1}{3} \int \frac {\left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\left (\frac {3}{4}-\frac {7 i}{12}\right ) \int \frac {x^2}{\sqrt [3]{1+x^3} \left ((-16-16 i)+8 x^3\right )} \, dx+\left (\frac {3}{4}+\frac {7 i}{12}\right ) \int \frac {x^2}{\sqrt [3]{1+x^3} \left ((-16+16 i)+8 x^3\right )} \, dx \\ & = \frac {1}{96} \left (1+x^3\right )^{2/3}-\frac {3 \left (1+x^3\right )^{2/3}}{16 x^2}-\frac {\left (1+x^3\right )^{5/3}}{10 x^5}-\left (\frac {7}{384}+\frac {3 i}{128}\right ) x^2 \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},-x^3,\left (\frac {1}{4}-\frac {i}{4}\right ) x^3\right )-\left (\frac {7}{384}-\frac {3 i}{128}\right ) x^2 \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},-x^3,\left (\frac {1}{4}+\frac {i}{4}\right ) x^3\right )-\frac {\arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{8 \sqrt {3}}+\frac {1}{8} \sqrt {3} \arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )+\frac {\left (\frac {1}{6}+\frac {i}{48}\right ) \arctan \left (\frac {1+\frac {\sqrt [3]{10-2 i} x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{10-2 i}}+\frac {\left (\frac {1}{6}-\frac {i}{48}\right ) \arctan \left (\frac {1+\frac {\sqrt [3]{10+2 i} x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{10+2 i}}+\frac {1}{48} x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-x^3\right )+\frac {\left (\frac {1}{36}+\frac {i}{288}\right ) \log \left ((-16-16 i)+8 x^3\right )}{\sqrt [3]{10-2 i}}+\frac {\left (\frac {1}{36}-\frac {i}{288}\right ) \log \left ((-16+16 i)+8 x^3\right )}{\sqrt [3]{10+2 i}}-\frac {\left (\frac {1}{12}+\frac {i}{96}\right ) \log \left (\frac {\sqrt [3]{5-i} x}{2^{2/3}}-\sqrt [3]{1+x^3}\right )}{\sqrt [3]{10-2 i}}-\frac {\left (\frac {1}{12}-\frac {i}{96}\right ) \log \left (\frac {\sqrt [3]{5+i} x}{2^{2/3}}-\sqrt [3]{1+x^3}\right )}{\sqrt [3]{10+2 i}}-\frac {1}{8} \log \left (-x+\sqrt [3]{1+x^3}\right )-\frac {1}{48} \int \frac {x^3 \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx-\frac {5}{24} \int \frac {x \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\left (\frac {1}{4}-\frac {7 i}{36}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{1+x} ((-16-16 i)+8 x)} \, dx,x,x^3\right )+\left (\frac {1}{4}+\frac {7 i}{36}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{1+x} ((-16+16 i)+8 x)} \, dx,x,x^3\right )+\frac {1}{3} \int \frac {\left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx \\ & = \frac {1}{96} \left (1+x^3\right )^{2/3}-\frac {3 \left (1+x^3\right )^{2/3}}{16 x^2}-\frac {\left (1+x^3\right )^{5/3}}{10 x^5}-\left (\frac {7}{384}+\frac {3 i}{128}\right ) x^2 \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},-x^3,\left (\frac {1}{4}-\frac {i}{4}\right ) x^3\right )-\left (\frac {7}{384}-\frac {3 i}{128}\right ) x^2 \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},-x^3,\left (\frac {1}{4}+\frac {i}{4}\right ) x^3\right )-\frac {\arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{8 \sqrt {3}}+\frac {1}{8} \sqrt {3} \arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )+\frac {\left (\frac {1}{6}+\frac {i}{48}\right ) \arctan \left (\frac {1+\frac {\sqrt [3]{10-2 i} x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{10-2 i}}+\frac {\left (\frac {1}{6}-\frac {i}{48}\right ) \arctan \left (\frac {1+\frac {\sqrt [3]{10+2 i} x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{10+2 i}}+\frac {1}{48} x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-x^3\right )-\left (\frac {1}{576}+\frac {i}{192}\right ) (3-2 i)^{2/3} \log \left ((2-2 i)-x^3\right )-\left (\frac {1}{576}-\frac {i}{192}\right ) (3+2 i)^{2/3} \log \left ((2+2 i)-x^3\right )+\frac {\left (\frac {1}{36}+\frac {i}{288}\right ) \log \left ((-16-16 i)+8 x^3\right )}{\sqrt [3]{10-2 i}}+\frac {\left (\frac {1}{36}-\frac {i}{288}\right ) \log \left ((-16+16 i)+8 x^3\right )}{\sqrt [3]{10+2 i}}-\frac {\left (\frac {1}{12}+\frac {i}{96}\right ) \log \left (\frac {\sqrt [3]{5-i} x}{2^{2/3}}-\sqrt [3]{1+x^3}\right )}{\sqrt [3]{10-2 i}}-\frac {\left (\frac {1}{12}-\frac {i}{96}\right ) \log \left (\frac {\sqrt [3]{5+i} x}{2^{2/3}}-\sqrt [3]{1+x^3}\right )}{\sqrt [3]{10+2 i}}-\frac {1}{8} \log \left (-x+\sqrt [3]{1+x^3}\right )-\frac {1}{48} \int \frac {x^3 \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\left (\frac {3}{64}-\frac {7 i}{192}\right ) \text {Subst}\left (\int \frac {1}{(3+2 i)^{2/3}+\sqrt [3]{3+2 i} x+x^2} \, dx,x,\sqrt [3]{1+x^3}\right )+\left (\frac {3}{64}+\frac {7 i}{192}\right ) \text {Subst}\left (\int \frac {1}{(3-2 i)^{2/3}+\sqrt [3]{3-2 i} x+x^2} \, dx,x,\sqrt [3]{1+x^3}\right )-\frac {5}{24} \int \frac {x \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\frac {1}{3} \int \frac {\left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\left (\left (-\frac {1}{192}-\frac {i}{64}\right ) (3-2 i)^{2/3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{3-2 i}-x} \, dx,x,\sqrt [3]{1+x^3}\right )+\left (\left (-\frac {1}{192}+\frac {i}{64}\right ) (3+2 i)^{2/3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{3+2 i}-x} \, dx,x,\sqrt [3]{1+x^3}\right ) \\ & = \frac {1}{96} \left (1+x^3\right )^{2/3}-\frac {3 \left (1+x^3\right )^{2/3}}{16 x^2}-\frac {\left (1+x^3\right )^{5/3}}{10 x^5}-\left (\frac {7}{384}+\frac {3 i}{128}\right ) x^2 \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},-x^3,\left (\frac {1}{4}-\frac {i}{4}\right ) x^3\right )-\left (\frac {7}{384}-\frac {3 i}{128}\right ) x^2 \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},-x^3,\left (\frac {1}{4}+\frac {i}{4}\right ) x^3\right )-\frac {\arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{8 \sqrt {3}}+\frac {1}{8} \sqrt {3} \arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )+\frac {\left (\frac {1}{6}+\frac {i}{48}\right ) \arctan \left (\frac {1+\frac {\sqrt [3]{10-2 i} x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{10-2 i}}+\frac {\left (\frac {1}{6}-\frac {i}{48}\right ) \arctan \left (\frac {1+\frac {\sqrt [3]{10+2 i} x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{10+2 i}}+\frac {1}{48} x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-x^3\right )-\left (\frac {1}{576}+\frac {i}{192}\right ) (3-2 i)^{2/3} \log \left ((2-2 i)-x^3\right )-\left (\frac {1}{576}-\frac {i}{192}\right ) (3+2 i)^{2/3} \log \left ((2+2 i)-x^3\right )+\frac {\left (\frac {1}{36}+\frac {i}{288}\right ) \log \left ((-16-16 i)+8 x^3\right )}{\sqrt [3]{10-2 i}}+\frac {\left (\frac {1}{36}-\frac {i}{288}\right ) \log \left ((-16+16 i)+8 x^3\right )}{\sqrt [3]{10+2 i}}+\left (\frac {1}{192}+\frac {i}{64}\right ) (3-2 i)^{2/3} \log \left (\sqrt [3]{3-2 i}-\sqrt [3]{1+x^3}\right )+\left (\frac {1}{192}-\frac {i}{64}\right ) (3+2 i)^{2/3} \log \left (\sqrt [3]{3+2 i}-\sqrt [3]{1+x^3}\right )-\frac {\left (\frac {1}{12}+\frac {i}{96}\right ) \log \left (\frac {\sqrt [3]{5-i} x}{2^{2/3}}-\sqrt [3]{1+x^3}\right )}{\sqrt [3]{10-2 i}}-\frac {\left (\frac {1}{12}-\frac {i}{96}\right ) \log \left (\frac {\sqrt [3]{5+i} x}{2^{2/3}}-\sqrt [3]{1+x^3}\right )}{\sqrt [3]{10+2 i}}-\frac {1}{8} \log \left (-x+\sqrt [3]{1+x^3}\right )-\frac {1}{48} \int \frac {x^3 \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx-\frac {5}{24} \int \frac {x \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\frac {1}{3} \int \frac {\left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\left (\left (-\frac {1}{96}-\frac {i}{32}\right ) (3-2 i)^{2/3}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{1+x^3}}{\sqrt [3]{3-2 i}}\right )+\left (\left (-\frac {1}{96}+\frac {i}{32}\right ) (3+2 i)^{2/3}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{1+x^3}}{\sqrt [3]{3+2 i}}\right ) \\ & = \frac {1}{96} \left (1+x^3\right )^{2/3}-\frac {3 \left (1+x^3\right )^{2/3}}{16 x^2}-\frac {\left (1+x^3\right )^{5/3}}{10 x^5}-\left (\frac {7}{384}+\frac {3 i}{128}\right ) x^2 \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},-x^3,\left (\frac {1}{4}-\frac {i}{4}\right ) x^3\right )-\left (\frac {7}{384}-\frac {3 i}{128}\right ) x^2 \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},-x^3,\left (\frac {1}{4}+\frac {i}{4}\right ) x^3\right )-\frac {\arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{8 \sqrt {3}}+\frac {1}{8} \sqrt {3} \arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )+\frac {\left (\frac {1}{6}+\frac {i}{48}\right ) \arctan \left (\frac {1+\frac {\sqrt [3]{10-2 i} x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{10-2 i}}+\frac {\left (\frac {1}{6}-\frac {i}{48}\right ) \arctan \left (\frac {1+\frac {\sqrt [3]{10+2 i} x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{10+2 i}}+\frac {\left (\frac {1}{96}+\frac {i}{32}\right ) (3-2 i)^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{1+x^3}}{\sqrt [3]{3-2 i}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\left (\frac {1}{96}-\frac {i}{32}\right ) (3+2 i)^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{1+x^3}}{\sqrt [3]{3+2 i}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{48} x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-x^3\right )-\left (\frac {1}{576}+\frac {i}{192}\right ) (3-2 i)^{2/3} \log \left ((2-2 i)-x^3\right )-\left (\frac {1}{576}-\frac {i}{192}\right ) (3+2 i)^{2/3} \log \left ((2+2 i)-x^3\right )+\frac {\left (\frac {1}{36}+\frac {i}{288}\right ) \log \left ((-16-16 i)+8 x^3\right )}{\sqrt [3]{10-2 i}}+\frac {\left (\frac {1}{36}-\frac {i}{288}\right ) \log \left ((-16+16 i)+8 x^3\right )}{\sqrt [3]{10+2 i}}+\left (\frac {1}{192}+\frac {i}{64}\right ) (3-2 i)^{2/3} \log \left (\sqrt [3]{3-2 i}-\sqrt [3]{1+x^3}\right )+\left (\frac {1}{192}-\frac {i}{64}\right ) (3+2 i)^{2/3} \log \left (\sqrt [3]{3+2 i}-\sqrt [3]{1+x^3}\right )-\frac {\left (\frac {1}{12}+\frac {i}{96}\right ) \log \left (\frac {\sqrt [3]{5-i} x}{2^{2/3}}-\sqrt [3]{1+x^3}\right )}{\sqrt [3]{10-2 i}}-\frac {\left (\frac {1}{12}-\frac {i}{96}\right ) \log \left (\frac {\sqrt [3]{5+i} x}{2^{2/3}}-\sqrt [3]{1+x^3}\right )}{\sqrt [3]{10+2 i}}-\frac {1}{8} \log \left (-x+\sqrt [3]{1+x^3}\right )-\frac {1}{48} \int \frac {x^3 \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx-\frac {5}{24} \int \frac {x \left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx+\frac {1}{3} \int \frac {\left (1+x^3\right )^{2/3}}{4-4 x+2 x^2-2 x^3+x^4} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (4+x^3\right )}{x^6 \left (8-4 x^3+x^6\right )} \, dx=\frac {\left (-8-23 x^3\right ) \left (1+x^3\right )^{2/3}}{80 x^5}+\frac {1}{96} \text {RootSum}\left [13-20 \text {$\#$1}^3+8 \text {$\#$1}^6\&,\frac {-39 \log (x)+39 \log \left (\sqrt [3]{1+x^3}-x \text {$\#$1}\right )+32 \log (x) \text {$\#$1}^3-32 \log \left (\sqrt [3]{1+x^3}-x \text {$\#$1}\right ) \text {$\#$1}^3}{-5 \text {$\#$1}+4 \text {$\#$1}^4}\&\right ] \]

[In]

Integrate[((1 + x^3)^(2/3)*(4 + x^3))/(x^6*(8 - 4*x^3 + x^6)),x]

[Out]

((-8 - 23*x^3)*(1 + x^3)^(2/3))/(80*x^5) + RootSum[13 - 20*#1^3 + 8*#1^6 & , (-39*Log[x] + 39*Log[(1 + x^3)^(1
/3) - x*#1] + 32*Log[x]*#1^3 - 32*Log[(1 + x^3)^(1/3) - x*#1]*#1^3)/(-5*#1 + 4*#1^4) & ]/96

Maple [N/A] (verified)

Time = 158.56 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.76

method	result	size
pseudoelliptic	\(\frac {-5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (8 \textit {\_Z}^{6}-20 \textit {\_Z}^{3}+13\right )}{\sum }\frac {\left (32 \textit {\_R}^{3}-39\right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{3}+1\right )^{\frac {1}{3}}}{x}\right )}{4 \textit {\_R}^{4}-5 \textit {\_R}}\right ) x^{5}-138 x^{3} \left (x^{3}+1\right )^{\frac {2}{3}}-48 \left (x^{3}+1\right )^{\frac {2}{3}}}{480 x^{5}}\)	\(85\)
risch	\(\text {Expression too large to display}\)	\(9664\)
trager	\(\text {Expression too large to display}\)	\(11243\)

[In]

int((x^3+1)^(2/3)*(x^3+4)/x^6/(x^6-4*x^3+8),x,method=_RETURNVERBOSE)

[Out]

1/480*(-5*sum((32*_R^3-39)*ln((-_R*x+(x^3+1)^(1/3))/x)/(4*_R^4-5*_R),_R=RootOf(8*_Z^6-20*_Z^3+13))*x^5-138*x^3
*(x^3+1)^(2/3)-48*(x^3+1)^(2/3))/x^5

Fricas [F(-2)]

Exception generated. \[ \int \frac {\left (1+x^3\right )^{2/3} \left (4+x^3\right )}{x^6 \left (8-4 x^3+x^6\right )} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((x^3+1)^(2/3)*(x^3+4)/x^6/(x^6-4*x^3+8),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (tr
ace 0)

Sympy [N/A]

Not integrable

Time = 54.70 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.44 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (4+x^3\right )}{x^6 \left (8-4 x^3+x^6\right )} \, dx=\int \frac {\left (\left (x + 1\right ) \left (x^{2} - x + 1\right )\right )^{\frac {2}{3}} \left (x^{3} + 4\right )}{x^{6} \left (x^{2} + 2 x + 2\right ) \left (x^{4} - 2 x^{3} + 2 x^{2} - 4 x + 4\right )}\, dx \]

[In]

integrate((x**3+1)**(2/3)*(x**3+4)/x**6/(x**6-4*x**3+8),x)

[Out]

Integral(((x + 1)*(x**2 - x + 1))**(2/3)*(x**3 + 4)/(x**6*(x**2 + 2*x + 2)*(x**4 - 2*x**3 + 2*x**2 - 4*x + 4))
, x)

Maxima [N/A]

Not integrable

Time = 0.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.27 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (4+x^3\right )}{x^6 \left (8-4 x^3+x^6\right )} \, dx=\int { \frac {{\left (x^{3} + 4\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{{\left (x^{6} - 4 \, x^{3} + 8\right )} x^{6}} \,d x } \]

[In]

integrate((x^3+1)^(2/3)*(x^3+4)/x^6/(x^6-4*x^3+8),x, algorithm="maxima")

[Out]

integrate((x^3 + 4)*(x^3 + 1)^(2/3)/((x^6 - 4*x^3 + 8)*x^6), x)

Giac [N/A]

Not integrable

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.27 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (4+x^3\right )}{x^6 \left (8-4 x^3+x^6\right )} \, dx=\int { \frac {{\left (x^{3} + 4\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{{\left (x^{6} - 4 \, x^{3} + 8\right )} x^{6}} \,d x } \]

[In]

integrate((x^3+1)^(2/3)*(x^3+4)/x^6/(x^6-4*x^3+8),x, algorithm="giac")

[Out]

integrate((x^3 + 4)*(x^3 + 1)^(2/3)/((x^6 - 4*x^3 + 8)*x^6), x)

Mupad [N/A]

Not integrable

Time = 5.89 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.27 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (4+x^3\right )}{x^6 \left (8-4 x^3+x^6\right )} \, dx=\int \frac {{\left (x^3+1\right )}^{2/3}\,\left (x^3+4\right )}{x^6\,\left (x^6-4\,x^3+8\right )} \,d x \]

[In]

int(((x^3 + 1)^(2/3)*(x^3 + 4))/(x^6*(x^6 - 4*x^3 + 8)),x)

[Out]

int(((x^3 + 1)^(2/3)*(x^3 + 4))/(x^6*(x^6 - 4*x^3 + 8)), x)