Integrand size = 32, antiderivative size = 149 \[ \int \frac {\sqrt [3]{-1+2 x^3+x^8} \left (3+5 x^8\right )}{x^2 \left (-1+x^8\right )} \, dx=\frac {3 \sqrt [3]{-1+2 x^3+x^8}}{x}+\sqrt [3]{2} \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{-1+2 x^3+x^8}}\right )+\sqrt [3]{2} \log \left (-2 x+2^{2/3} \sqrt [3]{-1+2 x^3+x^8}\right )-\frac {\log \left (2 x^2+2^{2/3} x \sqrt [3]{-1+2 x^3+x^8}+\sqrt [3]{2} \left (-1+2 x^3+x^8\right )^{2/3}\right )}{2^{2/3}} \]
3*(x^8+2*x^3-1)^(1/3)/x+2^(1/3)*3^(1/2)*arctan(3^(1/2)*x/(x+2^(2/3)*(x^8+2 *x^3-1)^(1/3)))+2^(1/3)*ln(-2*x+2^(2/3)*(x^8+2*x^3-1)^(1/3))-1/2*ln(2*x^2+ 2^(2/3)*x*(x^8+2*x^3-1)^(1/3)+2^(1/3)*(x^8+2*x^3-1)^(2/3))*2^(1/3)
Time = 1.50 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [3]{-1+2 x^3+x^8} \left (3+5 x^8\right )}{x^2 \left (-1+x^8\right )} \, dx=\frac {3 \sqrt [3]{-1+2 x^3+x^8}}{x}+\sqrt [3]{2} \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{-1+2 x^3+x^8}}\right )+\sqrt [3]{2} \log \left (-2 x+2^{2/3} \sqrt [3]{-1+2 x^3+x^8}\right )-\frac {\log \left (2 x^2+2^{2/3} x \sqrt [3]{-1+2 x^3+x^8}+\sqrt [3]{2} \left (-1+2 x^3+x^8\right )^{2/3}\right )}{2^{2/3}} \]
(3*(-1 + 2*x^3 + x^8)^(1/3))/x + 2^(1/3)*Sqrt[3]*ArcTan[(Sqrt[3]*x)/(x + 2 ^(2/3)*(-1 + 2*x^3 + x^8)^(1/3))] + 2^(1/3)*Log[-2*x + 2^(2/3)*(-1 + 2*x^3 + x^8)^(1/3)] - Log[2*x^2 + 2^(2/3)*x*(-1 + 2*x^3 + x^8)^(1/3) + 2^(1/3)* (-1 + 2*x^3 + x^8)^(2/3)]/2^(2/3)
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.
\(\displaystyle \int \frac {\sqrt [3]{x^8+2 x^3-1} \left (5 x^8+3\right )}{x^2 \left (x^8-1\right )} \, dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {\sqrt [3]{x^8+2 x^3-1}}{-x-1}+\frac {\sqrt [3]{x^8+2 x^3-1}}{x-1}+\frac {2 \sqrt [3]{x^8+2 x^3-1}}{x^2+1}-\frac {3 \sqrt [3]{x^8+2 x^3-1}}{x^2}+\frac {4 \sqrt [3]{x^8+2 x^3-1} x^2}{x^4+1}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {\sqrt [3]{x^8+2 x^3-1}}{-x-1}dx+i \int \frac {\sqrt [3]{x^8+2 x^3-1}}{i-x}dx+(-1)^{3/4} \int \frac {\sqrt [3]{x^8+2 x^3-1}}{\sqrt [4]{-1}-x}dx-\sqrt [4]{-1} \int \frac {\sqrt [3]{x^8+2 x^3-1}}{-x-(-1)^{3/4}}dx+\int \frac {\sqrt [3]{x^8+2 x^3-1}}{x-1}dx+i \int \frac {\sqrt [3]{x^8+2 x^3-1}}{x+i}dx+(-1)^{3/4} \int \frac {\sqrt [3]{x^8+2 x^3-1}}{x+\sqrt [4]{-1}}dx-\sqrt [4]{-1} \int \frac {\sqrt [3]{x^8+2 x^3-1}}{x-(-1)^{3/4}}dx-3 \int \frac {\sqrt [3]{x^8+2 x^3-1}}{x^2}dx\) |
3.21.67.3.1 Defintions of rubi rules used
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 10.47 (sec) , antiderivative size = 2496, normalized size of antiderivative = 16.75
\[\text {Expression too large to display}\]
3*(x^8+2*x^3-1)^(1/3)/x+(2*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_ Z^2)*ln(-(16*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*x^6+4*Roo tOf(_Z^3-2)*x^11-4*RootOf(_Z^3-2)*x^3-2*RootOf(_Z^3-2)*x^8+RootOf(_Z^3-2)* x^16-16*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*x^3+4*RootOf(_ Z^3-2)*x^6-8*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*x^8+4*Roo tOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)+RootOf(_Z^3-2)+8*RootOf(R ootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)^2*RootOf(_Z^3-2)^2*x^3+2*RootO f(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^3*x^3-8*Root Of(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)^2*RootOf(_Z^3-2)^2*x^11-2* RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^3*x^11- 4*RootOf(_Z^3-2)^3*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*x^6 -16*RootOf(_Z^3-2)^2*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)^2 *x^6+6*(x^16+4*x^11-2*x^8+4*x^6-4*x^3+1)^(1/3)*RootOf(_Z^3-2)^2*x^4-3*(x^1 6+4*x^11-2*x^8+4*x^6-4*x^3+1)^(1/3)*RootOf(_Z^3-2)^2*x+3*(x^16+4*x^11-2*x^ 8+4*x^6-4*x^3+1)^(1/3)*RootOf(_Z^3-2)^2*x^9+6*(x^16+4*x^11-2*x^8+4*x^6-4*x ^3+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^ 3-2)*x^9+6*(x^16+4*x^11-2*x^8+4*x^6-4*x^3+1)^(2/3)*RootOf(RootOf(_Z^3-2)^2 +2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^2*x^2+12*(x^16+4*x^11-2*x^8+4* x^6-4*x^3+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*Roo tOf(_Z^3-2)*x^4-6*(x^16+4*x^11-2*x^8+4*x^6-4*x^3+1)^(1/3)*RootOf(RootOf...
Timed out. \[ \int \frac {\sqrt [3]{-1+2 x^3+x^8} \left (3+5 x^8\right )}{x^2 \left (-1+x^8\right )} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt [3]{-1+2 x^3+x^8} \left (3+5 x^8\right )}{x^2 \left (-1+x^8\right )} \, dx=\int \frac {\left (5 x^{8} + 3\right ) \sqrt [3]{x^{8} + 2 x^{3} - 1}}{x^{2} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \left (x^{4} + 1\right )}\, dx \]
Integral((5*x**8 + 3)*(x**8 + 2*x**3 - 1)**(1/3)/(x**2*(x - 1)*(x + 1)*(x* *2 + 1)*(x**4 + 1)), x)
\[ \int \frac {\sqrt [3]{-1+2 x^3+x^8} \left (3+5 x^8\right )}{x^2 \left (-1+x^8\right )} \, dx=\int { \frac {{\left (5 \, x^{8} + 3\right )} {\left (x^{8} + 2 \, x^{3} - 1\right )}^{\frac {1}{3}}}{{\left (x^{8} - 1\right )} x^{2}} \,d x } \]
\[ \int \frac {\sqrt [3]{-1+2 x^3+x^8} \left (3+5 x^8\right )}{x^2 \left (-1+x^8\right )} \, dx=\int { \frac {{\left (5 \, x^{8} + 3\right )} {\left (x^{8} + 2 \, x^{3} - 1\right )}^{\frac {1}{3}}}{{\left (x^{8} - 1\right )} x^{2}} \,d x } \]
Timed out. \[ \int \frac {\sqrt [3]{-1+2 x^3+x^8} \left (3+5 x^8\right )}{x^2 \left (-1+x^8\right )} \, dx=\int \frac {\left (5\,x^8+3\right )\,{\left (x^8+2\,x^3-1\right )}^{1/3}}{x^2\,\left (x^8-1\right )} \,d x \]