\(\int \frac {\sqrt [3]{-1+2 x^3+x^8} (3+5 x^8)}{x^2 (-1+x^8)} \, dx\) [2067]

Optimal result

Integrand size = 32, antiderivative size = 149 \[ \int \frac {\sqrt [3]{-1+2 x^3+x^8} \left (3+5 x^8\right )}{x^2 \left (-1+x^8\right )} \, dx=\frac {3 \sqrt [3]{-1+2 x^3+x^8}}{x}+\sqrt [3]{2} \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{-1+2 x^3+x^8}}\right )+\sqrt [3]{2} \log \left (-2 x+2^{2/3} \sqrt [3]{-1+2 x^3+x^8}\right )-\frac {\log \left (2 x^2+2^{2/3} x \sqrt [3]{-1+2 x^3+x^8}+\sqrt [3]{2} \left (-1+2 x^3+x^8\right )^{2/3}\right )}{2^{2/3}} \] Output:

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)
 

Mathematica [A] (verified)

Time = 1.53 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [3]{-1+2 x^3+x^8} \left (3+5 x^8\right )}{x^2 \left (-1+x^8\right )} \, dx=\frac {3 \sqrt [3]{-1+2 x^3+x^8}}{x}+\sqrt [3]{2} \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{-1+2 x^3+x^8}}\right )+\sqrt [3]{2} \log \left (-2 x+2^{2/3} \sqrt [3]{-1+2 x^3+x^8}\right )-\frac {\log \left (2 x^2+2^{2/3} x \sqrt [3]{-1+2 x^3+x^8}+\sqrt [3]{2} \left (-1+2 x^3+x^8\right )^{2/3}\right )}{2^{2/3}} \] Input:

Integrate[((-1 + 2*x^3 + x^8)^(1/3)*(3 + 5*x^8))/(x^2*(-1 + x^8)),x]
 

Output:

(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)
 

Rubi [F]

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

\(\displaystyle \int \frac {\sqrt [3]{x^8+2 x^3-1}}{-x-1}dx+i \int \frac {\sqrt [3]{x^8+2 x^3-1}}{i-x}dx+(-1)^{3/4} \int \frac {\sqrt [3]{x^8+2 x^3-1}}{\sqrt [4]{-1}-x}dx-\sqrt [4]{-1} \int \frac {\sqrt [3]{x^8+2 x^3-1}}{-x-(-1)^{3/4}}dx+\int \frac {\sqrt [3]{x^8+2 x^3-1}}{x-1}dx+i \int \frac {\sqrt [3]{x^8+2 x^3-1}}{x+i}dx+(-1)^{3/4} \int \frac {\sqrt [3]{x^8+2 x^3-1}}{x+\sqrt [4]{-1}}dx-\sqrt [4]{-1} \int \frac {\sqrt [3]{x^8+2 x^3-1}}{x-(-1)^{3/4}}dx-3 \int \frac {\sqrt [3]{x^8+2 x^3-1}}{x^2}dx\)

Input:

Int[((-1 + 2*x^3 + x^8)^(1/3)*(3 + 5*x^8))/(x^2*(-1 + x^8)),x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 6.95 (sec) , antiderivative size = 2336, normalized size of antiderivative = 15.68

\[\text {Expression too large to display}\]

Input:

int((x^8+2*x^3-1)^(1/3)*(5*x^8+3)/x^2/(x^8-1),x)
 

Output:

3*(x^8+2*x^3-1)^(1/3)/x+(-ln(-(-RootOf(_Z^3-2)*x^16-RootOf(RootOf(_Z^3-2)^ 
2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*x^16+2*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_ 
Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^3*x^11+2*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf 
(_Z^3-2)+4*_Z^2)^2*RootOf(_Z^3-2)^2*x^11-3*(x^16+4*x^11-2*x^8+4*x^6-4*x^3+ 
1)^(1/3)*RootOf(_Z^3-2)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2 
)*x^9-4*RootOf(_Z^3-2)*x^11-4*x^11*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^ 
3-2)+4*_Z^2)+4*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf( 
_Z^3-2)^3*x^6+4*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)^2*Root 
Of(_Z^3-2)^2*x^6+2*RootOf(_Z^3-2)*x^8+2*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootO 
f(_Z^3-2)+4*_Z^2)*x^8-2*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2 
)*RootOf(_Z^3-2)^3*x^3-2*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^ 
2)^2*RootOf(_Z^3-2)^2*x^3+3*(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^2-6*(x^16 
+4*x^11-2*x^8+4*x^6-4*x^3+1)^(1/3)*RootOf(_Z^3-2)*RootOf(RootOf(_Z^3-2)^2+ 
2*_Z*RootOf(_Z^3-2)+4*_Z^2)*x^4-4*RootOf(_Z^3-2)*x^6-4*RootOf(RootOf(_Z^3- 
2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*x^6+3*(x^16+4*x^11-2*x^8+4*x^6-4*x^3+1)^( 
1/3)*RootOf(_Z^3-2)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*x+ 
4*RootOf(_Z^3-2)*x^3+4*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2) 
*x^3+3*(x^16+4*x^11-2*x^8+4*x^6-4*x^3+1)^(2/3)*x^2-RootOf(_Z^3-2)-RootOf(R 
ootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2))/(x^8+2*x^3-1)/(-1+x)/(1+x)...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{-1+2 x^3+x^8} \left (3+5 x^8\right )}{x^2 \left (-1+x^8\right )} \, dx=\text {Timed out} \] Input:

integrate((x^8+2*x^3-1)^(1/3)*(5*x^8+3)/x^2/(x^8-1),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\sqrt [3]{-1+2 x^3+x^8} \left (3+5 x^8\right )}{x^2 \left (-1+x^8\right )} \, dx=\int \frac {\left (5 x^{8} + 3\right ) \sqrt [3]{x^{8} + 2 x^{3} - 1}}{x^{2} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \left (x^{4} + 1\right )}\, dx \] Input:

integrate((x**8+2*x**3-1)**(1/3)*(5*x**8+3)/x**2/(x**8-1),x)
 

Output:

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)
 

Maxima [F]

\[ \int \frac {\sqrt [3]{-1+2 x^3+x^8} \left (3+5 x^8\right )}{x^2 \left (-1+x^8\right )} \, dx=\int { \frac {{\left (5 \, x^{8} + 3\right )} {\left (x^{8} + 2 \, x^{3} - 1\right )}^{\frac {1}{3}}}{{\left (x^{8} - 1\right )} x^{2}} \,d x } \] Input:

integrate((x^8+2*x^3-1)^(1/3)*(5*x^8+3)/x^2/(x^8-1),x, algorithm="maxima")
 

Output:

integrate((5*x^8 + 3)*(x^8 + 2*x^3 - 1)^(1/3)/((x^8 - 1)*x^2), x)
 

Giac [F]

\[ \int \frac {\sqrt [3]{-1+2 x^3+x^8} \left (3+5 x^8\right )}{x^2 \left (-1+x^8\right )} \, dx=\int { \frac {{\left (5 \, x^{8} + 3\right )} {\left (x^{8} + 2 \, x^{3} - 1\right )}^{\frac {1}{3}}}{{\left (x^{8} - 1\right )} x^{2}} \,d x } \] Input:

integrate((x^8+2*x^3-1)^(1/3)*(5*x^8+3)/x^2/(x^8-1),x, algorithm="giac")
 

Output:

integrate((5*x^8 + 3)*(x^8 + 2*x^3 - 1)^(1/3)/((x^8 - 1)*x^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{-1+2 x^3+x^8} \left (3+5 x^8\right )}{x^2 \left (-1+x^8\right )} \, dx=\int \frac {\left (5\,x^8+3\right )\,{\left (x^8+2\,x^3-1\right )}^{1/3}}{x^2\,\left (x^8-1\right )} \,d x \] Input:

int(((5*x^8 + 3)*(2*x^3 + x^8 - 1)^(1/3))/(x^2*(x^8 - 1)),x)
 

Output:

int(((5*x^8 + 3)*(2*x^3 + x^8 - 1)^(1/3))/(x^2*(x^8 - 1)), x)
 

Reduce [F]

\[ \int \frac {\sqrt [3]{-1+2 x^3+x^8} \left (3+5 x^8\right )}{x^2 \left (-1+x^8\right )} \, dx=3 \left (\int \frac {\left (x^{8}+2 x^{3}-1\right )^{\frac {1}{3}}}{x^{10}-x^{2}}d x \right )+5 \left (\int \frac {\left (x^{8}+2 x^{3}-1\right )^{\frac {1}{3}} x^{6}}{x^{8}-1}d x \right ) \] Input:

int((x^8+2*x^3-1)^(1/3)*(5*x^8+3)/x^2/(x^8-1),x)
 

Output:

3*int((x**8 + 2*x**3 - 1)**(1/3)/(x**10 - x**2),x) + 5*int(((x**8 + 2*x**3 
 - 1)**(1/3)*x**6)/(x**8 - 1),x)