\(\int \frac {x^2 (-2+x^4)}{\sqrt [3]{-x+x^5} (-1+x^4+x^8)} \, dx\) [1466]

Optimal result

Integrand size = 30, antiderivative size = 103 \[ \int \frac {x^2 \left (-2+x^4\right )}{\sqrt [3]{-x+x^5} \left (-1+x^4+x^8\right )} \, dx=\frac {1}{4} \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{-x+x^5}}{-2 x^3+\sqrt [3]{-x+x^5}}\right )+\frac {1}{4} \log \left (x^3+\sqrt [3]{-x+x^5}\right )-\frac {1}{8} \log \left (x^6-x^3 \sqrt [3]{-x+x^5}+\left (-x+x^5\right )^{2/3}\right ) \] Output:

1/4*3^(1/2)*arctan(3^(1/2)*(x^5-x)^(1/3)/(-2*x^3+(x^5-x)^(1/3)))+1/4*ln(x^ 
3+(x^5-x)^(1/3))-1/8*ln(x^6-x^3*(x^5-x)^(1/3)+(x^5-x)^(2/3))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 10.44 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 \left (-2+x^4\right )}{\sqrt [3]{-x+x^5} \left (-1+x^4+x^8\right )} \, dx=\frac {1}{4} \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{-x+x^5}}{-2 x^3+\sqrt [3]{-x+x^5}}\right )+\frac {1}{4} \log \left (x^3+\sqrt [3]{-x+x^5}\right )-\frac {1}{8} \log \left (x^6-x^3 \sqrt [3]{-x+x^5}+\left (-x+x^5\right )^{2/3}\right ) \] Input:

Integrate[(x^2*(-2 + x^4))/((-x + x^5)^(1/3)*(-1 + x^4 + x^8)),x]
 

Output:

(Sqrt[3]*ArcTan[(Sqrt[3]*(-x + x^5)^(1/3))/(-2*x^3 + (-x + x^5)^(1/3))])/4 
 + Log[x^3 + (-x + x^5)^(1/3)]/4 - Log[x^6 - x^3*(-x + x^5)^(1/3) + (-x + 
x^5)^(2/3)]/8
 

Rubi [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 1.16 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2467, 2035, 7279, 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.

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

\(\Big \downarrow \) 2467

\(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{x^4-1} \int \frac {x^{5/3} \left (2-x^4\right )}{\sqrt [3]{x^4-1} \left (-x^8-x^4+1\right )}dx}{\sqrt [3]{x^5-x}}\)

\(\Big \downarrow \) 2035

\(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{x^4-1} \int \frac {x^{7/3} \left (2-x^4\right )}{\sqrt [3]{x^4-1} \left (-x^8-x^4+1\right )}d\sqrt [3]{x}}{\sqrt [3]{x^5-x}}\)

\(\Big \downarrow \) 7279

\(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{x^4-1} \int \left (\frac {x^{19/3}}{\sqrt [3]{x^4-1} \left (x^8+x^4-1\right )}-\frac {2 x^{7/3}}{\sqrt [3]{x^4-1} \left (x^8+x^4-1\right )}\right )d\sqrt [3]{x}}{\sqrt [3]{x^5-x}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{x^4-1} \left (\frac {\sqrt [3]{1-x^4} x^{8/3} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},x^4,-\frac {2 x^4}{1+\sqrt {5}}\right )}{2 \sqrt {5} \left (1+\sqrt {5}\right ) \sqrt [3]{x^4-1}}+\frac {\sqrt [3]{1-x^4} x^{8/3} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},x^4,-\frac {2 x^4}{1+\sqrt {5}}\right )}{8 \sqrt {5} \sqrt [3]{x^4-1}}-\frac {\sqrt [3]{1-x^4} x^{8/3} \operatorname {AppellF1}\left (\frac {2}{3},1,\frac {1}{3},\frac {5}{3},-\frac {2 x^4}{1-\sqrt {5}},x^4\right )}{2 \sqrt {5} \left (1-\sqrt {5}\right ) \sqrt [3]{x^4-1}}-\frac {\sqrt [3]{1-x^4} x^{8/3} \operatorname {AppellF1}\left (\frac {2}{3},1,\frac {1}{3},\frac {5}{3},-\frac {2 x^4}{1-\sqrt {5}},x^4\right )}{8 \sqrt {5} \sqrt [3]{x^4-1}}\right )}{\sqrt [3]{x^5-x}}\)

Input:

Int[(x^2*(-2 + x^4))/((-x + x^5)^(1/3)*(-1 + x^4 + x^8)),x]
 

Output:

(3*x^(1/3)*(-1 + x^4)^(1/3)*((x^(8/3)*(1 - x^4)^(1/3)*AppellF1[2/3, 1/3, 1 
, 5/3, x^4, (-2*x^4)/(1 + Sqrt[5])])/(8*Sqrt[5]*(-1 + x^4)^(1/3)) + (x^(8/ 
3)*(1 - x^4)^(1/3)*AppellF1[2/3, 1/3, 1, 5/3, x^4, (-2*x^4)/(1 + Sqrt[5])] 
)/(2*Sqrt[5]*(1 + Sqrt[5])*(-1 + x^4)^(1/3)) - (x^(8/3)*(1 - x^4)^(1/3)*Ap 
pellF1[2/3, 1, 1/3, 5/3, (-2*x^4)/(1 - Sqrt[5]), x^4])/(8*Sqrt[5]*(-1 + x^ 
4)^(1/3)) - (x^(8/3)*(1 - x^4)^(1/3)*AppellF1[2/3, 1, 1/3, 5/3, (-2*x^4)/( 
1 - Sqrt[5]), x^4])/(2*Sqrt[5]*(1 - Sqrt[5])*(-1 + x^4)^(1/3))))/(-x + x^5 
)^(1/3)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

rule 7279

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ 
{v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su 
mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
 

Maple [C] (verified)

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

Time = 31.52 (sec) , antiderivative size = 626, normalized size of antiderivative = 6.08

method	result	size
trager	\(\text {Expression too large to display}\)	\(626\)

Input:

int(x^2*(x^4-2)/(x^5-x)^(1/3)/(x^8+x^4-1),x,method=_RETURNVERBOSE)
 

Output:

-1/4*ln(-(1027936778409280454041232991842640*RootOf(16*_Z^2+4*_Z+1)^2*x^8+ 
4529039623257700492643576695162024*RootOf(16*_Z^2+4*_Z+1)*x^8+610884417613 
243258599747967858176*x^8-6458608904506033300021952070673408*RootOf(16*_Z^ 
2+4*_Z+1)*(x^5-x)^(1/3)*x^5-17543454351518386415637043060781056*RootOf(16* 
_Z^2+4*_Z+1)^2*x^4-25262880761481365661024699935681*(x^5-x)^(1/3)*x^5+6458 
608904506033300021952070673408*RootOf(16*_Z^2+4*_Z+1)*(x^5-x)^(2/3)*x^2-65 
72417063730249524797944388667308*RootOf(16*_Z^2+4*_Z+1)*x^4+25262880761481 
365661024699935681*(x^5-x)^(2/3)*x^2-575090408768717286416168985366486*x^4 
+17543454351518386415637043060781056*RootOf(16*_Z^2+4*_Z+1)^2+657241706373 
0249524797944388667308*RootOf(16*_Z^2+4*_Z+1)+5750904087687172864161689853 
66486)/(x^8+x^4-1))-ln(-(1027936778409280454041232991842640*RootOf(16*_Z^2 
+4*_Z+1)^2*x^8+4529039623257700492643576695162024*RootOf(16*_Z^2+4*_Z+1)*x 
^8+610884417613243258599747967858176*x^8-645860890450603330002195207067340 
8*RootOf(16*_Z^2+4*_Z+1)*(x^5-x)^(1/3)*x^5-1754345435151838641563704306078 
1056*RootOf(16*_Z^2+4*_Z+1)^2*x^4-25262880761481365661024699935681*(x^5-x) 
^(1/3)*x^5+6458608904506033300021952070673408*RootOf(16*_Z^2+4*_Z+1)*(x^5- 
x)^(2/3)*x^2-6572417063730249524797944388667308*RootOf(16*_Z^2+4*_Z+1)*x^4 
+25262880761481365661024699935681*(x^5-x)^(2/3)*x^2-5750904087687172864161 
68985366486*x^4+17543454351518386415637043060781056*RootOf(16*_Z^2+4*_Z+1) 
^2+6572417063730249524797944388667308*RootOf(16*_Z^2+4*_Z+1)[phone]...
 

Fricas [A] (verification not implemented)

Time = 2.04 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.09 \[ \int \frac {x^2 \left (-2+x^4\right )}{\sqrt [3]{-x+x^5} \left (-1+x^4+x^8\right )} \, dx=-\frac {1}{4} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} x^{8} + 2 \, \sqrt {3} {\left (x^{5} - x\right )}^{\frac {1}{3}} x^{5} + 4 \, \sqrt {3} {\left (x^{5} - x\right )}^{\frac {2}{3}} x^{2}}{x^{8} - 8 \, x^{4} + 8}\right ) + \frac {1}{8} \, \log \left (\frac {x^{8} + 3 \, {\left (x^{5} - x\right )}^{\frac {1}{3}} x^{5} + x^{4} + 3 \, {\left (x^{5} - x\right )}^{\frac {2}{3}} x^{2} - 1}{x^{8} + x^{4} - 1}\right ) \] Input:

integrate(x^2*(x^4-2)/(x^5-x)^(1/3)/(x^8+x^4-1),x, algorithm="fricas")
 

Output:

-1/4*sqrt(3)*arctan((sqrt(3)*x^8 + 2*sqrt(3)*(x^5 - x)^(1/3)*x^5 + 4*sqrt( 
3)*(x^5 - x)^(2/3)*x^2)/(x^8 - 8*x^4 + 8)) + 1/8*log((x^8 + 3*(x^5 - x)^(1 
/3)*x^5 + x^4 + 3*(x^5 - x)^(2/3)*x^2 - 1)/(x^8 + x^4 - 1))
 

Sympy [F(-1)]

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

integrate(x**2*(x**4-2)/(x**5-x)**(1/3)/(x**8+x**4-1),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(x^2*(x^4-2)/(x^5-x)^(1/3)/(x^8+x^4-1),x, algorithm="maxima")
 

Output:

integrate((x^4 - 2)*x^2/((x^8 + x^4 - 1)*(x^5 - x)^(1/3)), x)
 

Giac [F]

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

integrate(x^2*(x^4-2)/(x^5-x)^(1/3)/(x^8+x^4-1),x, algorithm="giac")
 

Output:

integrate((x^4 - 2)*x^2/((x^8 + x^4 - 1)*(x^5 - x)^(1/3)), x)
 

Mupad [F(-1)]

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

int((x^2*(x^4 - 2))/((x^5 - x)^(1/3)*(x^4 + x^8 - 1)),x)
 

Output:

int((x^2*(x^4 - 2))/((x^5 - x)^(1/3)*(x^4 + x^8 - 1)), x)
 

Reduce [F]

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

int(x^2*(x^4-2)/(x^5-x)^(1/3)/(x^8+x^4-1),x)
                                                                                    
                                                                                    
 

Output:

int(x^2*(x^4-2)/(x^5-x)^(1/3)/(x^8+x^4-1),x)