Integrand size = 26, antiderivative size = 469 \[ \int \frac {-1+x^8}{\sqrt [4]{-x^2+x^6} \left (1+x^8\right )} \, dx=-\frac {1}{4} \sqrt [4]{-4+3 \sqrt {2}} \arctan \left (\frac {\sqrt {2-\sqrt {2}} x}{-\sqrt {2+\sqrt {2}} x+2^{7/8} \sqrt [4]{-x^2+x^6}}\right )-\frac {1}{4} \sqrt [4]{-4+3 \sqrt {2}} \arctan \left (\frac {\sqrt {2-\sqrt {2}} x}{\sqrt {2+\sqrt {2}} x+2^{7/8} \sqrt [4]{-x^2+x^6}}\right )-\frac {1}{4} \sqrt [4]{4+3 \sqrt {2}} \arctan \left (\frac {2^{7/8} \sqrt {2+\sqrt {2}} x \sqrt [4]{-x^2+x^6}}{-2 x^2+2^{3/4} \sqrt {-x^2+x^6}}\right )-\frac {1}{4} \sqrt [4]{-4+3 \sqrt {2}} \text {arctanh}\left (\frac {\frac {\sqrt [8]{2} x^2}{\sqrt {2-\sqrt {2}}}+\frac {\sqrt {-x^2+x^6}}{\sqrt [8]{2} \sqrt {2-\sqrt {2}}}}{x \sqrt [4]{-x^2+x^6}}\right )+\frac {1}{8} \sqrt [4]{4+3 \sqrt {2}} \log \left (-2 x^2+2^{7/8} \sqrt {2+\sqrt {2}} x \sqrt [4]{-x^2+x^6}-2^{3/4} \sqrt {-x^2+x^6}\right )-\frac {1}{8} \sqrt [4]{4+3 \sqrt {2}} \log \left (2 \sqrt {2-\sqrt {2}} x^2+2\ 2^{3/8} x \sqrt [4]{-x^2+x^6}+2^{3/4} \sqrt {2-\sqrt {2}} \sqrt {-x^2+x^6}\right ) \]
-1/4*(-4+3*2^(1/2))^(1/4)*arctan((2-2^(1/2))^(1/2)*x/(-(2+2^(1/2))^(1/2)*x +2^(7/8)*(x^6-x^2)^(1/4)))-1/4*(-4+3*2^(1/2))^(1/4)*arctan((2-2^(1/2))^(1/ 2)*x/((2+2^(1/2))^(1/2)*x+2^(7/8)*(x^6-x^2)^(1/4)))-1/4*(4+3*2^(1/2))^(1/4 )*arctan(2^(7/8)*(2+2^(1/2))^(1/2)*x*(x^6-x^2)^(1/4)/(-2*x^2+2^(3/4)*(x^6- x^2)^(1/2)))-1/4*(-4+3*2^(1/2))^(1/4)*arctanh((2^(1/8)*x^2/(2-2^(1/2))^(1/ 2)+1/2*(x^6-x^2)^(1/2)*2^(7/8)/(2-2^(1/2))^(1/2))/x/(x^6-x^2)^(1/4))+1/8*( 4+3*2^(1/2))^(1/4)*ln(-2*x^2+2^(7/8)*(2+2^(1/2))^(1/2)*x*(x^6-x^2)^(1/4)-2 ^(3/4)*(x^6-x^2)^(1/2))-1/8*(4+3*2^(1/2))^(1/4)*ln(2*(2-2^(1/2))^(1/2)*x^2 +2*2^(3/8)*x*(x^6-x^2)^(1/4)+2^(3/4)*(2-2^(1/2))^(1/2)*(x^6-x^2)^(1/2))
Time = 25.29 (sec) , antiderivative size = 342, normalized size of antiderivative = 0.73 \[ \int \frac {-1+x^8}{\sqrt [4]{-x^2+x^6} \left (1+x^8\right )} \, dx=\frac {\sqrt [4]{-1+\frac {1}{x^4}} x^{3/2} \left (2 \sqrt [4]{-4+3 \sqrt {2}} \arctan \left (\frac {\sqrt [4]{-8+6 \sqrt {2}} \sqrt [4]{-1+\frac {1}{x^4}} \sqrt {x}}{\sqrt [4]{2}-\sqrt {-1+\frac {1}{x^4}} x}\right )-2 \sqrt [4]{-4+3 \sqrt {2}} \text {arctanh}\left (\frac {2 \sqrt [4]{-4+3 \sqrt {2}} \sqrt [4]{-1+\frac {1}{x^4}} \sqrt {x}}{2+2^{3/4} \sqrt {-1+\frac {1}{x^4}} x}\right )+\sqrt [4]{4+3 \sqrt {2}} \left (2 \arctan \left (\frac {\sqrt [4]{8+6 \sqrt {2}} \sqrt [4]{-1+\frac {1}{x^4}} \sqrt {x}}{\sqrt [4]{2}-\sqrt {-1+\frac {1}{x^4}} x}\right )+\log \left (\frac {2-2 \sqrt [4]{4+3 \sqrt {2}} \sqrt [4]{-1+\frac {1}{x^4}} \sqrt {x}+2^{3/4} \sqrt {-1+\frac {1}{x^4}} x}{x}\right )-\log \left (\frac {\sqrt {2-\sqrt {2}}+2^{3/8} \sqrt [4]{-1+\frac {1}{x^4}} \sqrt {x}+\sqrt {-1+\sqrt {2}} \sqrt {-1+\frac {1}{x^4}} x}{x}\right )\right )\right )}{8 \sqrt [4]{x^2 \left (-1+x^4\right )}} \]
((-1 + x^(-4))^(1/4)*x^(3/2)*(2*(-4 + 3*Sqrt[2])^(1/4)*ArcTan[((-8 + 6*Sqr t[2])^(1/4)*(-1 + x^(-4))^(1/4)*Sqrt[x])/(2^(1/4) - Sqrt[-1 + x^(-4)]*x)] - 2*(-4 + 3*Sqrt[2])^(1/4)*ArcTanh[(2*(-4 + 3*Sqrt[2])^(1/4)*(-1 + x^(-4)) ^(1/4)*Sqrt[x])/(2 + 2^(3/4)*Sqrt[-1 + x^(-4)]*x)] + (4 + 3*Sqrt[2])^(1/4) *(2*ArcTan[((8 + 6*Sqrt[2])^(1/4)*(-1 + x^(-4))^(1/4)*Sqrt[x])/(2^(1/4) - Sqrt[-1 + x^(-4)]*x)] + Log[(2 - 2*(4 + 3*Sqrt[2])^(1/4)*(-1 + x^(-4))^(1/ 4)*Sqrt[x] + 2^(3/4)*Sqrt[-1 + x^(-4)]*x)/x] - Log[(Sqrt[2 - Sqrt[2]] + 2^ (3/8)*(-1 + x^(-4))^(1/4)*Sqrt[x] + Sqrt[-1 + Sqrt[2]]*Sqrt[-1 + x^(-4)]*x )/x])))/(8*(x^2*(-1 + x^4))^(1/4))
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 0.52 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.29, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {2467, 25, 1388, 2035, 25, 7276, 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.
| \(\displaystyle \int \frac {x^8-1}{\sqrt [4]{x^6-x^2} \left (x^8+1\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt {x} \sqrt [4]{x^4-1} \int -\frac {1-x^8}{\sqrt {x} \sqrt [4]{x^4-1} \left (x^8+1\right )}dx}{\sqrt [4]{x^6-x^2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt {x} \sqrt [4]{x^4-1} \int \frac {1-x^8}{\sqrt {x} \sqrt [4]{x^4-1} \left (x^8+1\right )}dx}{\sqrt [4]{x^6-x^2}}\) |
| \(\Big \downarrow \) 1388 |
| \(\displaystyle -\frac {\sqrt {x} \sqrt [4]{x^4-1} \int \frac {\left (-x^4-1\right ) \left (x^4-1\right )^{3/4}}{\sqrt {x} \left (x^8+1\right )}dx}{\sqrt [4]{x^6-x^2}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{x^4-1} \int -\frac {\left (x^4-1\right )^{3/4} \left (x^4+1\right )}{x^8+1}d\sqrt {x}}{\sqrt [4]{x^6-x^2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 \sqrt {x} \sqrt [4]{x^4-1} \int \frac {\left (x^4-1\right )^{3/4} \left (x^4+1\right )}{x^8+1}d\sqrt {x}}{\sqrt [4]{x^6-x^2}}\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle \frac {2 \sqrt {x} \sqrt [4]{x^4-1} \int \left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (x^4-1\right )^{3/4}}{x^4+i}-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (x^4-1\right )^{3/4}}{i-x^4}\right )d\sqrt {x}}{\sqrt [4]{x^6-x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{x^4-1} \left (-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {x} \left (x^4-1\right )^{3/4} \operatorname {AppellF1}\left (\frac {1}{8},-\frac {3}{4},1,\frac {9}{8},x^4,i x^4\right )}{\left (1-x^4\right )^{3/4}}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {x} \left (x^4-1\right )^{3/4} \operatorname {AppellF1}\left (\frac {1}{8},1,-\frac {3}{4},\frac {9}{8},-i x^4,x^4\right )}{\left (1-x^4\right )^{3/4}}\right )}{\sqrt [4]{x^6-x^2}}\) |
(-2*Sqrt[x]*(-1 + x^4)^(1/4)*(((-1/2 + I/2)*Sqrt[x]*(-1 + x^4)^(3/4)*Appel lF1[1/8, -3/4, 1, 9/8, x^4, I*x^4])/(1 - x^4)^(3/4) - ((1/2 + I/2)*Sqrt[x] *(-1 + x^4)^(3/4)*AppellF1[1/8, 1, -3/4, 9/8, (-I)*x^4, x^4])/(1 - x^4)^(3 /4)))/(-x^2 + x^6)^(1/4)
3.31.58.3.1 Defintions of rubi rules used
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
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]
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]
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]
Timed out.
\[\int \frac {x^{8}-1}{\left (x^{6}-x^{2}\right )^{\frac {1}{4}} \left (x^{8}+1\right )}d x\]
Timed out. \[ \int \frac {-1+x^8}{\sqrt [4]{-x^2+x^6} \left (1+x^8\right )} \, dx=\text {Timed out} \]
\[ \int \frac {-1+x^8}{\sqrt [4]{-x^2+x^6} \left (1+x^8\right )} \, dx=\int \frac {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \left (x^{4} + 1\right )}{\sqrt [4]{x^{2} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )} \left (x^{8} + 1\right )}\, dx \]
Integral((x - 1)*(x + 1)*(x**2 + 1)*(x**4 + 1)/((x**2*(x - 1)*(x + 1)*(x** 2 + 1))**(1/4)*(x**8 + 1)), x)
\[ \int \frac {-1+x^8}{\sqrt [4]{-x^2+x^6} \left (1+x^8\right )} \, dx=\int { \frac {x^{8} - 1}{{\left (x^{8} + 1\right )} {\left (x^{6} - x^{2}\right )}^{\frac {1}{4}}} \,d x } \]
\[ \int \frac {-1+x^8}{\sqrt [4]{-x^2+x^6} \left (1+x^8\right )} \, dx=\int { \frac {x^{8} - 1}{{\left (x^{8} + 1\right )} {\left (x^{6} - x^{2}\right )}^{\frac {1}{4}}} \,d x } \]
Timed out. \[ \int \frac {-1+x^8}{\sqrt [4]{-x^2+x^6} \left (1+x^8\right )} \, dx=\int \frac {x^8-1}{\left (x^8+1\right )\,{\left (x^6-x^2\right )}^{1/4}} \,d x \]