\(\int \frac {-1+x^8}{\sqrt [4]{-x^2+x^6} (1+x^8)} \, dx\) [3059]

Optimal result

Integrand size = 26, antiderivative size = 469 \[ \int \frac {-1+x^8}{\sqrt [4]{-x^2+x^6} \left (1+x^8\right )} \, dx=-\frac {1}{4} \sqrt [4]{-4+3 \sqrt {2}} \arctan \left (\frac {\sqrt {2-\sqrt {2}} x}{-\sqrt {2+\sqrt {2}} x+2^{7/8} \sqrt [4]{-x^2+x^6}}\right )-\frac {1}{4} \sqrt [4]{-4+3 \sqrt {2}} \arctan \left (\frac {\sqrt {2-\sqrt {2}} x}{\sqrt {2+\sqrt {2}} x+2^{7/8} \sqrt [4]{-x^2+x^6}}\right )-\frac {1}{4} \sqrt [4]{4+3 \sqrt {2}} \arctan \left (\frac {2^{7/8} \sqrt {2+\sqrt {2}} x \sqrt [4]{-x^2+x^6}}{-2 x^2+2^{3/4} \sqrt {-x^2+x^6}}\right )-\frac {1}{4} \sqrt [4]{-4+3 \sqrt {2}} \text {arctanh}\left (\frac {\frac {\sqrt [8]{2} x^2}{\sqrt {2-\sqrt {2}}}+\frac {\sqrt {-x^2+x^6}}{\sqrt [8]{2} \sqrt {2-\sqrt {2}}}}{x \sqrt [4]{-x^2+x^6}}\right )+\frac {1}{8} \sqrt [4]{4+3 \sqrt {2}} \log \left (-2 x^2+2^{7/8} \sqrt {2+\sqrt {2}} x \sqrt [4]{-x^2+x^6}-2^{3/4} \sqrt {-x^2+x^6}\right )-\frac {1}{8} \sqrt [4]{4+3 \sqrt {2}} \log \left (2 \sqrt {2-\sqrt {2}} x^2+2\ 2^{3/8} x \sqrt [4]{-x^2+x^6}+2^{3/4} \sqrt {2-\sqrt {2}} \sqrt {-x^2+x^6}\right ) \]

[Out]

-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))

Rubi [C] (verified)

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

Time = 0.31 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.22, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2081, 1600, 6847, 6857, 441, 440} \[ \int \frac {-1+x^8}{\sqrt [4]{-x^2+x^6} \left (1+x^8\right )} \, dx=-\frac {(1-i) x \sqrt [4]{1-x^4} \operatorname {AppellF1}\left (\frac {1}{8},-\frac {3}{4},1,\frac {9}{8},x^4,i x^4\right )}{\sqrt [4]{x^6-x^2}}-\frac {(1+i) x \sqrt [4]{1-x^4} \operatorname {AppellF1}\left (\frac {1}{8},1,-\frac {3}{4},\frac {9}{8},-i x^4,x^4\right )}{\sqrt [4]{x^6-x^2}} \]

[In]

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

[Out]

((-1 + I)*x*(1 - x^4)^(1/4)*AppellF1[1/8, -3/4, 1, 9/8, x^4, I*x^4])/(-x^2 + x^6)^(1/4) - ((1 + I)*x*(1 - x^4)
^(1/4)*AppellF1[1/8, 1, -3/4, 9/8, (-I)*x^4, x^4])/(-x^2 + x^6)^(1/4)

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 441

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^F
racPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rule 6847

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt [4]{-1+x^4}\right ) \int \frac {-1+x^8}{\sqrt {x} \sqrt [4]{-1+x^4} \left (1+x^8\right )} \, dx}{\sqrt [4]{-x^2+x^6}} \\ & = \frac {\left (\sqrt {x} \sqrt [4]{-1+x^4}\right ) \int \frac {\left (-1+x^4\right )^{3/4} \left (1+x^4\right )}{\sqrt {x} \left (1+x^8\right )} \, dx}{\sqrt [4]{-x^2+x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{-1+x^4}\right ) \text {Subst}\left (\int \frac {\left (-1+x^8\right )^{3/4} \left (1+x^8\right )}{1+x^{16}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{-1+x^4}\right ) \text {Subst}\left (\int \left (-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (-1+x^8\right )^{3/4}}{i-x^8}+\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (-1+x^8\right )^{3/4}}{i+x^8}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^6}} \\ & = -\frac {\left ((1-i) \sqrt {x} \sqrt [4]{-1+x^4}\right ) \text {Subst}\left (\int \frac {\left (-1+x^8\right )^{3/4}}{i-x^8} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^6}}+\frac {\left ((1+i) \sqrt {x} \sqrt [4]{-1+x^4}\right ) \text {Subst}\left (\int \frac {\left (-1+x^8\right )^{3/4}}{i+x^8} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^6}} \\ & = -\frac {\left ((1-i) \sqrt {x} \left (-1+x^4\right )\right ) \text {Subst}\left (\int \frac {\left (1-x^8\right )^{3/4}}{i-x^8} \, dx,x,\sqrt {x}\right )}{\left (1-x^4\right )^{3/4} \sqrt [4]{-x^2+x^6}}+\frac {\left ((1+i) \sqrt {x} \left (-1+x^4\right )\right ) \text {Subst}\left (\int \frac {\left (1-x^8\right )^{3/4}}{i+x^8} \, dx,x,\sqrt {x}\right )}{\left (1-x^4\right )^{3/4} \sqrt [4]{-x^2+x^6}} \\ & = -\frac {(1-i) x \sqrt [4]{1-x^4} \operatorname {AppellF1}\left (\frac {1}{8},-\frac {3}{4},1,\frac {9}{8},x^4,i x^4\right )}{\sqrt [4]{-x^2+x^6}}-\frac {(1+i) x \sqrt [4]{1-x^4} \operatorname {AppellF1}\left (\frac {1}{8},1,-\frac {3}{4},\frac {9}{8},-i x^4,x^4\right )}{\sqrt [4]{-x^2+x^6}} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 0.01 (sec) , antiderivative size = 342, normalized size of antiderivative = 0.73 \[ \int \frac {-1+x^8}{\sqrt [4]{-x^2+x^6} \left (1+x^8\right )} \, dx=\frac {\sqrt [4]{-1+\frac {1}{x^4}} x^{3/2} \left (2 \sqrt [4]{-4+3 \sqrt {2}} \arctan \left (\frac {\sqrt [4]{-8+6 \sqrt {2}} \sqrt [4]{-1+\frac {1}{x^4}} \sqrt {x}}{\sqrt [4]{2}-\sqrt {-1+\frac {1}{x^4}} x}\right )-2 \sqrt [4]{-4+3 \sqrt {2}} \text {arctanh}\left (\frac {2 \sqrt [4]{-4+3 \sqrt {2}} \sqrt [4]{-1+\frac {1}{x^4}} \sqrt {x}}{2+2^{3/4} \sqrt {-1+\frac {1}{x^4}} x}\right )+\sqrt [4]{4+3 \sqrt {2}} \left (2 \arctan \left (\frac {\sqrt [4]{8+6 \sqrt {2}} \sqrt [4]{-1+\frac {1}{x^4}} \sqrt {x}}{\sqrt [4]{2}-\sqrt {-1+\frac {1}{x^4}} x}\right )+\log \left (\frac {2-2 \sqrt [4]{4+3 \sqrt {2}} \sqrt [4]{-1+\frac {1}{x^4}} \sqrt {x}+2^{3/4} \sqrt {-1+\frac {1}{x^4}} x}{x}\right )-\log \left (\frac {\sqrt {2-\sqrt {2}}+2^{3/8} \sqrt [4]{-1+\frac {1}{x^4}} \sqrt {x}+\sqrt {-1+\sqrt {2}} \sqrt {-1+\frac {1}{x^4}} x}{x}\right )\right )\right )}{8 \sqrt [4]{x^2 \left (-1+x^4\right )}} \]

[In]

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

[Out]

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

Maple [F(-1)]

Timed out.

\[\int \frac {x^{8}-1}{\left (x^{6}-x^{2}\right )^{\frac {1}{4}} \left (x^{8}+1\right )}d x\]

[In]

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

[Out]

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

Fricas [F(-1)]

Timed out. \[ \int \frac {-1+x^8}{\sqrt [4]{-x^2+x^6} \left (1+x^8\right )} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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