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

Optimal. Leaf size=469 \[ \frac {1}{8} \sqrt [4]{4+3 \sqrt {2}} \log \left (-2 x^2+2^{7/8} \sqrt {2+\sqrt {2}} \sqrt [4]{x^6-x^2} x-2^{3/4} \sqrt {x^6-x^2}\right )-\frac {1}{8} \sqrt [4]{4+3 \sqrt {2}} \log \left (2 \sqrt {2-\sqrt {2}} x^2+2\ 2^{3/8} \sqrt [4]{x^6-x^2} x+2^{3/4} \sqrt {2-\sqrt {2}} \sqrt {x^6-x^2}\right )-\frac {1}{4} \sqrt [4]{3 \sqrt {2}-4} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}} x}{2^{7/8} \sqrt [4]{x^6-x^2}-\sqrt {2+\sqrt {2}} x}\right )-\frac {1}{4} \sqrt [4]{3 \sqrt {2}-4} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}} x}{2^{7/8} \sqrt [4]{x^6-x^2}+\sqrt {2+\sqrt {2}} x}\right )-\frac {1}{4} \sqrt [4]{4+3 \sqrt {2}} \tan ^{-1}\left (\frac {2^{7/8} \sqrt {2+\sqrt {2}} x \sqrt [4]{x^6-x^2}}{2^{3/4} \sqrt {x^6-x^2}-2 x^2}\right )-\frac {1}{4} \sqrt [4]{3 \sqrt {2}-4} \tanh ^{-1}\left (\frac {\frac {\sqrt [8]{2} x^2}{\sqrt {2-\sqrt {2}}}+\frac {\sqrt {x^6-x^2}}{\sqrt [8]{2} \sqrt {2-\sqrt {2}}}}{x \sqrt [4]{x^6-x^2}}\right ) \]

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Rubi [C]  time = 0.48, antiderivative size = 101, normalized size of antiderivative = 0.22, number of steps used = 9, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2056, 1586, 6715, 6725, 430, 429} \begin {gather*} -\frac {(1-i) x \sqrt [4]{1-x^4} F_1\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} F_1\left (\frac {1}{8};1,-\frac {3}{4};\frac {9}{8};-i x^4,x^4\right )}{\sqrt [4]{x^6-x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[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 429

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 430

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 1586

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 2056

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 6715

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 6725

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*} \int \frac {-1+x^8}{\sqrt [4]{-x^2+x^6} \left (1+x^8\right )} \, dx &=\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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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} F_1\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} F_1\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*}

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Mathematica [F]  time = 0.60, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-1+x^8}{\sqrt [4]{-x^2+x^6} \left (1+x^8\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.00, size = 394, normalized size = 0.84 \begin {gather*} \frac {1}{4} \sqrt [4]{-4+3 \sqrt {2}} \tan ^{-1}\left (\frac {\sqrt [4]{-8+6 \sqrt {2}} x \sqrt [4]{-x^2+x^6}}{\sqrt [4]{2} x^2-\sqrt {-x^2+x^6}}\right )+\frac {1}{4} \sqrt [4]{4+3 \sqrt {2}} \tan ^{-1}\left (\frac {\sqrt [4]{8+6 \sqrt {2}} x \sqrt [4]{-x^2+x^6}}{\sqrt [4]{2} x^2-\sqrt {-x^2+x^6}}\right )-\frac {1}{4} \sqrt [4]{-4+3 \sqrt {2}} \tanh ^{-1}\left (\frac {2 \sqrt [4]{\frac {1}{8}+\frac {3}{16 \sqrt {2}}} x^2+2^{3/4} \sqrt [4]{\frac {1}{8}+\frac {3}{16 \sqrt {2}}} \sqrt {-x^2+x^6}}{x \sqrt [4]{-x^2+x^6}}\right )+\frac {1}{8} \sqrt [4]{4+3 \sqrt {2}} \log \left (2 x^2-2 \sqrt [4]{4+3 \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 (\sqrt {2-\sqrt {2}} x^2+2^{3/8} x \sqrt [4]{-x^2+x^6}+\sqrt {-1+\sqrt {2}} \sqrt {-x^2+x^6}\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-4 + 3*Sqrt[2])^(1/4)*ArcTan[((-8 + 6*Sqrt[2])^(1/4)*x*(-x^2 + x^6)^(1/4))/(2^(1/4)*x^2 - Sqrt[-x^2 + x^6])]
)/4 + ((4 + 3*Sqrt[2])^(1/4)*ArcTan[((8 + 6*Sqrt[2])^(1/4)*x*(-x^2 + x^6)^(1/4))/(2^(1/4)*x^2 - Sqrt[-x^2 + x^
6])])/4 - ((-4 + 3*Sqrt[2])^(1/4)*ArcTanh[(2*(1/8 + 3/(16*Sqrt[2]))^(1/4)*x^2 + 2^(3/4)*(1/8 + 3/(16*Sqrt[2]))
^(1/4)*Sqrt[-x^2 + x^6])/(x*(-x^2 + x^6)^(1/4))])/4 + ((4 + 3*Sqrt[2])^(1/4)*Log[2*x^2 - 2*(4 + 3*Sqrt[2])^(1/
4)*x*(-x^2 + x^6)^(1/4) + 2^(3/4)*Sqrt[-x^2 + x^6]])/8 - ((4 + 3*Sqrt[2])^(1/4)*Log[Sqrt[2 - Sqrt[2]]*x^2 + 2^
(3/8)*x*(-x^2 + x^6)^(1/4) + Sqrt[-1 + Sqrt[2]]*Sqrt[-x^2 + x^6]])/8

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8} - 1}{{\left (x^{8} + 1\right )} {\left (x^{6} - x^{2}\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {x^{8}-1}{\left (x^{6}-x^{2}\right )^{\frac {1}{4}} \left (x^{8}+1\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8} - 1}{{\left (x^{8} + 1\right )} {\left (x^{6} - x^{2}\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^8-1}{\left (x^8+1\right )\,{\left (x^6-x^2\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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