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

Optimal. Leaf size=20 \[ \frac {2 \left (x^6-x^2\right )^{7/4}}{7 x^7} \]

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Rubi [B]  time = 0.16, antiderivative size = 41, normalized size of antiderivative = 2.05, number of steps used = 12, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {2052, 2025, 2011, 329, 246, 245, 2024} \begin {gather*} \frac {2 \left (x^6-x^2\right )^{3/4}}{7 x}-\frac {2 \left (x^6-x^2\right )^{3/4}}{7 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-x^2 + x^6)^(3/4))/(7*x^5) + (2*(-x^2 + x^6)^(3/4))/(7*x)

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
 && FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2011

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2024

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
 1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*p + 1)
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

Rule 2025

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
 1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rule 2052

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(c*x)
^m*Pq*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !In
tegerQ[p] && NeQ[n, j]

Rubi steps

\begin {align*} \int \frac {-1+x^8}{x^4 \sqrt [4]{-x^2+x^6}} \, dx &=\int \left (-\frac {1}{x^4 \sqrt [4]{-x^2+x^6}}+\frac {x^4}{\sqrt [4]{-x^2+x^6}}\right ) \, dx\\ &=-\int \frac {1}{x^4 \sqrt [4]{-x^2+x^6}} \, dx+\int \frac {x^4}{\sqrt [4]{-x^2+x^6}} \, dx\\ &=-\frac {2 \left (-x^2+x^6\right )^{3/4}}{7 x^5}+\frac {2 \left (-x^2+x^6\right )^{3/4}}{7 x}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 1.00 \begin {gather*} \frac {2 \left (x^2 \left (x^4-1\right )\right )^{7/4}}{7 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(x^2*(-1 + x^4))^(7/4))/(7*x^7)

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IntegrateAlgebraic [A]  time = 0.33, size = 20, normalized size = 1.00 \begin {gather*} \frac {2 \left (x^6-x^2\right )^{7/4}}{7 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-x^2 + x^6)^(7/4))/(7*x^7)

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fricas [A]  time = 0.40, size = 21, normalized size = 1.05 \begin {gather*} \frac {2 \, {\left (x^{6} - x^{2}\right )}^{\frac {3}{4}} {\left (x^{4} - 1\right )}}{7 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(x^6 - x^2)^(3/4)*(x^4 - 1)/x^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 33, normalized size = 1.65 \begin {gather*} \frac {2 \left (x^{4}-1\right ) \left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}{7 \left (x^{6}-x^{2}\right )^{\frac {1}{4}} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(x^4-1)*(-1+x)*(1+x)*(x^2+1)/(x^6-x^2)^(1/4)/x^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.26, size = 33, normalized size = 1.65 \begin {gather*} \frac {2\,{\left (x^6-x^2\right )}^{3/4}}{7\,x}-\frac {2\,{\left (x^6-x^2\right )}^{3/4}}{7\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(x^6 - x^2)^(3/4))/(7*x) - (2*(x^6 - x^2)^(3/4))/(7*x^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x - 1)*(x + 1)*(x**2 + 1)*(x**4 + 1)/(x**4*(x**2*(x - 1)*(x + 1)*(x**2 + 1))**(1/4)), x)

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