3.8.32 \(\int \frac {1}{x^3 \sqrt [8]{256-256 x^2+96 x^4-16 x^6+x^8}} \, dx\)

Optimal. Leaf size=56 \[ \frac {\left (\left (x^2-4\right )^4\right )^{7/8} \left (\frac {\sqrt {x^2-4}}{8 x^2}+\frac {1}{16} \tan ^{-1}\left (\frac {\sqrt {x^2-4}}{2}\right )\right )}{\left (x^2-4\right )^{7/2}} \]

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Rubi [A]  time = 0.15, antiderivative size = 64, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {6688, 6720, 266, 51, 63, 203} \begin {gather*} \frac {\sqrt {x^2-4} \tan ^{-1}\left (\frac {\sqrt {x^2-4}}{2}\right )}{16 \sqrt [8]{\left (x^2-4\right )^4}}-\frac {4-x^2}{8 x^2 \sqrt [8]{\left (x^2-4\right )^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(256 - 256*x^2 + 96*x^4 - 16*x^6 + x^8)^(1/8)),x]

[Out]

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

\begin {align*} \int \frac {1}{x^3 \sqrt [8]{256-256 x^2+96 x^4-16 x^6+x^8}} \, dx &=\int \frac {1}{x^3 \sqrt [8]{\left (-4+x^2\right )^4}} \, dx\\ &=\frac {\sqrt {-4+x^2} \int \frac {1}{x^3 \sqrt {-4+x^2}} \, dx}{\sqrt [8]{\left (-4+x^2\right )^4}}\\ &=\frac {\sqrt {-4+x^2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-4+x} x^2} \, dx,x,x^2\right )}{2 \sqrt [8]{\left (-4+x^2\right )^4}}\\ &=-\frac {4-x^2}{8 x^2 \sqrt [8]{\left (-4+x^2\right )^4}}+\frac {\sqrt {-4+x^2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-4+x} x} \, dx,x,x^2\right )}{16 \sqrt [8]{\left (-4+x^2\right )^4}}\\ &=-\frac {4-x^2}{8 x^2 \sqrt [8]{\left (-4+x^2\right )^4}}+\frac {\sqrt {-4+x^2} \operatorname {Subst}\left (\int \frac {1}{4+x^2} \, dx,x,\sqrt {-4+x^2}\right )}{8 \sqrt [8]{\left (-4+x^2\right )^4}}\\ &=-\frac {4-x^2}{8 x^2 \sqrt [8]{\left (-4+x^2\right )^4}}+\frac {\sqrt {-4+x^2} \tan ^{-1}\left (\frac {1}{2} \sqrt {-4+x^2}\right )}{16 \sqrt [8]{\left (-4+x^2\right )^4}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 52, normalized size = 0.93 \begin {gather*} \frac {\left (x^2-4\right ) \left (\frac {2}{x^2}+\frac {\tanh ^{-1}\left (\sqrt {1-\frac {x^2}{4}}\right )}{\sqrt {4-x^2}}\right )}{16 \sqrt [8]{\left (x^2-4\right )^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(256 - 256*x^2 + 96*x^4 - 16*x^6 + x^8)^(1/8)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 11.50, size = 56, normalized size = 1.00 \begin {gather*} \frac {\left (\left (-4+x^2\right )^4\right )^{7/8} \left (\frac {\sqrt {-4+x^2}}{8 x^2}+\frac {1}{16} \tan ^{-1}\left (\frac {1}{2} \sqrt {-4+x^2}\right )\right )}{\left (-4+x^2\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^3*(256 - 256*x^2 + 96*x^4 - 16*x^6 + x^8)^(1/8)),x]

[Out]

(((-4 + x^2)^4)^(7/8)*(Sqrt[-4 + x^2]/(8*x^2) + ArcTan[Sqrt[-4 + x^2]/2]/16))/(-4 + x^2)^(7/2)

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fricas [A]  time = 0.52, size = 61, normalized size = 1.09 \begin {gather*} \frac {x^{2} \arctan \left (-\frac {1}{2} \, x + \frac {1}{2} \, {\left (x^{8} - 16 \, x^{6} + 96 \, x^{4} - 256 \, x^{2} + 256\right )}^{\frac {1}{8}}\right ) + {\left (x^{8} - 16 \, x^{6} + 96 \, x^{4} - 256 \, x^{2} + 256\right )}^{\frac {1}{8}}}{8 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(x^8-16*x^6+96*x^4-256*x^2+256)^(1/8),x, algorithm="fricas")

[Out]

1/8*(x^2*arctan(-1/2*x + 1/2*(x^8 - 16*x^6 + 96*x^4 - 256*x^2 + 256)^(1/8)) + (x^8 - 16*x^6 + 96*x^4 - 256*x^2
+ 256)^(1/8))/x^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{8} - 16 \, x^{6} + 96 \, x^{4} - 256 \, x^{2} + 256\right )}^{\frac {1}{8}} x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(x^8-16*x^6+96*x^4-256*x^2+256)^(1/8),x, algorithm="giac")

[Out]

integrate(1/((x^8 - 16*x^6 + 96*x^4 - 256*x^2 + 256)^(1/8)*x^3), x)

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maple [A]  time = 0.07, size = 49, normalized size = 0.88

method	result	size
risch	\(\frac {x^{2}-4}{8 x^{2} \left (\left (x^{2}-4\right )^{4}\right )^{\frac {1}{8}}}-\frac {\arctan \left (\frac {2}{\sqrt {x^{2}-4}}\right ) \sqrt {x^{2}-4}}{16 \left (\left (x^{2}-4\right )^{4}\right )^{\frac {1}{8}}}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(x^8-16*x^6+96*x^4-256*x^2+256)^(1/8),x,method=_RETURNVERBOSE)

[Out]

1/8*(x^2-4)/x^2/((x^2-4)^4)^(1/8)-1/16*arctan(2/(x^2-4)^(1/2))/((x^2-4)^4)^(1/8)*(x^2-4)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{8} - 16 \, x^{6} + 96 \, x^{4} - 256 \, x^{2} + 256\right )}^{\frac {1}{8}} x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(x^8-16*x^6+96*x^4-256*x^2+256)^(1/8),x, algorithm="maxima")

[Out]

integrate(1/((x^8 - 16*x^6 + 96*x^4 - 256*x^2 + 256)^(1/8)*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{x^3\,{\left (x^8-16\,x^6+96\,x^4-256\,x^2+256\right )}^{1/8}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(96*x^4 - 256*x^2 - 16*x^6 + x^8 + 256)^(1/8)),x)

[Out]

int(1/(x^3*(96*x^4 - 256*x^2 - 16*x^6 + x^8 + 256)^(1/8)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{3} \sqrt [8]{\left (x - 2\right )^{4} \left (x + 2\right )^{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(x**8-16*x**6+96*x**4-256*x**2+256)**(1/8),x)

[Out]

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

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