3.8.32 \(\int \frac {1}{x^3 \sqrt [8]{256-256 x^2+96 x^4-16 x^6+x^8}} \, dx\) [732]

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

[Out]

((x^2-4)^4)^(7/8)*(1/8*(x^2-4)^(1/2)/x^2+1/16*arctan(1/2*(x^2-4)^(1/2)))/(x^2-4)^(7/2)

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Rubi [A]
time = 0.07, antiderivative size = 66, normalized size of antiderivative = 1.18, 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 = {6820, 1973, 272, 44, 65, 212} \begin {gather*} -\frac {4-x^2}{8 x^2 \sqrt [8]{\left (x^2-4\right )^4}}-\frac {\sqrt {4-x^2} \tanh ^{-1}\left (\sqrt {1-\frac {x^2}{4}}\right )}{16 \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]*ArcTanh[Sqrt[1 - x^2/4]])/(16*((-4 + x^2)^4)^(1/8))

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 212

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

Rule 272

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 1973

Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Dist[Simp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/
a))^(p*q)], Int[u*(1 + b*(x^n/a))^(p*q), x], x] /; FreeQ[{a, b, c, n, p, q}, x] &&  !GeQ[a, 0]

Rule 6820

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

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} \text {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} \text {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} \text {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 = 0.03, size = 52, normalized size = 0.93 \begin {gather*} \frac {-8+2 x^2+x^2 \sqrt {-4+x^2} \text {ArcTan}\left (\frac {1}{2} \sqrt {-4+x^2}\right )}{16 x^2 \sqrt [8]{\left (-4+x^2\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]

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

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Maple [A]
time = 0.14, 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*} \text {Failed to integrate} \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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Fricas [A]
time = 0.48, 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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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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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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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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