3.19.42 \(\int \frac {1}{\sqrt [6]{1+2 x-x^2-4 x^3-x^4+2 x^5+x^6}} \, dx\)

Optimal. Leaf size=125 \[ \frac {\sqrt [3]{x-1} (x+1)^{2/3} \left (-\log \left (\sqrt [3]{x-1}-\sqrt [3]{x+1}\right )+\frac {1}{2} \log \left ((x-1)^{2/3}+\sqrt [3]{x+1} \sqrt [3]{x-1}+(x+1)^{2/3}\right )+\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x+1}}{2 \sqrt [3]{x-1}+\sqrt [3]{x+1}}\right )\right )}{\sqrt [6]{(x-1)^2 (x+1)^4}} \]

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Rubi [A]  time = 0.04, antiderivative size = 159, normalized size of antiderivative = 1.27, number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {6688, 6719, 59} \begin {gather*} -\frac {\sqrt [3]{x-1} (x+1)^{2/3} \log (x+1)}{2 \sqrt [6]{(1-x)^2 (x+1)^4}}-\frac {3 \sqrt [3]{x-1} (x+1)^{2/3} \log \left (\frac {\sqrt [3]{x-1}}{\sqrt [3]{x+1}}-1\right )}{2 \sqrt [6]{(1-x)^2 (x+1)^4}}-\frac {\sqrt {3} \sqrt [3]{x-1} (x+1)^{2/3} \tan ^{-1}\left (\frac {2 \sqrt [3]{x-1}}{\sqrt {3} \sqrt [3]{x+1}}+\frac {1}{\sqrt {3}}\right )}{\sqrt [6]{(1-x)^2 (x+1)^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x - x^2 - 4*x^3 - x^4 + 2*x^5 + x^6)^(-1/6),x]

[Out]

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

Rule 59

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[d/b, 3]}, -Simp[(Sqrt
[3]*q*ArcTan[(2*q*(a + b*x)^(1/3))/(Sqrt[3]*(c + d*x)^(1/3)) + 1/Sqrt[3]])/d, x] + (-Simp[(3*q*Log[(q*(a + b*x
)^(1/3))/(c + d*x)^(1/3) - 1])/(2*d), x] - Simp[(q*Log[c + d*x])/(2*d), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[
b*c - a*d, 0] && PosQ[d/b]

Rule 6688

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

Rule 6719

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

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 53, normalized size = 0.42 \begin {gather*} \frac {3 (x-1) (x+1)^{2/3} \, _2F_1\left (\frac {2}{3},\frac {2}{3};\frac {5}{3};\frac {1-x}{2}\right )}{2\ 2^{2/3} \sqrt [6]{(x-1)^2 (x+1)^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x - x^2 - 4*x^3 - x^4 + 2*x^5 + x^6)^(-1/6),x]

[Out]

(3*(-1 + x)*(1 + x)^(2/3)*Hypergeometric2F1[2/3, 2/3, 5/3, (1 - x)/2])/(2*2^(2/3)*((-1 + x)^2*(1 + x)^4)^(1/6)
)

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IntegrateAlgebraic [A]  time = 11.34, size = 122, normalized size = 0.98 \begin {gather*} \frac {\sqrt [3]{-1+x} (1+x)^{2/3} \left (\sqrt {3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x}}{\sqrt {3} \sqrt [3]{-1+x}}\right )-\log \left (-1+\frac {\sqrt [3]{1+x}}{\sqrt [3]{-1+x}}\right )+\frac {1}{2} \log \left (1+\frac {\sqrt [3]{1+x}}{\sqrt [3]{-1+x}}+\frac {(1+x)^{2/3}}{(-1+x)^{2/3}}\right )\right )}{\sqrt [6]{(-1+x)^2 (1+x)^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + 2*x - x^2 - 4*x^3 - x^4 + 2*x^5 + x^6)^(-1/6),x]

[Out]

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

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fricas [A]  time = 0.59, size = 188, normalized size = 1.50 \begin {gather*} -\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (x + 1\right )} + 2 \, \sqrt {3} {\left (x^{6} + 2 \, x^{5} - x^{4} - 4 \, x^{3} - x^{2} + 2 \, x + 1\right )}^{\frac {1}{6}}}{3 \, {\left (x + 1\right )}}\right ) + \frac {1}{2} \, \log \left (\frac {x^{2} + {\left (x^{6} + 2 \, x^{5} - x^{4} - 4 \, x^{3} - x^{2} + 2 \, x + 1\right )}^{\frac {1}{6}} {\left (x + 1\right )} + 2 \, x + {\left (x^{6} + 2 \, x^{5} - x^{4} - 4 \, x^{3} - x^{2} + 2 \, x + 1\right )}^{\frac {1}{3}} + 1}{x^{2} + 2 \, x + 1}\right ) - \log \left (-\frac {x - {\left (x^{6} + 2 \, x^{5} - x^{4} - 4 \, x^{3} - x^{2} + 2 \, x + 1\right )}^{\frac {1}{6}} + 1}{x + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^6+2*x^5-x^4-4*x^3-x^2+2*x+1)^(1/6),x, algorithm="fricas")

[Out]

-sqrt(3)*arctan(1/3*(sqrt(3)*(x + 1) + 2*sqrt(3)*(x^6 + 2*x^5 - x^4 - 4*x^3 - x^2 + 2*x + 1)^(1/6))/(x + 1)) +
1/2*log((x^2 + (x^6 + 2*x^5 - x^4 - 4*x^3 - x^2 + 2*x + 1)^(1/6)*(x + 1) + 2*x + (x^6 + 2*x^5 - x^4 - 4*x^3 -
x^2 + 2*x + 1)^(1/3) + 1)/(x^2 + 2*x + 1)) - log(-(x - (x^6 + 2*x^5 - x^4 - 4*x^3 - x^2 + 2*x + 1)^(1/6) + 1)
/(x + 1))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{6} + 2 \, x^{5} - x^{4} - 4 \, x^{3} - x^{2} + 2 \, x + 1\right )}^{\frac {1}{6}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^6+2*x^5-x^4-4*x^3-x^2+2*x+1)^(1/6),x, algorithm="giac")

[Out]

integrate((x^6 + 2*x^5 - x^4 - 4*x^3 - x^2 + 2*x + 1)^(-1/6), x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (x^{6}+2 x^{5}-x^{4}-4 x^{3}-x^{2}+2 x +1\right )^{\frac {1}{6}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6+2*x^5-x^4-4*x^3-x^2+2*x+1)^(1/6),x)

[Out]

int(1/(x^6+2*x^5-x^4-4*x^3-x^2+2*x+1)^(1/6),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{6} + 2 \, x^{5} - x^{4} - 4 \, x^{3} - x^{2} + 2 \, x + 1\right )}^{\frac {1}{6}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^6+2*x^5-x^4-4*x^3-x^2+2*x+1)^(1/6),x, algorithm="maxima")

[Out]

integrate((x^6 + 2*x^5 - x^4 - 4*x^3 - x^2 + 2*x + 1)^(-1/6), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (x^6+2\,x^5-x^4-4\,x^3-x^2+2\,x+1\right )}^{1/6}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*x - x^2 - 4*x^3 - x^4 + 2*x^5 + x^6 + 1)^(1/6),x)

[Out]

int(1/(2*x - x^2 - 4*x^3 - x^4 + 2*x^5 + x^6 + 1)^(1/6), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [6]{x^{6} + 2 x^{5} - x^{4} - 4 x^{3} - x^{2} + 2 x + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**6+2*x**5-x**4-4*x**3-x**2+2*x+1)**(1/6),x)

[Out]

Integral((x**6 + 2*x**5 - x**4 - 4*x**3 - x**2 + 2*x + 1)**(-1/6), x)

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