\(\int \frac {1}{\sqrt [6]{1+2 x-x^2-4 x^3-x^4+2 x^5+x^6}} \, dx\) [1842]

Optimal result

Integrand size = 32, antiderivative size = 125 \[ \int \frac {1}{\sqrt [6]{1+2 x-x^2-4 x^3-x^4+2 x^5+x^6}} \, dx=\frac {\sqrt [3]{-1+x} (1+x)^{2/3} \left (\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{1+x}}{2 \sqrt [3]{-1+x}+\sqrt [3]{1+x}}\right )-\log \left (\sqrt [3]{-1+x}-\sqrt [3]{1+x}\right )+\frac {1}{2} \log \left ((-1+x)^{2/3}+\sqrt [3]{-1+x} \sqrt [3]{1+x}+(1+x)^{2/3}\right )\right )}{\sqrt [6]{(-1+x)^2 (1+x)^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.27, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {6820, 6851, 61} \[ \int \frac {1}{\sqrt [6]{1+2 x-x^2-4 x^3-x^4+2 x^5+x^6}} \, dx=-\frac {\sqrt {3} \sqrt [3]{x-1} (x+1)^{2/3} \arctan \left (\frac {2 \sqrt [3]{x-1}}{\sqrt {3} \sqrt [3]{x+1}}+\frac {1}{\sqrt {3}}\right )}{\sqrt [6]{(1-x)^2 (x+1)^4}}-\frac {\sqrt [3]{x-1} (x+1)^{2/3} \log (x+1)}{2 \sqrt [6]{(1-x)^2 (x+1)^4}}-\frac {3 \sqrt [3]{x-1} (x+1)^{2/3} \log \left (\frac {\sqrt [3]{x-1}}{\sqrt [3]{x+1}}-1\right )}{2 \sqrt [6]{(1-x)^2 (x+1)^4}} \]

[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 61

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

Rule 6820

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

Rule 6851

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a*v^m*w^n)^FracPart[p]/(v^(m*Fr
acPart[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*} \text {integral}& = \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} \arctan \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*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt [6]{1+2 x-x^2-4 x^3-x^4+2 x^5+x^6}} \, dx=\frac {\sqrt [3]{-1+x} (1+x)^{2/3} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{1+x}}{2 \sqrt [3]{-1+x}+\sqrt [3]{1+x}}\right )-2 \log \left (\sqrt [3]{-1+x}-\sqrt [3]{1+x}\right )+\log \left ((-1+x)^{2/3}+\sqrt [3]{-1+x} \sqrt [3]{1+x}+(1+x)^{2/3}\right )\right )}{2 \sqrt [6]{(-1+x)^2 (1+x)^4}} \]

[In]

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

Maple [F]

\[\int \frac {1}{\left (x^{6}+2 x^{5}-x^{4}-4 x^{3}-x^{2}+2 x +1\right )^{\frac {1}{6}}}d x\]

[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)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.50 \[ \int \frac {1}{\sqrt [6]{1+2 x-x^2-4 x^3-x^4+2 x^5+x^6}} \, dx=-\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (x + 1\right )} + 2 \, \sqrt {3} {\left (x^{6} + 2 \, x^{5} - x^{4} - 4 \, x^{3} - x^{2} + 2 \, x + 1\right )}^{\frac {1}{6}}}{3 \, {\left (x + 1\right )}}\right ) + \frac {1}{2} \, \log \left (\frac {x^{2} + {\left (x^{6} + 2 \, x^{5} - x^{4} - 4 \, x^{3} - x^{2} + 2 \, x + 1\right )}^{\frac {1}{6}} {\left (x + 1\right )} + 2 \, x + {\left (x^{6} + 2 \, x^{5} - x^{4} - 4 \, x^{3} - x^{2} + 2 \, x + 1\right )}^{\frac {1}{3}} + 1}{x^{2} + 2 \, x + 1}\right ) - \log \left (-\frac {x - {\left (x^{6} + 2 \, x^{5} - x^{4} - 4 \, x^{3} - x^{2} + 2 \, x + 1\right )}^{\frac {1}{6}} + 1}{x + 1}\right ) \]

[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))

Sympy [F]

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

[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)

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [6]{1+2 x-x^2-4 x^3-x^4+2 x^5+x^6}} \, dx=\int \frac {1}{{\left (x^6+2\,x^5-x^4-4\,x^3-x^2+2\,x+1\right )}^{1/6}} \,d x \]

[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)