\(\int \frac {(1+x^3)^{2/3} (2+x^3)}{x^6 (-2-x^3+x^6)} \, dx\) [2212]

Optimal result

Integrand size = 30, antiderivative size = 164 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (2+x^3\right )}{x^6 \left (-2-x^3+x^6\right )} \, dx=\frac {\left (1+x^3\right )^{5/3}}{5 x^5}-\frac {\arctan \left (\frac {3^{5/6} x}{\sqrt [3]{3} x+2 \sqrt [3]{2} \sqrt [3]{1+x^3}}\right )}{2^{2/3} 3^{5/6}}+\frac {\log \left (-3 x+\sqrt [3]{2} 3^{2/3} \sqrt [3]{1+x^3}\right )}{3\ 2^{2/3} \sqrt [3]{3}}-\frac {\log \left (3 x^2+\sqrt [3]{2} 3^{2/3} x \sqrt [3]{1+x^3}+2^{2/3} \sqrt [3]{3} \left (1+x^3\right )^{2/3}\right )}{6\ 2^{2/3} \sqrt [3]{3}} \]

[Out]

1/5*(x^3+1)^(5/3)/x^5-1/6*arctan(3^(5/6)*x/(3^(1/3)*x+2*2^(1/3)*(x^3+1)^(1/3)))*2^(1/3)*3^(1/6)+1/18*ln(-3*x+2
^(1/3)*3^(2/3)*(x^3+1)^(1/3))*2^(1/3)*3^(2/3)-1/36*ln(3*x^2+2^(1/3)*3^(2/3)*x*(x^3+1)^(1/3)+2^(2/3)*3^(1/3)*(x
^3+1)^(2/3))*2^(1/3)*3^(2/3)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.80, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1600, 597, 12, 384} \[ \int \frac {\left (1+x^3\right )^{2/3} \left (2+x^3\right )}{x^6 \left (-2-x^3+x^6\right )} \, dx=-\frac {\arctan \left (\frac {\frac {2^{2/3} \sqrt [3]{3} x}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )}{2^{2/3} 3^{5/6}}-\frac {\log \left (x^3-2\right )}{6\ 2^{2/3} \sqrt [3]{3}}+\frac {\log \left (\sqrt [3]{\frac {3}{2}} x-\sqrt [3]{x^3+1}\right )}{2\ 2^{2/3} \sqrt [3]{3}}+\frac {\left (x^3+1\right )^{2/3}}{5 x^5}+\frac {\left (x^3+1\right )^{2/3}}{5 x^2} \]

[In]

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

[Out]

(1 + x^3)^(2/3)/(5*x^5) + (1 + x^3)^(2/3)/(5*x^2) - ArcTan[(1 + (2^(2/3)*3^(1/3)*x)/(1 + x^3)^(1/3))/Sqrt[3]]/
(2^(2/3)*3^(5/6)) - Log[-2 + x^3]/(6*2^(2/3)*3^(1/3)) + Log[(3/2)^(1/3)*x - (1 + x^3)^(1/3)]/(2*2^(2/3)*3^(1/3
))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 384

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[
ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q
), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 597

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*g*(m + 1))), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {2+x^3}{x^6 \left (-2+x^3\right ) \sqrt [3]{1+x^3}} \, dx \\ & = \frac {\left (1+x^3\right )^{2/3}}{5 x^5}+\frac {1}{10} \int \frac {8+6 x^3}{x^3 \left (-2+x^3\right ) \sqrt [3]{1+x^3}} \, dx \\ & = \frac {\left (1+x^3\right )^{2/3}}{5 x^5}+\frac {\left (1+x^3\right )^{2/3}}{5 x^2}+\frac {1}{40} \int \frac {40}{\left (-2+x^3\right ) \sqrt [3]{1+x^3}} \, dx \\ & = \frac {\left (1+x^3\right )^{2/3}}{5 x^5}+\frac {\left (1+x^3\right )^{2/3}}{5 x^2}+\int \frac {1}{\left (-2+x^3\right ) \sqrt [3]{1+x^3}} \, dx \\ & = \frac {\left (1+x^3\right )^{2/3}}{5 x^5}+\frac {\left (1+x^3\right )^{2/3}}{5 x^2}-\frac {\arctan \left (\frac {1+\frac {2^{2/3} \sqrt [3]{3} x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{2^{2/3} 3^{5/6}}-\frac {\log \left (-2+x^3\right )}{6\ 2^{2/3} \sqrt [3]{3}}+\frac {\log \left (\sqrt [3]{\frac {3}{2}} x-\sqrt [3]{1+x^3}\right )}{2\ 2^{2/3} \sqrt [3]{3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (2+x^3\right )}{x^6 \left (-2-x^3+x^6\right )} \, dx=\frac {\left (1+x^3\right )^{5/3}}{5 x^5}-\frac {\arctan \left (\frac {3^{5/6} x}{\sqrt [3]{3} x+2 \sqrt [3]{2} \sqrt [3]{1+x^3}}\right )}{2^{2/3} 3^{5/6}}+\frac {\log \left (-3 x+\sqrt [3]{2} 3^{2/3} \sqrt [3]{1+x^3}\right )}{3\ 2^{2/3} \sqrt [3]{3}}-\frac {\log \left (3 x^2+\sqrt [3]{2} 3^{2/3} x \sqrt [3]{1+x^3}+2^{2/3} \sqrt [3]{3} \left (1+x^3\right )^{2/3}\right )}{6\ 2^{2/3} \sqrt [3]{3}} \]

[In]

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

[Out]

(1 + x^3)^(5/3)/(5*x^5) - ArcTan[(3^(5/6)*x)/(3^(1/3)*x + 2*2^(1/3)*(1 + x^3)^(1/3))]/(2^(2/3)*3^(5/6)) + Log[
-3*x + 2^(1/3)*3^(2/3)*(1 + x^3)^(1/3)]/(3*2^(2/3)*3^(1/3)) - Log[3*x^2 + 2^(1/3)*3^(2/3)*x*(1 + x^3)^(1/3) +
2^(2/3)*3^(1/3)*(1 + x^3)^(2/3)]/(6*2^(2/3)*3^(1/3))

Maple [A] (verified)

Time = 15.35 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.93

method	result	size
pseudoelliptic	\(\frac {-5 x^{5} \left (\left (\ln \left (\frac {2^{\frac {1}{3}} 3^{\frac {2}{3}} x^{2}+2^{\frac {2}{3}} 3^{\frac {1}{3}} {\left (\left (1+x \right ) \left (x^{2}-x +1\right )\right )}^{\frac {1}{3}} x +2 {\left (\left (1+x \right ) \left (x^{2}-x +1\right )\right )}^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {-2^{\frac {2}{3}} 3^{\frac {1}{3}} x +2 {\left (\left (1+x \right ) \left (x^{2}-x +1\right )\right )}^{\frac {1}{3}}}{x}\right )+\ln \left (2\right )\right ) 3^{\frac {2}{3}}-6 \arctan \left (\frac {\sqrt {3}\, \left (2 \,2^{\frac {1}{3}} 3^{\frac {2}{3}} \left (x^{3}+1\right )^{\frac {1}{3}}+3 x \right )}{9 x}\right ) 3^{\frac {1}{6}}\right ) 2^{\frac {1}{3}}+36 \left (x^{3}+1\right )^{\frac {5}{3}}}{180 x^{5}}\)	\(152\)
risch	\(\text {Expression too large to display}\)	\(606\)
trager	\(\text {Expression too large to display}\)	\(737\)

[In]

int((x^3+1)^(2/3)*(x^3+2)/x^6/(x^6-x^3-2),x,method=_RETURNVERBOSE)

[Out]

1/180*(-5*x^5*((ln((2^(1/3)*3^(2/3)*x^2+2^(2/3)*3^(1/3)*((1+x)*(x^2-x+1))^(1/3)*x+2*((1+x)*(x^2-x+1))^(2/3))/x
^2)-2*ln((-2^(2/3)*3^(1/3)*x+2*((1+x)*(x^2-x+1))^(1/3))/x)+ln(2))*3^(2/3)-6*arctan(1/9*3^(1/2)*(2*2^(1/3)*3^(2
/3)*(x^3+1)^(1/3)+3*x)/x)*3^(1/6))*2^(1/3)+36*(x^3+1)^(5/3))/x^5

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (118) = 236\).

Time = 1.70 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.54 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (2+x^3\right )}{x^6 \left (-2-x^3+x^6\right )} \, dx=\frac {10 \cdot 12^{\frac {2}{3}} x^{5} \log \left (\frac {18 \cdot 12^{\frac {1}{3}} {\left (x^{3} + 1\right )}^{\frac {1}{3}} x^{2} - 12^{\frac {2}{3}} {\left (x^{3} - 2\right )} - 36 \, {\left (x^{3} + 1\right )}^{\frac {2}{3}} x}{x^{3} - 2}\right ) - 5 \cdot 12^{\frac {2}{3}} x^{5} \log \left (\frac {6 \cdot 12^{\frac {2}{3}} {\left (4 \, x^{4} + x\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}} + 12^{\frac {1}{3}} {\left (55 \, x^{6} + 50 \, x^{3} + 4\right )} + 18 \, {\left (7 \, x^{5} + 4 \, x^{2}\right )} {\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x^{6} - 4 \, x^{3} + 4}\right ) - 60 \cdot 12^{\frac {1}{6}} x^{5} \arctan \left (\frac {12^{\frac {1}{6}} {\left (12 \cdot 12^{\frac {2}{3}} {\left (4 \, x^{7} - 7 \, x^{4} - 2 \, x\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}} - 12^{\frac {1}{3}} {\left (377 \, x^{9} + 600 \, x^{6} + 204 \, x^{3} + 8\right )} - 36 \, {\left (55 \, x^{8} + 50 \, x^{5} + 4 \, x^{2}\right )} {\left (x^{3} + 1\right )}^{\frac {1}{3}}\right )}}{6 \, {\left (487 \, x^{9} + 480 \, x^{6} + 12 \, x^{3} - 8\right )}}\right ) + 216 \, {\left (x^{3} + 1\right )}^{\frac {5}{3}}}{1080 \, x^{5}} \]

[In]

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

[Out]

1/1080*(10*12^(2/3)*x^5*log((18*12^(1/3)*(x^3 + 1)^(1/3)*x^2 - 12^(2/3)*(x^3 - 2) - 36*(x^3 + 1)^(2/3)*x)/(x^3
- 2)) - 5*12^(2/3)*x^5*log((6*12^(2/3)*(4*x^4 + x)*(x^3 + 1)^(2/3) + 12^(1/3)*(55*x^6 + 50*x^3 + 4) + 18*(7*x
^5 + 4*x^2)*(x^3 + 1)^(1/3))/(x^6 - 4*x^3 + 4)) - 60*12^(1/6)*x^5*arctan(1/6*12^(1/6)*(12*12^(2/3)*(4*x^7 - 7*
x^4 - 2*x)*(x^3 + 1)^(2/3) - 12^(1/3)*(377*x^9 + 600*x^6 + 204*x^3 + 8) - 36*(55*x^8 + 50*x^5 + 4*x^2)*(x^3 +
1)^(1/3))/(487*x^9 + 480*x^6 + 12*x^3 - 8)) + 216*(x^3 + 1)^(5/3))/x^5

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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