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

Optimal. Leaf size=164 \[ \frac {\left (1+x^3\right )^{5/3}}{5 x^5}-\frac {\text {ArcTan}\left (\frac {3^{5/6} x}{\sqrt [3]{3} x+2 \sqrt [3]{2} \sqrt [3]{1+x^3}}\right )}{2^{2/3} 3^{5/6}}+\frac {\log \left (-3 x+\sqrt [3]{2} 3^{2/3} \sqrt [3]{1+x^3}\right )}{3\ 2^{2/3} \sqrt [3]{3}}-\frac {\log \left (3 x^2+\sqrt [3]{2} 3^{2/3} x \sqrt [3]{1+x^3}+2^{2/3} \sqrt [3]{3} \left (1+x^3\right )^{2/3}\right )}{6\ 2^{2/3} \sqrt [3]{3}} \]

[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)

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Rubi [A]
time = 0.09, antiderivative size = 131, normalized size of antiderivative = 0.80, number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1600, 597, 12, 384} \begin {gather*} -\frac {\text {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} \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]
time = 0.34, size = 164, normalized size = 1.00 \begin {gather*} \frac {\left (1+x^3\right )^{5/3}}{5 x^5}-\frac {\text {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}} \end {gather*}

Antiderivative was successfully verified.

[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))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 10.43, size = 890, normalized size = 5.43

method	result	size
risch	\(\text {Expression too large to display}\)	\(890\)
trager	\(\text {Expression too large to display}\)	\(1088\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(x^6+2*x^3+1)/x^5/(x^3+1)^(1/3)+1/18*RootOf(_Z^3-18)*ln((135*RootOf(RootOf(_Z^3-18)^2+18*_Z*RootOf(_Z^3-18
)+324*_Z^2)^2*RootOf(_Z^3-18)^2*x^3+3*RootOf(RootOf(_Z^3-18)^2+18*_Z*RootOf(_Z^3-18)+324*_Z^2)*RootOf(_Z^3-18)
^3*x^3+21*(x^3+1)^(2/3)*RootOf(RootOf(_Z^3-18)^2+18*_Z*RootOf(_Z^3-18)+324*_Z^2)*RootOf(_Z^3-18)^2*x+9*(x^3+1)
^(1/3)*RootOf(RootOf(_Z^3-18)^2+18*_Z*RootOf(_Z^3-18)+324*_Z^2)*RootOf(_Z^3-18)*x^2+4*RootOf(_Z^3-18)^2*(x^3+1
)^(1/3)*x^2-90*RootOf(RootOf(_Z^3-18)^2+18*_Z*RootOf(_Z^3-18)+324*_Z^2)*x^3-2*RootOf(_Z^3-18)*x^3-3*x*(x^3+1)^
(2/3)-90*RootOf(RootOf(_Z^3-18)^2+18*_Z*RootOf(_Z^3-18)+324*_Z^2)-2*RootOf(_Z^3-18))/(x^3-2))-1/18*ln(-(162*Ro
otOf(RootOf(_Z^3-18)^2+18*_Z*RootOf(_Z^3-18)+324*_Z^2)^2*RootOf(_Z^3-18)^2*x^3-6*RootOf(RootOf(_Z^3-18)^2+18*_
Z*RootOf(_Z^3-18)+324*_Z^2)*RootOf(_Z^3-18)^3*x^3+42*(x^3+1)^(2/3)*RootOf(RootOf(_Z^3-18)^2+18*_Z*RootOf(_Z^3-
18)+324*_Z^2)*RootOf(_Z^3-18)^2*x-144*(x^3+1)^(1/3)*RootOf(RootOf(_Z^3-18)^2+18*_Z*RootOf(_Z^3-18)+324*_Z^2)*R
ootOf(_Z^3-18)*x^2-RootOf(_Z^3-18)^2*(x^3+1)^(1/3)*x^2+270*RootOf(RootOf(_Z^3-18)^2+18*_Z*RootOf(_Z^3-18)+324*
_Z^2)*x^3-10*RootOf(_Z^3-18)*x^3+48*x*(x^3+1)^(2/3)+108*RootOf(RootOf(_Z^3-18)^2+18*_Z*RootOf(_Z^3-18)+324*_Z^
2)-4*RootOf(_Z^3-18))/(x^3-2))*RootOf(_Z^3-18)-ln(-(162*RootOf(RootOf(_Z^3-18)^2+18*_Z*RootOf(_Z^3-18)+324*_Z^
2)^2*RootOf(_Z^3-18)^2*x^3-6*RootOf(RootOf(_Z^3-18)^2+18*_Z*RootOf(_Z^3-18)+324*_Z^2)*RootOf(_Z^3-18)^3*x^3+42
*(x^3+1)^(2/3)*RootOf(RootOf(_Z^3-18)^2+18*_Z*RootOf(_Z^3-18)+324*_Z^2)*RootOf(_Z^3-18)^2*x-144*(x^3+1)^(1/3)*
RootOf(RootOf(_Z^3-18)^2+18*_Z*RootOf(_Z^3-18)+324*_Z^2)*RootOf(_Z^3-18)*x^2-RootOf(_Z^3-18)^2*(x^3+1)^(1/3)*x
^2+270*RootOf(RootOf(_Z^3-18)^2+18*_Z*RootOf(_Z^3-18)+324*_Z^2)*x^3-10*RootOf(_Z^3-18)*x^3+48*x*(x^3+1)^(2/3)+
108*RootOf(RootOf(_Z^3-18)^2+18*_Z*RootOf(_Z^3-18)+324*_Z^2)-4*RootOf(_Z^3-18))/(x^3-2))*RootOf(RootOf(_Z^3-18
)^2+18*_Z*RootOf(_Z^3-18)+324*_Z^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((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)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (118) = 236\).
time = 1.87, size = 253, normalized size = 1.54 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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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((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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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