3.11.52 \(\int \frac {(-1+x^3)^{2/3} (2+x^3)}{x^6 (2+x^3+2 x^6)} \, dx\) [1052]

Optimal. Leaf size=79 \[ \frac {\left (-1+x^3\right )^{5/3}}{5 x^5}-\frac {2}{3} \text {RootSum}\left [5-5 \text {$\#$1}^3+2 \text {$\#$1}^6\& ,\frac {-\log (x) \text {$\#$1}^2+\log \left (\sqrt [3]{-1+x^3}-x \text {$\#$1}\right ) \text {$\#$1}^2}{-5+4 \text {$\#$1}^3}\& \right ] \]

[Out]

Unintegrable

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Rubi [C] Result contains complex when optimal does not.
time = 0.77, antiderivative size = 455, normalized size of antiderivative = 5.76, number of steps used = 10, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6860, 270, 1442, 399, 245, 384} \begin {gather*} \frac {2 i \text {ArcTan}\left (\frac {1+\frac {2 x}{\sqrt [3]{\frac {-\sqrt {15}+i}{-\sqrt {15}+5 i}} \sqrt [3]{x^3-1}}}{\sqrt {3}}\right )}{3 \sqrt {5} \left (\frac {-\sqrt {15}+i}{-\sqrt {15}+5 i}\right )^{2/3}}-\frac {2 i \text {ArcTan}\left (\frac {1+\frac {2 x}{\sqrt [3]{\frac {\sqrt {15}+i}{\sqrt {15}+5 i}} \sqrt [3]{x^3-1}}}{\sqrt {3}}\right )}{3 \sqrt {5} \left (\frac {\sqrt {15}+i}{\sqrt {15}+5 i}\right )^{2/3}}-\frac {i \log \left (4 x^3-i \sqrt {15}+1\right )}{3 \sqrt {15} \left (\frac {\sqrt {15}+i}{\sqrt {15}+5 i}\right )^{2/3}}+\frac {i \log \left (4 x^3+i \sqrt {15}+1\right )}{3 \sqrt {15} \left (\frac {-\sqrt {15}+i}{-\sqrt {15}+5 i}\right )^{2/3}}-\frac {i \log \left (-\sqrt [3]{x^3-1}+\frac {x}{\sqrt [3]{\frac {-\sqrt {15}+i}{-\sqrt {15}+5 i}}}\right )}{\sqrt {15} \left (\frac {-\sqrt {15}+i}{-\sqrt {15}+5 i}\right )^{2/3}}+\frac {i \log \left (-\sqrt [3]{x^3-1}+\frac {x}{\sqrt [3]{\frac {\sqrt {15}+i}{\sqrt {15}+5 i}}}\right )}{\sqrt {15} \left (\frac {\sqrt {15}+i}{\sqrt {15}+5 i}\right )^{2/3}}+\frac {\left (x^3-1\right )^{5/3}}{5 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-1 + x^3)^(5/3)/(5*x^5) + (((2*I)/3)*ArcTan[(1 + (2*x)/(((I - Sqrt[15])/(5*I - Sqrt[15]))^(1/3)*(-1 + x^3)^(1
/3)))/Sqrt[3]])/(Sqrt[5]*((I - Sqrt[15])/(5*I - Sqrt[15]))^(2/3)) - (((2*I)/3)*ArcTan[(1 + (2*x)/(((I + Sqrt[1
5])/(5*I + Sqrt[15]))^(1/3)*(-1 + x^3)^(1/3)))/Sqrt[3]])/(Sqrt[5]*((I + Sqrt[15])/(5*I + Sqrt[15]))^(2/3)) - (
(I/3)*Log[1 - I*Sqrt[15] + 4*x^3])/(Sqrt[15]*((I + Sqrt[15])/(5*I + Sqrt[15]))^(2/3)) + ((I/3)*Log[1 + I*Sqrt[
15] + 4*x^3])/(Sqrt[15]*((I - Sqrt[15])/(5*I - Sqrt[15]))^(2/3)) - (I*Log[x/((I - Sqrt[15])/(5*I - Sqrt[15]))^
(1/3) - (-1 + x^3)^(1/3)])/(Sqrt[15]*((I - Sqrt[15])/(5*I - Sqrt[15]))^(2/3)) + (I*Log[x/((I + Sqrt[15])/(5*I
+ Sqrt[15]))^(1/3) - (-1 + x^3)^(1/3)])/(Sqrt[15]*((I + Sqrt[15])/(5*I + Sqrt[15]))^(2/3))

Rule 245

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

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 399

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Dist[b/d, Int[(a + b*x^n)^(p - 1), x]
, x] - Dist[(b*c - a*d)/d, Int[(a + b*x^n)^(p - 1)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c
- a*d, 0] && EqQ[n*(p - 1) + 1, 0] && IntegerQ[n]

Rule 1442

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{r = Rt[b^2 -
4*a*c, 2]}, Dist[2*(c/r), Int[(d + e*x^n)^q/(b - r + 2*c*x^n), x], x] - Dist[2*(c/r), Int[(d + e*x^n)^q/(b +
r + 2*c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] &&  !IntegerQ[q]

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 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+2 x^6\right )} \, dx &=\int \left (\frac {\left (-1+x^3\right )^{2/3}}{x^6}-\frac {2 \left (-1+x^3\right )^{2/3}}{2+x^3+2 x^6}\right ) \, dx\\ &=-\left (2 \int \frac {\left (-1+x^3\right )^{2/3}}{2+x^3+2 x^6} \, dx\right )+\int \frac {\left (-1+x^3\right )^{2/3}}{x^6} \, dx\\ &=\frac {\left (-1+x^3\right )^{5/3}}{5 x^5}+\frac {(8 i) \int \frac {\left (-1+x^3\right )^{2/3}}{1-i \sqrt {15}+4 x^3} \, dx}{\sqrt {15}}-\frac {(8 i) \int \frac {\left (-1+x^3\right )^{2/3}}{1+i \sqrt {15}+4 x^3} \, dx}{\sqrt {15}}\\ &=\frac {\left (-1+x^3\right )^{5/3}}{5 x^5}+\frac {\left (8 i \left (-1+x^3\right )^{2/3}\right ) \int \frac {\left (1-x^3\right )^{2/3}}{1-i \sqrt {15}+4 x^3} \, dx}{\sqrt {15} \left (1-x^3\right )^{2/3}}-\frac {\left (8 i \left (-1+x^3\right )^{2/3}\right ) \int \frac {\left (1-x^3\right )^{2/3}}{1+i \sqrt {15}+4 x^3} \, dx}{\sqrt {15} \left (1-x^3\right )^{2/3}}\\ &=\frac {\left (-1+x^3\right )^{5/3}}{5 x^5}-\frac {8 x \left (-1+x^3\right )^{2/3} F_1\left (\frac {1}{3};-\frac {2}{3},1;\frac {4}{3};x^3,-\frac {4 x^3}{1-i \sqrt {15}}\right )}{\sqrt {15} \left (i+\sqrt {15}\right ) \left (1-x^3\right )^{2/3}}+\frac {8 x \left (-1+x^3\right )^{2/3} F_1\left (\frac {1}{3};-\frac {2}{3},1;\frac {4}{3};x^3,-\frac {4 x^3}{1+i \sqrt {15}}\right )}{\sqrt {15} \left (i-\sqrt {15}\right ) \left (1-x^3\right )^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 79, normalized size = 1.00 \begin {gather*} \frac {\left (-1+x^3\right )^{5/3}}{5 x^5}-\frac {2}{3} \text {RootSum}\left [5-5 \text {$\#$1}^3+2 \text {$\#$1}^6\&,\frac {-\log (x) \text {$\#$1}^2+\log \left (\sqrt [3]{-1+x^3}-x \text {$\#$1}\right ) \text {$\#$1}^2}{-5+4 \text {$\#$1}^3}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-1 + x^3)^(5/3)/(5*x^5) - (2*RootSum[5 - 5*#1^3 + 2*#1^6 & , (-(Log[x]*#1^2) + Log[(-1 + x^3)^(1/3) - x*#1]*#
1^2)/(-5 + 4*#1^3) & ])/3

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 1.
time = 192.95, size = 7735, normalized size = 97.91

method	result	size
risch	\(\text {Expression too large to display}\)	\(7735\)
trager	\(\text {Expression too large to display}\)	\(10913\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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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/(2*x^6+x^3+2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (tr
ace 0)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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/(2*x^6+x^3+2),x, algorithm="giac")

[Out]

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

[Out]

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

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