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

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

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Rubi [C]  time = 0.37, antiderivative size = 161, normalized size of antiderivative = 2.04, number of steps used = 8, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6728, 264, 1428, 430, 429} \begin {gather*} -\frac {8 x \left (x^3-1\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 (\sqrt {15}+i\right ) \left (1-x^3\right )^{2/3}}+\frac {8 x \left (x^3-1\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 (-\sqrt {15}+i\right ) \left (1-x^3\right )^{2/3}}+\frac {\left (x^3-1\right )^{5/3}}{5 x^5} \end {gather*}

Warning: Unable to verify antiderivative.

[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) - (8*x*(-1 + x^3)^(2/3)*AppellF1[1/3, -2/3, 1, 4/3, x^3, (-4*x^3)/(1 - I*Sqrt[15])])/
(Sqrt[15]*(I + Sqrt[15])*(1 - x^3)^(2/3)) + (8*x*(-1 + x^3)^(2/3)*AppellF1[1/3, -2/3, 1, 4/3, x^3, (-4*x^3)/(1
+ I*Sqrt[15])])/(Sqrt[15]*(I - Sqrt[15])*(1 - x^3)^(2/3))

Rule 264

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 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 430

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^F
racPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 1428

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 6728

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 [F]  time = 0.08, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.00, 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]

IntegrateAlgebraic[((-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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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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{3} + 2\right )} {\left (x^{3} - 1\right )}^{\frac {2}{3}}}{{\left (2 \, x^{6} + x^{3} + 2\right )} x^{6}}\,{d x} \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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maple [B]  time = 265.57, size = 6630, normalized size = 83.92

method	result	size
risch	\(\text {Expression too large to display}\)	\(6630\)
trager	\(\text {Expression too large to display}\)	\(7528\)

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*} \int \frac {{\left (x^{3} + 2\right )} {\left (x^{3} - 1\right )}^{\frac {2}{3}}}{{\left (2 \, x^{6} + x^{3} + 2\right )} x^{6}}\,{d x} \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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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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sympy [F(-1)]  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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