3.313 \(\int \frac {-2 x^5+3 x^8-x^2 (-1+3 x^3)^{2/3}}{(-1+3 x^3)^{3/4}} \, dx\)

Optimal. Leaf size=46 \[ \frac {4}{243} \left (3 x^3-1\right )^{9/4}-\frac {4}{33} \left (3 x^3-1\right )^{11/12}-\frac {4}{27} \sqrt [4]{3 x^3-1} \]

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Rubi [A]  time = 0.19, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6742, 266, 43, 261} \[ \frac {4}{243} \left (3 x^3-1\right )^{9/4}-\frac {4}{33} \left (3 x^3-1\right )^{11/12}-\frac {4}{27} \sqrt [4]{3 x^3-1} \]

Antiderivative was successfully verified.

[In]

Int[(-2*x^5 + 3*x^8 - x^2*(-1 + 3*x^3)^(2/3))/(-1 + 3*x^3)^(3/4),x]

[Out]

(-4*(-1 + 3*x^3)^(1/4))/27 - (4*(-1 + 3*x^3)^(11/12))/33 + (4*(-1 + 3*x^3)^(9/4))/243

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 261

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 6742

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

Rubi steps

\begin {align*} \int \frac {-2 x^5+3 x^8-x^2 \left (-1+3 x^3\right )^{2/3}}{\left (-1+3 x^3\right )^{3/4}} \, dx &=\int \left (-\frac {2 x^5}{\left (-1+3 x^3\right )^{3/4}}+\frac {3 x^8}{\left (-1+3 x^3\right )^{3/4}}-\frac {x^2}{\sqrt [12]{-1+3 x^3}}\right ) \, dx\\ &=-\left (2 \int \frac {x^5}{\left (-1+3 x^3\right )^{3/4}} \, dx\right )+3 \int \frac {x^8}{\left (-1+3 x^3\right )^{3/4}} \, dx-\int \frac {x^2}{\sqrt [12]{-1+3 x^3}} \, dx\\ &=-\frac {4}{33} \left (-1+3 x^3\right )^{11/12}-\frac {2}{3} \operatorname {Subst}\left (\int \frac {x}{(-1+3 x)^{3/4}} \, dx,x,x^3\right )+\operatorname {Subst}\left (\int \frac {x^2}{(-1+3 x)^{3/4}} \, dx,x,x^3\right )\\ &=-\frac {4}{33} \left (-1+3 x^3\right )^{11/12}-\frac {2}{3} \operatorname {Subst}\left (\int \left (\frac {1}{3 (-1+3 x)^{3/4}}+\frac {1}{3} \sqrt [4]{-1+3 x}\right ) \, dx,x,x^3\right )+\operatorname {Subst}\left (\int \left (\frac {1}{9 (-1+3 x)^{3/4}}+\frac {2}{9} \sqrt [4]{-1+3 x}+\frac {1}{9} (-1+3 x)^{5/4}\right ) \, dx,x,x^3\right )\\ &=-\frac {4}{27} \sqrt [4]{-1+3 x^3}-\frac {4}{33} \left (-1+3 x^3\right )^{11/12}+\frac {4}{243} \left (-1+3 x^3\right )^{9/4}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 40, normalized size = 0.87 \[ -\frac {4 \sqrt [4]{3 x^3-1} \left (-99 x^6+66 x^3+81 \left (3 x^3-1\right )^{2/3}+88\right )}{2673} \]

Antiderivative was successfully verified.

[In]

Integrate[(-2*x^5 + 3*x^8 - x^2*(-1 + 3*x^3)^(2/3))/(-1 + 3*x^3)^(3/4),x]

[Out]

(-4*(-1 + 3*x^3)^(1/4)*(88 + 66*x^3 - 99*x^6 + 81*(-1 + 3*x^3)^(2/3)))/2673

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IntegrateAlgebraic [A]  time = 0.04, size = 43, normalized size = 0.93 \[ \frac {4}{243} \sqrt [4]{3 x^3-1} \left (9 x^6-6 x^3-8\right )-\frac {4}{33} \left (3 x^3-1\right )^{11/12} \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-2*x^5 + 3*x^8 - x^2*(-1 + 3*x^3)^(2/3))/(-1 + 3*x^3)^(3/4),x]

[Out]

(-4*(-1 + 3*x^3)^(11/12))/33 + (4*(-1 + 3*x^3)^(1/4)*(-8 - 6*x^3 + 9*x^6))/243

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fricas [A]  time = 0.62, size = 35, normalized size = 0.76 \[ \frac {4}{243} \, {\left (9 \, x^{6} - 6 \, x^{3} - 8\right )} {\left (3 \, x^{3} - 1\right )}^{\frac {1}{4}} - \frac {4}{33} \, {\left (3 \, x^{3} - 1\right )}^{\frac {11}{12}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^5+3*x^8-x^2*(3*x^3-1)^(2/3))/(3*x^3-1)^(3/4),x, algorithm="fricas")

[Out]

4/243*(9*x^6 - 6*x^3 - 8)*(3*x^3 - 1)^(1/4) - 4/33*(3*x^3 - 1)^(11/12)

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giac [A]  time = 0.62, size = 34, normalized size = 0.74 \[ \frac {4}{243} \, {\left (3 \, x^{3} - 1\right )}^{\frac {9}{4}} - \frac {4}{33} \, {\left (3 \, x^{3} - 1\right )}^{\frac {11}{12}} - \frac {4}{27} \, {\left (3 \, x^{3} - 1\right )}^{\frac {1}{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^5+3*x^8-x^2*(3*x^3-1)^(2/3))/(3*x^3-1)^(3/4),x, algorithm="giac")

[Out]

4/243*(3*x^3 - 1)^(9/4) - 4/33*(3*x^3 - 1)^(11/12) - 4/27*(3*x^3 - 1)^(1/4)

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maple [C]  time = 0.35, size = 116, normalized size = 2.52

method	result	size
meijerg	\(-\frac {\left (-\mathrm {signum}\left (3 x^{3}-1\right )\right )^{\frac {3}{4}} x^{6} \hypergeom \left (\left [\frac {3}{4}, 2\right ], \relax [3], 3 x^{3}\right )}{3 \mathrm {signum}\left (3 x^{3}-1\right )^{\frac {3}{4}}}+\frac {\left (-\mathrm {signum}\left (3 x^{3}-1\right )\right )^{\frac {3}{4}} x^{9} \hypergeom \left (\left [\frac {3}{4}, 3\right ], \relax [4], 3 x^{3}\right )}{3 \mathrm {signum}\left (3 x^{3}-1\right )^{\frac {3}{4}}}-\frac {\left (-\mathrm {signum}\left (3 x^{3}-1\right )\right )^{\frac {1}{12}} x^{3} \hypergeom \left (\left [\frac {1}{12}, 1\right ], \relax [2], 3 x^{3}\right )}{3 \mathrm {signum}\left (3 x^{3}-1\right )^{\frac {1}{12}}}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x^5+3*x^8-x^2*(3*x^3-1)^(2/3))/(3*x^3-1)^(3/4),x,method=_RETURNVERBOSE)

[Out]

-1/3/signum(3*x^3-1)^(3/4)*(-signum(3*x^3-1))^(3/4)*x^6*hypergeom([3/4,2],[3],3*x^3)+1/3/signum(3*x^3-1)^(3/4)
*(-signum(3*x^3-1))^(3/4)*x^9*hypergeom([3/4,3],[4],3*x^3)-1/3/signum(3*x^3-1)^(1/12)*(-signum(3*x^3-1))^(1/12
)*x^3*hypergeom([1/12,1],[2],3*x^3)

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maxima [A]  time = 0.57, size = 34, normalized size = 0.74 \[ \frac {4}{243} \, {\left (3 \, x^{3} - 1\right )}^{\frac {9}{4}} - \frac {4}{33} \, {\left (3 \, x^{3} - 1\right )}^{\frac {11}{12}} - \frac {4}{27} \, {\left (3 \, x^{3} - 1\right )}^{\frac {1}{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^5+3*x^8-x^2*(3*x^3-1)^(2/3))/(3*x^3-1)^(3/4),x, algorithm="maxima")

[Out]

4/243*(3*x^3 - 1)^(9/4) - 4/33*(3*x^3 - 1)^(11/12) - 4/27*(3*x^3 - 1)^(1/4)

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mupad [B]  time = 0.37, size = 34, normalized size = 0.74 \[ -{\left (3\,x^3-1\right )}^{1/4}\,\left (\frac {8\,x^3}{81}-\frac {4\,x^6}{27}+\frac {4\,{\left (3\,x^3-1\right )}^{2/3}}{33}+\frac {32}{243}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^2*(3*x^3 - 1)^(2/3) + 2*x^5 - 3*x^8)/(3*x^3 - 1)^(3/4),x)

[Out]

-(3*x^3 - 1)^(1/4)*((8*x^3)/81 - (4*x^6)/27 + (4*(3*x^3 - 1)^(2/3))/33 + 32/243)

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sympy [C]  time = 3.94, size = 221, normalized size = 4.80 \[ - \frac {4 \left (3 x^{3} - 1\right )^{\frac {11}{12}}}{33} - 2 \left (\begin {cases} \frac {4 x^{3} \sqrt [4]{3 x^{3} - 1}}{45} + \frac {16 \sqrt [4]{3 x^{3} - 1}}{135} & \text {for}\: 3 \left |{x^{3}}\right | > 1 \\- \frac {4 x^{3} \sqrt [4]{1 - 3 x^{3}} e^{- \frac {3 i \pi }{4}}}{45} - \frac {16 \sqrt [4]{1 - 3 x^{3}} e^{- \frac {3 i \pi }{4}}}{135} & \text {otherwise} \end {cases}\right ) + 3 \left (\begin {cases} \frac {4 x^{6} \sqrt [4]{3 x^{3} - 1}}{81} + \frac {32 x^{3} \sqrt [4]{3 x^{3} - 1}}{1215} + \frac {128 \sqrt [4]{3 x^{3} - 1}}{3645} & \text {for}\: 3 \left |{x^{3}}\right | > 1 \\\frac {4 x^{6} \sqrt [4]{1 - 3 x^{3}} e^{\frac {i \pi }{4}}}{81} + \frac {32 x^{3} \sqrt [4]{1 - 3 x^{3}} e^{\frac {i \pi }{4}}}{1215} + \frac {128 \sqrt [4]{1 - 3 x^{3}} e^{\frac {i \pi }{4}}}{3645} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x**5+3*x**8-x**2*(3*x**3-1)**(2/3))/(3*x**3-1)**(3/4),x)

[Out]

-4*(3*x**3 - 1)**(11/12)/33 - 2*Piecewise((4*x**3*(3*x**3 - 1)**(1/4)/45 + 16*(3*x**3 - 1)**(1/4)/135, 3*Abs(x
**3) > 1), (-4*x**3*(1 - 3*x**3)**(1/4)*exp(-3*I*pi/4)/45 - 16*(1 - 3*x**3)**(1/4)*exp(-3*I*pi/4)/135, True))
+ 3*Piecewise((4*x**6*(3*x**3 - 1)**(1/4)/81 + 32*x**3*(3*x**3 - 1)**(1/4)/1215 + 128*(3*x**3 - 1)**(1/4)/3645
, 3*Abs(x**3) > 1), (4*x**6*(1 - 3*x**3)**(1/4)*exp(I*pi/4)/81 + 32*x**3*(1 - 3*x**3)**(1/4)*exp(I*pi/4)/1215
+ 128*(1 - 3*x**3)**(1/4)*exp(I*pi/4)/3645, True))

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