3.26.46 \(\int \frac {(1+x^3)^{2/3} (8-4 x^3+x^6)}{x^6 (2+x^3)} \, dx\)

Optimal. Leaf size=214 \[ -\frac {1}{3} \log \left (\sqrt [3]{x^3+1}-x\right )+\frac {5 \log \left (\sqrt [3]{2} \sqrt [3]{x^3+1}-x\right )}{3\ 2^{2/3}}+\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3+1}+x}\right )}{\sqrt {3}}-\frac {5 \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{2} \sqrt [3]{x^3+1}+x}\right )}{2^{2/3} \sqrt {3}}+\frac {2 \left (x^3+1\right )^{2/3} \left (3 x^3-2\right )}{5 x^5}+\frac {1}{6} \log \left (\sqrt [3]{x^3+1} x+\left (x^3+1\right )^{2/3}+x^2\right )-\frac {5 \log \left (\sqrt [3]{2} \sqrt [3]{x^3+1} x+2^{2/3} \left (x^3+1\right )^{2/3}+x^2\right )}{6\ 2^{2/3}} \]

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Rubi [C]  time = 0.39, antiderivative size = 103, normalized size of antiderivative = 0.48, number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6725, 264, 277, 239, 429} \begin {gather*} \frac {5}{2} x F_1\left (\frac {1}{3};-\frac {2}{3},1;\frac {4}{3};-x^3,-\frac {x^3}{2}\right )+2 \log \left (\sqrt [3]{x^3+1}-x\right )-\frac {4 \tan ^{-1}\left (\frac {\frac {2 x}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {4 \left (x^3+1\right )^{5/3}}{5 x^5}+\frac {2 \left (x^3+1\right )^{2/3}}{x^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[((1 + x^3)^(2/3)*(8 - 4*x^3 + x^6))/(x^6*(2 + x^3)),x]

[Out]

(2*(1 + x^3)^(2/3))/x^2 - (4*(1 + x^3)^(5/3))/(5*x^5) + (5*x*AppellF1[1/3, -2/3, 1, 4/3, -x^3, -1/2*x^3])/2 -
(4*ArcTan[(1 + (2*x)/(1 + x^3)^(1/3))/Sqrt[3]])/Sqrt[3] + 2*Log[-x + (1 + x^3)^(1/3)]

Rule 239

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 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 277

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

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 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (1+x^3\right )^{2/3} \left (8-4 x^3+x^6\right )}{x^6 \left (2+x^3\right )} \, dx &=\int \left (\frac {4 \left (1+x^3\right )^{2/3}}{x^6}-\frac {4 \left (1+x^3\right )^{2/3}}{x^3}+\frac {5 \left (1+x^3\right )^{2/3}}{2+x^3}\right ) \, dx\\ &=4 \int \frac {\left (1+x^3\right )^{2/3}}{x^6} \, dx-4 \int \frac {\left (1+x^3\right )^{2/3}}{x^3} \, dx+5 \int \frac {\left (1+x^3\right )^{2/3}}{2+x^3} \, dx\\ &=\frac {2 \left (1+x^3\right )^{2/3}}{x^2}-\frac {4 \left (1+x^3\right )^{5/3}}{5 x^5}+\frac {5}{2} x F_1\left (\frac {1}{3};-\frac {2}{3},1;\frac {4}{3};-x^3,-\frac {x^3}{2}\right )-4 \int \frac {1}{\sqrt [3]{1+x^3}} \, dx\\ &=\frac {2 \left (1+x^3\right )^{2/3}}{x^2}-\frac {4 \left (1+x^3\right )^{5/3}}{5 x^5}+\frac {5}{2} x F_1\left (\frac {1}{3};-\frac {2}{3},1;\frac {4}{3};-x^3,-\frac {x^3}{2}\right )-\frac {4 \tan ^{-1}\left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+2 \log \left (-x+\sqrt [3]{1+x^3}\right )\\ \end {align*}

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Mathematica [C]  time = 0.23, size = 154, normalized size = 0.72 \begin {gather*} \frac {1}{8} \left (x^4 F_1\left (\frac {4}{3};\frac {1}{3},1;\frac {7}{3};-x^3,-\frac {x^3}{2}\right )+\frac {16 \left (x^3+1\right )^{2/3} \left (3 x^3-2\right )}{5 x^5}-2 \sqrt [3]{2} \left (-2 \log \left (2-\frac {2^{2/3} x}{\sqrt [3]{x^3+1}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2^{2/3} x}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )+\log \left (\frac {2^{2/3} x}{\sqrt [3]{x^3+1}}+\frac {\sqrt [3]{2} x^2}{\left (x^3+1\right )^{2/3}}+2\right )\right )\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((1 + x^3)^(2/3)*(8 - 4*x^3 + x^6))/(x^6*(2 + x^3)),x]

[Out]

((16*(1 + x^3)^(2/3)*(-2 + 3*x^3))/(5*x^5) + x^4*AppellF1[4/3, 1/3, 1, 7/3, -x^3, -1/2*x^3] - 2*2^(1/3)*(2*Sqr
t[3]*ArcTan[(1 + (2^(2/3)*x)/(1 + x^3)^(1/3))/Sqrt[3]] - 2*Log[2 - (2^(2/3)*x)/(1 + x^3)^(1/3)] + Log[2 + (2^(
1/3)*x^2)/(1 + x^3)^(2/3) + (2^(2/3)*x)/(1 + x^3)^(1/3)]))/8

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IntegrateAlgebraic [A]  time = 0.47, size = 214, normalized size = 1.00 \begin {gather*} \frac {2 \left (1+x^3\right )^{2/3} \left (-2+3 x^3\right )}{5 x^5}+\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{1+x^3}}\right )}{\sqrt {3}}-\frac {5 \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{2} \sqrt [3]{1+x^3}}\right )}{2^{2/3} \sqrt {3}}-\frac {1}{3} \log \left (-x+\sqrt [3]{1+x^3}\right )+\frac {5 \log \left (-x+\sqrt [3]{2} \sqrt [3]{1+x^3}\right )}{3\ 2^{2/3}}+\frac {1}{6} \log \left (x^2+x \sqrt [3]{1+x^3}+\left (1+x^3\right )^{2/3}\right )-\frac {5 \log \left (x^2+\sqrt [3]{2} x \sqrt [3]{1+x^3}+2^{2/3} \left (1+x^3\right )^{2/3}\right )}{6\ 2^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((1 + x^3)^(2/3)*(8 - 4*x^3 + x^6))/(x^6*(2 + x^3)),x]

[Out]

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

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fricas [B]  time = 27.93, size = 361, normalized size = 1.69 \begin {gather*} \frac {100 \cdot 4^{\frac {1}{6}} \sqrt {3} x^{5} \arctan \left (\frac {4^{\frac {1}{6}} {\left (12 \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (2 \, x^{7} + 5 \, x^{4} + 2 \, x\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}} + 4^{\frac {1}{3}} \sqrt {3} {\left (91 \, x^{9} + 168 \, x^{6} + 84 \, x^{3} + 8\right )} + 12 \, \sqrt {3} {\left (19 \, x^{8} + 22 \, x^{5} + 4 \, x^{2}\right )} {\left (x^{3} + 1\right )}^{\frac {1}{3}}\right )}}{6 \, {\left (53 \, x^{9} + 48 \, x^{6} - 12 \, x^{3} - 8\right )}}\right ) + 50 \cdot 4^{\frac {2}{3}} x^{5} \log \left (-\frac {6 \cdot 4^{\frac {1}{3}} {\left (x^{3} + 1\right )}^{\frac {1}{3}} x^{2} + 4^{\frac {2}{3}} {\left (x^{3} + 2\right )} - 12 \, {\left (x^{3} + 1\right )}^{\frac {2}{3}} x}{x^{3} + 2}\right ) - 25 \cdot 4^{\frac {2}{3}} x^{5} \log \left (\frac {6 \cdot 4^{\frac {2}{3}} {\left (2 \, x^{4} + x\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (19 \, x^{6} + 22 \, x^{3} + 4\right )} + 6 \, {\left (5 \, x^{5} + 4 \, x^{2}\right )} {\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x^{6} + 4 \, x^{3} + 4}\right ) + 120 \, \sqrt {3} x^{5} \arctan \left (-\frac {25382 \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {1}{3}} x^{2} - 13720 \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {2}{3}} x + \sqrt {3} {\left (5831 \, x^{3} + 7200\right )}}{58653 \, x^{3} + 8000}\right ) - 60 \, x^{5} \log \left (3 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}} x^{2} - 3 \, {\left (x^{3} + 1\right )}^{\frac {2}{3}} x + 1\right ) + 144 \, {\left (3 \, x^{3} - 2\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{360 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+1)^(2/3)*(x^6-4*x^3+8)/x^6/(x^3+2),x, algorithm="fricas")

[Out]

1/360*(100*4^(1/6)*sqrt(3)*x^5*arctan(1/6*4^(1/6)*(12*4^(2/3)*sqrt(3)*(2*x^7 + 5*x^4 + 2*x)*(x^3 + 1)^(2/3) +
4^(1/3)*sqrt(3)*(91*x^9 + 168*x^6 + 84*x^3 + 8) + 12*sqrt(3)*(19*x^8 + 22*x^5 + 4*x^2)*(x^3 + 1)^(1/3))/(53*x^
9 + 48*x^6 - 12*x^3 - 8)) + 50*4^(2/3)*x^5*log(-(6*4^(1/3)*(x^3 + 1)^(1/3)*x^2 + 4^(2/3)*(x^3 + 2) - 12*(x^3 +
1)^(2/3)*x)/(x^3 + 2)) - 25*4^(2/3)*x^5*log((6*4^(2/3)*(2*x^4 + x)*(x^3 + 1)^(2/3) + 4^(1/3)*(19*x^6 + 22*x^3
+ 4) + 6*(5*x^5 + 4*x^2)*(x^3 + 1)^(1/3))/(x^6 + 4*x^3 + 4)) + 120*sqrt(3)*x^5*arctan(-(25382*sqrt(3)*(x^3 +
1)^(1/3)*x^2 - 13720*sqrt(3)*(x^3 + 1)^(2/3)*x + sqrt(3)*(5831*x^3 + 7200))/(58653*x^3 + 8000)) - 60*x^5*log(3
*(x^3 + 1)^(1/3)*x^2 - 3*(x^3 + 1)^(2/3)*x + 1) + 144*(3*x^3 - 2)*(x^3 + 1)^(2/3))/x^5

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{6} - 4 \, x^{3} + 8\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{{\left (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^6-4*x^3+8)/x^6/(x^3+2),x, algorithm="giac")

[Out]

integrate((x^6 - 4*x^3 + 8)*(x^3 + 1)^(2/3)/((x^3 + 2)*x^6), x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (x^{3}+1\right )^{\frac {2}{3}} \left (x^{6}-4 x^{3}+8\right )}{x^{6} \left (x^{3}+2\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3+1)^(2/3)*(x^6-4*x^3+8)/x^6/(x^3+2),x)

[Out]

int((x^3+1)^(2/3)*(x^6-4*x^3+8)/x^6/(x^3+2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{6} - 4 \, x^{3} + 8\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{{\left (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^6-4*x^3+8)/x^6/(x^3+2),x, algorithm="maxima")

[Out]

integrate((x^6 - 4*x^3 + 8)*(x^3 + 1)^(2/3)/((x^3 + 2)*x^6), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (x^3+1\right )}^{2/3}\,\left (x^6-4\,x^3+8\right )}{x^6\,\left (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^6 - 4*x^3 + 8))/(x^6*(x^3 + 2)),x)

[Out]

int(((x^3 + 1)^(2/3)*(x^6 - 4*x^3 + 8))/(x^6*(x^3 + 2)), x)

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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^{2} + 2 x + 2\right ) \left (x^{4} - 2 x^{3} + 2 x^{2} - 4 x + 4\right )}{x^{6} \left (x^{3} + 2\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3+1)**(2/3)*(x**6-4*x**3+8)/x**6/(x**3+2),x)

[Out]

Integral(((x + 1)*(x**2 - x + 1))**(2/3)*(x**2 + 2*x + 2)*(x**4 - 2*x**3 + 2*x**2 - 4*x + 4)/(x**6*(x**3 + 2))
, x)

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