∫ (1+x3)2∕3(2+x3+2x6)_______________

3.11.5 $\int \frac {(1+x^3)^{2/3} (2+x^3+2 x^6)}{x^6 (1+x^3+x^6)} \, dx$

Optimal. Leaf size=76 \[ \frac {\left (x^3-4\right ) \left (x^3+1\right )^{2/3}}{10 x^5}-\frac {1}{3} \text {RootSum}\left [\text {$\#$1}^6-\text {$\#$1}^3+1\& ,\frac {\log \left (\sqrt [3]{x^3+1}-\text {$\#$1} x\right )-\log (x)}{2 \text {$\#$1}^4-\text {$\#$1}}\& \right ] \]

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Rubi [C] time = 0.40, antiderivative size = 187, normalized size of antiderivative = 2.46, number of steps used = 8, number of rules used = 6, integrand size = 33, $\frac {\text {number of rules}}{\text {integrand size}}$ = 0.182, Rules used = {6728, 264, 277, 239, 1428, 429} \begin {gather*} \frac {2 x F_1\left (\frac {1}{3};-\frac {5}{3},1;\frac {4}{3};-x^3,-\frac {2 x^3}{1-i \sqrt {3}}\right )}{\sqrt {3} \left (\sqrt {3}+i\right )}-\frac {2 x F_1\left (\frac {1}{3};-\frac {5}{3},1;\frac {4}{3};-x^3,-\frac {2 x^3}{1+i \sqrt {3}}\right )}{\sqrt {3} \left (-\sqrt {3}+i\right )}+\frac {1}{2} \log \left (\sqrt [3]{x^3+1}-x\right )-\frac {\tan ^{-1}\left (\frac {\frac {2 x}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {2 \left (x^3+1\right )^{5/3}}{5 x^5}+\frac {\left (x^3+1\right )^{2/3}}{2 x^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(1 + x^3)^(2/3)/(2*x^2) - (2*(1 + x^3)^(5/3))/(5*x^5) + (2*x*AppellF1[1/3, -5/3, 1, 4/3, -x^3, (-2*x^3)/(1 - I

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

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

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

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Mathematica [C] time = 1.76, size = 411, normalized size = 5.41 \begin {gather*} \frac {\left (x^3+1\right )^{2/3} \left (x^3-4\right )}{10 x^5}+\frac {i \left (-\frac {2 \log \left (\sqrt [3]{\sqrt {3}+i}-\frac {\sqrt [3]{\sqrt {3}-i} x}{\sqrt [3]{x^3+1}}\right )}{\sqrt [3]{\frac {\sqrt {3}-i}{\sqrt {3}+i}}}+2 \sqrt [3]{\frac {\sqrt {3}-i}{\sqrt {3}+i}} \log \left (\sqrt [3]{\sqrt {3}-i}-\frac {\sqrt [3]{\sqrt {3}+i} x}{\sqrt [3]{x^3+1}}\right )+\frac {2 \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {\sqrt [3]{2} \left (\sqrt {3}-i\right )^{2/3} x}{\sqrt [3]{x^3+1}}}{\sqrt {3}}\right )+\log \left (\frac {2^{2/3} x}{\sqrt [3]{x^3+1}}+\frac {\left (\sqrt {3}-i\right )^{2/3} x^2}{\left (x^3+1\right )^{2/3}}+\left (\sqrt {3}+i\right )^{2/3}\right )}{\left (\frac {1}{2} \left (\sqrt {3}-i\right )\right )^{2/3}}-\frac {2 \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {\sqrt [3]{2} \left (\sqrt {3}+i\right )^{2/3} x}{\sqrt [3]{x^3+1}}}{\sqrt {3}}\right )+\log \left (\frac {2^{2/3} x}{\sqrt [3]{x^3+1}}+\frac {\left (\sqrt {3}+i\right )^{2/3} x^2}{\left (x^3+1\right )^{2/3}}+\left (\sqrt {3}-i\right )^{2/3}\right )}{\left (\frac {1}{2} \left (\sqrt {3}+i\right )\right )^{2/3}}\right )}{6 \sqrt {3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-4 + x^3)*(1 + x^3)^(2/3))/(10*x^5) + ((I/6)*((2*Sqrt[3]*ArcTan[(1 + (2^(1/3)*(-I + Sqrt[3])^(2/3)*x)/(1 + x

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

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

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

)])/((I + Sqrt[3])/2)^(2/3) - (2*Log[(I + Sqrt[3])^(1/3) - ((-I + Sqrt[3])^(1/3)*x)/(1 + x^3)^(1/3)])/((-I + S

qrt[3])/(I + Sqrt[3]))^(1/3) + 2*((-I + Sqrt[3])/(I + Sqrt[3]))^(1/3)*Log[(-I + Sqrt[3])^(1/3) - ((I + Sqrt[3]

)^(1/3)*x)/(1 + x^3)^(1/3)]))/Sqrt[3]

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IntegrateAlgebraic [A] time = 0.24, size = 76, normalized size = 1.00 \begin {gather*} \frac {\left (-4+x^3\right ) \left (1+x^3\right )^{2/3}}{10 x^5}-\frac {1}{3} \text {RootSum}\left [1-\text {$\#$1}^3+\text {$\#$1}^6\&,\frac {-\log (x)+\log \left (\sqrt [3]{1+x^3}-x \text {$\#$1}\right )}{-\text {$\#$1}+2 \text {$\#$1}^4}\&\right ] \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((-4 + x^3)*(1 + x^3)^(2/3))/(10*x^5) - RootSum[1 - #1^3 + #1^6 & , (-Log[x] + Log[(1 + x^3)^(1/3) - x*#1])/(-

#1 + 2*#1^4) & ]/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)*(2*x^6+x^3+2)/x^6/(x^6+x^3+1),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 (2 \, x^{6} + x^{3} + 2\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{{\left (x^{6} + x^{3} + 1\right )} x^{6}}\,{d x} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B] time = 22.32, size = 1837, normalized size = 24.17


method	result	size

risch	$\text {Expression too large to display}$	$1837$
trager	$\text {Expression too large to display}$	$4248$

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(x^6-3*x^3-4)/x^5/(x^3+1)^(1/3)+1458*ln((729*RootOf(19683*_Z^6+243*_Z^3+1)^4*x^3+243*(x^3+1)^(1/3)*RootOf

(19683*_Z^6+243*_Z^3+1)^3*x^2+27*RootOf(19683*_Z^6+243*_Z^3+1)^2*(x^3+1)^(2/3)*x+6*RootOf(19683*_Z^6+243*_Z^3+

1)*x^3+x^2*(x^3+1)^(1/3)+3*RootOf(19683*_Z^6+243*_Z^3+1))/(6561*RootOf(19683*_Z^6+243*_Z^3+1)^5*x+54*RootOf(19

683*_Z^6+243*_Z^3+1)^2*x-1)/(6561*RootOf(19683*_Z^6+243*_Z^3+1)^5*x+27*RootOf(19683*_Z^6+243*_Z^3+1)^2*x+1)/(2

7*RootOf(19683*_Z^6+243*_Z^3+1)^2*x+1))*RootOf(19683*_Z^6+243*_Z^3+1)^5-729*ln((-4374*RootOf(19683*_Z^6+243*_Z

^3+1)^5*x^3+81*(x^3+1)^(2/3)*RootOf(19683*_Z^6+243*_Z^3+1)^3*x-2187*RootOf(19683*_Z^6+243*_Z^3+1)^5-18*RootOf(

19683*_Z^6+243*_Z^3+1)^2*x^3+3*(x^3+1)^(1/3)*RootOf(19683*_Z^6+243*_Z^3+1)*x^2+x*(x^3+1)^(2/3)-9*RootOf(19683*

_Z^6+243*_Z^3+1)^2)/(6561*RootOf(19683*_Z^6+243*_Z^3+1)^5*x+54*RootOf(19683*_Z^6+243*_Z^3+1)^2*x-1)/(6561*Root

Of(19683*_Z^6+243*_Z^3+1)^5*x+27*RootOf(19683*_Z^6+243*_Z^3+1)^2*x+1)/(27*RootOf(19683*_Z^6+243*_Z^3+1)^2*x+1)

)*RootOf(19683*_Z^6+243*_Z^3+1)^5+243*ln(-(-59049*RootOf(19683*_Z^6+243*_Z^3+1)^7*x^3-4374*(x^3+1)^(1/3)*RootO

f(19683*_Z^6+243*_Z^3+1)^5*x^2-972*RootOf(19683*_Z^6+243*_Z^3+1)^4*x^3+81*(x^3+1)^(2/3)*RootOf(19683*_Z^6+243*

_Z^3+1)^3*x-27*(x^3+1)^(1/3)*RootOf(19683*_Z^6+243*_Z^3+1)^2*x^2+243*RootOf(19683*_Z^6+243*_Z^3+1)^4-3*RootOf(

19683*_Z^6+243*_Z^3+1)*x^3+x*(x^3+1)^(2/3)+3*RootOf(19683*_Z^6+243*_Z^3+1))/(729*RootOf(19683*_Z^6+243*_Z^3+1)

^4*x+9*RootOf(19683*_Z^6+243*_Z^3+1)*x-1)/(1458*RootOf(19683*_Z^6+243*_Z^3+1)^4*x+9*RootOf(19683*_Z^6+243*_Z^3

+1)*x+1)/(729*RootOf(19683*_Z^6+243*_Z^3+1)^4*x-1))*RootOf(19683*_Z^6+243*_Z^3+1)^4+9*ln((729*RootOf(19683*_Z^

6+243*_Z^3+1)^4*x^3+243*(x^3+1)^(1/3)*RootOf(19683*_Z^6+243*_Z^3+1)^3*x^2+27*RootOf(19683*_Z^6+243*_Z^3+1)^2*(

x^3+1)^(2/3)*x+6*RootOf(19683*_Z^6+243*_Z^3+1)*x^3+x^2*(x^3+1)^(1/3)+3*RootOf(19683*_Z^6+243*_Z^3+1))/(6561*Ro

otOf(19683*_Z^6+243*_Z^3+1)^5*x+54*RootOf(19683*_Z^6+243*_Z^3+1)^2*x-1)/(6561*RootOf(19683*_Z^6+243*_Z^3+1)^5*

x+27*RootOf(19683*_Z^6+243*_Z^3+1)^2*x+1)/(27*RootOf(19683*_Z^6+243*_Z^3+1)^2*x+1))*RootOf(19683*_Z^6+243*_Z^3

+1)^2-9*ln((-4374*RootOf(19683*_Z^6+243*_Z^3+1)^5*x^3+81*(x^3+1)^(2/3)*RootOf(19683*_Z^6+243*_Z^3+1)^3*x-2187*

RootOf(19683*_Z^6+243*_Z^3+1)^5-18*RootOf(19683*_Z^6+243*_Z^3+1)^2*x^3+3*(x^3+1)^(1/3)*RootOf(19683*_Z^6+243*_

Z^3+1)*x^2+x*(x^3+1)^(2/3)-9*RootOf(19683*_Z^6+243*_Z^3+1)^2)/(6561*RootOf(19683*_Z^6+243*_Z^3+1)^5*x+54*RootO

f(19683*_Z^6+243*_Z^3+1)^2*x-1)/(6561*RootOf(19683*_Z^6+243*_Z^3+1)^5*x+27*RootOf(19683*_Z^6+243*_Z^3+1)^2*x+1

)/(27*RootOf(19683*_Z^6+243*_Z^3+1)^2*x+1))*RootOf(19683*_Z^6+243*_Z^3+1)^2+ln(-(-59049*RootOf(19683*_Z^6+243*

_Z^3+1)^7*x^3-4374*(x^3+1)^(1/3)*RootOf(19683*_Z^6+243*_Z^3+1)^5*x^2-972*RootOf(19683*_Z^6+243*_Z^3+1)^4*x^3+8

1*(x^3+1)^(2/3)*RootOf(19683*_Z^6+243*_Z^3+1)^3*x-27*(x^3+1)^(1/3)*RootOf(19683*_Z^6+243*_Z^3+1)^2*x^2+243*Roo

tOf(19683*_Z^6+243*_Z^3+1)^4-3*RootOf(19683*_Z^6+243*_Z^3+1)*x^3+x*(x^3+1)^(2/3)+3*RootOf(19683*_Z^6+243*_Z^3+

1))/(729*RootOf(19683*_Z^6+243*_Z^3+1)^4*x+9*RootOf(19683*_Z^6+243*_Z^3+1)*x-1)/(1458*RootOf(19683*_Z^6+243*_Z

^3+1)^4*x+9*RootOf(19683*_Z^6+243*_Z^3+1)*x+1)/(729*RootOf(19683*_Z^6+243*_Z^3+1)^4*x-1))*RootOf(19683*_Z^6+24

3*_Z^3+1)+RootOf(19683*_Z^6+243*_Z^3+1)*ln(-(729*(x^3+1)^(1/3)*RootOf(19683*_Z^6+243*_Z^3+1)^4*x^2+324*x^3*Roo

tOf(19683*_Z^6+243*_Z^3+1)^3+27*RootOf(19683*_Z^6+243*_Z^3+1)^2*(x^3+1)^(2/3)*x+9*(x^3+1)^(1/3)*RootOf(19683*_

Z^6+243*_Z^3+1)*x^2+162*RootOf(19683*_Z^6+243*_Z^3+1)^3+2*x^3+1)/(729*RootOf(19683*_Z^6+243*_Z^3+1)^4*x+9*Root

Of(19683*_Z^6+243*_Z^3+1)*x-1)/(1458*RootOf(19683*_Z^6+243*_Z^3+1)^4*x+9*RootOf(19683*_Z^6+243*_Z^3+1)*x+1)/(7

29*RootOf(19683*_Z^6+243*_Z^3+1)^4*x-1))

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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, x^{6} + x^{3} + 2\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{{\left (x^{6} + x^{3} + 1\right )} x^{6}}\,{d x} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((2*x^6 + x^3 + 2)*(x^3 + 1)^(2/3)/((x^6 + x^3 + 1)*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 (2\,x^6+x^3+2\right )}{x^6\,\left (x^6+x^3+1\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 + 2))/(x^6*(x^3 + x^6 + 1)),x)

[Out]

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

[Out]

Timed out

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