3.16.77 \(\int \frac {-1+x^6}{x^6 \sqrt [3]{x+x^3}} \, dx\)

Optimal. Leaf size=108 \[ -\frac {1}{2} \log \left (\sqrt [3]{x^3+x}-x\right )+\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3+x}+x}\right )+\frac {1}{4} \log \left (\sqrt [3]{x^3+x} x+\left (x^3+x\right )^{2/3}+x^2\right )+\frac {3 \left (x^3+x\right )^{2/3} \left (9 x^4-6 x^2+5\right )}{80 x^6} \]

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Rubi [A]  time = 0.19, antiderivative size = 151, normalized size of antiderivative = 1.40, number of steps used = 9, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {2052, 2011, 329, 275, 239, 2016, 2014} \begin {gather*} \frac {3 \left (x^3+x\right )^{2/3}}{16 x^6}-\frac {9 \left (x^3+x\right )^{2/3}}{40 x^4}+\frac {27 \left (x^3+x\right )^{2/3}}{80 x^2}-\frac {3 \sqrt [3]{x} \sqrt [3]{x^2+1} \log \left (x^{2/3}-\sqrt [3]{x^2+1}\right )}{4 \sqrt [3]{x^3+x}}+\frac {\sqrt {3} \sqrt [3]{x} \sqrt [3]{x^2+1} \tan ^{-1}\left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{x^3+x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(x + x^3)^(2/3))/(16*x^6) - (9*(x + x^3)^(2/3))/(40*x^4) + (27*(x + x^3)^(2/3))/(80*x^2) + (Sqrt[3]*x^(1/3)
*(1 + x^2)^(1/3)*ArcTan[(1 + (2*x^(2/3))/(1 + x^2)^(1/3))/Sqrt[3]])/(2*(x + x^3)^(1/3)) - (3*x^(1/3)*(1 + x^2)
^(1/3)*Log[x^(2/3) - (1 + x^2)^(1/3)])/(4*(x + 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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2011

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2014

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

Rule 2016

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

Rule 2052

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(c*x)
^m*Pq*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !In
tegerQ[p] && NeQ[n, j]

Rubi steps

\begin {align*} \int \frac {-1+x^6}{x^6 \sqrt [3]{x+x^3}} \, dx &=\int \left (\frac {1}{\sqrt [3]{x+x^3}}-\frac {1}{x^6 \sqrt [3]{x+x^3}}\right ) \, dx\\ &=\int \frac {1}{\sqrt [3]{x+x^3}} \, dx-\int \frac {1}{x^6 \sqrt [3]{x+x^3}} \, dx\\ &=\frac {3 \left (x+x^3\right )^{2/3}}{16 x^6}+\frac {3}{4} \int \frac {1}{x^4 \sqrt [3]{x+x^3}} \, dx+\frac {\left (\sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \int \frac {1}{\sqrt [3]{x} \sqrt [3]{1+x^2}} \, dx}{\sqrt [3]{x+x^3}}\\ &=\frac {3 \left (x+x^3\right )^{2/3}}{16 x^6}-\frac {9 \left (x+x^3\right )^{2/3}}{40 x^4}-\frac {9}{20} \int \frac {1}{x^2 \sqrt [3]{x+x^3}} \, dx+\frac {\left (3 \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt [3]{1+x^6}} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x+x^3}}\\ &=\frac {3 \left (x+x^3\right )^{2/3}}{16 x^6}-\frac {9 \left (x+x^3\right )^{2/3}}{40 x^4}+\frac {27 \left (x+x^3\right )^{2/3}}{80 x^2}+\frac {\left (3 \sqrt [3]{x} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{1+x^3}} \, dx,x,x^{2/3}\right )}{2 \sqrt [3]{x+x^3}}\\ &=\frac {3 \left (x+x^3\right )^{2/3}}{16 x^6}-\frac {9 \left (x+x^3\right )^{2/3}}{40 x^4}+\frac {27 \left (x+x^3\right )^{2/3}}{80 x^2}+\frac {\sqrt {3} \sqrt [3]{x} \sqrt [3]{1+x^2} \tan ^{-1}\left (\frac {1+\frac {2 x^{2/3}}{\sqrt [3]{1+x^2}}}{\sqrt {3}}\right )}{2 \sqrt [3]{x+x^3}}-\frac {3 \sqrt [3]{x} \sqrt [3]{1+x^2} \log \left (x^{2/3}-\sqrt [3]{1+x^2}\right )}{4 \sqrt [3]{x+x^3}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 163, normalized size = 1.51 \begin {gather*} \frac {27 x^6+9 x^4-3 x^2-40 \sqrt [3]{x^2+1} x^{16/3} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{x^2+1}}\right )+20 \sqrt [3]{x^2+1} x^{16/3} \log \left (\frac {x^{4/3}}{\left (x^2+1\right )^{2/3}}+\frac {x^{2/3}}{\sqrt [3]{x^2+1}}+1\right )+40 \sqrt {3} \sqrt [3]{x^2+1} x^{16/3} \tan ^{-1}\left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )+15}{80 x^5 \sqrt [3]{x^3+x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15 - 3*x^2 + 9*x^4 + 27*x^6 + 40*Sqrt[3]*x^(16/3)*(1 + x^2)^(1/3)*ArcTan[(1 + (2*x^(2/3))/(1 + x^2)^(1/3))/Sq
rt[3]] - 40*x^(16/3)*(1 + x^2)^(1/3)*Log[1 - x^(2/3)/(1 + x^2)^(1/3)] + 20*x^(16/3)*(1 + x^2)^(1/3)*Log[1 + x^
(4/3)/(1 + x^2)^(2/3) + x^(2/3)/(1 + x^2)^(1/3)])/(80*x^5*(x + x^3)^(1/3))

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IntegrateAlgebraic [A]  time = 0.35, size = 108, normalized size = 1.00 \begin {gather*} \frac {3 \left (x+x^3\right )^{2/3} \left (5-6 x^2+9 x^4\right )}{80 x^6}+\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x+x^3}}\right )-\frac {1}{2} \log \left (-x+\sqrt [3]{x+x^3}\right )+\frac {1}{4} \log \left (x^2+x \sqrt [3]{x+x^3}+\left (x+x^3\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(x + x^3)^(2/3)*(5 - 6*x^2 + 9*x^4))/(80*x^6) + (Sqrt[3]*ArcTan[(Sqrt[3]*x)/(x + 2*(x + x^3)^(1/3))])/2 - L
og[-x + (x + x^3)^(1/3)]/2 + Log[x^2 + x*(x + x^3)^(1/3) + (x + x^3)^(2/3)]/4

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fricas [A]  time = 0.79, size = 112, normalized size = 1.04 \begin {gather*} \frac {40 \, \sqrt {3} x^{6} \arctan \left (-\frac {196 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {1}{3}} x - \sqrt {3} {\left (539 \, x^{2} + 507\right )} - 1274 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {2}{3}}}{2205 \, x^{2} + 2197}\right ) - 20 \, x^{6} \log \left (3 \, {\left (x^{3} + x\right )}^{\frac {1}{3}} x - 3 \, {\left (x^{3} + x\right )}^{\frac {2}{3}} + 1\right ) + 3 \, {\left (9 \, x^{4} - 6 \, x^{2} + 5\right )} {\left (x^{3} + x\right )}^{\frac {2}{3}}}{80 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*(40*sqrt(3)*x^6*arctan(-(196*sqrt(3)*(x^3 + x)^(1/3)*x - sqrt(3)*(539*x^2 + 507) - 1274*sqrt(3)*(x^3 + x)
^(2/3))/(2205*x^2 + 2197)) - 20*x^6*log(3*(x^3 + x)^(1/3)*x - 3*(x^3 + x)^(2/3) + 1) + 3*(9*x^4 - 6*x^2 + 5)*(
x^3 + x)^(2/3))/x^6

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giac [A]  time = 0.39, size = 82, normalized size = 0.76 \begin {gather*} \frac {3}{16} \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {8}{3}} - \frac {3}{5} \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {5}{3}} - \frac {1}{2} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {3}{4} \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + \frac {1}{4} \, \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {1}{2} \, \log \left ({\left | {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16*(1/x^2 + 1)^(8/3) - 3/5*(1/x^2 + 1)^(5/3) - 1/2*sqrt(3)*arctan(1/3*sqrt(3)*(2*(1/x^2 + 1)^(1/3) + 1)) + 3
/4*(1/x^2 + 1)^(2/3) + 1/4*log((1/x^2 + 1)^(2/3) + (1/x^2 + 1)^(1/3) + 1) - 1/2*log(abs((1/x^2 + 1)^(1/3) - 1)
)

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maple [C]  time = 1.66, size = 42, normalized size = 0.39

method	result	size
meijerg	\(\frac {3 \left (\frac {9}{5} x^{4}-\frac {6}{5} x^{2}+1\right ) \left (x^{2}+1\right )^{\frac {2}{3}}}{16 x^{\frac {16}{3}}}+\frac {3 x^{\frac {2}{3}} \hypergeom \left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x^{2}\right )}{2}\)	\(42\)
risch	\(\frac {-\frac {3}{80} x^{2}+\frac {3}{16}+\frac {9}{80} x^{4}+\frac {27}{80} x^{6}}{x^{5} \left (x \left (x^{2}+1\right )\right )^{\frac {1}{3}}}+\frac {3 x^{\frac {2}{3}} \hypergeom \left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x^{2}\right )}{2}\)	\(49\)
trager	\(\frac {3 \left (x^{3}+x \right )^{\frac {2}{3}} \left (9 x^{4}-6 x^{2}+5\right )}{80 x^{6}}+\frac {\ln \left (180 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2} x^{2}+144 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}+144 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x +114 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) x^{2}-180 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2}-9 \left (x^{3}+x \right )^{\frac {2}{3}}-9 x \left (x^{3}+x \right )^{\frac {1}{3}}-4 x^{2}+96 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )-3\right )}{2}-3 \ln \left (180 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2} x^{2}+144 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}+144 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x +114 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) x^{2}-180 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2}-9 \left (x^{3}+x \right )^{\frac {2}{3}}-9 x \left (x^{3}+x \right )^{\frac {1}{3}}-4 x^{2}+96 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )-3\right ) \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )+3 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \ln \left (180 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2} x^{2}-144 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}-144 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x -174 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right ) x^{2}-180 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )^{2}+15 \left (x^{3}+x \right )^{\frac {2}{3}}+15 x \left (x^{3}+x \right )^{\frac {1}{3}}+20 x^{2}-36 \RootOf \left (36 \textit {\_Z}^{2}-6 \textit {\_Z} +1\right )+8\right )\)	\(444\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16/x^(16/3)*(9/5*x^4-6/5*x^2+1)*(x^2+1)^(2/3)+3/2*x^(2/3)*hypergeom([1/3,1/3],[4/3],-x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.19, size = 52, normalized size = 0.48 \begin {gather*} \frac {3\,{\left (x^3+x\right )}^{2/3}\,\left (9\,x^4-6\,x^2+5\right )}{80\,x^6}+\frac {3\,x\,{\left (x^2+1\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ -x^2\right )}{2\,{\left (x^3+x\right )}^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*(x + x^3)^(2/3)*(9*x^4 - 6*x^2 + 5))/(80*x^6) + (3*x*(x^2 + 1)^(1/3)*hypergeom([1/3, 1/3], 4/3, -x^2))/(2*(
x + x^3)^(1/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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