3.1.61 \(\int \frac {2+x}{(1+x+x^2) \sqrt [3]{2+x^3}} \, dx\) [61]

Optimal. Leaf size=168 \[ -\frac {x^2 F_1\left (\frac {2}{3};1,\frac {1}{3};\frac {5}{3};x^3,-\frac {x^3}{2}\right )}{2 \sqrt [3]{2}}+\frac {2 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{3} x}{\sqrt [3]{2+x^3}}}{\sqrt {3}}\right )}{3^{5/6}}+\frac {\tan ^{-1}\left (\frac {\sqrt [3]{3}+2 \sqrt [3]{2+x^3}}{3^{5/6}}\right )}{3^{5/6}}+\frac {\log \left (1-x^3\right )}{6 \sqrt [3]{3}}+\frac {\log \left (\sqrt [3]{3}-\sqrt [3]{2+x^3}\right )}{2 \sqrt [3]{3}}-\frac {\log \left (\sqrt [3]{3} x-\sqrt [3]{2+x^3}\right )}{\sqrt [3]{3}} \]

[Out]

-1/4*x^2*AppellF1(2/3,1,1/3,5/3,x^3,-1/2*x^3)*2^(2/3)+1/3*arctan(1/3*(3^(1/3)+2*(x^3+2)^(1/3))*3^(1/6))*3^(1/6
)+2/3*arctan(1/3*(1+2*3^(1/3)*x/(x^3+2)^(1/3))*3^(1/2))*3^(1/6)+1/18*ln(-x^3+1)*3^(2/3)+1/6*ln(3^(1/3)-(x^3+2)
^(1/3))*3^(2/3)-1/3*ln(3^(1/3)*x-(x^3+2)^(1/3))*3^(2/3)

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Rubi [A]
time = 0.10, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {2183, 384, 524, 455, 57, 631, 210, 31} \begin {gather*} -\frac {x^2 F_1\left (\frac {2}{3};1,\frac {1}{3};\frac {5}{3};x^3,-\frac {x^3}{2}\right )}{2 \sqrt [3]{2}}+\frac {2 \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{3} x}{\sqrt [3]{x^3+2}}+1}{\sqrt {3}}\right )}{3^{5/6}}+\frac {\text {ArcTan}\left (\frac {2 \sqrt [3]{x^3+2}+\sqrt [3]{3}}{3^{5/6}}\right )}{3^{5/6}}+\frac {\log \left (1-x^3\right )}{6 \sqrt [3]{3}}+\frac {\log \left (\sqrt [3]{3}-\sqrt [3]{x^3+2}\right )}{2 \sqrt [3]{3}}-\frac {\log \left (\sqrt [3]{3} x-\sqrt [3]{x^3+2}\right )}{\sqrt [3]{3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(x^2*AppellF1[2/3, 1, 1/3, 5/3, x^3, -1/2*x^3])/2^(1/3) + (2*ArcTan[(1 + (2*3^(1/3)*x)/(2 + x^3)^(1/3))/S
qrt[3]])/3^(5/6) + ArcTan[(3^(1/3) + 2*(2 + x^3)^(1/3))/3^(5/6)]/3^(5/6) + Log[1 - x^3]/(6*3^(1/3)) + Log[3^(1
/3) - (2 + x^3)^(1/3)]/(2*3^(1/3)) - Log[3^(1/3)*x - (2 + x^3)^(1/3)]/3^(1/3)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 384

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[
ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q
), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 524

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

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2183

Int[(Px_.)*((c_) + (d_.)*(x_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^3)^(p_.), x_Symbol] :> Dist[1/c^q, Int[E
xpandIntegrand[(c^3 - d^3*x^3)^q*(a + b*x^3)^p, Px/(c - d*x)^q, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] &&
PolyQ[Px, x] && EqQ[d^2 - c*e, 0] && ILtQ[q, 0] && RationalQ[p] && EqQ[Denominator[p], 3]

Rubi steps

\begin {align*} \int \frac {2+x}{\left (1+x+x^2\right ) \sqrt [3]{2+x^3}} \, dx &=\int \left (\frac {1-i \sqrt {3}}{\left (1-i \sqrt {3}+2 x\right ) \sqrt [3]{2+x^3}}+\frac {1+i \sqrt {3}}{\left (1+i \sqrt {3}+2 x\right ) \sqrt [3]{2+x^3}}\right ) \, dx\\ &=\left (1-i \sqrt {3}\right ) \int \frac {1}{\left (1-i \sqrt {3}+2 x\right ) \sqrt [3]{2+x^3}} \, dx+\left (1+i \sqrt {3}\right ) \int \frac {1}{\left (1+i \sqrt {3}+2 x\right ) \sqrt [3]{2+x^3}} \, dx\\ \end {align*}

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Mathematica [F]
time = 10.10, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2+x}{\left (1+x+x^2\right ) \sqrt [3]{2+x^3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \frac {2+x}{\left (x^{2}+x +1\right ) \left (x^{3}+2\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+x)/(x^2+x+1)/(x^3+2)^(1/3),x, algorithm="fricas")

[Out]

integral((x^3 + 2)^(2/3)*(x + 2)/(x^5 + x^4 + x^3 + 2*x^2 + 2*x + 2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x+2}{{\left (x^3+2\right )}^{1/3}\,\left (x^2+x+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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