3.1.60 \(\int \frac {\sqrt [3]{1-x^3}}{2+x} \, dx\) [60]
Optimal. Leaf size=232 \[ \sqrt [3]{1-x^3}+\frac {1}{2} x F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};x^3,-\frac {x^3}{8}\right )-\frac {2 \tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\sqrt [6]{3} \tan ^{-1}\left (\frac {1-\frac {3^{2/3} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )-\sqrt [6]{3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1-x^3}}{3 \sqrt [6]{3}}\right )-\frac {\log \left (8+x^3\right )}{\sqrt [3]{3}}+\frac {1}{2} 3^{2/3} \log \left (3^{2/3}-\sqrt [3]{1-x^3}\right )-\log \left (-x-\sqrt [3]{1-x^3}\right )+\frac {1}{2} 3^{2/3} \log \left (-\frac {1}{2} 3^{2/3} x-\sqrt [3]{1-x^3}\right ) \]
[Out]
(-x^3+1)^(1/3)+1/2*x*AppellF1(1/3,-1/3,1,4/3,x^3,-1/8*x^3)-3^(1/6)*arctan(2/9*(-x^3+1)^(1/3)*3^(5/6)+1/3*3^(1/
2))+3^(1/6)*arctan(1/3*(1-3^(2/3)*x/(-x^3+1)^(1/3))*3^(1/2))-1/3*ln(x^3+8)*3^(2/3)+1/2*3^(2/3)*ln(3^(2/3)-(-x^
3+1)^(1/3))-ln(-x-(-x^3+1)^(1/3))+1/2*3^(2/3)*ln(-1/2*3^(2/3)*x-(-x^3+1)^(1/3))-2/3*arctan(1/3*(1-2*x/(-x^3+1)
^(1/3))*3^(1/2))*3^(1/2)
________________________________________________________________________________________
Rubi [A]
time = 0.12, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps
used = 12, number of rules used = 11, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.647, Rules used = {2181, 440,
495, 337, 503, 455, 52, 59, 631, 210, 31} \begin {gather*} \frac {1}{2} x F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};x^3,-\frac {x^3}{8}\right )-\frac {2 \text {ArcTan}\left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\sqrt [6]{3} \text {ArcTan}\left (\frac {1-\frac {3^{2/3} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )-\sqrt [6]{3} \text {ArcTan}\left (\frac {2 \sqrt [3]{1-x^3}}{3 \sqrt [6]{3}}+\frac {1}{\sqrt {3}}\right )+\sqrt [3]{1-x^3}-\frac {\log \left (x^3+8\right )}{\sqrt [3]{3}}+\frac {1}{2} 3^{2/3} \log \left (3^{2/3}-\sqrt [3]{1-x^3}\right )-\log \left (-\sqrt [3]{1-x^3}-x\right )+\frac {1}{2} 3^{2/3} \log \left (-\sqrt [3]{1-x^3}-\frac {1}{2} 3^{2/3} x\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(1 - x^3)^(1/3)/(2 + x),x]
[Out]
(1 - x^3)^(1/3) + (x*AppellF1[1/3, -1/3, 1, 4/3, x^3, -1/8*x^3])/2 - (2*ArcTan[(1 - (2*x)/(1 - x^3)^(1/3))/Sqr
t[3]])/Sqrt[3] + 3^(1/6)*ArcTan[(1 - (3^(2/3)*x)/(1 - x^3)^(1/3))/Sqrt[3]] - 3^(1/6)*ArcTan[1/Sqrt[3] + (2*(1
- x^3)^(1/3))/(3*3^(1/6))] - Log[8 + x^3]/3^(1/3) + (3^(2/3)*Log[3^(2/3) - (1 - x^3)^(1/3)])/2 - Log[-x - (1 -
x^3)^(1/3)] + (3^(2/3)*Log[-1/2*(3^(2/3)*x) - (1 - x^3)^(1/3)])/2
Rule 31
Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]
Rule 52
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]
Rule 59
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), 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 337
Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Simp[-ArcTan[(1 + 2*q*(x/(a + b*x^3)^(
1/3)))/Sqrt[3]]/(Sqrt[3]*q^2), x] - Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*q^2), x]] /; FreeQ[{a, b}, x]
Rule 440
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 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 495
Int[((x_)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Dist[b/d, Int[x*(a + b*x^n)^(p
- 1), x], x] - Dist[(b*c - a*d)/d, Int[x*((a + b*x^n)^(p - 1)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d}, x]
&& NeQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[p, 0] && IntBinomialQ[a, b, c, d, 1, 1, n, p, -1, x]
Rule 503
Int[(x_)/(((a_) + (b_.)*(x_)^3)^(2/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Si
mp[-ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q^2), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/
(2*c*q^2), x] + Simp[Log[c + d*x^3]/(6*c*q^2), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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 2181
Int[(Px_.)*((c_) + (d_.)*(x_))^(q_)*((a_) + (b_.)*(x_)^3)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c^3 + d^3*x
^3)^q*(a + b*x^3)^p, Px/(c^2 - c*d*x + d^2*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p}, x] && PolyQ[Px, x] && ILtQ
[q, 0] && RationalQ[p] && EqQ[Denominator[p], 3]
Rubi steps
\begin {align*} \int \frac {\sqrt [3]{1-x^3}}{2+x} \, dx &=\int \frac {\sqrt [3]{1-x^3}}{2+x} \, dx\\ \end {align*}
________________________________________________________________________________________
Mathematica [F]
time = 45.67, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [3]{1-x^3}}{2+x} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
Integrate[(1 - x^3)^(1/3)/(2 + x),x]
[Out]
Integrate[(1 - x^3)^(1/3)/(2 + x), x]
________________________________________________________________________________________
Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (-x^{3}+1\right )^{\frac {1}{3}}}{2+x}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((-x^3+1)^(1/3)/(2+x),x)
[Out]
int((-x^3+1)^(1/3)/(2+x),x)
________________________________________________________________________________________
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((-x^3+1)^(1/3)/(2+x),x, algorithm="maxima")
[Out]
integrate((-x^3 + 1)^(1/3)/(x + 2), x)
________________________________________________________________________________________
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)^(1/3)/(2+x),x, algorithm="fricas")
[Out]
Exception raised: TypeError >> Error detected within library code: integrate: implementation incomplete (re
sidue poly has multiple non-linear factors)
________________________________________________________________________________________
Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [3]{- \left (x - 1\right ) \left (x^{2} + x + 1\right )}}{x + 2}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-x**3+1)**(1/3)/(2+x),x)
[Out]
Integral((-(x - 1)*(x**2 + x + 1))**(1/3)/(x + 2), x)
________________________________________________________________________________________
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((-x^3+1)^(1/3)/(2+x),x, algorithm="giac")
[Out]
integrate((-x^3 + 1)^(1/3)/(x + 2), x)
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (1-x^3\right )}^{1/3}}{x+2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((1 - x^3)^(1/3)/(x + 2),x)
[Out]
int((1 - x^3)^(1/3)/(x + 2), x)
________________________________________________________________________________________