\(\int \frac {(1-x)^2}{(1-x^3)^{4/3}} \, dx\) [105]

Optimal result

Integrand size = 19, antiderivative size = 39 \[ \int \frac {(1-x)^2}{\left (1-x^3\right )^{4/3}} \, dx=\frac {1+(1-2 x) x}{\sqrt [3]{1-x^3}}+x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},x^3\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1868, 12, 371} \[ \int \frac {(1-x)^2}{\left (1-x^3\right )^{4/3}} \, dx=x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},x^3\right )+\frac {(1-2 x) x+1}{\sqrt [3]{1-x^3}} \]

[In]

Int[(1 - x)^2/(1 - x^3)^(4/3),x]

[Out]

(1 + (1 - 2*x)*x)/(1 - x^3)^(1/3) + x^2*Hypergeometric2F1[1/3, 2/3, 5/3, x^3]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1868

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], i}, Simp[(a*Coeff[Pq, x, q] -
b*x*ExpandToSum[Pq - Coeff[Pq, x, q]*x^q, x])*((a + b*x^n)^(p + 1)/(a*b*n*(p + 1))), x] + Dist[1/(a*n*(p + 1))
, Int[Sum[(n*(p + 1) + i + 1)*Coeff[Pq, x, i]*x^i, {i, 0, q - 1}]*(a + b*x^n)^(p + 1), x], x] /; q == n - 1] /
; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {1+(1-2 x) x}{\sqrt [3]{1-x^3}}-\int -\frac {2 x}{\sqrt [3]{1-x^3}} \, dx \\ & = \frac {1+(1-2 x) x}{\sqrt [3]{1-x^3}}+2 \int \frac {x}{\sqrt [3]{1-x^3}} \, dx \\ & = \frac {1+(1-2 x) x}{\sqrt [3]{1-x^3}}+x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},x^3\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 10.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.10 \[ \int \frac {(1-x)^2}{\left (1-x^3\right )^{4/3}} \, dx=\frac {1}{\sqrt [3]{1-x^3}}+\frac {x}{\sqrt [3]{1-x^3}}-x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {4}{3},\frac {5}{3},x^3\right ) \]

[In]

Integrate[(1 - x)^2/(1 - x^3)^(4/3),x]

[Out]

(1 - x^3)^(-1/3) + x/(1 - x^3)^(1/3) - x^2*Hypergeometric2F1[2/3, 4/3, 5/3, x^3]

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.87

method	result	size
risch	\(-\frac {\left (-1+x \right ) \left (1+2 x \right )}{\left (-x^{3}+1\right )^{\frac {1}{3}}}+x^{2} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};x^{3}\right )\)	\(34\)
meijerg	\(\frac {x}{\left (-x^{3}+1\right )^{\frac {1}{3}}}-x^{2} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {2}{3},\frac {4}{3};\frac {5}{3};x^{3}\right )+\frac {x^{3} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (1,\frac {4}{3};2;x^{3}\right )}{3}\)	\(41\)

[In]

int((1-x)^2/(-x^3+1)^(4/3),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int \frac {(1-x)^2}{\left (1-x^3\right )^{4/3}} \, dx=\int { \frac {{\left (x - 1\right )}^{2}}{{\left (-x^{3} + 1\right )}^{\frac {4}{3}}} \,d x } \]

[In]

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

[Out]

integral((-x^3 + 1)^(2/3)/(x^4 + 2*x^3 + 3*x^2 + 2*x + 1), x)

Sympy [F]

\[ \int \frac {(1-x)^2}{\left (1-x^3\right )^{4/3}} \, dx=\int \frac {\left (x - 1\right )^{2}}{\left (- \left (x - 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac {4}{3}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {(1-x)^2}{\left (1-x^3\right )^{4/3}} \, dx=\int { \frac {{\left (x - 1\right )}^{2}}{{\left (-x^{3} + 1\right )}^{\frac {4}{3}}} \,d x } \]

[In]

integrate((1-x)^2/(-x^3+1)^(4/3),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {(1-x)^2}{\left (1-x^3\right )^{4/3}} \, dx=\int { \frac {{\left (x - 1\right )}^{2}}{{\left (-x^{3} + 1\right )}^{\frac {4}{3}}} \,d x } \]

[In]

integrate((1-x)^2/(-x^3+1)^(4/3),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(1-x)^2}{\left (1-x^3\right )^{4/3}} \, dx=\int \frac {{\left (x-1\right )}^2}{{\left (1-x^3\right )}^{4/3}} \,d x \]

[In]

int((x - 1)^2/(1 - x^3)^(4/3),x)

[Out]

int((x - 1)^2/(1 - x^3)^(4/3), x)