3.7.56 \(\int \frac {1}{x^2 (1-x^3)^{4/3} (1+x^3)} \, dx\) [656]

Optimal. Leaf size=292 \[ \frac {1}{2 x \sqrt [3]{1-x^3}}-\frac {3 \left (1-x^3\right )^{2/3}}{2 x}-\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}-\frac {\tan ^{-1}\left (\frac {1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{4 \sqrt [3]{2} \sqrt {3}}-\frac {3}{4} x^2 \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};x^3\right )-\frac {\log \left ((1-x) (1+x)^2\right )}{24 \sqrt [3]{2}}-\frac {\log \left (1+\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{12 \sqrt [3]{2}}+\frac {\log \left (1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{6 \sqrt [3]{2}}+\frac {\log \left (-1+x+2^{2/3} \sqrt [3]{1-x^3}\right )}{8 \sqrt [3]{2}} \]

[Out]

1/2/x/(-x^3+1)^(1/3)-3/2*(-x^3+1)^(2/3)/x-3/4*x^2*hypergeom([1/3, 2/3],[5/3],x^3)-1/48*ln((1-x)*(1+x)^2)*2^(2/
3)-1/24*ln(1+2^(2/3)*(1-x)^2/(-x^3+1)^(2/3)-2^(1/3)*(1-x)/(-x^3+1)^(1/3))*2^(2/3)+1/12*ln(1+2^(1/3)*(1-x)/(-x^
3+1)^(1/3))*2^(2/3)+1/16*ln(-1+x+2^(2/3)*(-x^3+1)^(1/3))*2^(2/3)-1/12*arctan(1/3*(1-2*2^(1/3)*(1-x)/(-x^3+1)^(
1/3))*3^(1/2))*2^(2/3)*3^(1/2)-1/24*arctan(1/3*(1+2^(1/3)*(1-x)/(-x^3+1)^(1/3))*3^(1/2))*2^(2/3)*3^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 12, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {483, 597, 598, 371, 502, 2174, 206, 31, 648, 631, 210, 642} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}-\frac {\text {ArcTan}\left (\frac {\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1}{\sqrt {3}}\right )}{4 \sqrt [3]{2} \sqrt {3}}-\frac {3}{4} x^2 \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};x^3\right )-\frac {3 \left (1-x^3\right )^{2/3}}{2 x}+\frac {1}{2 \sqrt [3]{1-x^3} x}-\frac {\log \left (\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )}{12 \sqrt [3]{2}}+\frac {\log \left (\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )}{6 \sqrt [3]{2}}+\frac {\log \left (2^{2/3} \sqrt [3]{1-x^3}+x-1\right )}{8 \sqrt [3]{2}}-\frac {\log \left ((1-x) (x+1)^2\right )}{24 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1/(2*x*(1 - x^3)^(1/3)) - (3*(1 - x^3)^(2/3))/(2*x) - ArcTan[(1 - (2*2^(1/3)*(1 - x))/(1 - x^3)^(1/3))/Sqrt[3]
]/(2*2^(1/3)*Sqrt[3]) - ArcTan[(1 + (2^(1/3)*(1 - x))/(1 - x^3)^(1/3))/Sqrt[3]]/(4*2^(1/3)*Sqrt[3]) - (3*x^2*H
ypergeometric2F1[1/3, 2/3, 5/3, x^3])/4 - Log[(1 - x)*(1 + x)^2]/(24*2^(1/3)) - Log[1 + (2^(2/3)*(1 - x)^2)/(1
- x^3)^(2/3) - (2^(1/3)*(1 - x))/(1 - x^3)^(1/3)]/(12*2^(1/3)) + Log[1 + (2^(1/3)*(1 - x))/(1 - x^3)^(1/3)]/(
6*2^(1/3)) + Log[-1 + x + 2^(2/3)*(1 - x^3)^(1/3)]/(8*2^(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 206

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

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 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 483

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*(e*
x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*e*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a*d)
*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n
*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ
[p, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 502

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

Rule 597

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*g*(m + 1))), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 598

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Sy
mbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, p}, x] && IGtQ[n, 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 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 2174

Int[1/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^3)^(1/3)), x_Symbol] :> Simp[Sqrt[3]*(ArcTan[(1 - 2^(1/3)*Rt[b,
3]*((c - d*x)/(d*(a + b*x^3)^(1/3))))/Sqrt[3]]/(2^(4/3)*Rt[b, 3]*c)), x] + (Simp[Log[(c + d*x)^2*(c - d*x)]/(2
^(7/3)*Rt[b, 3]*c), x] - Simp[(3*Log[Rt[b, 3]*(c - d*x) + 2^(2/3)*d*(a + b*x^3)^(1/3)])/(2^(7/3)*Rt[b, 3]*c),
x]) /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^3 + a*d^3, 0]

Rubi steps

\begin {align*} \int \frac {1}{x^2 \left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx &=-\frac {F_1\left (-\frac {1}{3};\frac {4}{3},1;\frac {2}{3};x^3,-x^3\right )}{x}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 20.06, size = 76, normalized size = 0.26 \begin {gather*} \frac {-2+3 x^3}{2 x \sqrt [3]{1-x^3}}-x^2 F_1\left (\frac {2}{3};\frac {1}{3},1;\frac {5}{3};x^3,-x^3\right )-\frac {3}{10} x^5 F_1\left (\frac {5}{3};\frac {1}{3},1;\frac {8}{3};x^3,-x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2 + 3*x^3)/(2*x*(1 - x^3)^(1/3)) - x^2*AppellF1[2/3, 1/3, 1, 5/3, x^3, -x^3] - (3*x^5*AppellF1[5/3, 1/3, 1,
8/3, x^3, -x^3])/10

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{2} \left (-x^{3}+1\right )^{\frac {4}{3}} \left (x^{3}+1\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral((-x^3 + 1)^(2/3)/(x^11 - x^8 - x^5 + x^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{2} \left (- \left (x - 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac {4}{3}} \left (x + 1\right ) \left (x^{2} - x + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{x^2\,{\left (1-x^3\right )}^{4/3}\,\left (x^3+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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