3.21.90 \(\int \frac {(1+x^3)^{2/3} (2+x^6)}{x^6 (-1+x^3)^2} \, dx\) [2090]
Optimal. Leaf size=151 \[ \frac {\left (1+x^3\right )^{2/3} \left (2+10 x^3-17 x^6\right )}{5 x^5 \left (-1+x^3\right )}+\frac {5\ 2^{2/3} \text {ArcTan}\left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{1+x^3}}\right )}{\sqrt {3}}-\frac {5}{3} 2^{2/3} \log \left (-2 x+2^{2/3} \sqrt [3]{1+x^3}\right )+\frac {5 \log \left (2 x^2+2^{2/3} x \sqrt [3]{1+x^3}+\sqrt [3]{2} \left (1+x^3\right )^{2/3}\right )}{3 \sqrt [3]{2}} \]
[Out]
1/5*(x^3+1)^(2/3)*(-17*x^6+10*x^3+2)/x^5/(x^3-1)+5/3*2^(2/3)*arctan(3^(1/2)*x/(x+2^(2/3)*(x^3+1)^(1/3)))*3^(1/
2)-5/3*2^(2/3)*ln(-2*x+2^(2/3)*(x^3+1)^(1/3))+5/6*ln(2*x^2+2^(2/3)*x*(x^3+1)^(1/3)+2^(1/3)*(x^3+1)^(2/3))*2^(2
/3)
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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(545\) vs. \(2(151)=302\).
time = 1.36, antiderivative size = 545, normalized size of antiderivative = 3.61, number of steps
used = 93, number of rules used = 31, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.240, Rules used = {6874, 2181,
386, 384, 480, 371, 455, 43, 57, 631, 210, 31, 478, 544, 245, 598, 502, 2174, 206, 648, 642, 2178,
2177, 270, 283, 2183, 1600, 21, 399, 495, 52} \begin {gather*} 2^{2/3} \sqrt {3} \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{2} x}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )-\frac {2\ 2^{2/3} \text {ArcTan}\left (\frac {1-\frac {2 \sqrt [3]{2} (x+1)}{\sqrt [3]{x^3+1}}}{\sqrt {3}}\right )}{\sqrt {3}}+2^{2/3} \sqrt {3} \text {ArcTan}\left (\frac {\frac {\sqrt [3]{2} (x+1)}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )-\frac {2^{2/3} \text {ArcTan}\left (\frac {\frac {\sqrt [3]{2} (x+1)}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {2\ 2^{2/3} \text {ArcTan}\left (\frac {2^{2/3} \sqrt [3]{x^3+1}+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {x \left (x^3+1\right )^{2/3}}{1-x^3}+\frac {2}{9} 2^{2/3} \log \left (1-x^3\right )-\frac {2}{27} 2^{2/3} \log \left (x^3-1\right )+\frac {\log \left (x^3-1\right )}{27 \sqrt [3]{2}}-\frac {1}{3} 2^{2/3} \log \left (\frac {2^{2/3} (x+1)^2}{\left (x^3+1\right )^{2/3}}-\frac {\sqrt [3]{2} (x+1)}{\sqrt [3]{x^3+1}}+1\right )+\frac {2}{3} 2^{2/3} \log \left (\frac {\sqrt [3]{2} (x+1)}{\sqrt [3]{x^3+1}}+1\right )+2^{2/3} \log \left (\sqrt [3]{2}-\sqrt [3]{x^3+1}\right )+\frac {2}{9} 2^{2/3} \log \left (\sqrt [3]{2} x-\sqrt [3]{x^3+1}\right )-\frac {31 \log \left (\sqrt [3]{2} x-\sqrt [3]{x^3+1}\right )}{9 \sqrt [3]{2}}+\frac {\log \left (-2^{2/3} \sqrt [3]{x^3+1}+x+1\right )}{\sqrt [3]{2}}-\frac {3 \log \left (2^{2/3} \sqrt [3]{x^3+1}-x-1\right )}{\sqrt [3]{2}}-\frac {2 \left (x^3+1\right )^{5/3}}{5 x^5}-\frac {2 \left (x^3+1\right )^{2/3}}{x^2}+\frac {\log \left (-(1-x)^2 (x+1)\right )}{\sqrt [3]{2}}-\frac {\log \left ((1-x)^2 (x+1)\right )}{3 \sqrt [3]{2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[((1 + x^3)^(2/3)*(2 + x^6))/(x^6*(-1 + x^3)^2),x]
[Out]
(-2*(1 + x^3)^(2/3))/x^2 + (x*(1 + x^3)^(2/3))/(1 - x^3) - (2*(1 + x^3)^(5/3))/(5*x^5) + 2^(2/3)*Sqrt[3]*ArcTa
n[(1 + (2*2^(1/3)*x)/(1 + x^3)^(1/3))/Sqrt[3]] - (2*2^(2/3)*ArcTan[(1 - (2*2^(1/3)*(1 + x))/(1 + x^3)^(1/3))/S
qrt[3]])/Sqrt[3] - (2^(2/3)*ArcTan[(1 + (2^(1/3)*(1 + x))/(1 + x^3)^(1/3))/Sqrt[3]])/Sqrt[3] + 2^(2/3)*Sqrt[3]
*ArcTan[(1 + (2^(1/3)*(1 + x))/(1 + x^3)^(1/3))/Sqrt[3]] + (2*2^(2/3)*ArcTan[(1 + 2^(2/3)*(1 + x^3)^(1/3))/Sqr
t[3]])/Sqrt[3] + Log[-((1 - x)^2*(1 + x))]/2^(1/3) - Log[(1 - x)^2*(1 + x)]/(3*2^(1/3)) + (2*2^(2/3)*Log[1 - x
^3])/9 + Log[-1 + x^3]/(27*2^(1/3)) - (2*2^(2/3)*Log[-1 + x^3])/27 - (2^(2/3)*Log[1 + (2^(2/3)*(1 + x)^2)/(1 +
x^3)^(2/3) - (2^(1/3)*(1 + x))/(1 + x^3)^(1/3)])/3 + (2*2^(2/3)*Log[1 + (2^(1/3)*(1 + x))/(1 + x^3)^(1/3)])/3
+ 2^(2/3)*Log[2^(1/3) - (1 + x^3)^(1/3)] - (31*Log[2^(1/3)*x - (1 + x^3)^(1/3)])/(9*2^(1/3)) + (2*2^(2/3)*Log
[2^(1/3)*x - (1 + x^3)^(1/3)])/9 + Log[1 + x - 2^(2/3)*(1 + x^3)^(1/3)]/2^(1/3) - (3*Log[-1 - x + 2^(2/3)*(1 +
x^3)^(1/3)])/2^(1/3)
Rule 21
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])
Rule 31
Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]
Rule 43
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] && !IntegerQ[n] && GtQ[n, 0]
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 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 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 245
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 270
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]
Rule 283
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, 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 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 386
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[(-x)*(a + b*x^n)^(p + 1)*(
(c + d*x^n)^q/(a*n*(p + 1))), x] - Dist[c*(q/(a*(p + 1))), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 0] && NeQ[p, -1]
Rule 399
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Dist[b/d, Int[(a + b*x^n)^(p - 1), x]
, x] - Dist[(b*c - a*d)/d, Int[(a + b*x^n)^(p - 1)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c
- a*d, 0] && EqQ[n*(p - 1) + 1, 0] && IntegerQ[n]
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 478
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[e^(n - 1
)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*n*(p + 1))), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] &
& IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Rule 480
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-(e*x)^
(m + 1))*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*e*n*(p + 1))), x] + Dist[1/(a*n*(p + 1)), Int[(e*x)^m*(a + b*x^
n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m + n*(p + 1) + 1) + d*(m + n*(p + q + 1) + 1)*x^n, x], x], x] /; FreeQ
[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[0, q, 1] && IntBinomialQ[a, b,
c, d, e, m, n, p, q, x]
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 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 544
Int[(((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Dist[f/d,
Int[(a + b*x^n)^p, x], x] + Dist[(d*e - c*f)/d, Int[(a + b*x^n)^p/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,
f, p, n}, x]
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 1600
Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]
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]
Rule 2177
Int[((e_.) + (f_.)*(x_))/(((c_.) + (d_.)*(x_))*((a_) + (b_.)*(x_)^3)^(1/3)), x_Symbol] :> Dist[f/d, Int[1/(a +
b*x^3)^(1/3), x], x] + Dist[(d*e - c*f)/d, Int[1/((c + d*x)*(a + b*x^3)^(1/3)), x], x] /; FreeQ[{a, b, c, d,
e, f}, x]
Rule 2178
Int[((a_) + (b_.)*(x_)^3)^(2/3)/((c_) + (d_.)*(x_)), x_Symbol] :> Simp[(a + b*x^3)^(2/3)/(2*d), x] + (Dist[1/d
^2, Int[(a*d^2 + b*c^2*x)/((c + d*x)*(a + b*x^3)^(1/3)), x], x] - Dist[b*(c/d^2), Int[x/(a + b*x^3)^(1/3), x],
x]) /; FreeQ[{a, b, c, d}, x]
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]
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]
Rule 6874
Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]
Rubi steps
\begin {align*} \int \frac {\left (1+x^3\right )^{2/3} \left (2+x^6\right )}{x^6 \left (-1+x^3\right )^2} \, dx &=\int \left (\frac {\left (1+x^3\right )^{2/3}}{3 (-1+x)^2}-\frac {2 \left (1+x^3\right )^{2/3}}{-1+x}+\frac {2 \left (1+x^3\right )^{2/3}}{x^6}+\frac {4 \left (1+x^3\right )^{2/3}}{x^3}+\frac {(1+x) \left (1+x^3\right )^{2/3}}{\left (1+x+x^2\right )^2}+\frac {(11+6 x) \left (1+x^3\right )^{2/3}}{3 \left (1+x+x^2\right )}\right ) \, dx\\ &=\frac {1}{3} \int \frac {\left (1+x^3\right )^{2/3}}{(-1+x)^2} \, dx+\frac {1}{3} \int \frac {(11+6 x) \left (1+x^3\right )^{2/3}}{1+x+x^2} \, dx-2 \int \frac {\left (1+x^3\right )^{2/3}}{-1+x} \, dx+2 \int \frac {\left (1+x^3\right )^{2/3}}{x^6} \, dx+4 \int \frac {\left (1+x^3\right )^{2/3}}{x^3} \, dx+\int \frac {(1+x) \left (1+x^3\right )^{2/3}}{\left (1+x+x^2\right )^2} \, dx\\ &=-\frac {2 \left (1+x^3\right )^{2/3}}{x^2}-\frac {2 \left (1+x^3\right )^{5/3}}{5 x^5}+\frac {1}{3} \int \frac {\left (1+x^3\right )^{2/3}}{(-1+x)^2} \, dx+\frac {1}{3} \int \left (\frac {\left (6-\frac {16 i}{\sqrt {3}}\right ) \left (1+x^3\right )^{2/3}}{1-i \sqrt {3}+2 x}+\frac {\left (6+\frac {16 i}{\sqrt {3}}\right ) \left (1+x^3\right )^{2/3}}{1+i \sqrt {3}+2 x}\right ) \, dx-2 \int \frac {\left (1+x^3\right )^{2/3}}{-1+x} \, dx+4 \int \frac {1}{\sqrt [3]{1+x^3}} \, dx+\int \left (\frac {\left (1+x^3\right )^{2/3}}{\left (1+x+x^2\right )^2}+\frac {x \left (1+x^3\right )^{2/3}}{\left (1+x+x^2\right )^2}\right ) \, dx\\ &=-\frac {2 \left (1+x^3\right )^{2/3}}{x^2}-\frac {2 \left (1+x^3\right )^{5/3}}{5 x^5}+\frac {4 \tan ^{-1}\left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-2 \log \left (-x+\sqrt [3]{1+x^3}\right )+\frac {1}{3} \int \frac {\left (1+x^3\right )^{2/3}}{(-1+x)^2} \, dx-2 \int \frac {\left (1+x^3\right )^{2/3}}{-1+x} \, dx+\frac {1}{9} \left (2 \left (9-8 i \sqrt {3}\right )\right ) \int \frac {\left (1+x^3\right )^{2/3}}{1-i \sqrt {3}+2 x} \, dx+\frac {1}{9} \left (2 \left (9+8 i \sqrt {3}\right )\right ) \int \frac {\left (1+x^3\right )^{2/3}}{1+i \sqrt {3}+2 x} \, dx+\int \frac {\left (1+x^3\right )^{2/3}}{\left (1+x+x^2\right )^2} \, dx+\int \frac {x \left (1+x^3\right )^{2/3}}{\left (1+x+x^2\right )^2} \, dx\\ &=-\frac {2 \left (1+x^3\right )^{2/3}}{x^2}-\frac {2 \left (1+x^3\right )^{5/3}}{5 x^5}+\frac {4 \tan ^{-1}\left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-2 \log \left (-x+\sqrt [3]{1+x^3}\right )+\frac {1}{3} \int \frac {\left (1+x^3\right )^{2/3}}{(-1+x)^2} \, dx-2 \int \frac {\left (1+x^3\right )^{2/3}}{-1+x} \, dx+\frac {1}{9} \left (2 \left (9-8 i \sqrt {3}\right )\right ) \int \frac {\left (1+x^3\right )^{2/3}}{1-i \sqrt {3}+2 x} \, dx+\frac {1}{9} \left (2 \left (9+8 i \sqrt {3}\right )\right ) \int \frac {\left (1+x^3\right )^{2/3}}{1+i \sqrt {3}+2 x} \, dx+\int \left (-\frac {2 \left (-1+i \sqrt {3}\right ) \left (1+x^3\right )^{2/3}}{3 \left (-1+i \sqrt {3}-2 x\right )^2}-\frac {2 i \left (1+x^3\right )^{2/3}}{3 \sqrt {3} \left (-1+i \sqrt {3}-2 x\right )}-\frac {2 \left (-1-i \sqrt {3}\right ) \left (1+x^3\right )^{2/3}}{3 \left (1+i \sqrt {3}+2 x\right )^2}-\frac {2 i \left (1+x^3\right )^{2/3}}{3 \sqrt {3} \left (1+i \sqrt {3}+2 x\right )}\right ) \, dx+\int \left (-\frac {4 \left (1+x^3\right )^{2/3}}{3 \left (-1+i \sqrt {3}-2 x\right )^2}+\frac {4 i \left (1+x^3\right )^{2/3}}{3 \sqrt {3} \left (-1+i \sqrt {3}-2 x\right )}-\frac {4 \left (1+x^3\right )^{2/3}}{3 \left (1+i \sqrt {3}+2 x\right )^2}+\frac {4 i \left (1+x^3\right )^{2/3}}{3 \sqrt {3} \left (1+i \sqrt {3}+2 x\right )}\right ) \, dx\\ &=-\frac {2 \left (1+x^3\right )^{2/3}}{x^2}-\frac {2 \left (1+x^3\right )^{5/3}}{5 x^5}+\frac {4 \tan ^{-1}\left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-2 \log \left (-x+\sqrt [3]{1+x^3}\right )+\frac {1}{3} \int \frac {\left (1+x^3\right )^{2/3}}{(-1+x)^2} \, dx-\frac {4}{3} \int \frac {\left (1+x^3\right )^{2/3}}{\left (-1+i \sqrt {3}-2 x\right )^2} \, dx-\frac {4}{3} \int \frac {\left (1+x^3\right )^{2/3}}{\left (1+i \sqrt {3}+2 x\right )^2} \, dx-2 \int \frac {\left (1+x^3\right )^{2/3}}{-1+x} \, dx-\frac {(2 i) \int \frac {\left (1+x^3\right )^{2/3}}{-1+i \sqrt {3}-2 x} \, dx}{3 \sqrt {3}}-\frac {(2 i) \int \frac {\left (1+x^3\right )^{2/3}}{1+i \sqrt {3}+2 x} \, dx}{3 \sqrt {3}}+\frac {(4 i) \int \frac {\left (1+x^3\right )^{2/3}}{-1+i \sqrt {3}-2 x} \, dx}{3 \sqrt {3}}+\frac {(4 i) \int \frac {\left (1+x^3\right )^{2/3}}{1+i \sqrt {3}+2 x} \, dx}{3 \sqrt {3}}+\frac {1}{3} \left (2 \left (1-i \sqrt {3}\right )\right ) \int \frac {\left (1+x^3\right )^{2/3}}{\left (-1+i \sqrt {3}-2 x\right )^2} \, dx+\frac {1}{3} \left (2 \left (1+i \sqrt {3}\right )\right ) \int \frac {\left (1+x^3\right )^{2/3}}{\left (1+i \sqrt {3}+2 x\right )^2} \, dx+\frac {1}{9} \left (2 \left (9-8 i \sqrt {3}\right )\right ) \int \frac {\left (1+x^3\right )^{2/3}}{1-i \sqrt {3}+2 x} \, dx+\frac {1}{9} \left (2 \left (9+8 i \sqrt {3}\right )\right ) \int \frac {\left (1+x^3\right )^{2/3}}{1+i \sqrt {3}+2 x} \, dx\\ \end {align*}
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Mathematica [A]
time = 0.39, size = 151, normalized size = 1.00 \begin {gather*} \frac {\left (1+x^3\right )^{2/3} \left (2+10 x^3-17 x^6\right )}{5 x^5 \left (-1+x^3\right )}+\frac {5\ 2^{2/3} \text {ArcTan}\left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{1+x^3}}\right )}{\sqrt {3}}-\frac {5}{3} 2^{2/3} \log \left (-2 x+2^{2/3} \sqrt [3]{1+x^3}\right )+\frac {5 \log \left (2 x^2+2^{2/3} x \sqrt [3]{1+x^3}+\sqrt [3]{2} \left (1+x^3\right )^{2/3}\right )}{3 \sqrt [3]{2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[((1 + x^3)^(2/3)*(2 + x^6))/(x^6*(-1 + x^3)^2),x]
[Out]
((1 + x^3)^(2/3)*(2 + 10*x^3 - 17*x^6))/(5*x^5*(-1 + x^3)) + (5*2^(2/3)*ArcTan[(Sqrt[3]*x)/(x + 2^(2/3)*(1 + x
^3)^(1/3))])/Sqrt[3] - (5*2^(2/3)*Log[-2*x + 2^(2/3)*(1 + x^3)^(1/3)])/3 + (5*Log[2*x^2 + 2^(2/3)*x*(1 + x^3)^
(1/3) + 2^(1/3)*(1 + x^3)^(2/3)])/(3*2^(1/3))
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
3.
time = 9.71, size = 769, normalized size = 5.09
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| method |
result |
size |
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| trager |
\(\text {Expression too large to display}\) |
\(769\) |
| risch |
\(\text {Expression too large to display}\) |
\(939\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((x^3+1)^(2/3)*(x^6+2)/x^6/(x^3-1)^2,x,method=_RETURNVERBOSE)
[Out]
-1/5*(17*x^6-10*x^3-2)/(x^3-1)/x^5*(x^3+1)^(2/3)+2240*RootOf(RootOf(_Z^3+4)^2+1344*_Z*RootOf(_Z^3+4)+1806336*_
Z^2)*ln(-(1645572096*RootOf(RootOf(_Z^3+4)^2+1344*_Z*RootOf(_Z^3+4)+1806336*_Z^2)^2*RootOf(_Z^3+4)^2*x^3+66057
6*RootOf(RootOf(_Z^3+4)^2+1344*_Z*RootOf(_Z^3+4)+1806336*_Z^2)*RootOf(_Z^3+4)^3*x^3-27740160*(x^3+1)^(2/3)*Roo
tOf(RootOf(_Z^3+4)^2+1344*_Z*RootOf(_Z^3+4)+1806336*_Z^2)*RootOf(_Z^3+4)^2*x+37025856*RootOf(RootOf(_Z^3+4)^2+
1344*_Z*RootOf(_Z^3+4)+1806336*_Z^2)*(x^3+1)^(1/3)*RootOf(_Z^3+4)*x^2-41280*(x^3+1)^(1/3)*RootOf(_Z^3+4)^2*x^2
-13164576768*RootOf(RootOf(_Z^3+4)^2+1344*_Z*RootOf(_Z^3+4)+1806336*_Z^2)^2*RootOf(_Z^3+4)^2-5284608*RootOf(Ro
otOf(_Z^3+4)^2+1344*_Z*RootOf(_Z^3+4)+1806336*_Z^2)*RootOf(_Z^3+4)^3+139579776*RootOf(RootOf(_Z^3+4)^2+1344*_Z
*RootOf(_Z^3+4)+1806336*_Z^2)*x^3+56031*RootOf(_Z^3+4)*x^3+137658*x*(x^3+1)^(2/3)+61219200*RootOf(RootOf(_Z^3+
4)^2+1344*_Z*RootOf(_Z^3+4)+1806336*_Z^2)+24575*RootOf(_Z^3+4))/(-1+x)/(x^2+x+1))+5/3*RootOf(_Z^3+4)*ln(-(1374
442417152*RootOf(RootOf(_Z^3+4)^2+1344*_Z*RootOf(_Z^3+4)+1806336*_Z^2)^2*RootOf(_Z^3+4)^2*x^3-946523424*RootOf
(RootOf(_Z^3+4)^2+1344*_Z*RootOf(_Z^3+4)+1806336*_Z^2)*RootOf(_Z^3+4)^3*x^3+73577581056*(x^3+1)^(2/3)*RootOf(R
ootOf(_Z^3+4)^2+1344*_Z*RootOf(_Z^3+4)+1806336*_Z^2)*RootOf(_Z^3+4)^2*x+50046484320*RootOf(RootOf(_Z^3+4)^2+13
44*_Z*RootOf(_Z^3+4)+1806336*_Z^2)*(x^3+1)^(1/3)*RootOf(_Z^3+4)*x^2+109490448*(x^3+1)^(1/3)*RootOf(_Z^3+4)^2*x
^2-10995539337216*RootOf(RootOf(_Z^3+4)^2+1344*_Z*RootOf(_Z^3+4)+1806336*_Z^2)^2*RootOf(_Z^3+4)^2+7572187392*R
ootOf(RootOf(_Z^3+4)^2+1344*_Z*RootOf(_Z^3+4)+1806336*_Z^2)*RootOf(_Z^3+4)^3-120672771744*RootOf(RootOf(_Z^3+4
)^2+1344*_Z*RootOf(_Z^3+4)+1806336*_Z^2)*x^3+83102503*RootOf(_Z^3+4)*x^3-144506961*x*(x^3+1)^(2/3)-18407710944
*RootOf(RootOf(_Z^3+4)^2+1344*_Z*RootOf(_Z^3+4)+1806336*_Z^2)+12676653*RootOf(_Z^3+4))/(-1+x)/(x^2+x+1))
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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((x^3+1)^(2/3)*(x^6+2)/x^6/(x^3-1)^2,x, algorithm="maxima")
[Out]
integrate((x^6 + 2)*(x^3 + 1)^(2/3)/((x^3 - 1)^2*x^6), x)
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 297 vs.
\(2 (117) = 234\).
time = 2.10, size = 297, normalized size = 1.97 \begin {gather*} -\frac {50 \, \sqrt {3} \left (-4\right )^{\frac {1}{3}} {\left (x^{8} - x^{5}\right )} \arctan \left (\frac {3 \, \sqrt {3} \left (-4\right )^{\frac {2}{3}} {\left (5 \, x^{7} - 4 \, x^{4} - x\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}} + 6 \, \sqrt {3} \left (-4\right )^{\frac {1}{3}} {\left (19 \, x^{8} + 16 \, x^{5} + x^{2}\right )} {\left (x^{3} + 1\right )}^{\frac {1}{3}} - \sqrt {3} {\left (71 \, x^{9} + 111 \, x^{6} + 33 \, x^{3} + 1\right )}}{3 \, {\left (109 \, x^{9} + 105 \, x^{6} + 3 \, x^{3} - 1\right )}}\right ) - 50 \, \left (-4\right )^{\frac {1}{3}} {\left (x^{8} - x^{5}\right )} \log \left (\frac {3 \, \left (-4\right )^{\frac {2}{3}} {\left (x^{3} + 1\right )}^{\frac {1}{3}} x^{2} - 6 \, {\left (x^{3} + 1\right )}^{\frac {2}{3}} x + \left (-4\right )^{\frac {1}{3}} {\left (x^{3} - 1\right )}}{x^{3} - 1}\right ) + 25 \, \left (-4\right )^{\frac {1}{3}} {\left (x^{8} - x^{5}\right )} \log \left (-\frac {6 \, \left (-4\right )^{\frac {1}{3}} {\left (5 \, x^{4} + x\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}} - \left (-4\right )^{\frac {2}{3}} {\left (19 \, x^{6} + 16 \, x^{3} + 1\right )} - 24 \, {\left (2 \, x^{5} + x^{2}\right )} {\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x^{6} - 2 \, x^{3} + 1}\right ) + 18 \, {\left (17 \, x^{6} - 10 \, x^{3} - 2\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{90 \, {\left (x^{8} - x^{5}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((x^3+1)^(2/3)*(x^6+2)/x^6/(x^3-1)^2,x, algorithm="fricas")
[Out]
-1/90*(50*sqrt(3)*(-4)^(1/3)*(x^8 - x^5)*arctan(1/3*(3*sqrt(3)*(-4)^(2/3)*(5*x^7 - 4*x^4 - x)*(x^3 + 1)^(2/3)
+ 6*sqrt(3)*(-4)^(1/3)*(19*x^8 + 16*x^5 + x^2)*(x^3 + 1)^(1/3) - sqrt(3)*(71*x^9 + 111*x^6 + 33*x^3 + 1))/(109
*x^9 + 105*x^6 + 3*x^3 - 1)) - 50*(-4)^(1/3)*(x^8 - x^5)*log((3*(-4)^(2/3)*(x^3 + 1)^(1/3)*x^2 - 6*(x^3 + 1)^(
2/3)*x + (-4)^(1/3)*(x^3 - 1))/(x^3 - 1)) + 25*(-4)^(1/3)*(x^8 - x^5)*log(-(6*(-4)^(1/3)*(5*x^4 + x)*(x^3 + 1)
^(2/3) - (-4)^(2/3)*(19*x^6 + 16*x^3 + 1) - 24*(2*x^5 + x^2)*(x^3 + 1)^(1/3))/(x^6 - 2*x^3 + 1)) + 18*(17*x^6
- 10*x^3 - 2)*(x^3 + 1)^(2/3))/(x^8 - x^5)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (x + 1\right ) \left (x^{2} - x + 1\right )\right )^{\frac {2}{3}} \left (x^{6} + 2\right )}{x^{6} \left (x - 1\right )^{2} \left (x^{2} + x + 1\right )^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((x**3+1)**(2/3)*(x**6+2)/x**6/(x**3-1)**2,x)
[Out]
Integral(((x + 1)*(x**2 - x + 1))**(2/3)*(x**6 + 2)/(x**6*(x - 1)**2*(x**2 + x + 1)**2), 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((x^3+1)^(2/3)*(x^6+2)/x^6/(x^3-1)^2,x, algorithm="giac")
[Out]
integrate((x^6 + 2)*(x^3 + 1)^(2/3)/((x^3 - 1)^2*x^6), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (x^3+1\right )}^{2/3}\,\left (x^6+2\right )}{x^6\,{\left (x^3-1\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((x^3 + 1)^(2/3)*(x^6 + 2))/(x^6*(x^3 - 1)^2),x)
[Out]
int(((x^3 + 1)^(2/3)*(x^6 + 2))/(x^6*(x^3 - 1)^2), x)
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