\(\int \frac {(1+x^2) \sqrt [3]{-x+2 x^3}}{x^2 (1+x^4)} \, dx\) [1699]

Optimal result

Integrand size = 29, antiderivative size = 114 \[ \int \frac {\left (1+x^2\right ) \sqrt [3]{-x+2 x^3}}{x^2 \left (1+x^4\right )} \, dx=-\frac {3 \sqrt [3]{-x+2 x^3}}{2 x}+\frac {1}{4} \text {RootSum}\left [5-4 \text {$\#$1}^3+\text {$\#$1}^6\&,\frac {-5 \log (x)+5 \log \left (\sqrt [3]{-x+2 x^3}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^3-\log \left (\sqrt [3]{-x+2 x^3}-x \text {$\#$1}\right ) \text {$\#$1}^3}{-2 \text {$\#$1}^2+\text {$\#$1}^5}\&\right ] \]

[Out]

Unintegrable

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.51 (sec) , antiderivative size = 437, normalized size of antiderivative = 3.83, number of steps used = 13, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2081, 6857, 477, 476, 486, 12, 503} \[ \int \frac {\left (1+x^2\right ) \sqrt [3]{-x+2 x^3}}{x^2 \left (1+x^4\right )} \, dx=-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \sqrt [3]{2-i} \sqrt {3} \sqrt [3]{2 x^3-x} \arctan \left (\frac {1+\frac {2 \sqrt [3]{2-i} x^{2/3}}{\sqrt [3]{2 x^2-1}}}{\sqrt {3}}\right )}{\sqrt [3]{x} \sqrt [3]{2 x^2-1}}-\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \sqrt [3]{2+i} \sqrt {3} \sqrt [3]{2 x^3-x} \arctan \left (\frac {1+\frac {2 \sqrt [3]{2+i} x^{2/3}}{\sqrt [3]{2 x^2-1}}}{\sqrt {3}}\right )}{\sqrt [3]{x} \sqrt [3]{2 x^2-1}}-\frac {3 \sqrt [3]{2 x^3-x}}{2 x}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \sqrt [3]{2+i} \sqrt [3]{2 x^3-x} \log \left (-x^2+i\right )}{\sqrt [3]{x} \sqrt [3]{2 x^2-1}}+\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \sqrt [3]{2-i} \sqrt [3]{2 x^3-x} \log \left (x^2+i\right )}{\sqrt [3]{x} \sqrt [3]{2 x^2-1}}-\frac {\left (\frac {3}{8}-\frac {3 i}{8}\right ) \sqrt [3]{2-i} \sqrt [3]{2 x^3-x} \log \left (-\sqrt [3]{2 x^2-1}+\sqrt [3]{2-i} x^{2/3}\right )}{\sqrt [3]{x} \sqrt [3]{2 x^2-1}}-\frac {\left (\frac {3}{8}+\frac {3 i}{8}\right ) \sqrt [3]{2+i} \sqrt [3]{2 x^3-x} \log \left (-\sqrt [3]{2 x^2-1}+\sqrt [3]{2+i} x^{2/3}\right )}{\sqrt [3]{x} \sqrt [3]{2 x^2-1}} \]

[In]

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

[Out]

(-3*(-x + 2*x^3)^(1/3))/(2*x) - ((1/4 - I/4)*(2 - I)^(1/3)*Sqrt[3]*(-x + 2*x^3)^(1/3)*ArcTan[(1 + (2*(2 - I)^(
1/3)*x^(2/3))/(-1 + 2*x^2)^(1/3))/Sqrt[3]])/(x^(1/3)*(-1 + 2*x^2)^(1/3)) - ((1/4 + I/4)*(2 + I)^(1/3)*Sqrt[3]*
(-x + 2*x^3)^(1/3)*ArcTan[(1 + (2*(2 + I)^(1/3)*x^(2/3))/(-1 + 2*x^2)^(1/3))/Sqrt[3]])/(x^(1/3)*(-1 + 2*x^2)^(
1/3)) + ((1/8 + I/8)*(2 + I)^(1/3)*(-x + 2*x^3)^(1/3)*Log[I - x^2])/(x^(1/3)*(-1 + 2*x^2)^(1/3)) + ((1/8 - I/8
)*(2 - I)^(1/3)*(-x + 2*x^3)^(1/3)*Log[I + x^2])/(x^(1/3)*(-1 + 2*x^2)^(1/3)) - ((3/8 - (3*I)/8)*(2 - I)^(1/3)
*(-x + 2*x^3)^(1/3)*Log[(2 - I)^(1/3)*x^(2/3) - (-1 + 2*x^2)^(1/3)])/(x^(1/3)*(-1 + 2*x^2)^(1/3)) - ((3/8 + (3
*I)/8)*(2 + I)^(1/3)*(-x + 2*x^3)^(1/3)*Log[(2 + I)^(1/3)*x^(2/3) - (-1 + 2*x^2)^(1/3)])/(x^(1/3)*(-1 + 2*x^2)
^(1/3))

Rule 12

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

Rule 476

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

Rule 477

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 486

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*(m + 1))), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*
x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*b*(m + 1) + n*(b*c*(p + 1) + a*d*q) + d*(b*(m + 1) + b*n*(p + q + 1))*x^n, x
], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[0, q, 1] && LtQ[m, -1] &&
IntBinomialQ[a, b, c, d, e, m, n, p, q, 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 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [3]{-x+2 x^3} \int \frac {\left (1+x^2\right ) \sqrt [3]{-1+2 x^2}}{x^{5/3} \left (1+x^4\right )} \, dx}{\sqrt [3]{x} \sqrt [3]{-1+2 x^2}} \\ & = \frac {\sqrt [3]{-x+2 x^3} \int \left (-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt [3]{-1+2 x^2}}{x^{5/3} \left (i-x^2\right )}+\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [3]{-1+2 x^2}}{x^{5/3} \left (i+x^2\right )}\right ) \, dx}{\sqrt [3]{x} \sqrt [3]{-1+2 x^2}} \\ & = -\frac {\left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt [3]{-x+2 x^3}\right ) \int \frac {\sqrt [3]{-1+2 x^2}}{x^{5/3} \left (i-x^2\right )} \, dx}{\sqrt [3]{x} \sqrt [3]{-1+2 x^2}}+\frac {\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [3]{-x+2 x^3}\right ) \int \frac {\sqrt [3]{-1+2 x^2}}{x^{5/3} \left (i+x^2\right )} \, dx}{\sqrt [3]{x} \sqrt [3]{-1+2 x^2}} \\ & = -\frac {\left (\left (\frac {3}{2}-\frac {3 i}{2}\right ) \sqrt [3]{-x+2 x^3}\right ) \text {Subst}\left (\int \frac {\sqrt [3]{-1+2 x^6}}{x^3 \left (i-x^6\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x} \sqrt [3]{-1+2 x^2}}+\frac {\left (\left (\frac {3}{2}+\frac {3 i}{2}\right ) \sqrt [3]{-x+2 x^3}\right ) \text {Subst}\left (\int \frac {\sqrt [3]{-1+2 x^6}}{x^3 \left (i+x^6\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x} \sqrt [3]{-1+2 x^2}} \\ & = -\frac {\left (\left (\frac {3}{4}-\frac {3 i}{4}\right ) \sqrt [3]{-x+2 x^3}\right ) \text {Subst}\left (\int \frac {\sqrt [3]{-1+2 x^3}}{x^2 \left (i-x^3\right )} \, dx,x,x^{2/3}\right )}{\sqrt [3]{x} \sqrt [3]{-1+2 x^2}}+\frac {\left (\left (\frac {3}{4}+\frac {3 i}{4}\right ) \sqrt [3]{-x+2 x^3}\right ) \text {Subst}\left (\int \frac {\sqrt [3]{-1+2 x^3}}{x^2 \left (i+x^3\right )} \, dx,x,x^{2/3}\right )}{\sqrt [3]{x} \sqrt [3]{-1+2 x^2}} \\ & = -\frac {3 \sqrt [3]{-x+2 x^3}}{2 x}+\frac {\left (\left (\frac {3}{4}-\frac {3 i}{4}\right ) \sqrt [3]{-x+2 x^3}\right ) \text {Subst}\left (\int \frac {(1+2 i) x}{\left (i+x^3\right ) \left (-1+2 x^3\right )^{2/3}} \, dx,x,x^{2/3}\right )}{\sqrt [3]{x} \sqrt [3]{-1+2 x^2}}+\frac {\left (\left (\frac {3}{4}+\frac {3 i}{4}\right ) \sqrt [3]{-x+2 x^3}\right ) \text {Subst}\left (\int -\frac {(1-2 i) x}{\left (i-x^3\right ) \left (-1+2 x^3\right )^{2/3}} \, dx,x,x^{2/3}\right )}{\sqrt [3]{x} \sqrt [3]{-1+2 x^2}} \\ & = -\frac {3 \sqrt [3]{-x+2 x^3}}{2 x}+-\frac {\left (\left (\frac {9}{4}-\frac {3 i}{4}\right ) \sqrt [3]{-x+2 x^3}\right ) \text {Subst}\left (\int \frac {x}{\left (i-x^3\right ) \left (-1+2 x^3\right )^{2/3}} \, dx,x,x^{2/3}\right )}{\sqrt [3]{x} \sqrt [3]{-1+2 x^2}}+\frac {\left (\left (\frac {9}{4}+\frac {3 i}{4}\right ) \sqrt [3]{-x+2 x^3}\right ) \text {Subst}\left (\int \frac {x}{\left (i+x^3\right ) \left (-1+2 x^3\right )^{2/3}} \, dx,x,x^{2/3}\right )}{\sqrt [3]{x} \sqrt [3]{-1+2 x^2}} \\ & = -\frac {3 \sqrt [3]{-x+2 x^3}}{2 x}-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \sqrt [3]{2-i} \sqrt {3} \sqrt [3]{-x+2 x^3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{2-i} x^{2/3}}{\sqrt [3]{-1+2 x^2}}}{\sqrt {3}}\right )}{\sqrt [3]{x} \sqrt [3]{-1+2 x^2}}-\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \sqrt [3]{2+i} \sqrt {3} \sqrt [3]{-x+2 x^3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{2+i} x^{2/3}}{\sqrt [3]{-1+2 x^2}}}{\sqrt {3}}\right )}{\sqrt [3]{x} \sqrt [3]{-1+2 x^2}}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \sqrt [3]{2+i} \sqrt [3]{-x+2 x^3} \log \left (i-x^2\right )}{\sqrt [3]{x} \sqrt [3]{-1+2 x^2}}+\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \sqrt [3]{2-i} \sqrt [3]{-x+2 x^3} \log \left (i+x^2\right )}{\sqrt [3]{x} \sqrt [3]{-1+2 x^2}}-\frac {\left (\frac {3}{8}-\frac {3 i}{8}\right ) \sqrt [3]{2-i} \sqrt [3]{-x+2 x^3} \log \left (\sqrt [3]{2-i} x^{2/3}-\sqrt [3]{-1+2 x^2}\right )}{\sqrt [3]{x} \sqrt [3]{-1+2 x^2}}-\frac {\left (\frac {3}{8}+\frac {3 i}{8}\right ) \sqrt [3]{2+i} \sqrt [3]{-x+2 x^3} \log \left (\sqrt [3]{2+i} x^{2/3}-\sqrt [3]{-1+2 x^2}\right )}{\sqrt [3]{x} \sqrt [3]{-1+2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.78 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.18 \[ \int \frac {\left (1+x^2\right ) \sqrt [3]{-x+2 x^3}}{x^2 \left (1+x^4\right )} \, dx=\frac {18-36 x^2+x^{2/3} \left (-1+2 x^2\right )^{2/3} \text {RootSum}\left [5-4 \text {$\#$1}^3+\text {$\#$1}^6\&,\frac {-10 \log (x)+15 \log \left (\sqrt [3]{-1+2 x^2}-x^{2/3} \text {$\#$1}\right )+2 \log (x) \text {$\#$1}^3-3 \log \left (\sqrt [3]{-1+2 x^2}-x^{2/3} \text {$\#$1}\right ) \text {$\#$1}^3}{-2 \text {$\#$1}^2+\text {$\#$1}^5}\&\right ]}{12 \left (x \left (-1+2 x^2\right )\right )^{2/3}} \]

[In]

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

[Out]

(18 - 36*x^2 + x^(2/3)*(-1 + 2*x^2)^(2/3)*RootSum[5 - 4*#1^3 + #1^6 & , (-10*Log[x] + 15*Log[(-1 + 2*x^2)^(1/3
) - x^(2/3)*#1] + 2*Log[x]*#1^3 - 3*Log[(-1 + 2*x^2)^(1/3) - x^(2/3)*#1]*#1^3)/(-2*#1^2 + #1^5) & ])/(12*(x*(-
1 + 2*x^2))^(2/3))

Maple [N/A] (verified)

Time = 168.42 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.65

method	result	size
pseudoelliptic	\(\frac {-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-4 \textit {\_Z}^{3}+5\right )}{\sum }\frac {\left (\textit {\_R}^{3}-5\right ) \ln \left (\frac {-\textit {\_R} x +\left (2 x^{3}-x \right )^{\frac {1}{3}}}{x}\right )}{\textit {\_R}^{2} \left (\textit {\_R}^{3}-2\right )}\right ) x -6 \left (2 x^{3}-x \right )^{\frac {1}{3}}}{4 x}\)	\(74\)
trager	\(\text {Expression too large to display}\)	\(10722\)
risch	\(\text {Expression too large to display}\)	\(11848\)

[In]

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

[Out]

1/4*(-sum((_R^3-5)*ln((-_R*x+(2*x^3-x)^(1/3))/x)/_R^2/(_R^3-2),_R=RootOf(_Z^6-4*_Z^3+5))*x-6*(2*x^3-x)^(1/3))/
x

Fricas [F(-2)]

Exception generated. \[ \int \frac {\left (1+x^2\right ) \sqrt [3]{-x+2 x^3}}{x^2 \left (1+x^4\right )} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (tr
ace 0)

Sympy [N/A]

Not integrable

Time = 1.71 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.23 \[ \int \frac {\left (1+x^2\right ) \sqrt [3]{-x+2 x^3}}{x^2 \left (1+x^4\right )} \, dx=\int \frac {\sqrt [3]{x \left (2 x^{2} - 1\right )} \left (x^{2} + 1\right )}{x^{2} \left (x^{4} + 1\right )}\, dx \]

[In]

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

[Out]

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

Maxima [N/A]

Not integrable

Time = 0.33 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.25 \[ \int \frac {\left (1+x^2\right ) \sqrt [3]{-x+2 x^3}}{x^2 \left (1+x^4\right )} \, dx=\int { \frac {{\left (2 \, x^{3} - x\right )}^{\frac {1}{3}} {\left (x^{2} + 1\right )}}{{\left (x^{4} + 1\right )} x^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [N/A]

Not integrable

Time = 0.38 (sec) , antiderivative size = 1, normalized size of antiderivative = 0.01 \[ \int \frac {\left (1+x^2\right ) \sqrt [3]{-x+2 x^3}}{x^2 \left (1+x^4\right )} \, dx=\int { \frac {{\left (2 \, x^{3} - x\right )}^{\frac {1}{3}} {\left (x^{2} + 1\right )}}{{\left (x^{4} + 1\right )} x^{2}} \,d x } \]

[In]

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

[Out]

undef

Mupad [N/A]

Not integrable

Time = 5.65 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.25 \[ \int \frac {\left (1+x^2\right ) \sqrt [3]{-x+2 x^3}}{x^2 \left (1+x^4\right )} \, dx=\int \frac {{\left (2\,x^3-x\right )}^{1/3}\,\left (x^2+1\right )}{x^2\,\left (x^4+1\right )} \,d x \]

[In]

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

[Out]

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