3.16.54 \(\int \frac {(-2+x^2) \sqrt [3]{x+x^3}}{x^2 (4-2 x^2+x^4)} \, dx\) [1554]

Optimal. Leaf size=107 \[ \frac {3 \sqrt [3]{x+x^3}}{4 x}-\frac {1}{8} \text {RootSum}\left [7-10 \text {$\#$1}^3+4 \text {$\#$1}^6\& ,\frac {-7 \log (x)+7 \log \left (\sqrt [3]{x+x^3}-x \text {$\#$1}\right )+4 \log (x) \text {$\#$1}^3-4 \log \left (\sqrt [3]{x+x^3}-x \text {$\#$1}\right ) \text {$\#$1}^3}{-5 \text {$\#$1}^2+4 \text {$\#$1}^5}\& \right ] \]

[Out]

Unintegrable

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Rubi [C] Result contains complex when optimal does not.
time = 0.75, antiderivative size = 763, normalized size of antiderivative = 7.13, number of steps used = 13, number of rules used = 7, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {2081, 6860, 477, 476, 486, 12, 503} \begin {gather*} -\frac {\left (1-i \sqrt {3}\right ) \sqrt [3]{x^3+x} \text {ArcTan}\left (\frac {1+\frac {2 x^{2/3}}{\sqrt [3]{\frac {-\sqrt {3}+i}{-\sqrt {3}+2 i}} \sqrt [3]{x^2+1}}}{\sqrt {3}}\right )}{4 \left (-\sqrt {3}+i\right ) \sqrt [3]{\frac {-\sqrt {3}+i}{-\sqrt {3}+2 i}} \sqrt [3]{x} \sqrt [3]{x^2+1}}+\frac {\left (1+i \sqrt {3}\right ) \sqrt [3]{x^3+x} \text {ArcTan}\left (\frac {1+\frac {2 x^{2/3}}{\sqrt [3]{\frac {\sqrt {3}+i}{\sqrt {3}+2 i}} \sqrt [3]{x^2+1}}}{\sqrt {3}}\right )}{4 \left (\sqrt {3}+i\right ) \sqrt [3]{\frac {\sqrt {3}+i}{\sqrt {3}+2 i}} \sqrt [3]{x} \sqrt [3]{x^2+1}}+\frac {\left (\sqrt {3}+3 i\right ) \sqrt [3]{x^3+x}}{4 \left (\sqrt {3}+i\right ) x}+\frac {\left (-\sqrt {3}+3 i\right ) \sqrt [3]{x^3+x}}{4 \left (-\sqrt {3}+i\right ) x}-\frac {\left (\sqrt {3}+3 i\right ) \sqrt [3]{x^3+x} \log \left (-x^2-i \sqrt {3}+1\right )}{24 \left (\sqrt {3}+i\right ) \sqrt [3]{\frac {\sqrt {3}+i}{\sqrt {3}+2 i}} \sqrt [3]{x} \sqrt [3]{x^2+1}}-\frac {\left (-\sqrt {3}+3 i\right ) \sqrt [3]{x^3+x} \log \left (-x^2+i \sqrt {3}+1\right )}{24 \left (-\sqrt {3}+i\right ) \sqrt [3]{\frac {-\sqrt {3}+i}{-\sqrt {3}+2 i}} \sqrt [3]{x} \sqrt [3]{x^2+1}}+\frac {\left (-\sqrt {3}+3 i\right ) \sqrt [3]{x^3+x} \log \left (x^{2/3}-\sqrt [3]{\frac {-\sqrt {3}+i}{-\sqrt {3}+2 i}} \sqrt [3]{x^2+1}\right )}{8 \left (-\sqrt {3}+i\right ) \sqrt [3]{\frac {-\sqrt {3}+i}{-\sqrt {3}+2 i}} \sqrt [3]{x} \sqrt [3]{x^2+1}}+\frac {\left (\sqrt {3}+3 i\right ) \sqrt [3]{x^3+x} \log \left (x^{2/3}-\sqrt [3]{\frac {\sqrt {3}+i}{\sqrt {3}+2 i}} \sqrt [3]{x^2+1}\right )}{8 \left (\sqrt {3}+i\right ) \sqrt [3]{\frac {\sqrt {3}+i}{\sqrt {3}+2 i}} \sqrt [3]{x} \sqrt [3]{x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3*I - Sqrt[3])*(x + x^3)^(1/3))/(4*(I - Sqrt[3])*x) + ((3*I + Sqrt[3])*(x + x^3)^(1/3))/(4*(I + Sqrt[3])*x)
- ((1 - I*Sqrt[3])*(x + x^3)^(1/3)*ArcTan[(1 + (2*x^(2/3))/(((I - Sqrt[3])/(2*I - Sqrt[3]))^(1/3)*(1 + x^2)^(1
/3)))/Sqrt[3]])/(4*(I - Sqrt[3])*((I - Sqrt[3])/(2*I - Sqrt[3]))^(1/3)*x^(1/3)*(1 + x^2)^(1/3)) + ((1 + I*Sqrt
[3])*(x + x^3)^(1/3)*ArcTan[(1 + (2*x^(2/3))/(((I + Sqrt[3])/(2*I + Sqrt[3]))^(1/3)*(1 + x^2)^(1/3)))/Sqrt[3]]
)/(4*(I + Sqrt[3])*((I + Sqrt[3])/(2*I + Sqrt[3]))^(1/3)*x^(1/3)*(1 + x^2)^(1/3)) - ((3*I + Sqrt[3])*(x + x^3)
^(1/3)*Log[1 - I*Sqrt[3] - x^2])/(24*(I + Sqrt[3])*((I + Sqrt[3])/(2*I + Sqrt[3]))^(1/3)*x^(1/3)*(1 + x^2)^(1/
3)) - ((3*I - Sqrt[3])*(x + x^3)^(1/3)*Log[1 + I*Sqrt[3] - x^2])/(24*(I - Sqrt[3])*((I - Sqrt[3])/(2*I - Sqrt[
3]))^(1/3)*x^(1/3)*(1 + x^2)^(1/3)) + ((3*I - Sqrt[3])*(x + x^3)^(1/3)*Log[x^(2/3) - ((I - Sqrt[3])/(2*I - Sqr
t[3]))^(1/3)*(1 + x^2)^(1/3)])/(8*(I - Sqrt[3])*((I - Sqrt[3])/(2*I - Sqrt[3]))^(1/3)*x^(1/3)*(1 + x^2)^(1/3))
+ ((3*I + Sqrt[3])*(x + x^3)^(1/3)*Log[x^(2/3) - ((I + Sqrt[3])/(2*I + Sqrt[3]))^(1/3)*(1 + x^2)^(1/3)])/(8*(
I + Sqrt[3])*((I + Sqrt[3])/(2*I + Sqrt[3]))^(1/3)*x^(1/3)*(1 + 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 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (-2+x^2\right ) \sqrt [3]{x+x^3}}{x^2 \left (4-2 x^2+x^4\right )} \, dx &=\frac {\sqrt [3]{x+x^3} \int \frac {\left (-2+x^2\right ) \sqrt [3]{1+x^2}}{x^{5/3} \left (4-2 x^2+x^4\right )} \, dx}{\sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {\sqrt [3]{x+x^3} \int \left (\frac {\left (1+\frac {i}{\sqrt {3}}\right ) \sqrt [3]{1+x^2}}{x^{5/3} \left (-2-2 i \sqrt {3}+2 x^2\right )}+\frac {\left (1-\frac {i}{\sqrt {3}}\right ) \sqrt [3]{1+x^2}}{x^{5/3} \left (-2+2 i \sqrt {3}+2 x^2\right )}\right ) \, dx}{\sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {\left (\left (3-i \sqrt {3}\right ) \sqrt [3]{x+x^3}\right ) \int \frac {\sqrt [3]{1+x^2}}{x^{5/3} \left (-2+2 i \sqrt {3}+2 x^2\right )} \, dx}{3 \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {\left (\left (3+i \sqrt {3}\right ) \sqrt [3]{x+x^3}\right ) \int \frac {\sqrt [3]{1+x^2}}{x^{5/3} \left (-2-2 i \sqrt {3}+2 x^2\right )} \, dx}{3 \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {\left (\left (3-i \sqrt {3}\right ) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {\sqrt [3]{1+x^6}}{x^3 \left (-2+2 i \sqrt {3}+2 x^6\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {\left (\left (3+i \sqrt {3}\right ) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {\sqrt [3]{1+x^6}}{x^3 \left (-2-2 i \sqrt {3}+2 x^6\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {\left (\left (3-i \sqrt {3}\right ) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {\sqrt [3]{1+x^3}}{x^2 \left (-2+2 i \sqrt {3}+2 x^3\right )} \, dx,x,x^{2/3}\right )}{2 \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {\left (\left (3+i \sqrt {3}\right ) \sqrt [3]{x+x^3}\right ) \text {Subst}\left (\int \frac {\sqrt [3]{1+x^3}}{x^2 \left (-2-2 i \sqrt {3}+2 x^3\right )} \, dx,x,x^{2/3}\right )}{2 \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {\left (3 i+\sqrt {3}\right ) \sqrt [3]{1-\frac {x^2}{1-i \sqrt {3}}} \sqrt [3]{x+x^3} \, _2F_1\left (-\frac {1}{3},-\frac {1}{3};\frac {2}{3};-\frac {x^2+\frac {x^2}{1-i \sqrt {3}}}{1-\frac {x^2}{1-i \sqrt {3}}}\right )}{4 \left (i+\sqrt {3}\right ) x \sqrt [3]{1+x^2}}+\frac {\left (3 i-\sqrt {3}\right ) \sqrt [3]{1-\frac {x^2}{1+i \sqrt {3}}} \sqrt [3]{x+x^3} \, _2F_1\left (-\frac {1}{3},-\frac {1}{3};\frac {2}{3};-\frac {x^2+\frac {x^2}{1+i \sqrt {3}}}{1-\frac {x^2}{1+i \sqrt {3}}}\right )}{4 \left (i-\sqrt {3}\right ) x \sqrt [3]{1+x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 140, normalized size = 1.31 \begin {gather*} \frac {\sqrt [3]{x+x^3} \left (6 \sqrt [3]{1+x^2}-\frac {1}{3} x^{2/3} \text {RootSum}\left [7-10 \text {$\#$1}^3+4 \text {$\#$1}^6\&,\frac {-14 \log (x)+21 \log \left (\sqrt [3]{1+x^2}-x^{2/3} \text {$\#$1}\right )+8 \log (x) \text {$\#$1}^3-12 \log \left (\sqrt [3]{1+x^2}-x^{2/3} \text {$\#$1}\right ) \text {$\#$1}^3}{-5 \text {$\#$1}^2+4 \text {$\#$1}^5}\&\right ]\right )}{8 x \sqrt [3]{1+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((x + x^3)^(1/3)*(6*(1 + x^2)^(1/3) - (x^(2/3)*RootSum[7 - 10*#1^3 + 4*#1^6 & , (-14*Log[x] + 21*Log[(1 + x^2)
^(1/3) - x^(2/3)*#1] + 8*Log[x]*#1^3 - 12*Log[(1 + x^2)^(1/3) - x^(2/3)*#1]*#1^3)/(-5*#1^2 + 4*#1^5) & ])/3))/
(8*x*(1 + x^2)^(1/3))

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (x^{2}-2\right ) \left (x^{3}+x \right )^{\frac {1}{3}}}{x^{2} \left (x^{4}-2 x^{2}+4\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

-3/8*((x^3 + x)*x^2 - 2*x^3 - 2*x)*(x^2 + 1)^(1/3)/(x^(17/3) - 2*x^(11/3) + 4*x^(5/3)) + integrate(6*((x^2 + 1
)*x^2 - x^2 - 1)*(x^2 + 1)^(1/3)/(x^(29/3) - 4*x^(23/3) + 12*x^(17/3) - 16*x^(11/3) + 16*x^(5/3)), x)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (x^2-2\right )\,{\left (x^3+x\right )}^{1/3}}{x^2\,\left (x^4-2\,x^2+4\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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