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

Optimal. Leaf size=114 \[ \frac {1}{4} \text {RootSum}\left [\text {$\#$1}^6-4 \text {$\#$1}^3+5\& ,\frac {-\text {$\#$1}^3 \log \left (\sqrt [3]{2 x^3-x}-\text {$\#$1} x\right )+\text {$\#$1}^3 \log (x)+5 \log \left (\sqrt [3]{2 x^3-x}-\text {$\#$1} x\right )-5 \log (x)}{\text {$\#$1}^5-2 \text {$\#$1}^2}\& \right ]-\frac {3 \sqrt [3]{2 x^3-x}}{2 x} \]

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Rubi [C]  time = 0.66, antiderivative size = 153, normalized size of antiderivative = 1.34, number of steps used = 11, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2056, 6725, 466, 465, 511, 510} \begin {gather*} -\frac {\left (\frac {3}{4}-\frac {3 i}{4}\right ) \sqrt [3]{1-i x^2} \sqrt [3]{2 x^3-x} \, _2F_1\left (-\frac {1}{3},-\frac {1}{3};\frac {2}{3};\frac {(2-i) x^2}{1-i x^2}\right )}{x \sqrt [3]{1-2 x^2}}-\frac {\left (\frac {3}{4}+\frac {3 i}{4}\right ) \sqrt [3]{1+i x^2} \sqrt [3]{2 x^3-x} \, _2F_1\left (-\frac {1}{3},-\frac {1}{3};\frac {2}{3};\frac {(2+i) x^2}{i x^2+1}\right )}{x \sqrt [3]{1-2 x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

Rule 465

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 466

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 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 2056

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 6725

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

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Mathematica [F]  time = 2.95, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (1+x^2\right ) \sqrt [3]{-x+2 x^3}}{x^2 \left (1+x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.26, size = 114, normalized size = 1.00 \begin {gather*} -\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 ] \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((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) + RootSum[5 - 4*#1^3 + #1^6 & , (-5*Log[x] + 5*Log[(-x + 2*x^3)^(1/3) - x*#1] +
Log[x]*#1^3 - Log[(-x + 2*x^3)^(1/3) - x*#1]*#1^3)/(-2*#1^2 + #1^5) & ]/4

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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+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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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maple [B]  time = 234.18, size = 10723, normalized size = 94.06

method	result	size
trager	\(\text {Expression too large to display}\)	\(10723\)
risch	\(\text {Expression too large to display}\)	\(19101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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