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

Optimal. Leaf size=254 \[ \frac {\log \left (2^{2/3} \sqrt [3]{x^4+x^2}-2 x\right )}{4 \sqrt [3]{2}}-\frac {3 \log \left (2^{2/3} \sqrt [3]{x^4+x^2}+2 x\right )}{4 \sqrt [3]{2}}+\frac {3 \log \left (-2 x^2+2^{2/3} \sqrt [3]{x^4+x^2} x-\sqrt [3]{2} \left (x^4+x^2\right )^{2/3}\right )}{8 \sqrt [3]{2}}-\frac {\log \left (2 x^2+2^{2/3} \sqrt [3]{x^4+x^2} x+\sqrt [3]{2} \left (x^4+x^2\right )^{2/3}\right )}{8 \sqrt [3]{2}}-\frac {3 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2^{2/3} \sqrt [3]{x^4+x^2}-x}\right )}{4 \sqrt [3]{2}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2^{2/3} \sqrt [3]{x^4+x^2}+x}\right )}{4 \sqrt [3]{2}} \]

________________________________________________________________________________________

Rubi [C]  time = 0.87, antiderivative size = 127, normalized size of antiderivative = 0.50, number of steps used = 18, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2056, 6725, 364, 959, 466, 429, 465, 510} \begin {gather*} \frac {3 \sqrt [3]{x^2+1} x^2 F_1\left (\frac {2}{3};1,\frac {1}{3};\frac {5}{3};x^2,-x^2\right )}{4 \sqrt [3]{x^4+x^2}}-\frac {6 \sqrt [3]{x^2+1} x F_1\left (\frac {1}{6};1,\frac {1}{3};\frac {7}{6};x^2,-x^2\right )}{\sqrt [3]{x^4+x^2}}+\frac {3 \sqrt [3]{x^2+1} x \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^2\right )}{\sqrt [3]{x^4+x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-6*x*(1 + x^2)^(1/3)*AppellF1[1/6, 1, 1/3, 7/6, x^2, -x^2])/(x^2 + x^4)^(1/3) + (3*x^2*(1 + x^2)^(1/3)*Appell
F1[2/3, 1, 1/3, 5/3, x^2, -x^2])/(4*(x^2 + x^4)^(1/3)) + (3*x*(1 + x^2)^(1/3)*Hypergeometric2F1[1/6, 1/3, 7/6,
-x^2])/(x^2 + x^4)^(1/3)

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 429

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

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 959

Int[(((g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[(d*(g*x)^n)/x^n, In
t[(x^n*(a + c*x^2)^p)/(d^2 - e^2*x^2), x], x] - Dist[(e*(g*x)^n)/x^n, Int[(x^(n + 1)*(a + c*x^2)^p)/(d^2 - e^2
*x^2), x], x] /; FreeQ[{a, c, d, e, g, n, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] &&  !IntegersQ[n, 2
*p]

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 {1-x+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx &=\frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \int \frac {1-x+x^2}{x^{2/3} \left (-1+x^2\right ) \sqrt [3]{1+x^2}} \, dx}{\sqrt [3]{x^2+x^4}}\\ &=\frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \int \left (\frac {1}{x^{2/3} \sqrt [3]{1+x^2}}+\frac {2-x}{x^{2/3} \left (-1+x^2\right ) \sqrt [3]{1+x^2}}\right ) \, dx}{\sqrt [3]{x^2+x^4}}\\ &=\frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \int \frac {1}{x^{2/3} \sqrt [3]{1+x^2}} \, dx}{\sqrt [3]{x^2+x^4}}+\frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \int \frac {2-x}{x^{2/3} \left (-1+x^2\right ) \sqrt [3]{1+x^2}} \, dx}{\sqrt [3]{x^2+x^4}}\\ &=\frac {3 x \sqrt [3]{1+x^2} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^2\right )}{\sqrt [3]{x^2+x^4}}+\frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \int \left (-\frac {1}{2 (1-x) x^{2/3} \sqrt [3]{1+x^2}}-\frac {3}{2 x^{2/3} (1+x) \sqrt [3]{1+x^2}}\right ) \, dx}{\sqrt [3]{x^2+x^4}}\\ &=\frac {3 x \sqrt [3]{1+x^2} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^2\right )}{\sqrt [3]{x^2+x^4}}-\frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \int \frac {1}{(1-x) x^{2/3} \sqrt [3]{1+x^2}} \, dx}{2 \sqrt [3]{x^2+x^4}}-\frac {\left (3 x^{2/3} \sqrt [3]{1+x^2}\right ) \int \frac {1}{x^{2/3} (1+x) \sqrt [3]{1+x^2}} \, dx}{2 \sqrt [3]{x^2+x^4}}\\ &=\frac {3 x \sqrt [3]{1+x^2} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^2\right )}{\sqrt [3]{x^2+x^4}}-\frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \int \frac {1}{x^{2/3} \left (1-x^2\right ) \sqrt [3]{1+x^2}} \, dx}{2 \sqrt [3]{x^2+x^4}}-\frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \int \frac {\sqrt [3]{x}}{\left (1-x^2\right ) \sqrt [3]{1+x^2}} \, dx}{2 \sqrt [3]{x^2+x^4}}-\frac {\left (3 x^{2/3} \sqrt [3]{1+x^2}\right ) \int \frac {1}{x^{2/3} \left (1-x^2\right ) \sqrt [3]{1+x^2}} \, dx}{2 \sqrt [3]{x^2+x^4}}+\frac {\left (3 x^{2/3} \sqrt [3]{1+x^2}\right ) \int \frac {\sqrt [3]{x}}{\left (1-x^2\right ) \sqrt [3]{1+x^2}} \, dx}{2 \sqrt [3]{x^2+x^4}}\\ &=\frac {3 x \sqrt [3]{1+x^2} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^2\right )}{\sqrt [3]{x^2+x^4}}-\frac {\left (3 x^{2/3} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-x^6\right ) \sqrt [3]{1+x^6}} \, dx,x,\sqrt [3]{x}\right )}{2 \sqrt [3]{x^2+x^4}}-\frac {\left (3 x^{2/3} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {x^3}{\left (1-x^6\right ) \sqrt [3]{1+x^6}} \, dx,x,\sqrt [3]{x}\right )}{2 \sqrt [3]{x^2+x^4}}-\frac {\left (9 x^{2/3} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-x^6\right ) \sqrt [3]{1+x^6}} \, dx,x,\sqrt [3]{x}\right )}{2 \sqrt [3]{x^2+x^4}}+\frac {\left (9 x^{2/3} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {x^3}{\left (1-x^6\right ) \sqrt [3]{1+x^6}} \, dx,x,\sqrt [3]{x}\right )}{2 \sqrt [3]{x^2+x^4}}\\ &=-\frac {6 x \sqrt [3]{1+x^2} F_1\left (\frac {1}{6};1,\frac {1}{3};\frac {7}{6};x^2,-x^2\right )}{\sqrt [3]{x^2+x^4}}+\frac {3 x \sqrt [3]{1+x^2} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^2\right )}{\sqrt [3]{x^2+x^4}}-\frac {\left (3 x^{2/3} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (1-x^3\right ) \sqrt [3]{1+x^3}} \, dx,x,x^{2/3}\right )}{4 \sqrt [3]{x^2+x^4}}+\frac {\left (9 x^{2/3} \sqrt [3]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (1-x^3\right ) \sqrt [3]{1+x^3}} \, dx,x,x^{2/3}\right )}{4 \sqrt [3]{x^2+x^4}}\\ &=-\frac {6 x \sqrt [3]{1+x^2} F_1\left (\frac {1}{6};1,\frac {1}{3};\frac {7}{6};x^2,-x^2\right )}{\sqrt [3]{x^2+x^4}}+\frac {3 x^2 \sqrt [3]{1+x^2} F_1\left (\frac {2}{3};1,\frac {1}{3};\frac {5}{3};x^2,-x^2\right )}{4 \sqrt [3]{x^2+x^4}}+\frac {3 x \sqrt [3]{1+x^2} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^2\right )}{\sqrt [3]{x^2+x^4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 0.33, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1-x+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.69, size = 254, normalized size = 1.00 \begin {gather*} -\frac {3 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{-x+2^{2/3} \sqrt [3]{x^2+x^4}}\right )}{4 \sqrt [3]{2}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{x^2+x^4}}\right )}{4 \sqrt [3]{2}}+\frac {\log \left (-2 x+2^{2/3} \sqrt [3]{x^2+x^4}\right )}{4 \sqrt [3]{2}}-\frac {3 \log \left (2 x+2^{2/3} \sqrt [3]{x^2+x^4}\right )}{4 \sqrt [3]{2}}+\frac {3 \log \left (-2 x^2+2^{2/3} x \sqrt [3]{x^2+x^4}-\sqrt [3]{2} \left (x^2+x^4\right )^{2/3}\right )}{8 \sqrt [3]{2}}-\frac {\log \left (2 x^2+2^{2/3} x \sqrt [3]{x^2+x^4}+\sqrt [3]{2} \left (x^2+x^4\right )^{2/3}\right )}{8 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 6.34, size = 0, normalized size = 0.00 \begin {gather*} {\rm integral}\left (\frac {{\left (x^{4} + x^{2}\right )}^{\frac {2}{3}} {\left (x^{2} - x + 1\right )}}{x^{6} - x^{2}}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((x^4 + x^2)^(2/3)*(x^2 - x + 1)/(x^6 - x^2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 74.65, size = 8780, normalized size = 34.57

method	result	size
trager	\(\text {Expression too large to display}\)	\(8780\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________