Integrand size = 26, antiderivative size = 157 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [3]{-x^2+x^4}} \, dx=-\frac {\arctan \left (\frac {\sqrt [3]{2} x}{\sqrt [3]{-x^2+x^4}}\right )}{\sqrt [3]{2}}-\frac {\arctan \left (\frac {2^{2/3} x \sqrt [3]{-x^2+x^4}}{-2 x^2+\sqrt [3]{2} \left (-x^2+x^4\right )^{2/3}}\right )}{2 \sqrt [3]{2}}-\frac {\sqrt {3} \text {arctanh}\left (\frac {\frac {\sqrt [3]{2} x^2}{\sqrt {3}}+\frac {\left (-x^2+x^4\right )^{2/3}}{\sqrt [3]{2} \sqrt {3}}}{x \sqrt [3]{-x^2+x^4}}\right )}{2 \sqrt [3]{2}} \]
-1/2*arctan(2^(1/3)*x/(x^4-x^2)^(1/3))*2^(2/3)-1/4*arctan(2^(2/3)*x*(x^4-x ^2)^(1/3)/(-2*x^2+2^(1/3)*(x^4-x^2)^(2/3)))*2^(2/3)-1/4*3^(1/2)*arctanh((1 /3*2^(1/3)*x^2*3^(1/2)+1/6*(x^4-x^2)^(2/3)*2^(2/3)*3^(1/2))/x/(x^4-x^2)^(1 /3))*2^(2/3)
Time = 0.47 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.04 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [3]{-x^2+x^4}} \, dx=-\frac {x^{2/3} \sqrt [3]{-1+x^2} \left (2 \arctan \left (\frac {\sqrt [3]{2} \sqrt [3]{x}}{\sqrt [3]{-1+x^2}}\right )+\arctan \left (\frac {2^{2/3} \sqrt [3]{x} \sqrt [3]{-1+x^2}}{-2 x^{2/3}+\sqrt [3]{2} \left (-1+x^2\right )^{2/3}}\right )+\sqrt {3} \text {arctanh}\left (\frac {2^{2/3} \sqrt {3} \sqrt [3]{x} \sqrt [3]{-1+x^2}}{2 x^{2/3}+\sqrt [3]{2} \left (-1+x^2\right )^{2/3}}\right )\right )}{2 \sqrt [3]{2} \sqrt [3]{x^2 \left (-1+x^2\right )}} \]
-1/2*(x^(2/3)*(-1 + x^2)^(1/3)*(2*ArcTan[(2^(1/3)*x^(1/3))/(-1 + x^2)^(1/3 )] + ArcTan[(2^(2/3)*x^(1/3)*(-1 + x^2)^(1/3))/(-2*x^(2/3) + 2^(1/3)*(-1 + x^2)^(2/3))] + Sqrt[3]*ArcTanh[(2^(2/3)*Sqrt[3]*x^(1/3)*(-1 + x^2)^(1/3)) /(2*x^(2/3) + 2^(1/3)*(-1 + x^2)^(2/3))]))/(2^(1/3)*(x^2*(-1 + x^2))^(1/3) )
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.32, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2467, 368, 937, 936}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^2-1}{\left (x^2+1\right ) \sqrt [3]{x^4-x^2}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {x^{2/3} \sqrt [3]{x^2-1} \int \frac {\left (x^2-1\right )^{2/3}}{x^{2/3} \left (x^2+1\right )}dx}{\sqrt [3]{x^4-x^2}}\) |
\(\Big \downarrow \) 368 |
\(\displaystyle \frac {3 x^{2/3} \sqrt [3]{x^2-1} \int \frac {\left (x^2-1\right )^{2/3}}{x^2+1}d\sqrt [3]{x}}{\sqrt [3]{x^4-x^2}}\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \frac {3 x^{2/3} \left (x^2-1\right ) \int \frac {\left (1-x^2\right )^{2/3}}{x^2+1}d\sqrt [3]{x}}{\left (1-x^2\right )^{2/3} \sqrt [3]{x^4-x^2}}\) |
\(\Big \downarrow \) 936 |
\(\displaystyle \frac {3 x \left (x^2-1\right ) \operatorname {AppellF1}\left (\frac {1}{6},-\frac {2}{3},1,\frac {7}{6},x^2,-x^2\right )}{\left (1-x^2\right )^{2/3} \sqrt [3]{x^4-x^2}}\) |
(3*x*(-1 + x^2)*AppellF1[1/6, -2/3, 1, 7/6, x^2, -x^2])/((1 - x^2)^(2/3)*( -x^2 + x^4)^(1/3))
3.22.47.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) , x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*2)/e^2))^p*(c + d*(x^(k*2)/e^2))^q, x], x, (e*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && FractionQ[m ] && IntegerQ[p]
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])
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]) Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, p, q }, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] && !(IntegerQ[p] || GtQ[a, 0])
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Time = 6.80 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.15
method | result | size |
pseudoelliptic | \(-\frac {2^{\frac {2}{3}} \left (\left (\ln \left (\frac {\sqrt {3}\, 2^{\frac {1}{3}} \left (x^{4}-x^{2}\right )^{\frac {1}{3}} x +2^{\frac {2}{3}} x^{2}+\left (x^{4}-x^{2}\right )^{\frac {2}{3}}}{x^{2}}\right )-\ln \left (\frac {-\sqrt {3}\, 2^{\frac {1}{3}} \left (x^{4}-x^{2}\right )^{\frac {1}{3}} x +2^{\frac {2}{3}} x^{2}+\left (x^{4}-x^{2}\right )^{\frac {2}{3}}}{x^{2}}\right )\right ) \sqrt {3}-4 \arctan \left (\frac {2^{\frac {2}{3}} \left (x^{4}-x^{2}\right )^{\frac {1}{3}}}{2 x}\right )+2 \arctan \left (\frac {-2^{\frac {2}{3}} \left (x^{4}-x^{2}\right )^{\frac {1}{3}}+x \sqrt {3}}{x}\right )-2 \arctan \left (\frac {2^{\frac {2}{3}} \left (x^{4}-x^{2}\right )^{\frac {1}{3}}+x \sqrt {3}}{x}\right )\right )}{8}\) | \(180\) |
-1/8*2^(2/3)*((ln((3^(1/2)*2^(1/3)*(x^4-x^2)^(1/3)*x+2^(2/3)*x^2+(x^4-x^2) ^(2/3))/x^2)-ln((-3^(1/2)*2^(1/3)*(x^4-x^2)^(1/3)*x+2^(2/3)*x^2+(x^4-x^2)^ (2/3))/x^2))*3^(1/2)-4*arctan(1/2*2^(2/3)/x*(x^4-x^2)^(1/3))+2*arctan((-2^ (2/3)*(x^4-x^2)^(1/3)+x*3^(1/2))/x)-2*arctan((2^(2/3)*(x^4-x^2)^(1/3)+x*3^ (1/2))/x))
Result contains complex when optimal does not.
Time = 1.64 (sec) , antiderivative size = 1380, normalized size of antiderivative = 8.79 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [3]{-x^2+x^4}} \, dx=\text {Too large to display} \]
1/16*2^(2/3)*(-1)^(1/6)*(sqrt(-3) - 1)*log(-2*(8*(x^4 - x^2)^(2/3)*((512*I - 59)*x^2 - (118*I + 1024)*x - 512*I + 59) + 4*2^(1/3)*(x^4 - x^2)^(1/3)* ((-1)^(5/6)*(59*x^3 + 1024*x^2 + sqrt(-3)*(59*x^3 + 1024*x^2 - 59*x) - 59* x) + 2*(-1)^(1/3)*(256*x^3 - 59*x^2 + sqrt(-3)*(256*x^3 - 59*x^2 - 256*x) - 256*x)) + 2^(2/3)*(4*(-1)^(2/3)*(128*x^5 - 59*x^4 - 768*x^3 + 59*x^2 - s qrt(-3)*(128*x^5 - 59*x^4 - 768*x^3 + 59*x^2 + 128*x) + 128*x) - (-1)^(1/6 )*(59*x^5 + 2048*x^4 - 354*x^3 - 2048*x^2 - sqrt(-3)*(59*x^5 + 2048*x^4 - 354*x^3 - 2048*x^2 + 59*x) + 59*x)))/(x^5 + 2*x^3 + x)) - 1/16*2^(2/3)*(-1 )^(1/6)*(sqrt(-3) + 1)*log(-2*(8*(x^4 - x^2)^(2/3)*((512*I - 59)*x^2 - (11 8*I + 1024)*x - 512*I + 59) + 4*2^(1/3)*(x^4 - x^2)^(1/3)*((-1)^(5/6)*(59* x^3 + 1024*x^2 - sqrt(-3)*(59*x^3 + 1024*x^2 - 59*x) - 59*x) + 2*(-1)^(1/3 )*(256*x^3 - 59*x^2 - sqrt(-3)*(256*x^3 - 59*x^2 - 256*x) - 256*x)) + 2^(2 /3)*(4*(-1)^(2/3)*(128*x^5 - 59*x^4 - 768*x^3 + 59*x^2 + sqrt(-3)*(128*x^5 - 59*x^4 - 768*x^3 + 59*x^2 + 128*x) + 128*x) - (-1)^(1/6)*(59*x^5 + 2048 *x^4 - 354*x^3 - 2048*x^2 + sqrt(-3)*(59*x^5 + 2048*x^4 - 354*x^3 - 2048*x ^2 + 59*x) + 59*x)))/(x^5 + 2*x^3 + x)) - 1/16*2^(2/3)*(-1)^(1/6)*(sqrt(-3 ) - 1)*log(-2*(8*(x^4 - x^2)^(2/3)*(-(512*I + 59)*x^2 + (118*I - 1024)*x + 512*I + 59) - 4*2^(1/3)*(x^4 - x^2)^(1/3)*((-1)^(5/6)*(59*x^3 + 1024*x^2 + sqrt(-3)*(59*x^3 + 1024*x^2 - 59*x) - 59*x) - 2*(-1)^(1/3)*(256*x^3 - 59 *x^2 + sqrt(-3)*(256*x^3 - 59*x^2 - 256*x) - 256*x)) + 2^(2/3)*(4*(-1)^...
\[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [3]{-x^2+x^4}} \, dx=\int \frac {\left (x - 1\right ) \left (x + 1\right )}{\sqrt [3]{x^{2} \left (x - 1\right ) \left (x + 1\right )} \left (x^{2} + 1\right )}\, dx \]
\[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [3]{-x^2+x^4}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{4} - x^{2}\right )}^{\frac {1}{3}} {\left (x^{2} + 1\right )}} \,d x } \]
\[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [3]{-x^2+x^4}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{4} - x^{2}\right )}^{\frac {1}{3}} {\left (x^{2} + 1\right )}} \,d x } \]
Timed out. \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [3]{-x^2+x^4}} \, dx=\int \frac {x^2-1}{\left (x^2+1\right )\,{\left (x^4-x^2\right )}^{1/3}} \,d x \]