Integrand size = 23, antiderivative size = 199 \[ \int \frac {1}{\left (-b+a x^2\right ) \sqrt [3]{-x+x^3}} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{a-b} x}{\sqrt [3]{a-b} x-2 \sqrt [3]{b} \sqrt [3]{-x+x^3}}\right )}{2 \sqrt [3]{a-b} b^{2/3}}-\frac {\log \left (\sqrt [3]{a-b} x+\sqrt [3]{b} \sqrt [3]{-x+x^3}\right )}{2 \sqrt [3]{a-b} b^{2/3}}+\frac {\log \left ((a-b)^{2/3} x^2-\sqrt [3]{a-b} \sqrt [3]{b} x \sqrt [3]{-x+x^3}+b^{2/3} \left (-x+x^3\right )^{2/3}\right )}{4 \sqrt [3]{a-b} b^{2/3}} \]
1/2*3^(1/2)*arctan(3^(1/2)*(a-b)^(1/3)*x/((a-b)^(1/3)*x-2*b^(1/3)*(x^3-x)^ (1/3)))/(a-b)^(1/3)/b^(2/3)-1/2*ln((a-b)^(1/3)*x+b^(1/3)*(x^3-x)^(1/3))/(a -b)^(1/3)/b^(2/3)+1/4*ln((a-b)^(2/3)*x^2-(a-b)^(1/3)*b^(1/3)*x*(x^3-x)^(1/ 3)+b^(2/3)*(x^3-x)^(2/3))/(a-b)^(1/3)/b^(2/3)
Time = 5.02 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\left (-b+a x^2\right ) \sqrt [3]{-x+x^3}} \, dx=\frac {\sqrt [3]{x} \sqrt [3]{-1+x^2} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{a-b} x^{2/3}}{\sqrt [3]{a-b} x^{2/3}-2 \sqrt [3]{b} \sqrt [3]{-1+x^2}}\right )-2 \log \left (\sqrt [3]{a-b} x^{2/3}+\sqrt [3]{b} \sqrt [3]{-1+x^2}\right )+\log \left ((a-b)^{2/3} x^{4/3}-\sqrt [3]{a-b} \sqrt [3]{b} x^{2/3} \sqrt [3]{-1+x^2}+b^{2/3} \left (-1+x^2\right )^{2/3}\right )\right )}{4 \sqrt [3]{a-b} b^{2/3} \sqrt [3]{x \left (-1+x^2\right )}} \]
(x^(1/3)*(-1 + x^2)^(1/3)*(2*Sqrt[3]*ArcTan[(Sqrt[3]*(a - b)^(1/3)*x^(2/3) )/((a - b)^(1/3)*x^(2/3) - 2*b^(1/3)*(-1 + x^2)^(1/3))] - 2*Log[(a - b)^(1 /3)*x^(2/3) + b^(1/3)*(-1 + x^2)^(1/3)] + Log[(a - b)^(2/3)*x^(4/3) - (a - b)^(1/3)*b^(1/3)*x^(2/3)*(-1 + x^2)^(1/3) + b^(2/3)*(-1 + x^2)^(2/3)]))/( 4*(a - b)^(1/3)*b^(2/3)*(x*(-1 + x^2))^(1/3))
Time = 0.33 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.82, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2467, 25, 368, 965, 901}
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.
\(\displaystyle \int \frac {1}{\sqrt [3]{x^3-x} \left (a x^2-b\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{x^2-1} \int -\frac {1}{\sqrt [3]{x} \sqrt [3]{x^2-1} \left (b-a x^2\right )}dx}{\sqrt [3]{x^3-x}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt [3]{x} \sqrt [3]{x^2-1} \int \frac {1}{\sqrt [3]{x} \sqrt [3]{x^2-1} \left (b-a x^2\right )}dx}{\sqrt [3]{x^3-x}}\) |
\(\Big \downarrow \) 368 |
\(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{x^2-1} \int \frac {\sqrt [3]{x}}{\sqrt [3]{x^2-1} \left (b-a x^2\right )}d\sqrt [3]{x}}{\sqrt [3]{x^3-x}}\) |
\(\Big \downarrow \) 965 |
\(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{x^2-1} \int \frac {1}{\sqrt [3]{x-1} (b-a x)}dx^{2/3}}{2 \sqrt [3]{x^3-x}}\) |
\(\Big \downarrow \) 901 |
\(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{x^2-1} \left (-\frac {\arctan \left (\frac {1-\frac {2 x^{2/3} \sqrt [3]{a-b}}{\sqrt [3]{b} \sqrt [3]{x-1}}}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3} \sqrt [3]{a-b}}+\frac {\log \left (-\frac {x^{2/3} \sqrt [3]{a-b}}{\sqrt [3]{b}}-\sqrt [3]{x-1}\right )}{2 b^{2/3} \sqrt [3]{a-b}}-\frac {\log (b-a x)}{6 b^{2/3} \sqrt [3]{a-b}}\right )}{2 \sqrt [3]{x^3-x}}\) |
(-3*x^(1/3)*(-1 + x^2)^(1/3)*(-(ArcTan[(1 - (2*(a - b)^(1/3)*x^(2/3))/(b^( 1/3)*(-1 + x)^(1/3)))/Sqrt[3]]/(Sqrt[3]*(a - b)^(1/3)*b^(2/3))) + Log[-(-1 + x)^(1/3) - ((a - b)^(1/3)*x^(2/3))/b^(1/3)]/(2*(a - b)^(1/3)*b^(2/3)) - Log[b - a*x]/(6*(a - b)^(1/3)*b^(2/3))))/(2*(-x + x^3)^(1/3))
3.25.49.3.1 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[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> Wit h[{q = Rt[(b*c - a*d)/c, 3]}, Simp[ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/S qrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[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] /; Free Q[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]
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 = 1.04 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.77
method | result | size |
pseudoelliptic | \(-\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a -b}{b}\right )^{\frac {1}{3}} x -2 \left (x^{3}-x \right )^{\frac {1}{3}}\right )}{3 \left (\frac {a -b}{b}\right )^{\frac {1}{3}} x}\right )+2 \ln \left (\frac {\left (\frac {a -b}{b}\right )^{\frac {1}{3}} x +\left (x^{3}-x \right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {\left (\frac {a -b}{b}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a -b}{b}\right )^{\frac {1}{3}} \left (x^{3}-x \right )^{\frac {1}{3}} x +\left (x^{3}-x \right )^{\frac {2}{3}}}{x^{2}}\right )}{4 \left (\frac {a -b}{b}\right )^{\frac {1}{3}} b}\) | \(153\) |
-1/4/((a-b)/b)^(1/3)*(2*3^(1/2)*arctan(1/3*3^(1/2)*(((a-b)/b)^(1/3)*x-2*(x ^3-x)^(1/3))/((a-b)/b)^(1/3)/x)+2*ln((((a-b)/b)^(1/3)*x+(x^3-x)^(1/3))/x)- ln((((a-b)/b)^(2/3)*x^2-((a-b)/b)^(1/3)*(x^3-x)^(1/3)*x+(x^3-x)^(2/3))/x^2 ))/b
Timed out. \[ \int \frac {1}{\left (-b+a x^2\right ) \sqrt [3]{-x+x^3}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{\left (-b+a x^2\right ) \sqrt [3]{-x+x^3}} \, dx=\int \frac {1}{\sqrt [3]{x \left (x - 1\right ) \left (x + 1\right )} \left (a x^{2} - b\right )}\, dx \]
\[ \int \frac {1}{\left (-b+a x^2\right ) \sqrt [3]{-x+x^3}} \, dx=\int { \frac {1}{{\left (a x^{2} - b\right )} {\left (x^{3} - x\right )}^{\frac {1}{3}}} \,d x } \]
Time = 0.30 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\left (-b+a x^2\right ) \sqrt [3]{-x+x^3}} \, dx=-\frac {3 \, {\left (-a b^{2} + b^{3}\right )}^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {a - b}{b}\right )^{\frac {1}{3}} + 2 \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a - b}{b}\right )^{\frac {1}{3}}}\right )}{2 \, {\left (\sqrt {3} a b^{2} - \sqrt {3} b^{3}\right )}} + \frac {{\left (-a b^{2} + b^{3}\right )}^{\frac {2}{3}} \log \left (\left (-\frac {a - b}{b}\right )^{\frac {2}{3}} + \left (-\frac {a - b}{b}\right )^{\frac {1}{3}} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}}\right )}{4 \, {\left (a b^{2} - b^{3}\right )}} - \frac {\left (-\frac {a - b}{b}\right )^{\frac {2}{3}} \log \left ({\left | -\left (-\frac {a - b}{b}\right )^{\frac {1}{3}} + {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} \right |}\right )}{2 \, {\left (a - b\right )}} \]
-3/2*(-a*b^2 + b^3)^(2/3)*arctan(1/3*sqrt(3)*((-(a - b)/b)^(1/3) + 2*(-1/x ^2 + 1)^(1/3))/(-(a - b)/b)^(1/3))/(sqrt(3)*a*b^2 - sqrt(3)*b^3) + 1/4*(-a *b^2 + b^3)^(2/3)*log((-(a - b)/b)^(2/3) + (-(a - b)/b)^(1/3)*(-1/x^2 + 1) ^(1/3) + (-1/x^2 + 1)^(2/3))/(a*b^2 - b^3) - 1/2*(-(a - b)/b)^(2/3)*log(ab s(-(-(a - b)/b)^(1/3) + (-1/x^2 + 1)^(1/3)))/(a - b)
Timed out. \[ \int \frac {1}{\left (-b+a x^2\right ) \sqrt [3]{-x+x^3}} \, dx=-\int \frac {1}{{\left (x^3-x\right )}^{1/3}\,\left (b-a\,x^2\right )} \,d x \]