Integrand size = 13, antiderivative size = 66 \[ \int \frac {1}{\sqrt [3]{x \left (-q+x^2\right )}} \, dx=\frac {1}{2} \sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 x}{\sqrt {3} \sqrt [3]{x \left (-q+x^2\right )}}\right )+\frac {\log (x)}{4}-\frac {3}{4} \log \left (-x+\sqrt [3]{x \left (-q+x^2\right )}\right ) \]
1/4*ln(x)-3/4*ln(-x+(x*(x^2-q))^(1/3))+1/2*arctan(1/3*3^(1/2)+2/3*x/(x*(x^ 2-q))^(1/3)*3^(1/2))*3^(1/2)
Time = 0.82 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.97 \[ \int \frac {1}{\sqrt [3]{x \left (-q+x^2\right )}} \, dx=\frac {\sqrt [3]{x} \sqrt [3]{-q+x^2} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x^{2/3}}{x^{2/3}+2 \sqrt [3]{-q+x^2}}\right )-2 \log \left (-x^{2/3}+\sqrt [3]{-q+x^2}\right )+\log \left (x^{4/3}+x^{2/3} \sqrt [3]{-q+x^2}+\left (-q+x^2\right )^{2/3}\right )\right )}{4 \sqrt [3]{-q x+x^3}} \]
(x^(1/3)*(-q + x^2)^(1/3)*(2*Sqrt[3]*ArcTan[(Sqrt[3]*x^(2/3))/(x^(2/3) + 2 *(-q + x^2)^(1/3))] - 2*Log[-x^(2/3) + (-q + x^2)^(1/3)] + Log[x^(4/3) + x ^(2/3)*(-q + x^2)^(1/3) + (-q + x^2)^(2/3)]))/(4*(-(q*x) + x^3)^(1/3))
Time = 0.21 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.30, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2078, 1917, 266, 807, 769}
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.
\(\displaystyle \int \frac {1}{\sqrt [3]{x \left (x^2-q\right )}} \, dx\) |
\(\Big \downarrow \) 2078 |
\(\displaystyle \int \frac {1}{\sqrt [3]{x^3-q x}}dx\) |
\(\Big \downarrow \) 1917 |
\(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{x^2-q} \int \frac {1}{\sqrt [3]{x} \sqrt [3]{x^2-q}}dx}{\sqrt [3]{x^3-q x}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{x^2-q} \int \frac {\sqrt [3]{x}}{\sqrt [3]{x^2-q}}d\sqrt [3]{x}}{\sqrt [3]{x^3-q x}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{x^2-q} \int \frac {1}{\sqrt [3]{x-q}}dx^{2/3}}{2 \sqrt [3]{x^3-q x}}\) |
\(\Big \downarrow \) 769 |
\(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{x^2-q} \left (\frac {\arctan \left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x-q}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (\sqrt [3]{x-q}-x^{2/3}\right )\right )}{2 \sqrt [3]{x^3-q x}}\) |
(3*x^(1/3)*(-q + x^2)^(1/3)*(ArcTan[(1 + (2*x^(2/3))/(-q + x)^(1/3))/Sqrt[ 3]]/Sqrt[3] - Log[-x^(2/3) + (-q + x)^(1/3)]/2))/(2*(-(q*x) + x^3)^(1/3))
3.1.41.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]* (x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^ 3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]) Int[ x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] && !Integ erQ[p] && NeQ[n, j] && PosQ[n - j]
Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && G eneralizedBinomialQ[u, x] && !GeneralizedBinomialMatchQ[u, x]
Time = 0.85 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.39
method | result | size |
pseudoelliptic | \(\frac {\ln \left (\frac {{\left (-x \left (-x^{2}+q \right )\right )}^{\frac {2}{3}}+{\left (-x \left (-x^{2}+q \right )\right )}^{\frac {1}{3}} x +x^{2}}{x^{2}}\right )}{4}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (x +2 {\left (-x \left (-x^{2}+q \right )\right )}^{\frac {1}{3}}\right ) \sqrt {3}}{3 x}\right )}{2}-\frac {\ln \left (\frac {{\left (-x \left (-x^{2}+q \right )\right )}^{\frac {1}{3}}-x}{x}\right )}{2}\) | \(92\) |
1/4*ln(((-x*(-x^2+q))^(2/3)+(-x*(-x^2+q))^(1/3)*x+x^2)/x^2)-1/2*3^(1/2)*ar ctan(1/3*(x+2*(-x*(-x^2+q))^(1/3))*3^(1/2)/x)-1/2*ln(((-x*(-x^2+q))^(1/3)- x)/x)
Leaf count of result is larger than twice the leaf count of optimal. 415 vs. \(2 (52) = 104\).
Time = 0.90 (sec) , antiderivative size = 415, normalized size of antiderivative = 6.29 \[ \int \frac {1}{\sqrt [3]{x \left (-q+x^2\right )}} \, dx=\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {4 \, \sqrt {3} {\left (q^{12} - 15 \, q^{10} + 90 \, q^{8} - 351 \, q^{6} + 810 \, q^{4} - 1215 \, q^{2} + 729\right )} {\left (x^{3} - q x\right )}^{\frac {1}{3}} x - 2 \, \sqrt {3} {\left (q^{12} + 6 \, q^{11} - 15 \, q^{10} - 54 \, q^{9} + 90 \, q^{8} + 270 \, q^{7} - 351 \, q^{6} - 810 \, q^{5} + 810 \, q^{4} + 1458 \, q^{3} - 1215 \, q^{2} - 1458 \, q + 729\right )} {\left (x^{3} - q x\right )}^{\frac {2}{3}} - \sqrt {3} {\left (q^{13} + 10 \, q^{12} - 15 \, q^{11} - 282 \, q^{10} + 90 \, q^{9} + 2178 \, q^{8} - 351 \, q^{7} - 6534 \, q^{6} + 810 \, q^{5} + 7614 \, q^{4} - 1215 \, q^{3} - {\left (q^{12} - 6 \, q^{11} - 15 \, q^{10} + 54 \, q^{9} + 90 \, q^{8} - 270 \, q^{7} - 351 \, q^{6} + 810 \, q^{5} + 810 \, q^{4} - 1458 \, q^{3} - 1215 \, q^{2} + 1458 \, q + 729\right )} x^{2} - 2430 \, q^{2} + 729 \, q\right )}}{q^{13} + 18 \, q^{12} + 81 \, q^{11} - 162 \, q^{10} - 1350 \, q^{9} + 810 \, q^{8} + 6561 \, q^{7} - 2430 \, q^{6} - 12150 \, q^{5} + 4374 \, q^{4} + 6561 \, q^{3} - 9 \, {\left (q^{12} + 2 \, q^{11} - 15 \, q^{10} - 18 \, q^{9} + 90 \, q^{8} + 90 \, q^{7} - 351 \, q^{6} - 270 \, q^{5} + 810 \, q^{4} + 486 \, q^{3} - 1215 \, q^{2} - 486 \, q + 729\right )} x^{2} - 4374 \, q^{2} + 729 \, q}\right ) - \frac {1}{4} \, \log \left (-3 \, {\left (x^{3} - q x\right )}^{\frac {1}{3}} x + q + 3 \, {\left (x^{3} - q x\right )}^{\frac {2}{3}}\right ) \]
1/2*sqrt(3)*arctan((4*sqrt(3)*(q^12 - 15*q^10 + 90*q^8 - 351*q^6 + 810*q^4 - 1215*q^2 + 729)*(x^3 - q*x)^(1/3)*x - 2*sqrt(3)*(q^12 + 6*q^11 - 15*q^1 0 - 54*q^9 + 90*q^8 + 270*q^7 - 351*q^6 - 810*q^5 + 810*q^4 + 1458*q^3 - 1 215*q^2 - 1458*q + 729)*(x^3 - q*x)^(2/3) - sqrt(3)*(q^13 + 10*q^12 - 15*q ^11 - 282*q^10 + 90*q^9 + 2178*q^8 - 351*q^7 - 6534*q^6 + 810*q^5 + 7614*q ^4 - 1215*q^3 - (q^12 - 6*q^11 - 15*q^10 + 54*q^9 + 90*q^8 - 270*q^7 - 351 *q^6 + 810*q^5 + 810*q^4 - 1458*q^3 - 1215*q^2 + 1458*q + 729)*x^2 - 2430* q^2 + 729*q))/(q^13 + 18*q^12 + 81*q^11 - 162*q^10 - 1350*q^9 + 810*q^8 + 6561*q^7 - 2430*q^6 - 12150*q^5 + 4374*q^4 + 6561*q^3 - 9*(q^12 + 2*q^11 - 15*q^10 - 18*q^9 + 90*q^8 + 90*q^7 - 351*q^6 - 270*q^5 + 810*q^4 + 486*q^ 3 - 1215*q^2 - 486*q + 729)*x^2 - 4374*q^2 + 729*q)) - 1/4*log(-3*(x^3 - q *x)^(1/3)*x + q + 3*(x^3 - q*x)^(2/3))
\[ \int \frac {1}{\sqrt [3]{x \left (-q+x^2\right )}} \, dx=\int \frac {1}{\sqrt [3]{x \left (- q + x^{2}\right )}}\, dx \]
\[ \int \frac {1}{\sqrt [3]{x \left (-q+x^2\right )}} \, dx=\int { \frac {1}{\left ({\left (x^{2} - q\right )} x\right )^{\frac {1}{3}}} \,d x } \]
Time = 0.32 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\sqrt [3]{x \left (-q+x^2\right )}} \, dx=-\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-\frac {q}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {1}{4} \, \log \left ({\left (-\frac {q}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (-\frac {q}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {1}{2} \, \log \left ({\left | {\left (-\frac {q}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]
-1/2*sqrt(3)*arctan(1/3*sqrt(3)*(2*(-q/x^2 + 1)^(1/3) + 1)) + 1/4*log((-q/ x^2 + 1)^(2/3) + (-q/x^2 + 1)^(1/3) + 1) - 1/2*log(abs((-q/x^2 + 1)^(1/3) - 1))
Time = 0.48 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.56 \[ \int \frac {1}{\sqrt [3]{x \left (-q+x^2\right )}} \, dx=\frac {3\,x\,{\left (1-\frac {x^2}{q}\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ \frac {x^2}{q}\right )}{2\,{\left (x^3-q\,x\right )}^{1/3}} \]