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

Optimal. Leaf size=66 \[ -\frac{3}{4} \log \left (\sqrt [3]{x \left (x^2-q\right )}-x\right )+\frac{1}{2} \sqrt{3} \tan ^{-1}\left (\frac{2 x}{\sqrt{3} \sqrt [3]{x \left (x^2-q\right )}}+\frac{1}{\sqrt{3}}\right )+\frac{\log (x)}{4} \]

[Out]

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

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Rubi [A]  time = 0.0564275, antiderivative size = 117, normalized size of antiderivative = 1.77, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {1979, 2011, 329, 275, 239} \[ \frac{\sqrt{3} \sqrt [3]{x} \sqrt [3]{x^2-q} \tan ^{-1}\left (\frac{\frac{2 x^{2/3}}{\sqrt [3]{x^2-q}}+1}{\sqrt{3}}\right )}{2 \sqrt [3]{x^3-q x}}-\frac{3 \sqrt [3]{x} \sqrt [3]{x^2-q} \log \left (x^{2/3}-\sqrt [3]{x^2-q}\right )}{4 \sqrt [3]{x^3-q x}} \]

Antiderivative was successfully verified.

[In]

Int[(x*(-q + x^2))^(-1/3),x]

[Out]

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

Rule 1979

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && GeneralizedBinomialQ[u, x] &&  !Gene
ralizedBinomialMatchQ[u, x]

Rule 2011

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(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] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[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]

Rule 239

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]]/(Sq
rt[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]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt [3]{x \left (-q+x^2\right )}} \, dx &=\int \frac{1}{\sqrt [3]{-q x+x^3}} \, dx\\ &=\frac{\left (\sqrt [3]{x} \sqrt [3]{-q+x^2}\right ) \int \frac{1}{\sqrt [3]{x} \sqrt [3]{-q+x^2}} \, dx}{\sqrt [3]{-q x+x^3}}\\ &=\frac{\left (3 \sqrt [3]{x} \sqrt [3]{-q+x^2}\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt [3]{-q+x^6}} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{-q x+x^3}}\\ &=\frac{\left (3 \sqrt [3]{x} \sqrt [3]{-q+x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{-q+x^3}} \, dx,x,x^{2/3}\right )}{2 \sqrt [3]{-q x+x^3}}\\ &=\frac{\sqrt{3} \sqrt [3]{x} \sqrt [3]{-q+x^2} \tan ^{-1}\left (\frac{1+\frac{2 x^{2/3}}{\sqrt [3]{-q+x^2}}}{\sqrt{3}}\right )}{2 \sqrt [3]{-q x+x^3}}-\frac{3 \sqrt [3]{x} \sqrt [3]{-q+x^2} \log \left (x^{2/3}-\sqrt [3]{-q+x^2}\right )}{4 \sqrt [3]{-q x+x^3}}\\ \end{align*}

Mathematica [A]  time = 0.0822427, size = 127, normalized size = 1.92 \[ \frac{\sqrt [3]{x} \sqrt [3]{x^2-q} \left (-2 \log \left (1-\frac{x^{2/3}}{\sqrt [3]{x^2-q}}\right )+\log \left (\frac{x^{4/3}}{\left (x^2-q\right )^{2/3}}+\frac{x^{2/3}}{\sqrt [3]{x^2-q}}+1\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{\frac{2 x^{2/3}}{\sqrt [3]{x^2-q}}+1}{\sqrt{3}}\right )\right )}{4 \sqrt [3]{x^3-q x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(-q + x^2))^(-1/3),x]

[Out]

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

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Maple [F]  time = 0.042, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{\sqrt [3]{x \left ({x}^{2}-q \right ) }}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left ({\left (x^{2} - q\right )} x\right )^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x*(x^2-q))^(1/3),x, algorithm="maxima")

[Out]

integrate(((x^2 - q)*x)^(-1/3), x)

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Fricas [B]  time = 13.5156, size = 1153, normalized size = 17.47 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x*(x^2-q))^(1/3),x, algorithm="fricas")

[Out]

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^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^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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x*(x**2-q))**(1/3),x)

[Out]

Integral((x*(-q + x**2))**(-1/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left ({\left (x^{2} - q\right )} x\right )^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(((x^2 - q)*x)^(-1/3), x)