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]

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)

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Rubi [A]  time = 0.06, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.08, 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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fricas [B]  time = 3.86, size = 415, normalized size = 6.29 \[ \frac {1}{2} \, \sqrt {3} \arctan \left (\frac {4 \, \sqrt {3} {\left (q^{12} - 15 \, q^{10} + 90 \, q^{8} - 351 \, q^{6} + 810 \, q^{4} - 1215 \, q^{2} + 729\right )} {\left (x^{3} - q x\right )}^{\frac {1}{3}} x - 2 \, \sqrt {3} {\left (q^{12} + 6 \, q^{11} - 15 \, q^{10} - 54 \, q^{9} + 90 \, q^{8} + 270 \, q^{7} - 351 \, q^{6} - 810 \, q^{5} + 810 \, q^{4} + 1458 \, q^{3} - 1215 \, q^{2} - 1458 \, q + 729\right )} {\left (x^{3} - q x\right )}^{\frac {2}{3}} - \sqrt {3} {\left (q^{13} + 10 \, q^{12} - 15 \, q^{11} - 282 \, q^{10} + 90 \, q^{9} + 2178 \, q^{8} - 351 \, q^{7} - 6534 \, q^{6} + 810 \, q^{5} + 7614 \, q^{4} - 1215 \, q^{3} - {\left (q^{12} - 6 \, q^{11} - 15 \, q^{10} + 54 \, q^{9} + 90 \, q^{8} - 270 \, q^{7} - 351 \, q^{6} + 810 \, q^{5} + 810 \, q^{4} - 1458 \, q^{3} - 1215 \, q^{2} + 1458 \, q + 729\right )} x^{2} - 2430 \, q^{2} + 729 \, q\right )}}{q^{13} + 18 \, q^{12} + 81 \, q^{11} - 162 \, q^{10} - 1350 \, q^{9} + 810 \, q^{8} + 6561 \, q^{7} - 2430 \, q^{6} - 12150 \, q^{5} + 4374 \, q^{4} + 6561 \, q^{3} - 9 \, {\left (q^{12} + 2 \, q^{11} - 15 \, q^{10} - 18 \, q^{9} + 90 \, q^{8} + 90 \, q^{7} - 351 \, q^{6} - 270 \, q^{5} + 810 \, q^{4} + 486 \, q^{3} - 1215 \, q^{2} - 486 \, q + 729\right )} x^{2} - 4374 \, q^{2} + 729 \, q}\right ) - \frac {1}{4} \, \log \left (-3 \, {\left (x^{3} - q x\right )}^{\frac {1}{3}} x + q + 3 \, {\left (x^{3} - q x\right )}^{\frac {2}{3}}\right ) \]

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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giac [A]  time = 1.05, size = 67, normalized size = 1.02 \[ -\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-\frac {q}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {1}{4} \, \log \left ({\left (-\frac {q}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (-\frac {q}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {1}{2} \, \log \left ({\left | {\left (-\frac {q}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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))

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (\left (x^{2}-q \right ) x \right )^{\frac {1}{3}}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left ({\left (x^{2} - q\right )} x\right )^{\frac {1}{3}}}\,{d x} \]

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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mupad [B]  time = 0.39, size = 37, normalized size = 0.56 \[ \frac {3\,x\,{\left (1-\frac {x^2}{q}\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ \frac {x^2}{q}\right )}{2\,{\left (x^3-q\,x\right )}^{1/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-x*(q - x^2))^(1/3),x)

[Out]

(3*x*(1 - x^2/q)^(1/3)*hypergeom([1/3, 1/3], 4/3, x^2/q))/(2*(x^3 - q*x)^(1/3))

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

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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