3.25.49 \(\int \frac {1}{(-b+a x^2) \sqrt [3]{-x+x^3}} \, dx\)

Optimal. Leaf size=199 \[ -\frac {\log \left (x \sqrt [3]{a-b}+\sqrt [3]{b} \sqrt [3]{x^3-x}\right )}{2 b^{2/3} \sqrt [3]{a-b}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x \sqrt [3]{a-b}}{x \sqrt [3]{a-b}-2 \sqrt [3]{b} \sqrt [3]{x^3-x}}\right )}{2 b^{2/3} \sqrt [3]{a-b}}+\frac {\log \left (-\sqrt [3]{b} \sqrt [3]{x^3-x} x \sqrt [3]{a-b}+x^2 (a-b)^{2/3}+b^{2/3} \left (x^3-x\right )^{2/3}\right )}{4 b^{2/3} \sqrt [3]{a-b}} \]

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Rubi [A]  time = 0.34, antiderivative size = 270, normalized size of antiderivative = 1.36, number of steps used = 10, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {2056, 466, 465, 377, 200, 31, 634, 617, 204, 628} \begin {gather*} -\frac {\sqrt [3]{x} \sqrt [3]{x^2-1} \log \left (\frac {x^{2/3} \sqrt [3]{a-b}}{\sqrt [3]{x^2-1}}+\sqrt [3]{b}\right )}{2 b^{2/3} \sqrt [3]{x^3-x} \sqrt [3]{a-b}}+\frac {\sqrt [3]{x} \sqrt [3]{x^2-1} \log \left (\frac {x^{4/3} (a-b)^{2/3}}{\left (x^2-1\right )^{2/3}}-\frac {\sqrt [3]{b} x^{2/3} \sqrt [3]{a-b}}{\sqrt [3]{x^2-1}}+b^{2/3}\right )}{4 b^{2/3} \sqrt [3]{x^3-x} \sqrt [3]{a-b}}+\frac {\sqrt {3} \sqrt [3]{x} \sqrt [3]{x^2-1} \tan ^{-1}\left (\frac {\sqrt [3]{b}-\frac {2 x^{2/3} \sqrt [3]{a-b}}{\sqrt [3]{x^2-1}}}{\sqrt {3} \sqrt [3]{b}}\right )}{2 b^{2/3} \sqrt [3]{x^3-x} \sqrt [3]{a-b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((-b + a*x^2)*(-x + x^3)^(1/3)),x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 465

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1,
n]}, Dist[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] /;
FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 466

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/e^n)^p*(c + (d*x^(k*n))/e^n)^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps

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

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Mathematica [C]  time = 0.03, size = 75, normalized size = 0.38 \begin {gather*} -\frac {3 x \sqrt [3]{1-x^2} \, _2F_1\left (\frac {1}{3},\frac {1}{3};\frac {4}{3};\frac {(a-b) x^2}{a x^2-b}\right )}{2 b \sqrt [3]{x \left (x^2-1\right )} \sqrt [3]{1-\frac {a x^2}{b}}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((-b + a*x^2)*(-x + x^3)^(1/3)),x]

[Out]

(-3*x*(1 - x^2)^(1/3)*Hypergeometric2F1[1/3, 1/3, 4/3, ((a - b)*x^2)/(-b + a*x^2)])/(2*b*(x*(-1 + x^2))^(1/3)*
(1 - (a*x^2)/b)^(1/3))

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IntegrateAlgebraic [A]  time = 0.50, size = 199, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((-b + a*x^2)*(-x + x^3)^(1/3)),x]

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [A]  time = 0.52, size = 195, normalized size = 0.98 \begin {gather*} -\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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(abs(-(-(a - b)/b)^(1/3) + (-1/x^2 +
1)^(1/3)))/(a - b)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a \,x^{2}-b \right ) \left (x^{3}-x \right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (a x^{2} - b\right )} {\left (x^{3} - x\right )}^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((a*x^2 - b)*(x^3 - x)^(1/3)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {1}{{\left (x^3-x\right )}^{1/3}\,\left (b-a\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x**2-b)/(x**3-x)**(1/3),x)

[Out]

Integral(1/((x*(x - 1)*(x + 1))**(1/3)*(a*x**2 - b)), x)

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