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

Optimal. Leaf size=331 \[ \frac {(b c-a d) \log \left (x \sqrt [3]{c-d}+\sqrt [3]{d} \sqrt [3]{x^3-x}\right )}{2 c d^{2/3} \sqrt [3]{c-d}}-\frac {\sqrt {3} (b c-a d) \tan ^{-1}\left (\frac {\sqrt {3} x \sqrt [3]{c-d}}{x \sqrt [3]{c-d}-2 \sqrt [3]{d} \sqrt [3]{x^3-x}}\right )}{2 c d^{2/3} \sqrt [3]{c-d}}+\frac {(a d-b c) \log \left (-\sqrt [3]{d} \sqrt [3]{x^3-x} x \sqrt [3]{c-d}+x^2 (c-d)^{2/3}+d^{2/3} \left (x^3-x\right )^{2/3}\right )}{4 c d^{2/3} \sqrt [3]{c-d}}-\frac {a \log \left (\sqrt [3]{x^3-x}-x\right )}{2 c}+\frac {\sqrt {3} a \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3-x}+x}\right )}{2 c}+\frac {a \log \left (\sqrt [3]{x^3-x} x+\left (x^3-x\right )^{2/3}+x^2\right )}{4 c} \]

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Rubi [A]  time = 0.55, antiderivative size = 417, normalized size of antiderivative = 1.26, number of steps used = 15, number of rules used = 14, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {2056, 584, 329, 275, 239, 466, 465, 377, 200, 31, 634, 617, 204, 628} \begin {gather*} \frac {\sqrt [3]{x} \sqrt [3]{x^2-1} (b c-a d) \log \left (\frac {x^{2/3} \sqrt [3]{c-d}}{\sqrt [3]{x^2-1}}+\sqrt [3]{d}\right )}{2 c d^{2/3} \sqrt [3]{x^3-x} \sqrt [3]{c-d}}-\frac {\sqrt [3]{x} \sqrt [3]{x^2-1} (b c-a d) \log \left (\frac {x^{4/3} (c-d)^{2/3}}{\left (x^2-1\right )^{2/3}}-\frac {\sqrt [3]{d} x^{2/3} \sqrt [3]{c-d}}{\sqrt [3]{x^2-1}}+d^{2/3}\right )}{4 c d^{2/3} \sqrt [3]{x^3-x} \sqrt [3]{c-d}}-\frac {\sqrt {3} \sqrt [3]{x} \sqrt [3]{x^2-1} (b c-a d) \tan ^{-1}\left (\frac {\sqrt [3]{d}-\frac {2 x^{2/3} \sqrt [3]{c-d}}{\sqrt [3]{x^2-1}}}{\sqrt {3} \sqrt [3]{d}}\right )}{2 c d^{2/3} \sqrt [3]{x^3-x} \sqrt [3]{c-d}}-\frac {3 a \sqrt [3]{x} \sqrt [3]{x^2-1} \log \left (x^{2/3}-\sqrt [3]{x^2-1}\right )}{4 c \sqrt [3]{x^3-x}}+\frac {\sqrt {3} a \sqrt [3]{x} \sqrt [3]{x^2-1} \tan ^{-1}\left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2-1}}+1}{\sqrt {3}}\right )}{2 c \sqrt [3]{x^3-x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3]*a*x^(1/3)*(-1 + x^2)^(1/3)*ArcTan[(1 + (2*x^(2/3))/(-1 + x^2)^(1/3))/Sqrt[3]])/(2*c*(-x + x^3)^(1/3))
- (Sqrt[3]*(b*c - a*d)*x^(1/3)*(-1 + x^2)^(1/3)*ArcTan[(d^(1/3) - (2*(c - d)^(1/3)*x^(2/3))/(-1 + x^2)^(1/3))
/(Sqrt[3]*d^(1/3))])/(2*c*(c - d)^(1/3)*d^(2/3)*(-x + x^3)^(1/3)) + ((b*c - a*d)*x^(1/3)*(-1 + x^2)^(1/3)*Log[
d^(1/3) + ((c - d)^(1/3)*x^(2/3))/(-1 + x^2)^(1/3)])/(2*c*(c - d)^(1/3)*d^(2/3)*(-x + x^3)^(1/3)) - ((b*c - a*
d)*x^(1/3)*(-1 + x^2)^(1/3)*Log[d^(2/3) + ((c - d)^(2/3)*x^(4/3))/(-1 + x^2)^(2/3) - ((c - d)^(1/3)*d^(1/3)*x^
(2/3))/(-1 + x^2)^(1/3)])/(4*c*(c - d)^(1/3)*d^(2/3)*(-x + x^3)^(1/3)) - (3*a*x^(1/3)*(-1 + x^2)^(1/3)*Log[x^(
2/3) - (-1 + x^2)^(1/3)])/(4*c*(-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 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 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 584

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Sy
mbol] :> Int[ExpandIntegrand[((g*x)^m*(a + b*x^n)^p*(e + f*x^n))/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, p}, x] && IGtQ[n, 0]

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

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Mathematica [A]  time = 0.60, size = 310, normalized size = 0.94 \begin {gather*} \frac {\sqrt [3]{x} \sqrt [3]{x^2-1} \left (\frac {2 (b c-a d) \log \left (\frac {x^{2/3} \sqrt [3]{c-d}}{\sqrt [3]{x^2-1}}+\sqrt [3]{d}\right )}{d^{2/3} \sqrt [3]{c-d}}-\frac {(b c-a d) \left (\log \left (\frac {x^{4/3} (c-d)^{2/3}}{\left (x^2-1\right )^{2/3}}-\frac {\sqrt [3]{d} x^{2/3} \sqrt [3]{c-d}}{\sqrt [3]{x^2-1}}+d^{2/3}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {1-\frac {2 x^{2/3} \sqrt [3]{c-d}}{\sqrt [3]{d} \sqrt [3]{x^2-1}}}{\sqrt {3}}\right )\right )}{d^{2/3} \sqrt [3]{c-d}}-2 a \log \left (1-\frac {x^{2/3}}{\sqrt [3]{x^2-1}}\right )+a \left (\log \left (\frac {x^{4/3}}{\left (x^2-1\right )^{2/3}}+\frac {x^{2/3}}{\sqrt [3]{x^2-1}}+1\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2-1}}+1}{\sqrt {3}}\right )\right )\right )}{4 c \sqrt [3]{x \left (x^2-1\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 3.01, size = 334, normalized size = 1.01 \begin {gather*} \frac {\sqrt {3} a \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-x+x^3}}\right )}{2 c}-\frac {\sqrt {3} (b c-a d) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{c-d} x}{\sqrt [3]{c-d} x-2 \sqrt [3]{d} \sqrt [3]{-x+x^3}}\right )}{2 c \sqrt [3]{c-d} d^{2/3}}-\frac {a \log \left (-c x+c \sqrt [3]{-x+x^3}\right )}{2 c}+\frac {(b c-a d) \log \left (\sqrt [3]{c-d} x+\sqrt [3]{d} \sqrt [3]{-x+x^3}\right )}{2 c \sqrt [3]{c-d} d^{2/3}}+\frac {a \log \left (x^2+x \sqrt [3]{-x+x^3}+\left (-x+x^3\right )^{2/3}\right )}{4 c}+\frac {(-b c+a d) \log \left ((c-d)^{2/3} x^2-\sqrt [3]{c-d} \sqrt [3]{d} x \sqrt [3]{-x+x^3}+d^{2/3} \left (-x+x^3\right )^{2/3}\right )}{4 c \sqrt [3]{c-d} d^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3]*a*ArcTan[(Sqrt[3]*x)/(x + 2*(-x + x^3)^(1/3))])/(2*c) - (Sqrt[3]*(b*c - a*d)*ArcTan[(Sqrt[3]*(c - d)^
(1/3)*x)/((c - d)^(1/3)*x - 2*d^(1/3)*(-x + x^3)^(1/3))])/(2*c*(c - d)^(1/3)*d^(2/3)) - (a*Log[-(c*x) + c*(-x
+ x^3)^(1/3)])/(2*c) + ((b*c - a*d)*Log[(c - d)^(1/3)*x + d^(1/3)*(-x + x^3)^(1/3)])/(2*c*(c - d)^(1/3)*d^(2/3
)) + (a*Log[x^2 + x*(-x + x^3)^(1/3) + (-x + x^3)^(2/3)])/(4*c) + ((-(b*c) + a*d)*Log[(c - d)^(2/3)*x^2 - (c -
d)^(1/3)*d^(1/3)*x*(-x + x^3)^(1/3) + d^(2/3)*(-x + x^3)^(2/3)])/(4*c*(c - d)^(1/3)*d^(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((a*x^2-b)/(c*x^2-d)/(x^3-x)^(1/3),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.27, size = 300, normalized size = 0.91 \begin {gather*} -\frac {\sqrt {3} a \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right )}{2 \, c} + \frac {{\left (b c \left (-\frac {c - d}{d}\right )^{\frac {1}{3}} - a d \left (-\frac {c - d}{d}\right )^{\frac {1}{3}}\right )} \left (-\frac {c - d}{d}\right )^{\frac {1}{3}} \log \left ({\left | -\left (-\frac {c - d}{d}\right )^{\frac {1}{3}} + {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} \right |}\right )}{2 \, {\left (c^{2} - c d\right )}} + \frac {a \log \left ({\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}{4 \, c} - \frac {a \log \left ({\left | {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right )}{2 \, c} - \frac {{\left (\sqrt {3} b c - \sqrt {3} a d\right )} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {c - d}{d}\right )^{\frac {1}{3}} + 2 \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {c - d}{d}\right )^{\frac {1}{3}}}\right )}{2 \, {\left (-c d^{2} + d^{3}\right )}^{\frac {1}{3}} c} + \frac {{\left (b c - a d\right )} \log \left (\left (-\frac {c - d}{d}\right )^{\frac {2}{3}} + \left (-\frac {c - d}{d}\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 (-c d^{2} + d^{3}\right )}^{\frac {1}{3}} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(3)*a*arctan(1/3*sqrt(3)*(2*(-1/x^2 + 1)^(1/3) + 1))/c + 1/2*(b*c*(-(c - d)/d)^(1/3) - a*d*(-(c - d)/
d)^(1/3))*(-(c - d)/d)^(1/3)*log(abs(-(-(c - d)/d)^(1/3) + (-1/x^2 + 1)^(1/3)))/(c^2 - c*d) + 1/4*a*log((-1/x^
2 + 1)^(2/3) + (-1/x^2 + 1)^(1/3) + 1)/c - 1/2*a*log(abs((-1/x^2 + 1)^(1/3) - 1))/c - 1/2*(sqrt(3)*b*c - sqrt(
3)*a*d)*arctan(1/3*sqrt(3)*((-(c - d)/d)^(1/3) + 2*(-1/x^2 + 1)^(1/3))/(-(c - d)/d)^(1/3))/((-c*d^2 + d^3)^(1/
3)*c) + 1/4*(b*c - a*d)*log((-(c - d)/d)^(2/3) + (-(c - d)/d)^(1/3)*(-1/x^2 + 1)^(1/3) + (-1/x^2 + 1)^(2/3))/(
(-c*d^2 + d^3)^(1/3)*c)

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{2}-b}{\left (c \,x^{2}-d \right ) \left (x^{3}-x \right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a*x^2-b)/(c*x^2-d)/(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 {a x^{2} - b}{{\left (c x^{2} - d\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((a*x^2-b)/(c*x^2-d)/(x^3-x)^(1/3),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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