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

Optimal. Leaf size=79 \[ -\frac{3}{4} \log \left (\sqrt [3]{(x-1) \left (q+x^2-2 x\right )}-x+1\right )+\frac{1}{2} \sqrt{3} \tan ^{-1}\left (\frac{2 (x-1)}{\sqrt{3} \sqrt [3]{(x-1) \left (q+x^2-2 x\right )}}+\frac{1}{\sqrt{3}}\right )+\frac{1}{4} \log (1-x) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0978022, antiderivative size = 145, normalized size of antiderivative = 1.84, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {2067, 2011, 329, 275, 239} \[ \frac{\sqrt{3} \sqrt [3]{x-1} \sqrt [3]{q+(x-1)^2-1} \tan ^{-1}\left (\frac{\frac{2 (x-1)^{2/3}}{\sqrt [3]{q+(x-1)^2-1}}+1}{\sqrt{3}}\right )}{2 \sqrt [3]{(x-1)^3-(1-q) (x-1)}}-\frac{3 \sqrt [3]{x-1} \sqrt [3]{q+(x-1)^2-1} \log \left ((x-1)^{2/3}-\sqrt [3]{q+(x-1)^2-1}\right )}{4 \sqrt [3]{(x-1)^3-(1-q) (x-1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2067

Int[(P3_)^(p_), x_Symbol] :> With[{a = Coeff[P3, x, 0], b = Coeff[P3, x, 1], c = Coeff[P3, x, 2], d = Coeff[P3
, x, 3]}, Subst[Int[Simp[(2*c^3 - 9*b*c*d + 27*a*d^2)/(27*d^2) - ((c^2 - 3*b*d)*x)/(3*d) + d*x^3, x]^p, x], x,
x + c/(3*d)] /; NeQ[c, 0]] /; FreeQ[p, x] && PolyQ[P3, x, 3]

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

Mathematica [A]  time = 0.166759, size = 140, normalized size = 1.77 \[ \frac{\sqrt [3]{x-1} \sqrt [3]{q+(x-2) x} \left (-2 \log \left (1-\frac{(x-1)^{2/3}}{\sqrt [3]{q+(x-2) x}}\right )+\log \left (\frac{(x-1)^{4/3}}{(q+(x-2) x)^{2/3}}+\frac{(x-1)^{2/3}}{\sqrt [3]{q+(x-2) x}}+1\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{\frac{2 (x-1)^{2/3}}{\sqrt [3]{q+(x-2) x}}+1}{\sqrt{3}}\right )\right )}{4 \sqrt [3]{(x-1) (q+(x-2) x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.012, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{\sqrt [3]{ \left ( -1+x \right ) \left ({x}^{2}+q-2\,x \right ) }}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 12.361, size = 1866, normalized size = 23.62 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(3)*arctan((2*sqrt(3)*(q^12 - 18*q^11 + 117*q^10 - 346*q^9 + 414*q^8 - 18*q^7 + 69*q^6 - 774*q^5 - 234
*q^4 + 1058*q^3 + 621*q^2 + 378*q - 539)*(x^3 + (q + 2)*x - 3*x^2 - q)^(2/3) + 4*sqrt(3)*(q^12 - 12*q^11 + 51*
q^10 - 70*q^9 - 90*q^8 + 288*q^7 - 57*q^6 + 54*q^5 - 810*q^4 + 320*q^3 + 291*q^2 - (q^12 - 12*q^11 + 51*q^10 -
70*q^9 - 90*q^8 + 288*q^7 - 57*q^6 + 54*q^5 - 810*q^4 + 320*q^3 + 291*q^2 + 714*q + 49)*x + 714*q + 49)*(x^3
+ (q + 2)*x - 3*x^2 - q)^(1/3) - sqrt(3)*(q^13 - 22*q^12 + 177*q^11 - 514*q^10 - 434*q^9 + 5346*q^8 - 8247*q^7
- 4542*q^6 + 19638*q^5 - 8050*q^4 - 10343*q^3 + (q^12 - 6*q^11 - 15*q^10 + 206*q^9 - 594*q^8 + 594*q^7 - 183*
q^6 + 882*q^5 - 1386*q^4 - 418*q^3 - 39*q^2 + 1050*q + 637)*x^2 + 6186*q^2 - 2*(q^12 - 6*q^11 - 15*q^10 + 206*
q^9 - 594*q^8 + 594*q^7 - 183*q^6 + 882*q^5 - 1386*q^4 - 418*q^3 - 39*q^2 + 1050*q + 637)*x + 1501*q + 32))/(q
^13 - 22*q^12 + 249*q^11 - 1546*q^10 + 4702*q^9 - 4230*q^8 - 10623*q^7 + 25338*q^6 - 3546*q^5 - 31306*q^4 + 18
817*q^3 + 9*(q^12 - 14*q^11 + 73*q^10 - 162*q^9 + 78*q^8 + 186*q^7 - 15*q^6 - 222*q^5 - 618*q^4 + 566*q^3 + 40
1*q^2 + 602*q - 147)*x^2 + 9714*q^2 - 18*(q^12 - 14*q^11 + 73*q^10 - 162*q^9 + 78*q^8 + 186*q^7 - 15*q^6 - 222
*q^5 - 618*q^4 + 566*q^3 + 401*q^2 + 602*q - 147)*x - 995*q + 8)) - 1/4*log(3*(x^3 + (q + 2)*x - 3*x^2 - q)^(1
/3)*(x - 1) + q - 3*(x^3 + (q + 2)*x - 3*x^2 - q)^(2/3) - 1)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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