Integrand size = 77, antiderivative size = 276 \[ \int \frac {x^2 (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x^2}{\sqrt [3]{b} x^2+2 \left (x+(-1-k) x^2+k x^3\right )^{2/3}}\right )}{2 b^{2/3}}+\frac {\log \left (-\sqrt [6]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{2/3}}+\frac {\log \left (\sqrt [6]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{2/3}}-\frac {\log \left (\sqrt [3]{b} x^2-\sqrt [6]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{2/3}}-\frac {\log \left (\sqrt [3]{b} x^2+\sqrt [6]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{2/3}} \]
1/2*3^(1/2)*arctan(3^(1/2)*b^(1/3)*x^2/(b^(1/3)*x^2+2*(x+(-1-k)*x^2+k*x^3) ^(2/3)))/b^(2/3)+1/2*ln(-b^(1/6)*x+(x+(-1-k)*x^2+k*x^3)^(1/3))/b^(2/3)+1/2 *ln(b^(1/6)*x+(x+(-1-k)*x^2+k*x^3)^(1/3))/b^(2/3)-1/4*ln(b^(1/3)*x^2-b^(1/ 6)*x*(x+(-1-k)*x^2+k*x^3)^(1/3)+(x+(-1-k)*x^2+k*x^3)^(2/3))/b^(2/3)-1/4*ln (b^(1/3)*x^2+b^(1/6)*x*(x+(-1-k)*x^2+k*x^3)^(1/3)+(x+(-1-k)*x^2+k*x^3)^(2/ 3))/b^(2/3)
Time = 12.43 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.70 \[ \int \frac {x^2 (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=-\frac {2 \sqrt {3} \arctan \left (\frac {1+\frac {2 ((-1+x) x (-1+k x))^{2/3}}{\sqrt [3]{b} x^2}}{\sqrt {3}}\right )-2 \log \left (-\sqrt [6]{b} x+\sqrt [3]{(-1+x) x (-1+k x)}\right )-2 \log \left (\sqrt [6]{b} x+\sqrt [3]{(-1+x) x (-1+k x)}\right )+\log \left (\sqrt [3]{b} x^2-\sqrt [6]{b} x \sqrt [3]{(-1+x) x (-1+k x)}+((-1+x) x (-1+k x))^{2/3}\right )+\log \left (\sqrt [3]{b} x^2+\sqrt [6]{b} x \sqrt [3]{(-1+x) x (-1+k x)}+((-1+x) x (-1+k x))^{2/3}\right )}{4 b^{2/3}} \]
Integrate[(x^2*(-2 + (1 + k)*x))/(((1 - x)*x*(1 - k*x))^(1/3)*(1 - (2 + 2* k)*x + (1 + 4*k + k^2)*x^2 - (2*k + 2*k^2)*x^3 + (-b + k^2)*x^4)),x]
-1/4*(2*Sqrt[3]*ArcTan[(1 + (2*((-1 + x)*x*(-1 + k*x))^(2/3))/(b^(1/3)*x^2 ))/Sqrt[3]] - 2*Log[-(b^(1/6)*x) + ((-1 + x)*x*(-1 + k*x))^(1/3)] - 2*Log[ b^(1/6)*x + ((-1 + x)*x*(-1 + k*x))^(1/3)] + Log[b^(1/3)*x^2 - b^(1/6)*x*( (-1 + x)*x*(-1 + k*x))^(1/3) + ((-1 + x)*x*(-1 + k*x))^(2/3)] + Log[b^(1/3 )*x^2 + b^(1/6)*x*((-1 + x)*x*(-1 + k*x))^(1/3) + ((-1 + x)*x*(-1 + k*x))^ (2/3)])/b^(2/3)
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {x^2 ((k+1) x-2)}{\sqrt [3]{(1-x) x (1-k x)} \left (x^4 \left (k^2-b\right )-\left (2 k^2+2 k\right ) x^3+\left (k^2+4 k+1\right ) x^2-(2 k+2) x+1\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int -\frac {x^{5/3} (2-(k+1) x)}{\sqrt [3]{k x^2-(k+1) x+1} \left (-\left (\left (b-k^2\right ) x^4\right )-2 k (k+1) x^3+\left (k^2+4 k+1\right ) x^2-2 (k+1) x+1\right )}dx}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {x^{5/3} (2-(k+1) x)}{\sqrt [3]{k x^2-(k+1) x+1} \left (-\left (\left (b-k^2\right ) x^4\right )-2 k (k+1) x^3+\left (k^2+4 k+1\right ) x^2-2 (k+1) x+1\right )}dx}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {x^{7/3} (2-(k+1) x)}{\sqrt [3]{k x^2-(k+1) x+1} \left (-\left (\left (b-k^2\right ) x^4\right )-2 k (k+1) x^3+\left (k^2+4 k+1\right ) x^2-2 (k+1) x+1\right )}d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \left (\frac {(-k-1) x^{10/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b \left (1-\frac {k^2}{b}\right ) x^4-2 k (k+1) x^3+(k (k+4)+1) x^2-2 (k+1) x+1\right )}+\frac {2 x^{7/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b \left (1-\frac {k^2}{b}\right ) x^4-2 k (k+1) x^3+(k (k+4)+1) x^2-2 (k+1) x+1\right )}\right )d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \left (2 \int \frac {x^{7/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b \left (1-\frac {k^2}{b}\right ) x^4-2 k (k+1) x^3+(k (k+4)+1) x^2-2 (k+1) x+1\right )}d\sqrt [3]{x}-(k+1) \int \frac {x^{10/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b \left (1-\frac {k^2}{b}\right ) x^4-2 k (k+1) x^3+(k (k+4)+1) x^2-2 (k+1) x+1\right )}d\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\) |
Int[(x^2*(-2 + (1 + k)*x))/(((1 - x)*x*(1 - k*x))^(1/3)*(1 - (2 + 2*k)*x + (1 + 4*k + k^2)*x^2 - (2*k + 2*k^2)*x^3 + (-b + k^2)*x^4)),x]
3.29.13.3.1 Defintions of rubi rules used
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Time = 0.56 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.46
| method | result | size |
| pseudoelliptic | \(\frac {-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x^{2}+2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}\right )}{3 b^{\frac {1}{3}} x^{2}}\right )+2 \ln \left (\frac {-b^{\frac {1}{3}} x^{2}+\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )-\ln \left (\frac {b^{\frac {1}{3}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}} x +\left (-1+x \right ) \left (k x -1\right ) \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}+b^{\frac {2}{3}} x^{3}}{x^{3}}\right )}{4 b^{\frac {2}{3}}}\) | \(127\) |
int(x^2*(-2+(1+k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(1-(2+2*k)*x+(k^2+4*k+1)*x^2 -(2*k^2+2*k)*x^3+(k^2-b)*x^4),x,method=_RETURNVERBOSE)
1/4*(-2*3^(1/2)*arctan(1/3*3^(1/2)*(b^(1/3)*x^2+2*((-1+x)*x*(k*x-1))^(2/3) )/b^(1/3)/x^2)+2*ln((-b^(1/3)*x^2+((-1+x)*x*(k*x-1))^(2/3))/x^2)-ln((b^(1/ 3)*((-1+x)*x*(k*x-1))^(2/3)*x+(-1+x)*(k*x-1)*((-1+x)*x*(k*x-1))^(1/3)+b^(2 /3)*x^3)/x^3))/b^(2/3)
Timed out. \[ \int \frac {x^2 (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\text {Timed out} \]
integrate(x^2*(-2+(1+k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(1-(2+2*k)*x+(k^2+4*k+ 1)*x^2-(2*k^2+2*k)*x^3+(k^2-b)*x^4),x, algorithm="fricas")
Timed out. \[ \int \frac {x^2 (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\text {Timed out} \]
integrate(x**2*(-2+(1+k)*x)/((1-x)*x*(-k*x+1))**(1/3)/(1-(2+2*k)*x+(k**2+4 *k+1)*x**2-(2*k**2+2*k)*x**3+(k**2-b)*x**4),x)
\[ \int \frac {x^2 (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\int { \frac {{\left ({\left (k + 1\right )} x - 2\right )} x^{2}}{{\left ({\left (k^{2} - b\right )} x^{4} - 2 \, {\left (k^{2} + k\right )} x^{3} + {\left (k^{2} + 4 \, k + 1\right )} x^{2} - 2 \, {\left (k + 1\right )} x + 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}} \,d x } \]
integrate(x^2*(-2+(1+k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(1-(2+2*k)*x+(k^2+4*k+ 1)*x^2-(2*k^2+2*k)*x^3+(k^2-b)*x^4),x, algorithm="maxima")
integrate(((k + 1)*x - 2)*x^2/(((k^2 - b)*x^4 - 2*(k^2 + k)*x^3 + (k^2 + 4 *k + 1)*x^2 - 2*(k + 1)*x + 1)*((k*x - 1)*(x - 1)*x)^(1/3)), x)
Time = 0.41 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.45 \[ \int \frac {x^2 (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + b^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{2 \, b^{\frac {2}{3}}} - \frac {\log \left ({\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {4}{3}} + b^{\frac {1}{3}} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + b^{\frac {2}{3}}\right )}{4 \, b^{\frac {2}{3}}} + \frac {\log \left ({\left | {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} - b^{\frac {1}{3}} \right |}\right )}{2 \, b^{\frac {2}{3}}} \]
integrate(x^2*(-2+(1+k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(1-(2+2*k)*x+(k^2+4*k+ 1)*x^2-(2*k^2+2*k)*x^3+(k^2-b)*x^4),x, algorithm="giac")
-1/2*sqrt(3)*arctan(1/3*sqrt(3)*(2*(k - k/x - 1/x + 1/x^2)^(2/3) + b^(1/3) )/b^(1/3))/b^(2/3) - 1/4*log((k - k/x - 1/x + 1/x^2)^(4/3) + b^(1/3)*(k - k/x - 1/x + 1/x^2)^(2/3) + b^(2/3))/b^(2/3) + 1/2*log(abs((k - k/x - 1/x + 1/x^2)^(2/3) - b^(1/3)))/b^(2/3)
Timed out. \[ \int \frac {x^2 (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\int -\frac {x^2\,\left (x\,\left (k+1\right )-2\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (\left (b-k^2\right )\,x^4+\left (2\,k^2+2\,k\right )\,x^3+\left (-k^2-4\,k-1\right )\,x^2+\left (2\,k+2\right )\,x-1\right )} \,d x \]
int(-(x^2*(x*(k + 1) - 2))/((x*(k*x - 1)*(x - 1))^(1/3)*(x*(2*k + 2) + x^4 *(b - k^2) - x^2*(4*k + k^2 + 1) + x^3*(2*k + 2*k^2) - 1)),x)