Integrand size = 77, antiderivative size = 224 \[ \int \frac {x^3 (-2+(1+k) x)}{((1-x) x (1-k x))^{2/3} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [6]{b} x}{\sqrt [6]{b} x-2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{5/6}}+\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [6]{b} x}{\sqrt [6]{b} x+2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{5/6}}-\frac {\text {arctanh}\left (\frac {\sqrt [6]{b} x}{\sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{5/6}}-\frac {\text {arctanh}\left (\frac {\sqrt [6]{b} x^2+\frac {\left (x+(-1-k) x^2+k x^3\right )^{2/3}}{\sqrt [6]{b}}}{x \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{5/6}} \]
-1/2*3^(1/2)*arctan(3^(1/2)*b^(1/6)*x/(b^(1/6)*x-2*(x+(-1-k)*x^2+k*x^3)^(1 /3)))/b^(5/6)+1/2*3^(1/2)*arctan(3^(1/2)*b^(1/6)*x/(b^(1/6)*x+2*(x+(-1-k)* x^2+k*x^3)^(1/3)))/b^(5/6)-arctanh(b^(1/6)*x/(x+(-1-k)*x^2+k*x^3)^(1/3))/b ^(5/6)-1/2*arctanh((b^(1/6)*x^2+(x+(-1-k)*x^2+k*x^3)^(2/3)/b^(1/6))/x/(x+( -1-k)*x^2+k*x^3)^(1/3))/b^(5/6)
Time = 14.25 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.76 \[ \int \frac {x^3 (-2+(1+k) x)}{((1-x) x (1-k x))^{2/3} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\frac {\sqrt {3} \left (-\arctan \left (\frac {\sqrt {3} \sqrt [6]{b} x}{\sqrt [6]{b} x-2 \sqrt [3]{(-1+x) x (-1+k x)}}\right )+\arctan \left (\frac {\sqrt {3} \sqrt [6]{b} x}{\sqrt [6]{b} x+2 \sqrt [3]{(-1+x) x (-1+k x)}}\right )\right )-2 \text {arctanh}\left (\frac {\sqrt [6]{b} x}{\sqrt [3]{(-1+x) x (-1+k x)}}\right )-\text {arctanh}\left (\frac {\sqrt [3]{b} x^2+((-1+x) x (-1+k x))^{2/3}}{\sqrt [6]{b} x \sqrt [3]{(-1+x) x (-1+k x)}}\right )}{2 b^{5/6}} \]
Integrate[(x^3*(-2 + (1 + k)*x))/(((1 - x)*x*(1 - k*x))^(2/3)*(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]
(Sqrt[3]*(-ArcTan[(Sqrt[3]*b^(1/6)*x)/(b^(1/6)*x - 2*((-1 + x)*x*(-1 + k*x ))^(1/3))] + ArcTan[(Sqrt[3]*b^(1/6)*x)/(b^(1/6)*x + 2*((-1 + x)*x*(-1 + k *x))^(1/3))]) - 2*ArcTanh[(b^(1/6)*x)/((-1 + x)*x*(-1 + k*x))^(1/3)] - Arc Tanh[(b^(1/3)*x^2 + ((-1 + x)*x*(-1 + k*x))^(2/3))/(b^(1/6)*x*((-1 + x)*x* (-1 + k*x))^(1/3))])/(2*b^(5/6))
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.
| \(\displaystyle \int \frac {x^3 ((k+1) x-2)}{((1-x) x (1-k x))^{2/3} \left (x^4 \left (k^2-b\right )-\left (2 k^2+2 k\right ) x^3+\left (k^2+4 k+1\right ) x^2-(2 k+2) x+1\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int -\frac {x^{7/3} (2-(k+1) x)}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-\left (\left (b-k^2\right ) x^4\right )-2 k (k+1) x^3+\left (k^2+4 k+1\right ) x^2-2 (k+1) x+1\right )}dx}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \frac {x^{7/3} (2-(k+1) x)}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-\left (\left (b-k^2\right ) x^4\right )-2 k (k+1) x^3+\left (k^2+4 k+1\right ) x^2-2 (k+1) x+1\right )}dx}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \frac {x^3 (2-(k+1) x)}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-\left (\left (b-k^2\right ) x^4\right )-2 k (k+1) x^3+\left (k^2+4 k+1\right ) x^2-2 (k+1) x+1\right )}d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \left (\frac {k+1}{\left (b-k^2\right ) \left (k x^2-(k+1) x+1\right )^{2/3}}-\frac {-2 \left (k^3+k^2+k+b\right ) x^3+(k+1) \left (k^2+4 k+1\right ) x^2-2 (k+1)^2 x+k+1}{\left (b-k^2\right ) \left (k x^2-(k+1) x+1\right )^{2/3} \left (-\left (\left (b-k^2\right ) x^4\right )-2 k (k+1) x^3+\left (k^2+4 k+1\right ) x^2-2 (k+1) x+1\right )}\right )d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \left (-\frac {2 (k+1)^2 \int \frac {x}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (b \left (1-\frac {k^2}{b}\right ) x^4+2 k (k+1) x^3-(k (k+4)+1) x^2+2 (k+1) x-1\right )}d\sqrt [3]{x}}{b-k^2}-\frac {(k+1) \int \frac {1}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-b \left (1-\frac {k^2}{b}\right ) x^4-2 k (k+1) x^3+(k (k+4)+1) x^2-2 (k+1) x+1\right )}d\sqrt [3]{x}}{b-k^2}-\frac {\left (k^2+4 k+1\right ) (k+1) \int \frac {x^2}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-b \left (1-\frac {k^2}{b}\right ) x^4-2 k (k+1) x^3+(k (k+4)+1) x^2-2 (k+1) x+1\right )}d\sqrt [3]{x}}{b-k^2}+\frac {2 \left (b+k^3+k^2+k\right ) \int \frac {x^3}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-b \left (1-\frac {k^2}{b}\right ) x^4-2 k (k+1) x^3+(k (k+4)+1) x^2-2 (k+1) x+1\right )}d\sqrt [3]{x}}{b-k^2}+\frac {(k+1) (1-x)^{2/3} \sqrt [3]{x} (1-k x)^{2/3} \sqrt [3]{1-\frac {2 k x}{-\sqrt {k^2-2 k+1}+k+1}} \left (\frac {1-\frac {2 k x}{\sqrt {k^2-2 k+1}+k+1}}{1-\frac {2 k x}{-\sqrt {k^2-2 k+1}+k+1}}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {\sqrt {k^2-2 k+1} x}{1-\frac {2 k x}{k-\sqrt {k^2-2 k+1}+1}}\right )}{\left (b-k^2\right ) \left (1-\frac {2 k x}{\sqrt {k^2-2 k+1}+k+1}\right )^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3}}\right )}{((1-x) x (1-k x))^{2/3}}\) |
Int[(x^3*(-2 + (1 + k)*x))/(((1 - x)*x*(1 - k*x))^(2/3)*(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]
3.26.88.3.1 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 = 1.10 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.99
| method | result | size |
| pseudoelliptic | \(\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{6}} x -2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{6}} x}\right )-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{6}} x +2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{6}} x}\right )+2 \ln \left (\frac {b^{\frac {1}{6}} x -\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}}{x}\right )+\ln \left (\frac {b^{\frac {1}{6}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}} x -b^{\frac {1}{3}} x^{2}-\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {b^{\frac {1}{6}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {b^{\frac {1}{6}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}} x +b^{\frac {1}{3}} x^{2}+\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{4 b^{\frac {5}{6}}}\) | \(222\) |
int(x^3*(-2+(1+k)*x)/((1-x)*x*(-k*x+1))^(2/3)/(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^(1/6)*x-2*((-1+x)*x*(k*x-1))^(1/3))/b ^(1/6)/x)-2*3^(1/2)*arctan(1/3*3^(1/2)*(b^(1/6)*x+2*((-1+x)*x*(k*x-1))^(1/ 3))/b^(1/6)/x)+2*ln((b^(1/6)*x-((-1+x)*x*(k*x-1))^(1/3))/x)+ln((b^(1/6)*(( -1+x)*x*(k*x-1))^(1/3)*x-b^(1/3)*x^2-((-1+x)*x*(k*x-1))^(2/3))/x^2)-2*ln(( b^(1/6)*x+((-1+x)*x*(k*x-1))^(1/3))/x)-ln((b^(1/6)*((-1+x)*x*(k*x-1))^(1/3 )*x+b^(1/3)*x^2+((-1+x)*x*(k*x-1))^(2/3))/x^2))/b^(5/6)
Timed out. \[ \int \frac {x^3 (-2+(1+k) x)}{((1-x) x (1-k x))^{2/3} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\text {Timed out} \]
integrate(x^3*(-2+(1+k)*x)/((1-x)*x*(-k*x+1))^(2/3)/(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. \[ \int \frac {x^3 (-2+(1+k) x)}{((1-x) x (1-k x))^{2/3} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\text {Timed out} \]
integrate(x**3*(-2+(1+k)*x)/((1-x)*x*(-k*x+1))**(2/3)/(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)
\[ \int \frac {x^3 (-2+(1+k) x)}{((1-x) x (1-k x))^{2/3} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\int { \frac {{\left ({\left (k + 1\right )} x - 2\right )} x^{3}}{{\left ({\left (k^{2} - b\right )} x^{4} - 2 \, {\left (k^{2} + k\right )} x^{3} + {\left (k^{2} + 4 \, k + 1\right )} x^{2} - 2 \, {\left (k + 1\right )} x + 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}}} \,d x } \]
integrate(x^3*(-2+(1+k)*x)/((1-x)*x*(-k*x+1))^(2/3)/(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^3/(((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)^(2/3)), x)
Time = 0.57 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.17 \[ \int \frac {x^3 (-2+(1+k) x)}{((1-x) x (1-k x))^{2/3} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=-\frac {\sqrt {3} \left (-b\right )^{\frac {1}{6}} \log \left (\sqrt {3} \left (-b\right )^{\frac {1}{6}} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \left (-b\right )^{\frac {1}{3}}\right )}{4 \, b} + \frac {\sqrt {3} \left (-b\right )^{\frac {1}{6}} \log \left (-\sqrt {3} \left (-b\right )^{\frac {1}{6}} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \left (-b\right )^{\frac {1}{3}}\right )}{4 \, b} - \frac {\left (-b\right )^{\frac {1}{6}} \arctan \left (\frac {\sqrt {3} \left (-b\right )^{\frac {1}{6}} + 2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{\left (-b\right )^{\frac {1}{6}}}\right )}{2 \, b} - \frac {\left (-b\right )^{\frac {1}{6}} \arctan \left (-\frac {\sqrt {3} \left (-b\right )^{\frac {1}{6}} - 2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{\left (-b\right )^{\frac {1}{6}}}\right )}{2 \, b} - \frac {\left (-b\right )^{\frac {1}{6}} \arctan \left (\frac {{\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{\left (-b\right )^{\frac {1}{6}}}\right )}{b} \]
integrate(x^3*(-2+(1+k)*x)/((1-x)*x*(-k*x+1))^(2/3)/(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")
-1/4*sqrt(3)*(-b)^(1/6)*log(sqrt(3)*(-b)^(1/6)*(k - k/x - 1/x + 1/x^2)^(1/ 3) + (k - k/x - 1/x + 1/x^2)^(2/3) + (-b)^(1/3))/b + 1/4*sqrt(3)*(-b)^(1/6 )*log(-sqrt(3)*(-b)^(1/6)*(k - k/x - 1/x + 1/x^2)^(1/3) + (k - k/x - 1/x + 1/x^2)^(2/3) + (-b)^(1/3))/b - 1/2*(-b)^(1/6)*arctan((sqrt(3)*(-b)^(1/6) + 2*(k - k/x - 1/x + 1/x^2)^(1/3))/(-b)^(1/6))/b - 1/2*(-b)^(1/6)*arctan(- (sqrt(3)*(-b)^(1/6) - 2*(k - k/x - 1/x + 1/x^2)^(1/3))/(-b)^(1/6))/b - (-b )^(1/6)*arctan((k - k/x - 1/x + 1/x^2)^(1/3)/(-b)^(1/6))/b
Timed out. \[ \int \frac {x^3 (-2+(1+k) x)}{((1-x) x (1-k x))^{2/3} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\int -\frac {x^3\,\left (x\,\left (k+1\right )-2\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}\,\left (\left (b-k^2\right )\,x^4+\left (2\,k^2+2\,k\right )\,x^3+\left (-k^2-4\,k-1\right )\,x^2+\left (2\,k+2\right )\,x-1\right )} \,d x \]
int(-(x^3*(x*(k + 1) - 2))/((x*(k*x - 1)*(x - 1))^(2/3)*(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)