\(\int \frac {-1+2 x+(-2 k+k^2) x^2}{((1-x) x (1-k x))^{2/3} (1-(b+2 k) x+(b+k^2) x^2)} \, dx\) [2580]

   Optimal result
   Rubi [F]
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F(-1)]
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 55, antiderivative size = 223 \[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(b+2 k) x+\left (b+k^2\right ) x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}{2 \sqrt [3]{b} x-2 \sqrt [3]{b} x^2+\left (x+(-1-k) x^2+k x^3\right )^{2/3}}\right )}{\sqrt [3]{b}}+\frac {\log \left (-\sqrt {b} x+\sqrt {b} x^2+\sqrt [6]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{\sqrt [3]{b}}-\frac {\log \left (b x^2-2 b x^3+b x^4+\left (b^{2/3} x-b^{2/3} x^2\right ) \left (x+(-1-k) x^2+k x^3\right )^{2/3}+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{4/3}\right )}{2 \sqrt [3]{b}} \]

[Out]

3^(1/2)*arctan(3^(1/2)*(x+(-1-k)*x^2+k*x^3)^(2/3)/(2*b^(1/3)*x-2*b^(1/3)*x^2+(x+(-1-k)*x^2+k*x^3)^(2/3)))/b^(1
/3)+ln(-x*b^(1/2)+b^(1/2)*x^2+b^(1/6)*(x+(-1-k)*x^2+k*x^3)^(2/3))/b^(1/3)-1/2*ln(b*x^2-2*b*x^3+b*x^4+(b^(2/3)*
x-b^(2/3)*x^2)*(x+(-1-k)*x^2+k*x^3)^(2/3)+b^(1/3)*(x+(-1-k)*x^2+k*x^3)^(4/3))/b^(1/3)

Rubi [F]

\[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(b+2 k) x+\left (b+k^2\right ) x^2\right )} \, dx=\int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(b+2 k) x+\left (b+k^2\right ) x^2\right )} \, dx \]

[In]

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

[Out]

((2 - k*(1 + Sqrt[-4 + b + 4*k]/Sqrt[b]))*(1 - x)^(2/3)*x^(2/3)*(1 - k*x)^(2/3)*Defer[Int][(1 - k*x)^(1/3)/((1
 - x)^(2/3)*x^(2/3)*(-b - 2*k - Sqrt[b]*Sqrt[-4 + b + 4*k] + 2*(b + k^2)*x)), x])/((1 - x)*x*(1 - k*x))^(2/3)
+ ((2 - k*(1 - Sqrt[-4 + b + 4*k]/Sqrt[b]))*(1 - x)^(2/3)*x^(2/3)*(1 - k*x)^(2/3)*Defer[Int][(1 - k*x)^(1/3)/(
(1 - x)^(2/3)*x^(2/3)*(-b - 2*k + Sqrt[b]*Sqrt[-4 + b + 4*k] + 2*(b + k^2)*x)), x])/((1 - x)*x*(1 - k*x))^(2/3
)

Rubi steps \begin{align*} \text {integral}& = \frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{(1-x)^{2/3} x^{2/3} (1-k x)^{2/3} \left (1-(b+2 k) x+\left (b+k^2\right ) x^2\right )} \, dx}{((1-x) x (1-k x))^{2/3}} \\ & = \frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {(-1+(2-k) x) \sqrt [3]{1-k x}}{(1-x)^{2/3} x^{2/3} \left (1-(b+2 k) x+\left (b+k^2\right ) x^2\right )} \, dx}{((1-x) x (1-k x))^{2/3}} \\ & = \frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \left (\frac {\left (2-k-\frac {k \sqrt {-4+b+4 k}}{\sqrt {b}}\right ) \sqrt [3]{1-k x}}{(1-x)^{2/3} x^{2/3} \left (-b-2 k-\sqrt {b} \sqrt {-4+b+4 k}+2 \left (b+k^2\right ) x\right )}+\frac {\left (2-k+\frac {k \sqrt {-4+b+4 k}}{\sqrt {b}}\right ) \sqrt [3]{1-k x}}{(1-x)^{2/3} x^{2/3} \left (-b-2 k+\sqrt {b} \sqrt {-4+b+4 k}+2 \left (b+k^2\right ) x\right )}\right ) \, dx}{((1-x) x (1-k x))^{2/3}} \\ & = \frac {\left (\left (2-k \left (1-\frac {\sqrt {-4+b+4 k}}{\sqrt {b}}\right )\right ) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {\sqrt [3]{1-k x}}{(1-x)^{2/3} x^{2/3} \left (-b-2 k+\sqrt {b} \sqrt {-4+b+4 k}+2 \left (b+k^2\right ) x\right )} \, dx}{((1-x) x (1-k x))^{2/3}}+\frac {\left (\left (2-k \left (1+\frac {\sqrt {-4+b+4 k}}{\sqrt {b}}\right )\right ) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {\sqrt [3]{1-k x}}{(1-x)^{2/3} x^{2/3} \left (-b-2 k-\sqrt {b} \sqrt {-4+b+4 k}+2 \left (b+k^2\right ) x\right )} \, dx}{((1-x) x (1-k x))^{2/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 15.69 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.74 \[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(b+2 k) x+\left (b+k^2\right ) x^2\right )} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} ((-1+x) x (-1+k x))^{2/3}}{-2 \sqrt [3]{b} (-1+x) x+((-1+x) x (-1+k x))^{2/3}}\right )+2 \log \left (\sqrt [6]{b} \left (\sqrt [3]{b} (-1+x) x+((-1+x) x (-1+k x))^{2/3}\right )\right )-\log \left (b x^2-2 b x^3+b x^4-b^{2/3} (-1+x) x ((-1+x) x (-1+k x))^{2/3}+\sqrt [3]{b} ((-1+x) x (-1+k x))^{4/3}\right )}{2 \sqrt [3]{b}} \]

[In]

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

[Out]

(2*Sqrt[3]*ArcTan[(Sqrt[3]*((-1 + x)*x*(-1 + k*x))^(2/3))/(-2*b^(1/3)*(-1 + x)*x + ((-1 + x)*x*(-1 + k*x))^(2/
3))] + 2*Log[b^(1/6)*(b^(1/3)*(-1 + x)*x + ((-1 + x)*x*(-1 + k*x))^(2/3))] - Log[b*x^2 - 2*b*x^3 + b*x^4 - b^(
2/3)*(-1 + x)*x*((-1 + x)*x*(-1 + k*x))^(2/3) + b^(1/3)*((-1 + x)*x*(-1 + k*x))^(4/3)])/(2*b^(1/3))

Maple [F]

\[\int \frac {-1+2 x +\left (k^{2}-2 k \right ) x^{2}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (1-\left (b +2 k \right ) x +\left (k^{2}+b \right ) x^{2}\right )}d x\]

[In]

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

[Out]

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

Fricas [F(-1)]

Timed out. \[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(b+2 k) x+\left (b+k^2\right ) x^2\right )} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(b+2 k) x+\left (b+k^2\right ) x^2\right )} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(b+2 k) x+\left (b+k^2\right ) x^2\right )} \, dx=\int { \frac {{\left (k^{2} - 2 \, k\right )} x^{2} + 2 \, x - 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}} {\left ({\left (k^{2} + b\right )} x^{2} - {\left (b + 2 \, k\right )} x + 1\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(b+2 k) x+\left (b+k^2\right ) x^2\right )} \, dx=\int { \frac {{\left (k^{2} - 2 \, k\right )} x^{2} + 2 \, x - 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}} {\left ({\left (k^{2} + b\right )} x^{2} - {\left (b + 2 \, k\right )} x + 1\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(b+2 k) x+\left (b+k^2\right ) x^2\right )} \, dx=\int -\frac {\left (2\,k-k^2\right )\,x^2-2\,x+1}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}\,\left (\left (k^2+b\right )\,x^2+\left (-b-2\,k\right )\,x+1\right )} \,d x \]

[In]

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

[Out]

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