\(\int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} (b-b (1+k) x+(-1+b k) x^2)} \, dx\) [2133]

   Optimal result
   Rubi [F]
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [F(-1)]
   Sympy [F(-1)]
   Maxima [F]
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 45, antiderivative size = 155 \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{2/3}}+\frac {\log \left (x-\sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{b^{2/3}}-\frac {\log \left (x^2+\sqrt [3]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+b^{2/3} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 b^{2/3}} \]

[Out]

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

Rubi [F]

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

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 15.27 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.79 \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=-\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{b} \sqrt [3]{(-1+x) x (-1+k x)}}\right )-2 \log \left (x-\sqrt [3]{b} \sqrt [3]{(-1+x) x (-1+k x)}\right )+\log \left (x^2+\sqrt [3]{b} x \sqrt [3]{(-1+x) x (-1+k x)}+b^{2/3} ((-1+x) x (-1+k x))^{2/3}\right )}{2 b^{2/3}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.83

method result size
pseudoelliptic \(\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {1}{b}\right )^{\frac {1}{3}} x +2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}\right )}{3 \left (\frac {1}{b}\right )^{\frac {1}{3}} x}\right )+2 \ln \left (\frac {-\left (\frac {1}{b}\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {\left (\frac {1}{b}\right )^{\frac {2}{3}} x^{2}+\left (\frac {1}{b}\right )^{\frac {1}{3}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2 \left (\frac {1}{b}\right )^{\frac {1}{3}} b}\) \(129\)

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.83 \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=\frac {\sqrt {3} {\left | b \right |}^{\frac {4}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} b^{\frac {1}{3}} {\left (2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} + \frac {1}{b^{\frac {1}{3}}}\right )}\right )}{b^{2}} - \frac {{\left | b \right |}^{\frac {4}{3}} \log \left ({\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \frac {{\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{b^{\frac {1}{3}}} + \frac {1}{b^{\frac {2}{3}}}\right )}{2 \, b^{2}} + \frac {\log \left ({\left | {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} - \frac {1}{b^{\frac {1}{3}}} \right |}\right )}{b^{\frac {2}{3}}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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