\(\int \frac {-1+(2-k) x}{\sqrt [3]{(1-x) x (1-k x)} (1-(b+2 k) x+(b+k^2) x^2)} \, dx\) [2469]

Optimal result

Integrand size = 48, antiderivative size = 201 \[ \int \frac {-1+(2-k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(b+2 k) x+\left (b+k^2\right ) x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}{2-2 k x+\sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{2/3}}+\frac {\log \left (-1+k x+\sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{b^{2/3}}-\frac {\log \left (1-2 k x+k^2 x^2+\left (\sqrt [3]{b}-\sqrt [3]{b} k x\right ) \sqrt [3]{x+(-1-k) x^2+k x^3}+b^{2/3} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 b^{2/3}} \]

[Out]

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

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

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 15.49 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.80 \[ \int \frac {-1+(2-k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(b+2 k) x+\left (b+k^2\right ) x^2\right )} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{(-1+x) x (-1+k x)}}{2-2 k x+\sqrt [3]{b} \sqrt [3]{(-1+x) x (-1+k x)}}\right )+2 \log \left (-1+k x+\sqrt [3]{b} \sqrt [3]{(-1+x) x (-1+k x)}\right )-\log \left (1-2 k x+k^2 x^2+\sqrt [3]{b} (1-k x) \sqrt [3]{(-1+x) x (-1+k x)}+b^{2/3} ((-1+x) x (-1+k x))^{2/3}\right )}{2 b^{2/3}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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