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

Optimal. Leaf size=192 \[ \frac {\log \left (\sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}+x-1\right )}{b^{2/3}}-\frac {\log \left (b^{2/3} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+\left (\sqrt [3]{b}-\sqrt [3]{b} x\right ) \sqrt [3]{k x^3+(-k-1) x^2+x}+x^2-2 x+1\right )}{2 b^{2/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}}{\sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}-2 x+2}\right )}{b^{2/3}} \]

________________________________________________________________________________________

Rubi [F]  time = 2.70, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-1+(-1+2 k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+b) x+(1+b k) x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [F]  time = 7.52, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-1+(-1+2 k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+b) x+(1+b k) x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.64, size = 192, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}{2-2 x+\sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{2/3}}+\frac {\log \left (-1+x+\sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{b^{2/3}}-\frac {\log \left (1-2 x+x^2+\left (\sqrt [3]{b}-\sqrt [3]{b} 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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, k - 1\right )} x - 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}} {\left ({\left (b k + 1\right )} x^{2} - {\left (b + 2\right )} x + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {-1+\left (-1+2 k \right ) x}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (1-\left (2+b \right ) x +\left (b k +1\right ) x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, k - 1\right )} x - 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}} {\left ({\left (b k + 1\right )} x^{2} - {\left (b + 2\right )} x + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,\left (2\,k-1\right )-1}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (\left (b\,k+1\right )\,x^2+\left (-b-2\right )\,x+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________