3.24.90 \(\int \frac {-1+(-1+2 k) x}{\sqrt [3]{(1-x) x (1-k x)} (1-(2+b) x+(1+b k) x^2)} \, dx\) [2390]

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

[Out]

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

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Rubi [F]
time = 1.81, 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*}

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Mathematica [A]
time = 30.03, size = 297, normalized size = 1.55 \begin {gather*} \frac {x \sqrt [3]{\frac {-1+k x}{-1+x}} \left (2 \sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{\frac {x}{-1+x}} \sqrt [3]{\frac {-1+k x}{-1+x}}}{-2+\sqrt [3]{b} \sqrt [3]{\frac {x}{-1+x}} \sqrt [3]{\frac {-1+k x}{-1+x}}}\right )-2 \log \left (\sqrt [3]{\frac {x}{-1+x}} \sqrt [3]{\frac {-1+k x}{-1+x}}\right )+\log \left (\left (\frac {x}{-1+x}\right )^{2/3} \left (\frac {-1+k x}{-1+x}\right )^{2/3}\right )+2 \log \left (1+\sqrt [3]{b} \sqrt [3]{\frac {x}{-1+x}} \sqrt [3]{\frac {-1+k x}{-1+x}}\right )-\log \left (1-\sqrt [3]{b} \sqrt [3]{\frac {x}{-1+x}} \sqrt [3]{\frac {-1+k x}{-1+x}}+b^{2/3} \left (\frac {x}{-1+x}\right )^{2/3} \left (\frac {-1+k x}{-1+x}\right )^{2/3}\right )\right )}{2 b^{2/3} \left (\frac {x}{-1+x}\right )^{2/3} \sqrt [3]{(-1+x) x (-1+k x)}} \end {gather*}

Antiderivative was successfully verified.

[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]

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

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Maple [F]
time = 0.12, 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)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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)

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Fricas [F(-1)] Timed out
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

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Sympy [F(-1)] Timed out
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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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)

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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)

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