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

Optimal. Leaf size=160 \[ \frac {\log \left (\sqrt [3]{k x^3+(-k-1) x^2+x}-\sqrt [3]{b} x\right )}{b^{2/3}}-\frac {\log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{k x^3+(-k-1) x^2+x}+\left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{2 b^{2/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{b^{2/3}} \]

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Rubi [F]  time = 3.88, 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 {-2 x+(1+k) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 7.32, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-2 x+(1+k) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.36, size = 160, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{2/3}}+\frac {\log \left (-\sqrt [3]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{b^{2/3}}-\frac {\log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{x+(-1-k) x^2+k x^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[(-2*x + (1 + k)*x^2)/(((1 - x)*x*(1 - k*x))^(2/3)*(1 - (1 + k)*x + (-b + k)*x^2)),x]

[Out]

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

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

[Out]

Timed out

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giac [A]  time = 0.53, size = 122, normalized size = 0.76 \begin {gather*} -\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + 2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {2}{3}}} - \frac {\log \left (b^{\frac {2}{3}} + b^{\frac {1}{3}} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}}\right )}{2 \, b^{\frac {2}{3}}} + \frac {\log \left ({\left | -b^{\frac {1}{3}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} \right |}\right )}{b^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[\int \frac {-2 x +\left (1+k \right ) x^{2}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (1-\left (1+k \right ) x +\left (-b +k \right ) x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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