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

Optimal. Leaf size=208 \[ \frac {\log \left (\sqrt [3]{b} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+x^2-x\right )}{b^{2/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} \left (k x^3+(-k-1) x^2+x\right )^{2/3}}{\sqrt [3]{b} \left (k x^3+(-k-1) x^2+x\right )^{2/3}-2 x^2+2 x}\right )}{b^{2/3}}-\frac {\log \left (b^{2/3} \left (k x^3+(-k-1) x^2+x\right )^{4/3}+\left (\sqrt [3]{b} x-\sqrt [3]{b} x^2\right ) \left (k x^3+(-k-1) x^2+x\right )^{2/3}+x^4-2 x^3+x^2\right )}{2 b^{2/3}} \]

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Rubi [F]  time = 3.51, 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+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 5.50, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.94, size = 208, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}{2 x-2 x^2+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}\right )}{b^{2/3}}+\frac {\log \left (-x+x^2+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{b^{2/3}}-\frac {\log \left (x^2-2 x^3+x^4+\left (\sqrt [3]{b} x-\sqrt [3]{b} x^2\right ) \left (x+(-1-k) x^2+k x^3\right )^{2/3}+b^{2/3} \left (x+(-1-k) x^2+k x^3\right )^{4/3}\right )}{2 b^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3]*ArcTan[(Sqrt[3]*b^(1/3)*(x + (-1 - k)*x^2 + k*x^3)^(2/3))/(2*x - 2*x^2 + b^(1/3)*(x + (-1 - k)*x^2 +
k*x^3)^(2/3))])/b^(2/3) + Log[-x + x^2 + b^(1/3)*(x + (-1 - k)*x^2 + k*x^3)^(2/3)]/b^(2/3) - Log[x^2 - 2*x^3 +
x^4 + (b^(1/3)*x - b^(1/3)*x^2)*(x + (-1 - k)*x^2 + k*x^3)^(2/3) + b^(2/3)*(x + (-1 - k)*x^2 + k*x^3)^(4/3)]/
(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((-1+2*x+(k^2-2*k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(b-(2*b*k+1)*x+(b*k^2+1)*x^2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[\int \frac {-1+2 x +\left (k^{2}-2 k \right ) x^{2}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (b -\left (2 b k +1\right ) x +\left (b \,k^{2}+1\right ) x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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