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

Optimal. Leaf size=223 \[ -\frac {\log \left (\left (b^{2/3} x-b^{2/3} x^2\right ) \left (k x^3+(-k-1) x^2+x\right )^{2/3}+\sqrt [3]{b} \left (k x^3+(-k-1) x^2+x\right )^{4/3}+b x^4-2 b x^3+b x^2\right )}{2 \sqrt [3]{b}}+\frac {\log \left (\sqrt [6]{b} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+\sqrt {b} x^2-\sqrt {b} x\right )}{\sqrt [3]{b}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \left (k x^3+(-k-1) x^2+x\right )^{2/3}}{-2 \sqrt [3]{b} x^2+2 \sqrt [3]{b} x+\left (k x^3+(-k-1) x^2+x\right )^{2/3}}\right )}{\sqrt [3]{b}} \]

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Rubi [F]  time = 3.60, 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 (1-(b+2 k) x+\left (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)*(1 - (b + 2*k)*x + (b + k^2)*x^2)),x]

[Out]

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

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Mathematica [F]  time = 4.67, 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 (1-(b+2 k) x+\left (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)*(1 - (b + 2*k)*x + (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)*(1 - (b + 2*k)*x + (b + k^2)*x^2)), x]

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IntegrateAlgebraic [A]  time = 0.95, size = 223, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}{2 \sqrt [3]{b} x-2 \sqrt [3]{b} x^2+\left (x+(-1-k) x^2+k x^3\right )^{2/3}}\right )}{\sqrt [3]{b}}+\frac {\log \left (-\sqrt {b} x+\sqrt {b} x^2+\sqrt [6]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{\sqrt [3]{b}}-\frac {\log \left (b x^2-2 b x^3+b x^4+\left (b^{2/3} x-b^{2/3} x^2\right ) \left (x+(-1-k) x^2+k x^3\right )^{2/3}+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{4/3}\right )}{2 \sqrt [3]{b}} \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)*(1 - (b + 2*k)*x + (b + k^2)*x^2
)),x]

[Out]

(Sqrt[3]*ArcTan[(Sqrt[3]*(x + (-1 - k)*x^2 + k*x^3)^(2/3))/(2*b^(1/3)*x - 2*b^(1/3)*x^2 + (x + (-1 - k)*x^2 +
k*x^3)^(2/3))])/b^(1/3) + Log[-(Sqrt[b]*x) + Sqrt[b]*x^2 + b^(1/6)*(x + (-1 - k)*x^2 + k*x^3)^(2/3)]/b^(1/3) -
Log[b*x^2 - 2*b*x^3 + b*x^4 + (b^(2/3)*x - b^(2/3)*x^2)*(x + (-1 - k)*x^2 + k*x^3)^(2/3) + b^(1/3)*(x + (-1 -
k)*x^2 + k*x^3)^(4/3)]/(2*b^(1/3))

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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)/(1-(b+2*k)*x+(k^2+b)*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 (k^{2} + b\right )} x^{2} - {\left (b + 2 \, k\right )} x + 1\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)/(1-(b+2*k)*x+(k^2+b)*x^2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[\int \frac {-1+2 x +\left (k^{2}-2 k \right ) x^{2}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (1-\left (b +2 k \right ) x +\left (k^{2}+b \right ) x^{2}\right )}\, dx\]

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

[Out]

int((-1+2*x+(k^2-2*k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(1-(b+2*k)*x+(k^2+b)*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 (k^{2} + b\right )} x^{2} - {\left (b + 2 \, k\right )} x + 1\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)/(1-(b+2*k)*x+(k^2+b)*x^2),x, algorithm="maxima")

[Out]

integrate(((k^2 - 2*k)*x^2 + 2*x - 1)/(((k*x - 1)*(x - 1)*x)^(2/3)*((k^2 + b)*x^2 - (b + 2*k)*x + 1)), 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 (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}\,\left (\left (k^2+b\right )\,x^2+\left (-b-2\,k\right )\,x+1\right )} \,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)/((x*(k*x - 1)*(x - 1))^(2/3)*(x^2*(b + k^2) - x*(b + 2*k) + 1)),x)

[Out]

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

[Out]

Timed out

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