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

Optimal. Leaf size=230 \[ -\frac {\log \left (\left (k x^3+(-k-1) x^2+x\right )^{2/3} \left (b^{2/3} x-b^{2/3} k x^2\right )+b k^2 x^4-2 b k x^3+\sqrt [3]{b} \left (k x^3+(-k-1) x^2+x\right )^{4/3}+b x^2\right )}{2 \sqrt [3]{b}}+\frac {\log \left (\sqrt {b} k x^2+\sqrt [6]{b} \left (k x^3+(-k-1) x^2+x\right )^{2/3}-\sqrt {b} x\right )}{\sqrt [3]{b}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \left (k x^3+(-k-1) x^2+x\right )^{2/3}}{-2 \sqrt [3]{b} k x^2+2 \sqrt [3]{b} x+\left (k x^3+(-k-1) x^2+x\right )^{2/3}}\right )}{\sqrt [3]{b}} \]

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Rubi [F]  time = 4.18, 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 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(2+b) x+(1+b k) x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 8.57, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(2+b) x+(1+b k) x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.87, size = 230, 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} k 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} k 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 k x^3+b k^2 x^4+\left (b^{2/3} x-b^{2/3} k 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*k*x + (1 - 2*k)*x^2)/(((1 - x)*x*(1 - k*x))^(2/3)*(1 - (2 + b)*x + (1 + b*k)*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)*k*x^2 + (x + (-1 - k)*x^2
+ k*x^3)^(2/3))])/b^(1/3) + Log[-(Sqrt[b]*x) + Sqrt[b]*k*x^2 + b^(1/6)*(x + (-1 - k)*x^2 + k*x^3)^(2/3)]/b^(1/
3) - Log[b*x^2 - 2*b*k*x^3 + b*k^2*x^4 + (b^(2/3)*x - b^(2/3)*k*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*k*x+(1-2*k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(1-(2+b)*x+(b*k+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 (2 \, k - 1\right )} x^{2} - 2 \, k x + 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{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+2*k*x+(1-2*k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(1-(2+b)*x+(b*k+1)*x^2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.17, size = 0, normalized size = 0.00 \[\int \frac {-1+2 k x +\left (1-2 k \right ) x^{2}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (1-\left (2+b \right ) x +\left (b k +1\right ) x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

[Out]

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

[Out]

Timed out

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