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

Optimal. Leaf size=155 \[ \frac {\log \left (x-\sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{b^{2/3}}-\frac {\log \left (b^{2/3} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+\sqrt [3]{b} x \sqrt [3]{k x^3+(-k-1) x^2+x}+x^2\right )}{2 b^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}+x}\right )}{b^{2/3}} \]

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Rubi [F]  time = 4.43, 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+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 4.45, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.29, size = 155, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{2/3}}+\frac {\log \left (x-\sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{b^{2/3}}-\frac {\log \left (x^2+\sqrt [3]{b} x \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

[Out]

Timed out

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giac [A]  time = 0.59, size = 129, normalized size = 0.83 \begin {gather*} \frac {\sqrt {3} {\left | b \right |}^{\frac {4}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} b^{\frac {1}{3}} {\left (2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} + \frac {1}{b^{\frac {1}{3}}}\right )}\right )}{b^{2}} - \frac {{\left | b \right |}^{\frac {4}{3}} \log \left ({\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \frac {{\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{b^{\frac {1}{3}}} + \frac {1}{b^{\frac {2}{3}}}\right )}{2 \, b^{2}} + \frac {\log \left ({\left | {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} - \frac {1}{b^{\frac {1}{3}}} \right |}\right )}{b^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(3)*abs(b)^(4/3)*arctan(1/3*sqrt(3)*b^(1/3)*(2*(k - k/x - 1/x + 1/x^2)^(1/3) + 1/b^(1/3)))/b^2 - 1/2*abs(b
)^(4/3)*log((k - k/x - 1/x + 1/x^2)^(2/3) + (k - k/x - 1/x + 1/x^2)^(1/3)/b^(1/3) + 1/b^(2/3))/b^2 + log(abs((
k - k/x - 1/x + 1/x^2)^(1/3) - 1/b^(1/3)))/b^(2/3)

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maple [F]  time = 0.13, size = 0, normalized size = 0.00 \[\int \frac {-2+\left (1+k \right ) x}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (b -b \left (1+k \right ) x +\left (b k -1\right ) x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((-2+(1+k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(b-b*(1+k)*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 (k + 1\right )} x - 2}{{\left (b {\left (k + 1\right )} x - {\left (b k - 1\right )} x^{2} - b\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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