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

Optimal. Leaf size=161 \[ -\frac {\log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{k x^3+(-k-1) x^2+x}+\left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{2 \sqrt [3]{b}}+\frac {\log \left (\sqrt [3]{k x^3+(-k-1) x^2+x}-\sqrt [3]{b} x\right )}{\sqrt [3]{b}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{\sqrt [3]{b}} \]

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Rubi [F]  time = 3.49, 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 (1-(1+k) x+(-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)*(1 - (1 + k)*x + (-b + k)*x^2)),x]

[Out]

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

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Mathematica [F]  time = 6.57, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-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)*(1 - (1 + k)*x + (-b + k)*x^2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.28, size = 161, 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 )}{\sqrt [3]{b}}+\frac {\log \left (-\sqrt [3]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{\sqrt [3]{b}}-\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 \sqrt [3]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Timed out

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giac [A]  time = 0.30, size = 121, normalized size = 0.75 \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 {1}{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 {1}{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 {1}{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)/(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^(1/3) - 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^(1/3) + log(abs(-b^(1/3) + (k - k/x - 1
/x + 1/x^2)^(1/3)))/b^(1/3)

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[\int \frac {-2+\left (1+k \right ) x}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (1-\left (1+k \right ) x +\left (-b +k \right ) x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

[Out]

Timed out

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