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

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

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Rubi [F]  time = 0.63, 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)} (1-(1+k) x)} \, 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)),x]

[Out]

(3*(1 - x)^(1/3)*x*(1 - k*x)^(1/3)*AppellF1[2/3, 1/3, 1/3, 5/3, x, k*x])/(2*((1 - x)*x*(1 - k*x))^(1/3)) + ((1
- x)^(1/3)*x^(1/3)*(1 - k*x)^(1/3)*Defer[Int][1/((1 - x)^(1/3)*x^(1/3)*(1 + (-1 - k)*x)*(1 - k*x)^(1/3)), 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)} (1-(1+k) x)} \, 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} (1-(1+k) x)} \, 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 \frac {1}{\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}} \, 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 \frac {1}{\sqrt [3]{1-x} \sqrt [3]{x} (1+(-1-k) x) \sqrt [3]{1-k x}} \, dx}{\sqrt [3]{(1-x) x (1-k x)}}\\ &=\frac {3 \sqrt [3]{1-x} x \sqrt [3]{1-k x} F_1\left (\frac {2}{3};\frac {1}{3},\frac {1}{3};\frac {5}{3};x,k x\right )}{2 \sqrt [3]{(1-x) x (1-k x)}}+\frac {\left (\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{x} (1+(-1-k) x) \sqrt [3]{1-k x}} \, dx}{\sqrt [3]{(1-x) x (1-k x)}}\\ \end {align*}

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Mathematica [F]  time = 1.60, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2-(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} (1-(1+k) x)} \, 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)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.25, size = 161, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{k} x}{\sqrt [3]{k} x+2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{\sqrt [3]{k}}-\frac {\log \left (-\sqrt [3]{k} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{\sqrt [3]{k}}+\frac {\log \left (k^{2/3} x^2+\sqrt [3]{k} 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]{k}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3]*ArcTan[(Sqrt[3]*k^(1/3)*x)/(k^(1/3)*x + 2*(x + (-1 - k)*x^2 + k*x^3)^(1/3))])/k^(1/3) - Log[-(k^(1/3)
*x) + (x + (-1 - k)*x^2 + k*x^3)^(1/3)]/k^(1/3) + Log[k^(2/3)*x^2 + k^(1/3)*x*(x + (-1 - k)*x^2 + k*x^3)^(1/3)
+ (x + (-1 - k)*x^2 + k*x^3)^(2/3)]/(2*k^(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),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 + 1\right )} x - 2}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}} {\left ({\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),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {2-\left (1+k \right ) x}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (1-\left (1+k \right ) x \right )}\, dx\]

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),x)

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Integral((k*x + x - 2)/((x*(x - 1)*(k*x - 1))**(1/3)*(k*x + x - 1)), x)

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