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

Optimal. Leaf size=214 \[ -\frac {\log \left (b^{2/3} k^2 x^2-2 b^{2/3} k x+b^{2/3}+\sqrt [3]{k x^3+(-k-1) x^2+x} \left (\sqrt [3]{b}-\sqrt [3]{b} k x\right )+\left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{2 \sqrt [3]{b}}+\frac {\log \left (\sqrt [3]{b} k x-\sqrt [3]{b}+\sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{\sqrt [3]{b}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{k x^3+(-k-1) x^2+x}}{-2 \sqrt [3]{b} k x+2 \sqrt [3]{b}+\sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{\sqrt [3]{b}} \]

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Rubi [F]  time = 4.14, 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}{\sqrt [3]{(1-x) x (1-k x)} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 5.75, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-1+(2-k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.61, size = 214, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x+(-1-k) x^2+k x^3}}{2 \sqrt [3]{b}-2 \sqrt [3]{b} k x+\sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{\sqrt [3]{b}}+\frac {\log \left (-\sqrt [3]{b}+\sqrt [3]{b} k x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{\sqrt [3]{b}}-\frac {\log \left (b^{2/3}-2 b^{2/3} k x+b^{2/3} k^2 x^2+\left (\sqrt [3]{b}-\sqrt [3]{b} k x\right ) \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[(-1 + (2 - k)*x)/(((1 - x)*x*(1 - k*x))^(1/3)*(b - (1 + 2*b*k)*x + (1 + b*k^2)*x^2)),x]

[Out]

(Sqrt[3]*ArcTan[(Sqrt[3]*(x + (-1 - k)*x^2 + k*x^3)^(1/3))/(2*b^(1/3) - 2*b^(1/3)*k*x + (x + (-1 - k)*x^2 + k*
x^3)^(1/3))])/b^(1/3) + Log[-b^(1/3) + b^(1/3)*k*x + (x + (-1 - k)*x^2 + k*x^3)^(1/3)]/b^(1/3) - Log[b^(2/3) -
2*b^(2/3)*k*x + b^(2/3)*k^2*x^2 + (b^(1/3) - b^(1/3)*k*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((-1+(2-k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(b-(2*b*k+1)*x+(b*k^2+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 (k - 2\right )} x + 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}} {\left ({\left (b k^{2} + 1\right )} x^{2} - {\left (2 \, b k + 1\right )} x + b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[\int \frac {-1+\left (2-k \right ) x}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (b -\left (2 b k +1\right ) x +\left (b \,k^{2}+1\right ) x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

Timed out

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