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

Optimal. Leaf size=207 \[ -\frac {\log \left (b^{2/3} x^2-2 b^{2/3} x+b^{2/3}+\left (\sqrt [3]{b}-\sqrt [3]{b} x\right ) \sqrt [3]{k x^3+(-k-1) x^2+x}+\left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{2 \sqrt [3]{b}}+\frac {\log \left (\sqrt [3]{b} x-\sqrt [3]{b}+\sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{\sqrt [3]{b}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{k x^3+(-k-1) x^2+x}}{-2 \sqrt [3]{b} x+2 \sqrt [3]{b}+\sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{\sqrt [3]{b}} \]

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Rubi [F]  time = 2.65, 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+(-1+2 k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-(1+2 b) x+(b+k) x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 5.66, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-1+(-1+2 k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-(1+2 b) x+(b+k) x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.66, size = 207, 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} x+\sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{\sqrt [3]{b}}+\frac {\log \left (-\sqrt [3]{b}+\sqrt [3]{b} 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} x+b^{2/3} x^2+\left (\sqrt [3]{b}-\sqrt [3]{b} 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 + (-1 + 2*k)*x)/(((1 - x)*x*(1 - k*x))^(1/3)*(b - (1 + 2*b)*x + (b + k)*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)*x + (x + (-1 - k)*x^2 + k*x^
3)^(1/3))])/b^(1/3) + Log[-b^(1/3) + b^(1/3)*x + (x + (-1 - k)*x^2 + k*x^3)^(1/3)]/b^(1/3) - Log[b^(2/3) - 2*b
^(2/3)*x + b^(2/3)*x^2 + (b^(1/3) - 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((-1+(-1+2*k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(b-(1+2*b)*x+(b+k)*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 - 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}} {\left ({\left (b + k\right )} x^{2} - {\left (2 \, b + 1\right )} x + b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {-1+\left (-1+2 k \right ) x}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (b -\left (1+2 b \right ) x +\left (b +k \right ) x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

Timed out

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