3.25.95 \(\int \frac {-1+(-1+2 k) x}{\sqrt [3]{(1-x) x (1-k x)} (b-(1+2 b) x+(b+k) x^2)} \, dx\) [2495]

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

[Out]

3^(1/2)*arctan(3^(1/2)*(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)+
ln(-b^(1/3)+b^(1/3)*x+(x+(-1-k)*x^2+k*x^3)^(1/3))/b^(1/3)-1/2*ln(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))/b^(1/3)

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Rubi [F]
time = 1.85, 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 [A]
time = 28.97, size = 263, normalized size = 1.27 \begin {gather*} \frac {x \sqrt [3]{\frac {-1+k x}{-1+x}} \left (2 \sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} \sqrt [3]{\frac {x}{-1+x}} \sqrt [3]{\frac {-1+k x}{-1+x}}}{-2 \sqrt [3]{b}+\sqrt [3]{\frac {x}{-1+x}} \sqrt [3]{\frac {-1+k x}{-1+x}}}\right )+4 \tanh ^{-1}\left (1+\frac {2 \sqrt [3]{\frac {x}{-1+x}} \sqrt [3]{\frac {-1+k x}{-1+x}}}{\sqrt [3]{b}}\right )+\log \left (\left (\frac {x}{-1+x}\right )^{2/3} \left (\frac {-1+k x}{-1+x}\right )^{2/3}\right )-\log \left (b^{2/3}-\sqrt [3]{b} \sqrt [3]{\frac {x}{-1+x}} \sqrt [3]{\frac {-1+k x}{-1+x}}+\left (\frac {x}{-1+x}\right )^{2/3} \left (\frac {-1+k x}{-1+x}\right )^{2/3}\right )\right )}{2 \sqrt [3]{b} \left (\frac {x}{-1+x}\right )^{2/3} \sqrt [3]{(-1+x) x (-1+k x)}} \end {gather*}

Antiderivative was successfully verified.

[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]

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

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Maple [F]
time = 0.12, 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*} \text {Failed to integrate} \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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Fricas [F(-1)] Timed out
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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Sympy [F(-1)] Timed out
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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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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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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