3.29.20 \(\int \frac {x (-1+k x) (-1+(-1+2 k) x)}{\sqrt [3]{(1-x) x (1-k x)} (-1+4 x+(-6+b) x^2+(4-2 b k) x^3+(-1+b k^2) x^4)} \, dx\)

Optimal. Leaf size=279 \[ \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}}{\sqrt [6]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}-2 x+2}\right )}{2 b^{5/6}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}}{\sqrt [6]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}+2 x-2}\right )}{2 b^{5/6}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [6]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}}{x-1}\right )}{b^{5/6}}+\frac {\tanh ^{-1}\left (\frac {\left (\sqrt [6]{b} x-\sqrt [6]{b}\right ) \sqrt [3]{k x^3+(-k-1) x^2+x}}{\sqrt [3]{b} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+x^2-2 x+1}\right )}{2 b^{5/6}} \]

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Rubi [F]  time = 11.77, 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 {x (-1+k x) (-1+(-1+2 k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+4 x+(-6+b) x^2+(4-2 b k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x*(-1 + k*x)*(-1 + (-1 + 2*k)*x))/(((1 - x)*x*(1 - k*x))^(1/3)*(-1 + 4*x + (-6 + b)*x^2 + (4 - 2*b*k)*x^3
+ (-1 + b*k^2)*x^4)),x]

[Out]

(-3*(1 - x)^(1/3)*x^(1/3)*(1 - k*x)^(1/3)*Defer[Subst][Defer[Int][(x^4*(1 - k*x^3)^(2/3))/((1 - x^3)^(1/3)*(1
- 4*x^3 + 6*(1 - b/6)*x^6 - 4*(1 - (b*k)/2)*x^9 + (1 - b*k^2)*x^12)), x], x, x^(1/3)])/((1 - x)*x*(1 - k*x))^(
1/3) - (3*(1 - 2*k)*(1 - x)^(1/3)*x^(1/3)*(1 - k*x)^(1/3)*Defer[Subst][Defer[Int][(x^7*(1 - k*x^3)^(2/3))/((1
- x^3)^(1/3)*(1 - 4*x^3 + 6*(1 - b/6)*x^6 - 4*(1 - (b*k)/2)*x^9 + (1 - b*k^2)*x^12)), x], x, x^(1/3)])/((1 - x
)*x*(1 - k*x))^(1/3)

Rubi steps

\begin {align*} \int \frac {x (-1+k x) (-1+(-1+2 k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+4 x+(-6+b) x^2+(4-2 b k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx &=\frac {\left (\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {x^{2/3} (-1+k x) (-1+(-1+2 k) x)}{\sqrt [3]{1-x} \sqrt [3]{1-k x} \left (-1+4 x+(-6+b) x^2+(4-2 b k) x^3+\left (-1+b k^2\right ) x^4\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 {x^{2/3} (1-k x)^{2/3} (-1+(-1+2 k) x)}{\sqrt [3]{1-x} \left (-1+4 x+(-6+b) x^2+(4-2 b k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx}{\sqrt [3]{(1-x) x (1-k x)}}\\ &=-\frac {\left (3 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (1-k x^3\right )^{2/3} \left (-1+(-1+2 k) x^3\right )}{\sqrt [3]{1-x^3} \left (-1+4 x^3+(-6+b) x^6+(4-2 b k) x^9+\left (-1+b k^2\right ) x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\\ &=-\frac {\left (3 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \left (\frac {x^4 \left (1-k x^3\right )^{2/3}}{\sqrt [3]{1-x^3} \left (1-4 x^3+6 \left (1-\frac {b}{6}\right ) x^6-4 \left (1-\frac {b k}{2}\right ) x^9+\left (1-b k^2\right ) x^{12}\right )}+\frac {(1-2 k) x^7 \left (1-k x^3\right )^{2/3}}{\sqrt [3]{1-x^3} \left (1-4 x^3+6 \left (1-\frac {b}{6}\right ) x^6-4 \left (1-\frac {b k}{2}\right ) x^9+\left (1-b k^2\right ) x^{12}\right )}\right ) \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\\ &=-\frac {\left (3 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (1-k x^3\right )^{2/3}}{\sqrt [3]{1-x^3} \left (1-4 x^3+6 \left (1-\frac {b}{6}\right ) x^6-4 \left (1-\frac {b k}{2}\right ) x^9+\left (1-b k^2\right ) x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}-\frac {\left (3 (1-2 k) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^7 \left (1-k x^3\right )^{2/3}}{\sqrt [3]{1-x^3} \left (1-4 x^3+6 \left (1-\frac {b}{6}\right ) x^6-4 \left (1-\frac {b k}{2}\right ) x^9+\left (1-b k^2\right ) x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\\ \end {align*}

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Mathematica [F]  time = 4.41, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x (-1+k x) (-1+(-1+2 k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+4 x+(-6+b) x^2+(4-2 b k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(x*(-1 + k*x)*(-1 + (-1 + 2*k)*x))/(((1 - x)*x*(1 - k*x))^(1/3)*(-1 + 4*x + (-6 + b)*x^2 + (4 - 2*b*
k)*x^3 + (-1 + b*k^2)*x^4)),x]

[Out]

Integrate[(x*(-1 + k*x)*(-1 + (-1 + 2*k)*x))/(((1 - x)*x*(1 - k*x))^(1/3)*(-1 + 4*x + (-6 + b)*x^2 + (4 - 2*b*
k)*x^3 + (-1 + b*k^2)*x^4)), x]

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IntegrateAlgebraic [A]  time = 3.47, size = 279, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}{2-2 x+\sqrt [6]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{5/6}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}{-2+2 x+\sqrt [6]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{5/6}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [6]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}{-1+x}\right )}{b^{5/6}}+\frac {\tanh ^{-1}\left (\frac {\left (-\sqrt [6]{b}+\sqrt [6]{b} x\right ) \sqrt [3]{x+(-1-k) x^2+k x^3}}{1-2 x+x^2+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}\right )}{2 b^{5/6}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x*(-1 + k*x)*(-1 + (-1 + 2*k)*x))/(((1 - x)*x*(1 - k*x))^(1/3)*(-1 + 4*x + (-6 + b)*x^2 +
(4 - 2*b*k)*x^3 + (-1 + b*k^2)*x^4)),x]

[Out]

(Sqrt[3]*ArcTan[(Sqrt[3]*b^(1/6)*(x + (-1 - k)*x^2 + k*x^3)^(1/3))/(2 - 2*x + b^(1/6)*(x + (-1 - k)*x^2 + k*x^
3)^(1/3))])/(2*b^(5/6)) - (Sqrt[3]*ArcTan[(Sqrt[3]*b^(1/6)*(x + (-1 - k)*x^2 + k*x^3)^(1/3))/(-2 + 2*x + b^(1/
6)*(x + (-1 - k)*x^2 + k*x^3)^(1/3))])/(2*b^(5/6)) + ArcTanh[(b^(1/6)*(x + (-1 - k)*x^2 + k*x^3)^(1/3))/(-1 +
x)]/b^(5/6) + ArcTanh[((-b^(1/6) + b^(1/6)*x)*(x + (-1 - k)*x^2 + k*x^3)^(1/3))/(1 - 2*x + x^2 + b^(1/3)*(x +
(-1 - k)*x^2 + k*x^3)^(2/3))]/(2*b^(5/6))

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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(x*(k*x-1)*(-1+(-1+2*k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(-1+4*x+(-6+b)*x^2+(-2*b*k+4)*x^3+(b*k^2-1)*x^4),
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 ({\left (2 \, k - 1\right )} x - 1\right )} {\left (k x - 1\right )} x}{{\left ({\left (b k^{2} - 1\right )} x^{4} - 2 \, {\left (b k - 2\right )} x^{3} + {\left (b - 6\right )} x^{2} + 4 \, x - 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(k*x-1)*(-1+(-1+2*k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(-1+4*x+(-6+b)*x^2+(-2*b*k+4)*x^3+(b*k^2-1)*x^4),
x, algorithm="giac")

[Out]

integrate(((2*k - 1)*x - 1)*(k*x - 1)*x/(((b*k^2 - 1)*x^4 - 2*(b*k - 2)*x^3 + (b - 6)*x^2 + 4*x - 1)*((k*x - 1
)*(x - 1)*x)^(1/3)), x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {x \left (k x -1\right ) \left (-1+\left (-1+2 k \right ) x \right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (-1+4 x +\left (-6+b \right ) x^{2}+\left (-2 b k +4\right ) x^{3}+\left (b \,k^{2}-1\right ) x^{4}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(k*x-1)*(-1+(-1+2*k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(-1+4*x+(-6+b)*x^2+(-2*b*k+4)*x^3+(b*k^2-1)*x^4),x)

[Out]

int(x*(k*x-1)*(-1+(-1+2*k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(-1+4*x+(-6+b)*x^2+(-2*b*k+4)*x^3+(b*k^2-1)*x^4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left ({\left (2 \, k - 1\right )} x - 1\right )} {\left (k x - 1\right )} x}{{\left ({\left (b k^{2} - 1\right )} x^{4} - 2 \, {\left (b k - 2\right )} x^{3} + {\left (b - 6\right )} x^{2} + 4 \, x - 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(k*x-1)*(-1+(-1+2*k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(-1+4*x+(-6+b)*x^2+(-2*b*k+4)*x^3+(b*k^2-1)*x^4),
x, algorithm="maxima")

[Out]

integrate(((2*k - 1)*x - 1)*(k*x - 1)*x/(((b*k^2 - 1)*x^4 - 2*(b*k - 2)*x^3 + (b - 6)*x^2 + 4*x - 1)*((k*x - 1
)*(x - 1)*x)^(1/3)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x\,\left (x\,\left (2\,k-1\right )-1\right )\,\left (k\,x-1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (\left (b\,k^2-1\right )\,x^4+\left (4-2\,b\,k\right )\,x^3+\left (b-6\right )\,x^2+4\,x-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(x*(2*k - 1) - 1)*(k*x - 1))/((x*(k*x - 1)*(x - 1))^(1/3)*(4*x + x^4*(b*k^2 - 1) - x^3*(2*b*k - 4) + x^
2*(b - 6) - 1)),x)

[Out]

int((x*(x*(2*k - 1) - 1)*(k*x - 1))/((x*(k*x - 1)*(x - 1))^(1/3)*(4*x + x^4*(b*k^2 - 1) - x^3*(2*b*k - 4) + x^
2*(b - 6) - 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(x*(k*x-1)*(-1+(-1+2*k)*x)/((1-x)*x*(-k*x+1))**(1/3)/(-1+4*x+(-6+b)*x**2+(-2*b*k+4)*x**3+(b*k**2-1)*x
**4),x)

[Out]

Timed out

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