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

Optimal. Leaf size=289 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt [3]{k x^3+(-k-1) x^2+x} \left (\sqrt [6]{b} k x-\sqrt [6]{b}\right )}{\sqrt [3]{b} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+k^2 x^2-2 k x+1}\right )}{2 b^{5/6}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}}{\sqrt [6]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}-2 k x+2}\right )}{2 b^{5/6}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}}{\sqrt [6]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}+2 k x-2}\right )}{2 b^{5/6}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [6]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}}{k x-1}\right )}{b^{5/6}} \]

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Rubi [F]  time = 12.31, 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+x) x (1+(-2+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-4 k x+\left (-b+6 k^2\right ) x^2+\left (2 b-4 k^3\right ) x^3+\left (-b+k^4\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {(-1+x) x (1+(-2+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-4 k x+\left (-b+6 k^2\right ) x^2+\left (2 b-4 k^3\right ) x^3+\left (-b+k^4\right ) x^4\right )} \, dx &=\frac {\left (\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {(-1+x) x^{2/3} (1+(-2+k) x)}{\sqrt [3]{1-x} \sqrt [3]{1-k x} \left (1-4 k x+\left (-b+6 k^2\right ) x^2+\left (2 b-4 k^3\right ) x^3+\left (-b+k^4\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 {(1-x)^{2/3} x^{2/3} (1+(-2+k) x)}{\sqrt [3]{1-k x} \left (1-4 k x+\left (-b+6 k^2\right ) x^2+\left (2 b-4 k^3\right ) x^3+\left (-b+k^4\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-x^3\right )^{2/3} \left (1+(-2+k) x^3\right )}{\sqrt [3]{1-k x^3} \left (1-4 k x^3+\left (-b+6 k^2\right ) x^6+\left (2 b-4 k^3\right ) x^9+\left (-b+k^4\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-x^3\right )^{2/3}}{\sqrt [3]{1-k x^3} \left (1-4 k x^3-b \left (1-\frac {6 k^2}{b}\right ) x^6+2 b \left (1-\frac {2 k^3}{b}\right ) x^9-b \left (1-\frac {k^4}{b}\right ) x^{12}\right )}+\frac {(-2+k) x^7 \left (1-x^3\right )^{2/3}}{\sqrt [3]{1-k x^3} \left (1-4 k x^3-b \left (1-\frac {6 k^2}{b}\right ) x^6+2 b \left (1-\frac {2 k^3}{b}\right ) x^9-b \left (1-\frac {k^4}{b}\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-x^3\right )^{2/3}}{\sqrt [3]{1-k x^3} \left (1-4 k x^3-b \left (1-\frac {6 k^2}{b}\right ) x^6+2 b \left (1-\frac {2 k^3}{b}\right ) x^9-b \left (1-\frac {k^4}{b}\right ) x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}-\frac {\left (3 (-2+k) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^7 \left (1-x^3\right )^{2/3}}{\sqrt [3]{1-k x^3} \left (1-4 k x^3-b \left (1-\frac {6 k^2}{b}\right ) x^6+2 b \left (1-\frac {2 k^3}{b}\right ) x^9-b \left (1-\frac {k^4}{b}\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 = 4.86, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(-1+x) x (1+(-2+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-4 k x+\left (-b+6 k^2\right ) x^2+\left (2 b-4 k^3\right ) x^3+\left (-b+k^4\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 3.79, size = 289, 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 k 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 k 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+k x}\right )}{b^{5/6}}+\frac {\tanh ^{-1}\left (\frac {\left (-\sqrt [6]{b}+\sqrt [6]{b} k x\right ) \sqrt [3]{x+(-1-k) x^2+k x^3}}{1-2 k x+k^2 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[((-1 + x)*x*(1 + (-2 + k)*x))/(((1 - x)*x*(1 - k*x))^(1/3)*(1 - 4*k*x + (-b + 6*k^2)*x^2 +
(2*b - 4*k^3)*x^3 + (-b + k^4)*x^4)),x]

[Out]

(Sqrt[3]*ArcTan[(Sqrt[3]*b^(1/6)*(x + (-1 - k)*x^2 + k*x^3)^(1/3))/(2 - 2*k*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*k*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 + k*x)]/b^(5/6) + ArcTanh[((-b^(1/6) + b^(1/6)*k*x)*(x + (-1 - k)*x^2 + k*x^3)^(1/3))/(1 - 2*k*x + k^2*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((-1+x)*x*(1+(-2+k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(1-4*k*x+(6*k^2-b)*x^2+(-4*k^3+2*b)*x^3+(k^4-b)*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 (k - 2\right )} x + 1\right )} {\left (x - 1\right )} x}{{\left ({\left (k^{4} - b\right )} x^{4} - 2 \, {\left (2 \, k^{3} - b\right )} x^{3} + {\left (6 \, k^{2} - b\right )} x^{2} - 4 \, k 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((-1+x)*x*(1+(-2+k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(1-4*k*x+(6*k^2-b)*x^2+(-4*k^3+2*b)*x^3+(k^4-b)*x^4),
x, algorithm="giac")

[Out]

integrate(((k - 2)*x + 1)*(x - 1)*x/(((k^4 - b)*x^4 - 2*(2*k^3 - b)*x^3 + (6*k^2 - b)*x^2 - 4*k*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 \[\int \frac {\left (-1+x \right ) x \left (1+\left (-2+k \right ) x \right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (1-4 k x +\left (6 k^{2}-b \right ) x^{2}+\left (-4 k^{3}+2 b \right ) x^{3}+\left (k^{4}-b \right ) x^{4}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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