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

Optimal. Leaf size=212 \[ \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{x-2 \sqrt [6]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{2 b^{5/6}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [6]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}+x}\right )}{2 b^{5/6}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt [6]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{b^{5/6}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [6]{b} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+\frac {x^2}{\sqrt [6]{b}}}{x \sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{2 b^{5/6}} \]

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Rubi [F]  time = 18.57, 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) (-1+k x) (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (b-2 (b+b k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 4.55, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(-1+x) (-1+k x) (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (b-2 (b+b k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.00, size = 212, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{x-2 \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} x}{x+2 \sqrt [6]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{5/6}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt [6]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{5/6}}-\frac {\tanh ^{-1}\left (\frac {\frac {x^2}{\sqrt [6]{b}}+\sqrt [6]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}{x \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{5/6}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3]*ArcTan[(Sqrt[3]*x)/(x - 2*b^(1/6)*(x + (-1 - k)*x^2 + k*x^3)^(1/3))])/(2*b^(5/6)) - (Sqrt[3]*ArcTan[(
Sqrt[3]*x)/(x + 2*b^(1/6)*(x + (-1 - k)*x^2 + k*x^3)^(1/3))])/(2*b^(5/6)) - ArcTanh[x/(b^(1/6)*(x + (-1 - k)*x
^2 + k*x^3)^(1/3))]/b^(5/6) - ArcTanh[(x^2/b^(1/6) + b^(1/6)*(x + (-1 - k)*x^2 + k*x^3)^(2/3))/(x*(x + (-1 - k
)*x^2 + k*x^3)^(1/3))]/(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)*(k*x-1)*(-2+(1+k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(b-2*(b*k+b)*x+(b*k^2+4*b*k+b)*x^2-2*b*k*(1+k)*
x^3+(b*k^2-1)*x^4),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.48, size = 290, normalized size = 1.37 \begin {gather*} \frac {\sqrt {3} \left (-b^{5}\right )^{\frac {5}{6}} \log \left (\sqrt {3} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} \left (-\frac {1}{b}\right )^{\frac {1}{6}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \left (-\frac {1}{b}\right )^{\frac {1}{3}}\right )}{4 \, b^{5}} - \frac {\sqrt {3} \left (-b^{5}\right )^{\frac {5}{6}} \log \left (-\sqrt {3} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} \left (-\frac {1}{b}\right )^{\frac {1}{6}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \left (-\frac {1}{b}\right )^{\frac {1}{3}}\right )}{4 \, b^{5}} - \frac {\left (-b^{5}\right )^{\frac {5}{6}} \arctan \left (\frac {\sqrt {3} \left (-\frac {1}{b}\right )^{\frac {1}{6}} + 2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{\left (-\frac {1}{b}\right )^{\frac {1}{6}}}\right )}{2 \, b^{5}} - \frac {\left (-b^{5}\right )^{\frac {5}{6}} \arctan \left (-\frac {\sqrt {3} \left (-\frac {1}{b}\right )^{\frac {1}{6}} - 2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{\left (-\frac {1}{b}\right )^{\frac {1}{6}}}\right )}{2 \, b^{5}} - \frac {\left (-b^{5}\right )^{\frac {5}{6}} \arctan \left (\frac {{\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{\left (-\frac {1}{b}\right )^{\frac {1}{6}}}\right )}{b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(3)*(-b^5)^(5/6)*log(sqrt(3)*(k - k/x - 1/x + 1/x^2)^(1/3)*(-1/b)^(1/6) + (k - k/x - 1/x + 1/x^2)^(2/3
) + (-1/b)^(1/3))/b^5 - 1/4*sqrt(3)*(-b^5)^(5/6)*log(-sqrt(3)*(k - k/x - 1/x + 1/x^2)^(1/3)*(-1/b)^(1/6) + (k
- k/x - 1/x + 1/x^2)^(2/3) + (-1/b)^(1/3))/b^5 - 1/2*(-b^5)^(5/6)*arctan((sqrt(3)*(-1/b)^(1/6) + 2*(k - k/x -
1/x + 1/x^2)^(1/3))/(-1/b)^(1/6))/b^5 - 1/2*(-b^5)^(5/6)*arctan(-(sqrt(3)*(-1/b)^(1/6) - 2*(k - k/x - 1/x + 1/
x^2)^(1/3))/(-1/b)^(1/6))/b^5 - (-b^5)^(5/6)*arctan((k - k/x - 1/x + 1/x^2)^(1/3)/(-1/b)^(1/6))/b^5

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\left (-1+x \right ) \left (k x -1\right ) \left (-2+\left (1+k \right ) x \right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (b -2 \left (b k +b \right ) x +\left (b \,k^{2}+4 b k +b \right ) x^{2}-2 b k \left (1+k \right ) x^{3}+\left (b \,k^{2}-1\right ) x^{4}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((-1+x)*(k*x-1)*(-2+(1+k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(b-2*(b*k+b)*x+(b*k^2+4*b*k+b)*x^2-2*b*k*(1+k)*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 (k + 1\right )} x - 2\right )} {\left (k x - 1\right )} {\left (x - 1\right )}}{{\left (2 \, b {\left (k + 1\right )} k x^{3} - {\left (b k^{2} - 1\right )} x^{4} - {\left (b k^{2} + 4 \, b k + b\right )} x^{2} + 2 \, {\left (b k + b\right )} x - b\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)*(k*x-1)*(-2+(1+k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(b-2*(b*k+b)*x+(b*k^2+4*b*k+b)*x^2-2*b*k*(1+k)*
x^3+(b*k^2-1)*x^4),x, algorithm="maxima")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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