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

Optimal. Leaf size=276 \[ \frac {\log \left (\sqrt [3]{k x^3+(-k-1) x^2+x}-\sqrt [6]{b} x\right )}{2 b^{2/3}}+\frac {\log \left (\sqrt [6]{b} x+\sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{2 b^{2/3}}-\frac {\log \left (-\sqrt [6]{b} x \sqrt [3]{k x^3+(-k-1) x^2+x}+\sqrt [3]{b} x^2+\left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{4 b^{2/3}}-\frac {\log \left (\sqrt [6]{b} x \sqrt [3]{k x^3+(-k-1) x^2+x}+\sqrt [3]{b} x^2+\left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{4 b^{2/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x^2}{\sqrt [3]{b} x^2+2 \left (k x^3+(-k-1) x^2+x\right )^{2/3}}\right )}{2 b^{2/3}} \]

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Rubi [F]  time = 15.64, 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^2 (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(3*(1 + k)*(1 - x)^(1/3)*x^(1/3)*(1 - k*x)^(1/3)*Defer[Subst][Defer[Int][x^10/((1 - x^3)^(1/3)*(1 - k*x^3)^(1/
3)*(1 - 2*(1 + k)*x^3 + (1 + k*(4 + k))*x^6 - 2*k*(1 + k)*x^9 - b*(1 - k^2/b)*x^12)), x], x, x^(1/3)])/((1 - x
)*x*(1 - k*x))^(1/3) + (6*(1 - x)^(1/3)*x^(1/3)*(1 - k*x)^(1/3)*Defer[Subst][Defer[Int][x^7/((1 - x^3)^(1/3)*(
1 - k*x^3)^(1/3)*(-1 + 2*(1 + k)*x^3 - (1 + k*(4 + k))*x^6 + 2*k*(1 + k)*x^9 + b*(1 - k^2/b)*x^12)), x], x, x^
(1/3)])/((1 - x)*x*(1 - k*x))^(1/3)

Rubi steps

\begin {align*} \int \frac {x^2 (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-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^{5/3} (-2+(1+k) x)}{\sqrt [3]{1-x} \sqrt [3]{1-k x} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-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^7 \left (-2+(1+k) x^3\right )}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (1-(2+2 k) x^3+\left (1+4 k+k^2\right ) x^6-\left (2 k+2 k^2\right ) x^9+\left (-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 {(1+k) x^{10}}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (1-2 (1+k) x^3+(1+k (4+k)) x^6-2 k (1+k) x^9-b \left (1-\frac {k^2}{b}\right ) x^{12}\right )}+\frac {2 x^7}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (-1+2 (1+k) x^3-(1+k (4+k)) x^6+2 k (1+k) x^9+b \left (1-\frac {k^2}{b}\right ) x^{12}\right )}\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^7}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (-1+2 (1+k) x^3-(1+k (4+k)) x^6+2 k (1+k) x^9+b \left (1-\frac {k^2}{b}\right ) x^{12}\right )} \, 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^{10}}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (1-2 (1+k) x^3+(1+k (4+k)) x^6-2 k (1+k) x^9-b \left (1-\frac {k^2}{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 = 5.42, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^2 (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.63, size = 274, normalized size = 0.99 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {\frac {x^2}{\sqrt {3}}+\frac {2 \left (x+(-1-k) x^2+k x^3\right )^{2/3}}{\sqrt {3} \sqrt [3]{b}}}{x^2}\right )}{2 b^{2/3}}+\frac {\log \left (-\sqrt [6]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{2/3}}+\frac {\log \left (\sqrt [6]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{2/3}}-\frac {\log \left (\sqrt [3]{b} x^2-\sqrt [6]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{2/3}}-\frac {\log \left (\sqrt [3]{b} x^2+\sqrt [6]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Sqrt[3]*ArcTan[(x^2/Sqrt[3] + (2*(x + (-1 - k)*x^2 + k*x^3)^(2/3))/(Sqrt[3]*b^(1/3)))/x^2])/b^(2/3) + Lo
g[-(b^(1/6)*x) + (x + (-1 - k)*x^2 + k*x^3)^(1/3)]/(2*b^(2/3)) + Log[b^(1/6)*x + (x + (-1 - k)*x^2 + k*x^3)^(1
/3)]/(2*b^(2/3)) - Log[b^(1/3)*x^2 - b^(1/6)*x*(x + (-1 - k)*x^2 + k*x^3)^(1/3) + (x + (-1 - k)*x^2 + k*x^3)^(
2/3)]/(4*b^(2/3)) - Log[b^(1/3)*x^2 + b^(1/6)*x*(x + (-1 - k)*x^2 + k*x^3)^(1/3) + (x + (-1 - k)*x^2 + k*x^3)^
(2/3)]/(4*b^(2/3))

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

[Out]

Timed out

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giac [A]  time = 0.37, size = 123, normalized size = 0.45 \begin {gather*} -\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + b^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{2 \, b^{\frac {2}{3}}} - \frac {\log \left ({\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {4}{3}} + b^{\frac {1}{3}} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + b^{\frac {2}{3}}\right )}{4 \, b^{\frac {2}{3}}} + \frac {\log \left ({\left | {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} - b^{\frac {1}{3}} \right |}\right )}{2 \, b^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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