3.26.88 \(\int \frac {x^3 (-2+(1+k) x)}{((1-x) x (1-k x))^{2/3} (1-(2+2 k) x+(1+4 k+k^2) x^2-(2 k+2 k^2) x^3+(-b+k^2) x^4)} \, dx\)

Optimal. Leaf size=224 \[ -\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{b} x}{\sqrt [6]{b} x-2 \sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{2 b^{5/6}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{b} x}{\sqrt [6]{b} x+2 \sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{2 b^{5/6}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [6]{b} x}{\sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{b^{5/6}}-\frac {\tanh ^{-1}\left (\frac {\frac {\left (k x^3+(-k-1) x^2+x\right )^{2/3}}{\sqrt [6]{b}}+\sqrt [6]{b} x^2}{x \sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{2 b^{5/6}} \]

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Rubi [F]  time = 20.16, 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^3 (-2+(1+k) x)}{((1-x) x (1-k x))^{2/3} \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^3*(-2 + (1 + k)*x))/(((1 - x)*x*(1 - k*x))^(2/3)*(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)*x*((1 - x)/(1 - k*x))^(2/3)*(1 - k*x)*Hypergeometric2F1[1/3, 2/3, 4/3, ((1 - k)*x)/(1 - k*x)])/((b
- k^2)*((1 - x)*x*(1 - k*x))^(2/3)) + (3*(1 + k)*(1 - x)^(2/3)*x^(2/3)*(1 - k*x)^(2/3)*Defer[Subst][Defer[Int
][1/((1 - x^3)^(2/3)*(1 - k*x^3)^(2/3)*(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)])/((b - k^2)*((1 - x)*x*(1 - k*x))^(2/3)) + (3*(1 + k)*(1 + 4*k + k^2)*(1 - x)^(2/3
)*x^(2/3)*(1 - k*x)^(2/3)*Defer[Subst][Defer[Int][x^6/((1 - x^3)^(2/3)*(1 - k*x^3)^(2/3)*(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)])/((b - k^2)*((1 - x)*x*(1 - k*x)
)^(2/3)) - (6*(b + k + k^2 + k^3)*(1 - x)^(2/3)*x^(2/3)*(1 - k*x)^(2/3)*Defer[Subst][Defer[Int][x^9/((1 - x^3)
^(2/3)*(1 - k*x^3)^(2/3)*(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)])/((b - k^2)*((1 - x)*x*(1 - k*x))^(2/3)) + (6*(1 + k)^2*(1 - x)^(2/3)*x^(2/3)*(1 - k*x)^(2/3)*De
fer[Subst][Defer[Int][x^3/((1 - x^3)^(2/3)*(1 - k*x^3)^(2/3)*(-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)])/((b - k^2)*((1 - x)*x*(1 - k*x))^(2/3))

Rubi steps

\begin {align*} \int \frac {x^3 (-2+(1+k) x)}{((1-x) x (1-k x))^{2/3} \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 ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {x^{7/3} (-2+(1+k) x)}{(1-x)^{2/3} (1-k x)^{2/3} \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}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {\left (3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^9 \left (-2+(1+k) x^3\right )}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \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 )}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {\left (3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \left (-\frac {1+k}{\left (b-k^2\right ) \left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3}}+\frac {1+k-2 (1+k)^2 x^3+(1+k) \left (1+4 k+k^2\right ) x^6-2 \left (b+k+k^2+k^3\right ) x^9}{\left (b-k^2\right ) \left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \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 )}\right ) \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {\left (3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1+k-2 (1+k)^2 x^3+(1+k) \left (1+4 k+k^2\right ) x^6-2 \left (b+k+k^2+k^3\right ) x^9}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \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 )}{\left (b-k^2\right ) ((1-x) x (1-k x))^{2/3}}-\frac {\left (3 (1+k) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3}} \, dx,x,\sqrt [3]{x}\right )}{\left (b-k^2\right ) ((1-x) x (1-k x))^{2/3}}\\ &=-\frac {3 (1+k) x \left (\frac {1-x}{1-k x}\right )^{2/3} (1-k x) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};\frac {(1-k) x}{1-k x}\right )}{\left (b-k^2\right ) ((1-x) x (1-k x))^{2/3}}+\frac {\left (3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \left (\frac {1+k}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \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 {(1+k) \left (1+4 k+k^2\right ) x^6}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \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 \left (-b-k-k^2-k^3\right ) x^9}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \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 (1+k)^2 x^3}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \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 )}{\left (b-k^2\right ) ((1-x) x (1-k x))^{2/3}}\\ &=-\frac {3 (1+k) x \left (\frac {1-x}{1-k x}\right )^{2/3} (1-k x) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};\frac {(1-k) x}{1-k x}\right )}{\left (b-k^2\right ) ((1-x) x (1-k x))^{2/3}}+\frac {\left (3 (1+k) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \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 )}{\left (b-k^2\right ) ((1-x) x (1-k x))^{2/3}}+\frac {\left (6 (1+k)^2 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^3}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \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 )}{\left (b-k^2\right ) ((1-x) x (1-k x))^{2/3}}+\frac {\left (3 (1+k) \left (1+4 k+k^2\right ) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \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 )}{\left (b-k^2\right ) ((1-x) x (1-k x))^{2/3}}-\frac {\left (6 \left (b+k+k^2+k^3\right ) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^9}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \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 )}{\left (b-k^2\right ) ((1-x) x (1-k x))^{2/3}}\\ \end {align*}

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Mathematica [F]  time = 1.96, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^3 (-2+(1+k) x)}{((1-x) x (1-k x))^{2/3} \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^3*(-2 + (1 + k)*x))/(((1 - x)*x*(1 - k*x))^(2/3)*(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^3*(-2 + (1 + k)*x))/(((1 - x)*x*(1 - k*x))^(2/3)*(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 = 1.34, size = 224, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{b} x}{\sqrt [6]{b} x-2 \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} x}{\sqrt [6]{b} x+2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{5/6}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [6]{b} x}{\sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{5/6}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [6]{b} x^2+\frac {\left (x+(-1-k) x^2+k x^3\right )^{2/3}}{\sqrt [6]{b}}}{x \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{5/6}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^3*(-2 + (1 + k)*x))/(((1 - x)*x*(1 - k*x))^(2/3)*(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[(Sqrt[3]*b^(1/6)*x)/(b^(1/6)*x - 2*(x + (-1 - k)*x^2 + k*x^3)^(1/3))])/b^(5/6) + (Sqrt[3]
*ArcTan[(Sqrt[3]*b^(1/6)*x)/(b^(1/6)*x + 2*(x + (-1 - k)*x^2 + k*x^3)^(1/3))])/(2*b^(5/6)) - ArcTanh[(b^(1/6)*
x)/(x + (-1 - k)*x^2 + k*x^3)^(1/3)]/b^(5/6) - ArcTanh[(b^(1/6)*x^2 + (x + (-1 - k)*x^2 + k*x^3)^(2/3)/b^(1/6)
)/(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(x^3*(-2+(1+k)*x)/((1-x)*x*(-k*x+1))^(2/3)/(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 = 0.76, size = 262, normalized size = 1.17 \begin {gather*} -\frac {\sqrt {3} \left (-b\right )^{\frac {1}{6}} \log \left (\sqrt {3} \left (-b\right )^{\frac {1}{6}} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \left (-b\right )^{\frac {1}{3}}\right )}{4 \, b} + \frac {\sqrt {3} \left (-b\right )^{\frac {1}{6}} \log \left (-\sqrt {3} \left (-b\right )^{\frac {1}{6}} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \left (-b\right )^{\frac {1}{3}}\right )}{4 \, b} - \frac {\left (-b\right )^{\frac {1}{6}} \arctan \left (\frac {\sqrt {3} \left (-b\right )^{\frac {1}{6}} + 2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{\left (-b\right )^{\frac {1}{6}}}\right )}{2 \, b} - \frac {\left (-b\right )^{\frac {1}{6}} \arctan \left (-\frac {\sqrt {3} \left (-b\right )^{\frac {1}{6}} - 2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{\left (-b\right )^{\frac {1}{6}}}\right )}{2 \, b} - \frac {\left (-b\right )^{\frac {1}{6}} \arctan \left (\frac {{\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{\left (-b\right )^{\frac {1}{6}}}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-2+(1+k)*x)/((1-x)*x*(-k*x+1))^(2/3)/(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]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {x^{3} \left (-2+\left (1+k \right ) x \right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (1-\left (2+2 k \right ) x +\left (k^{2}+4 k +1\right ) x^{2}-\left (2 k^{2}+2 k \right ) x^{3}+\left (k^{2}-b \right ) x^{4}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(-2+(1+k)*x)/((1-x)*x*(-k*x+1))^(2/3)/(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^3*(-2+(1+k)*x)/((1-x)*x*(-k*x+1))^(2/3)/(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^{3}}{{\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 {2}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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