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

Optimal. Leaf size=83 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{k x^3+(-k-1) x^2+x}}{k x-1}\right )}{d^{3/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{k x^3+(-k-1) x^2+x}}{k x-1}\right )}{d^{3/4}} \]

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Rubi [F]  time = 15.01, 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 {-1-2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(4*(1 - x)^(1/4)*x^(1/4)*(1 - k*x)^(1/4)*Defer[Subst][Defer[Int][x^2/((1 - x^4)^(1/4)*(1 - k*x^4)^(1/4)*(-(-1
+ k*x^4)^3 - d*(x^4 - x^8))), x], x, x^(1/4)])/((1 - x)*x*(1 - k*x))^(1/4) - (8*(1 - k)*(1 - x)^(1/4)*x^(1/4)*
(1 - k*x)^(1/4)*Defer[Subst][Defer[Int][x^6/((1 - x^4)^(1/4)*(1 - k*x^4)^(1/4)*(-(-1 + k*x^4)^3 - d*(x^4 - x^8
))), x], x, x^(1/4)])/((1 - x)*x*(1 - k*x))^(1/4) + (4*k*(1 - x)^(1/4)*x^(1/4)*(1 - k*x)^(1/4)*Defer[Subst][De
fer[Int][x^10/((1 - x^4)^(1/4)*(1 - k*x^4)^(1/4)*((-1 + k*x^4)^3 + d*(x^4 - x^8))), x], x, x^(1/4)])/((1 - x)*
x*(1 - k*x))^(1/4)

Rubi steps

\begin {align*} \int \frac {-1-2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 x^3\right )} \, dx &=\frac {\left (\sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \int \frac {-1-2 (-1+k) x+k x^2}{\sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 x^3\right )} \, dx}{\sqrt [4]{(1-x) x (1-k x)}}\\ &=\frac {\left (4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (-1-2 (-1+k) x^4+k x^8\right )}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (-1+(d+3 k) x^4-\left (d+3 k^2\right ) x^8+k^3 x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}\\ &=\frac {\left (4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (-1-2 (-1+k) x^4+k x^8\right )}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (\left (-1+k x^4\right )^3+d \left (x^4-x^8\right )\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}\\ &=\frac {\left (4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \left (\frac {x^2}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (1-d \left (1+\frac {3 k}{d}\right ) x^4+d \left (1+\frac {3 k^2}{d}\right ) x^8-k^3 x^{12}\right )}+\frac {2 (-1+k) x^6}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (1-d \left (1+\frac {3 k}{d}\right ) x^4+d \left (1+\frac {3 k^2}{d}\right ) x^8-k^3 x^{12}\right )}+\frac {k x^{10}}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (-1+d \left (1+\frac {3 k}{d}\right ) x^4-d \left (1+\frac {3 k^2}{d}\right ) x^8+k^3 x^{12}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}\\ &=\frac {\left (4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (1-d \left (1+\frac {3 k}{d}\right ) x^4+d \left (1+\frac {3 k^2}{d}\right ) x^8-k^3 x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}-\frac {\left (8 (1-k) \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (1-d \left (1+\frac {3 k}{d}\right ) x^4+d \left (1+\frac {3 k^2}{d}\right ) x^8-k^3 x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}+\frac {\left (4 k \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^{10}}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (-1+d \left (1+\frac {3 k}{d}\right ) x^4-d \left (1+\frac {3 k^2}{d}\right ) x^8+k^3 x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}\\ &=\frac {\left (4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (-\left (-1+k x^4\right )^3-d \left (x^4-x^8\right )\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}-\frac {\left (8 (1-k) \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (-\left (-1+k x^4\right )^3-d \left (x^4-x^8\right )\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}+\frac {\left (4 k \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^{10}}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (\left (-1+k x^4\right )^3+d \left (x^4-x^8\right )\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}\\ \end {align*}

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Mathematica [F]  time = 3.75, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-1-2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.25, size = 83, normalized size = 1.00 \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{-1+k x}\right )}{d^{3/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{-1+k x}\right )}{d^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTan[(d^(1/4)*(x + (-1 - k)*x^2 + k*x^3)^(1/4))/(-1 + k*x)])/d^(3/4) - (2*ArcTanh[(d^(1/4)*(x + (-1 - k)*
x^2 + k*x^3)^(1/4))/(-1 + k*x)])/d^(3/4)

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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((-1-2*(-1+k)*x+k*x^2)/((1-x)*x*(-k*x+1))^(1/4)/(-1+(d+3*k)*x-(3*k^2+d)*x^2+k^3*x^3),x, algorithm="fr
icas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1-2*(-1+k)*x+k*x^2)/((1-x)*x*(-k*x+1))^(1/4)/(-1+(d+3*k)*x-(3*k^2+d)*x^2+k^3*x^3),x, algorithm="gi
ac")

[Out]

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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {-1-2 \left (-1+k \right ) x +k \,x^{2}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{4}} \left (-1+\left (d +3 k \right ) x -\left (3 k^{2}+d \right ) x^{2}+k^{3} x^{3}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1-2*(-1+k)*x+k*x^2)/((1-x)*x*(-k*x+1))^(1/4)/(-1+(d+3*k)*x-(3*k^2+d)*x^2+k^3*x^3),x, algorithm="ma
xima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {2\,x\,\left (k-1\right )-k\,x^2+1}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/4}\,\left (k^3\,x^3-x^2\,\left (3\,k^2+d\right )+x\,\left (d+3\,k\right )-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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