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

Optimal. Leaf size=285 \[ -\frac {(-1)^{2/3} \log \left (k^2 x^2+2 (-1)^{2/3} \sqrt [3]{2} (k-1)^{2/3} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+\left (\sqrt [3]{-1} 2^{2/3} \sqrt [3]{k-1}-\sqrt [3]{-1} 2^{2/3} \sqrt [3]{k-1} k x\right ) \sqrt [3]{k x^3+(-k-1) x^2+x}-2 k x+1\right )}{2\ 2^{2/3} \sqrt [3]{k-1}}+\frac {(-1)^{2/3} \log \left (\sqrt [3]{-1} 2^{2/3} \sqrt [3]{k-1} \sqrt [3]{k x^3+(-k-1) x^2+x}+k x-1\right )}{2^{2/3} \sqrt [3]{k-1}}+\frac {(-1)^{2/3} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} k x-\sqrt {3}}{-2 \sqrt [3]{-1} 2^{2/3} \sqrt [3]{k-1} \sqrt [3]{k x^3+(-k-1) x^2+x}+k x-1}\right )}{2^{2/3} \sqrt [3]{k-1}} \]

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Rubi [F]  time = 0.50, 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-k x}{(1+(-2+k) x) ((1-x) x (1-k x))^{2/3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 0.66, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1-k x}{(1+(-2+k) x) ((1-x) x (1-k x))^{2/3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 2.92, size = 285, normalized size = 1.00 \begin {gather*} \frac {(-1)^{2/3} \sqrt {3} \tan ^{-1}\left (\frac {-\sqrt {3}+\sqrt {3} k x}{-1+k x-2 \sqrt [3]{-1} 2^{2/3} \sqrt [3]{-1+k} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2^{2/3} \sqrt [3]{-1+k}}+\frac {(-1)^{2/3} \log \left (-1+k x+\sqrt [3]{-1} 2^{2/3} \sqrt [3]{-1+k} \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2^{2/3} \sqrt [3]{-1+k}}-\frac {(-1)^{2/3} \log \left (1-2 k x+k^2 x^2+\left (\sqrt [3]{-1} 2^{2/3} \sqrt [3]{-1+k}-\sqrt [3]{-1} 2^{2/3} \sqrt [3]{-1+k} k x\right ) \sqrt [3]{x+(-1-k) x^2+k x^3}+2 (-1)^{2/3} \sqrt [3]{2} (-1+k)^{2/3} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2\ 2^{2/3} \sqrt [3]{-1+k}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1)^(2/3)*Sqrt[3]*ArcTan[(-Sqrt[3] + Sqrt[3]*k*x)/(-1 + k*x - 2*(-1)^(1/3)*2^(2/3)*(-1 + k)^(1/3)*(x + (-1 -
k)*x^2 + k*x^3)^(1/3))])/(2^(2/3)*(-1 + k)^(1/3)) + ((-1)^(2/3)*Log[-1 + k*x + (-1)^(1/3)*2^(2/3)*(-1 + k)^(1
/3)*(x + (-1 - k)*x^2 + k*x^3)^(1/3)])/(2^(2/3)*(-1 + k)^(1/3)) - ((-1)^(2/3)*Log[1 - 2*k*x + k^2*x^2 + ((-1)^
(1/3)*2^(2/3)*(-1 + k)^(1/3) - (-1)^(1/3)*2^(2/3)*(-1 + k)^(1/3)*k*x)*(x + (-1 - k)*x^2 + k*x^3)^(1/3) + 2*(-1
)^(2/3)*2^(1/3)*(-1 + k)^(2/3)*(x + (-1 - k)*x^2 + k*x^3)^(2/3)])/(2*2^(2/3)*(-1 + k)^(1/3))

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fricas [B]  time = 93.99, size = 932, normalized size = 3.27

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(3)*2^(1/3)*arctan(1/3*(24*sqrt(3)*2^(1/3)*((k^5 - 3*k^4 - 4*k^3 + 22*k^2 - 24*k + 8)*x^4 - 2*(k^4 - 1
0*k^3 + 27*k^2 - 26*k + 8)*x^3 - 6*(k^3 - 4*k^2 + 4*k - 1)*x^2 - 2*(k^2 - 1)*x + k - 1)*(k*x^3 - (k + 1)*x^2 +
x)^(2/3)/(k - 1)^(1/3) - 6*sqrt(3)*2^(2/3)*((k^6 + 27*k^5 - 40*k^4 - 20*k^3 + 48*k^2 - 16*k)*x^5 - (33*k^5 +
55*k^4 - 220*k^3 + 132*k^2 + 16*k - 16)*x^4 + 2*(55*k^4 - 55*k^3 - 66*k^2 + 82*k - 16)*x^3 - 2*(55*k^3 - 99*k^
2 + 38*k + 6)*x^2 + (33*k^2 - 61*k + 28)*x - k + 1)*(k*x^3 - (k + 1)*x^2 + x)^(1/3)/(k - 1)^(2/3) + sqrt(3)*((
k^6 - 48*k^5 - 192*k^4 + 416*k^3 - 48*k^2 - 192*k + 64)*x^6 + 6*(7*k^5 + 104*k^4 - 80*k^3 - 176*k^2 + 176*k -
32)*x^5 - 3*(139*k^4 + 256*k^3 - 768*k^2 + 352*k + 16)*x^4 + 4*(203*k^3 - 192*k^2 - 120*k + 104)*x^3 - 3*(139*
k^2 - 208*k + 64)*x^2 + 6*(7*k - 8)*x + 1))/((k^6 + 96*k^5 - 48*k^4 - 160*k^3 + 240*k^2 - 192*k + 64)*x^6 - 6*
(17*k^5 + 64*k^4 - 112*k^3 + 80*k^2 - 80*k + 32)*x^5 + 3*(149*k^4 + 32*k^3 - 96*k^2 - 160*k + 80)*x^4 - 4*(157
*k^3 - 24*k^2 - 168*k + 40)*x^3 + 3*(149*k^2 - 128*k - 16)*x^2 - 6*(17*k - 16)*x + 1))/(k - 1)^(1/3) - 1/12*2^
(1/3)*log((12*2^(2/3)*(k*x^3 - (k + 1)*x^2 + x)^(2/3)*((k^3 + k^2 - 4*k + 2)*x^2 - 2*(2*k^2 - 3*k + 1)*x + k -
1)/(k - 1)^(2/3) + 6*((k^3 + 8*k^2 - 8*k)*x^3 - (11*k^2 - 8)*x^2 + (11*k - 8)*x - 1)*(k*x^3 - (k + 1)*x^2 + x
)^(1/3) + 2^(1/3)*((k^4 + 28*k^3 - 12*k^2 - 32*k + 16)*x^4 - 4*(8*k^3 + 15*k^2 - 30*k + 8)*x^3 + 6*(13*k^2 - 1
0*k - 2)*x^2 - 4*(8*k - 7)*x + 1)/(k - 1)^(1/3))/((k^4 - 8*k^3 + 24*k^2 - 32*k + 16)*x^4 + 4*(k^3 - 6*k^2 + 12
*k - 8)*x^3 + 6*(k^2 - 4*k + 4)*x^2 + 4*(k - 2)*x + 1))/(k - 1)^(1/3) + 1/6*2^(1/3)*log((6*2^(1/3)*(k*x^3 - (k
+ 1)*x^2 + x)^(1/3)*(k*x - 1)/(k - 1)^(1/3) - 2^(2/3)*((k^2 - 4*k + 4)*x^2 + 2*(k - 2)*x + 1)/(k - 1)^(2/3) -
12*(k*x^3 - (k + 1)*x^2 + x)^(2/3))/((k^2 - 4*k + 4)*x^2 + 2*(k - 2)*x + 1))/(k - 1)^(1/3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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