3.12.88 \(\int \frac {x (-1+k x) (-1+(-1+2 k) x)}{\sqrt [3]{(1-x) x (1-k x)} (-1+(4-c) x+(-6+b+2 c+c k) x^2+(4-c-2 b k-2 c k) x^3+(-1+c k+b k^2) x^4)} \, dx\)

Optimal. Leaf size=87 \[ \frac {1}{3} \text {RootSum}\left [-\text {$\#$1}^6-\text {$\#$1}^3 c+b\& ,\frac {3 \log \left (\text {$\#$1} \sqrt [3]{k x^3+(-k-1) x^2+x}+x-1\right )-\log \left (k x^3+(-k-1) x^2+x\right )}{2 \text {$\#$1}^5+\text {$\#$1}^2 c}\& \right ] \]

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Rubi [F]  time = 26.02, 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 (-1+k x) (-1+(-1+2 k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(4-c) x+(-6+b+2 c+c k) x^2+(4-c-2 b k-2 c k) x^3+\left (-1+c k+b k^2\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x*(-1 + k*x)*(-1 + (-1 + 2*k)*x))/(((1 - x)*x*(1 - k*x))^(1/3)*(-1 + (4 - c)*x + (-6 + b + 2*c + c*k)*x^2
+ (4 - c - 2*b*k - 2*c*k)*x^3 + (-1 + c*k + b*k^2)*x^4)),x]

[Out]

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

Rubi steps

\begin {align*} \int \frac {x (-1+k x) (-1+(-1+2 k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(4-c) x+(-6+b+2 c+c k) x^2+(4-c-2 b k-2 c k) x^3+\left (-1+c k+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^{2/3} (-1+k x) (-1+(-1+2 k) x)}{\sqrt [3]{1-x} \sqrt [3]{1-k x} \left (-1+(4-c) x+(-6+b+2 c+c k) x^2+(4-c-2 b k-2 c k) x^3+\left (-1+c k+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 {x^{2/3} (1-k x)^{2/3} (-1+(-1+2 k) x)}{\sqrt [3]{1-x} \left (-1+(4-c) x+(-6+b+2 c+c k) x^2+(4-c-2 b k-2 c k) x^3+\left (-1+c k+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^4 \left (1-k x^3\right )^{2/3} \left (-1+(-1+2 k) x^3\right )}{\sqrt [3]{1-x^3} \left (-1+(4-c) x^3+(-6+b+2 c+c k) x^6+(4-c-2 b k-2 c k) x^9+\left (-1+c k+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 {x^4 \left (1-k x^3\right )^{2/3}}{\sqrt [3]{1-x^3} \left (1-4 \left (1-\frac {c}{4}\right ) x^3+6 \left (1+\frac {1}{6} (-b-c (2+k))\right ) x^6-4 \left (1-\frac {c}{4}-\frac {1}{2} (b+c) k\right ) x^9+(1-k (c+b k)) x^{12}\right )}+\frac {(1-2 k) x^7 \left (1-k x^3\right )^{2/3}}{\sqrt [3]{1-x^3} \left (1-4 \left (1-\frac {c}{4}\right ) x^3+6 \left (1+\frac {1}{6} (-b-c (2+k))\right ) x^6-4 \left (1-\frac {c}{4}-\frac {1}{2} (b+c) k\right ) x^9+(1-k (c+b k)) x^{12}\right )}\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 \frac {x^4 \left (1-k x^3\right )^{2/3}}{\sqrt [3]{1-x^3} \left (1-4 \left (1-\frac {c}{4}\right ) x^3+6 \left (1+\frac {1}{6} (-b-c (2+k))\right ) x^6-4 \left (1-\frac {c}{4}-\frac {1}{2} (b+c) k\right ) x^9+(1-k (c+b k)) x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}-\frac {\left (3 (1-2 k) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^7 \left (1-k x^3\right )^{2/3}}{\sqrt [3]{1-x^3} \left (1-4 \left (1-\frac {c}{4}\right ) x^3+6 \left (1+\frac {1}{6} (-b-c (2+k))\right ) x^6-4 \left (1-\frac {c}{4}-\frac {1}{2} (b+c) k\right ) x^9+(1-k (c+b k)) 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 = 2.98, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x (-1+k x) (-1+(-1+2 k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(4-c) x+(-6+b+2 c+c k) x^2+(4-c-2 b k-2 c k) x^3+\left (-1+c k+b k^2\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(x*(-1 + k*x)*(-1 + (-1 + 2*k)*x))/(((1 - x)*x*(1 - k*x))^(1/3)*(-1 + (4 - c)*x + (-6 + b + 2*c + c*
k)*x^2 + (4 - c - 2*b*k - 2*c*k)*x^3 + (-1 + c*k + b*k^2)*x^4)),x]

[Out]

Integrate[(x*(-1 + k*x)*(-1 + (-1 + 2*k)*x))/(((1 - x)*x*(1 - k*x))^(1/3)*(-1 + (4 - c)*x + (-6 + b + 2*c + c*
k)*x^2 + (4 - c - 2*b*k - 2*c*k)*x^3 + (-1 + c*k + b*k^2)*x^4)), x]

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IntegrateAlgebraic [A]  time = 3.05, size = 87, normalized size = 1.00 \begin {gather*} \frac {1}{3} \text {RootSum}\left [b-c \text {$\#$1}^3-\text {$\#$1}^6\&,\frac {-\log \left (x+(-1-k) x^2+k x^3\right )+3 \log \left (-1+x+\sqrt [3]{x+(-1-k) x^2+k x^3} \text {$\#$1}\right )}{c \text {$\#$1}^2+2 \text {$\#$1}^5}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x*(-1 + k*x)*(-1 + (-1 + 2*k)*x))/(((1 - x)*x*(1 - k*x))^(1/3)*(-1 + (4 - c)*x + (-6 + b +
2*c + c*k)*x^2 + (4 - c - 2*b*k - 2*c*k)*x^3 + (-1 + c*k + b*k^2)*x^4)),x]

[Out]

RootSum[b - c*#1^3 - #1^6 & , (-Log[x + (-1 - k)*x^2 + k*x^3] + 3*Log[-1 + x + (x + (-1 - k)*x^2 + k*x^3)^(1/3
)*#1])/(c*#1^2 + 2*#1^5) & ]/3

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(k*x-1)*(-1+(-1+2*k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(-1+(4-c)*x+(c*k+b+2*c-6)*x^2+(-2*b*k-2*c*k-c+4)*
x^3+(b*k^2+c*k-1)*x^4),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (tr
ace 0)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left ({\left (2 \, k - 1\right )} x - 1\right )} {\left (k x - 1\right )} x}{{\left ({\left (b k^{2} + c k - 1\right )} x^{4} - {\left (2 \, b k + 2 \, c k + c - 4\right )} x^{3} + {\left (c k + b + 2 \, c - 6\right )} x^{2} - {\left (c - 4\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*(k*x-1)*(-1+(-1+2*k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(-1+(4-c)*x+(c*k+b+2*c-6)*x^2+(-2*b*k-2*c*k-c+4)*
x^3+(b*k^2+c*k-1)*x^4),x, algorithm="giac")

[Out]

integrate(((2*k - 1)*x - 1)*(k*x - 1)*x/(((b*k^2 + c*k - 1)*x^4 - (2*b*k + 2*c*k + c - 4)*x^3 + (c*k + b + 2*c
- 6)*x^2 - (c - 4)*x - 1)*((k*x - 1)*(x - 1)*x)^(1/3)), x)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {x \left (k x -1\right ) \left (-1+\left (-1+2 k \right ) x \right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (-1+\left (4-c \right ) x +\left (c k +b +2 c -6\right ) x^{2}+\left (-2 b k -2 c k -c +4\right ) x^{3}+\left (b \,k^{2}+c k -1\right ) x^{4}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(k*x-1)*(-1+(-1+2*k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(-1+(4-c)*x+(c*k+b+2*c-6)*x^2+(-2*b*k-2*c*k-c+4)*x^3+(b
*k^2+c*k-1)*x^4),x)

[Out]

int(x*(k*x-1)*(-1+(-1+2*k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(-1+(4-c)*x+(c*k+b+2*c-6)*x^2+(-2*b*k-2*c*k-c+4)*x^3+(b
*k^2+c*k-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 (2 \, k - 1\right )} x - 1\right )} {\left (k x - 1\right )} x}{{\left ({\left (b k^{2} + c k - 1\right )} x^{4} - {\left (2 \, b k + 2 \, c k + c - 4\right )} x^{3} + {\left (c k + b + 2 \, c - 6\right )} x^{2} - {\left (c - 4\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*(k*x-1)*(-1+(-1+2*k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(-1+(4-c)*x+(c*k+b+2*c-6)*x^2+(-2*b*k-2*c*k-c+4)*
x^3+(b*k^2+c*k-1)*x^4),x, algorithm="maxima")

[Out]

integrate(((2*k - 1)*x - 1)*(k*x - 1)*x/(((b*k^2 + c*k - 1)*x^4 - (2*b*k + 2*c*k + c - 4)*x^3 + (c*k + b + 2*c
- 6)*x^2 - (c - 4)*x - 1)*((k*x - 1)*(x - 1)*x)^(1/3)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {x\,\left (x\,\left (2\,k-1\right )-1\right )\,\left (k\,x-1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (\left (-b\,k^2-c\,k+1\right )\,x^4+\left (c+2\,b\,k+2\,c\,k-4\right )\,x^3+\left (6-2\,c-c\,k-b\right )\,x^2+\left (c-4\right )\,x+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x*(x*(2*k - 1) - 1)*(k*x - 1))/((x*(k*x - 1)*(x - 1))^(1/3)*(x^3*(c + 2*b*k + 2*c*k - 4) - x^4*(c*k + b*
k^2 - 1) + x*(c - 4) - x^2*(b + 2*c + c*k - 6) + 1)),x)

[Out]

-int((x*(x*(2*k - 1) - 1)*(k*x - 1))/((x*(k*x - 1)*(x - 1))^(1/3)*(x^3*(c + 2*b*k + 2*c*k - 4) - x^4*(c*k + b*
k^2 - 1) + x*(c - 4) - x^2*(b + 2*c + c*k - 6) + 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*(k*x-1)*(-1+(-1+2*k)*x)/((1-x)*x*(-k*x+1))**(1/3)/(-1+(4-c)*x+(c*k+b+2*c-6)*x**2+(-2*b*k-2*c*k-c+4
)*x**3+(b*k**2+c*k-1)*x**4),x)

[Out]

Timed out

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