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

Optimal. Leaf size=109 \[ -\text {RootSum}\left [-\text {$\#$1}^6-\text {$\#$1}^3 c+b\& ,\frac {-\text {$\#$1}^3 \log \left (\sqrt [3]{k x^3+(-k-1) x^2+x}-\text {$\#$1} x\right )+\text {$\#$1}^3 \log (x)-a \log \left (\sqrt [3]{k x^3+(-k-1) x^2+x}-\text {$\#$1} x\right )+a \log (x)}{2 \text {$\#$1}^4+\text {$\#$1} c}\& \right ] \]

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Rubi [F]  time = 52.69, 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 {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (1-2 (1+k) x+\left (1+c+4 k+k^2\right ) x^2-\left (c+2 k+c k+2 k^2\right ) x^3+\left (-b+c k+k^2\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (1-2 (1+k) x+\left (1+c+4 k+k^2\right ) x^2-\left (c+2 k+c k+2 k^2\right ) x^3+\left (-b+c k+k^2\right ) x^4\right )} \, dx &=\frac {\left (\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x} \left (1-2 (1+k) x+\left (1+c+4 k+k^2\right ) x^2-\left (c+2 k+c k+2 k^2\right ) x^3+\left (-b+c k+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 \left (-2+(1+k) x^3\right ) \left (1-(1+k) x^3+(a+k) x^6\right )}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (1-2 (1+k) x^3+\left (1+c+4 k+k^2\right ) x^6-\left (c+2 k+c k+2 k^2\right ) x^9+\left (-b+c k+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 {3 (1+k) x^4}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (1-2 (1+k) x^3+(1+c+k (4+k)) x^6-c \left (1+\frac {k (2+c+2 k)}{c}\right ) x^9-b \left (1-\frac {k (c+k)}{b}\right ) x^{12}\right )}+\frac {2 (-a-k) \left (1+\frac {(1+k)^2}{2 (a+k)}\right ) x^7}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (1-2 (1+k) x^3+(1+c+k (4+k)) x^6-c \left (1+\frac {k (2+c+2 k)}{c}\right ) x^9-b \left (1-\frac {k (c+k)}{b}\right ) x^{12}\right )}+\frac {(1+k) (a+k) x^{10}}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (1-2 (1+k) x^3+(1+c+k (4+k)) x^6-c \left (1+\frac {k (2+c+2 k)}{c}\right ) x^9-b \left (1-\frac {k (c+k)}{b}\right ) x^{12}\right )}+\frac {2 x}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (-1+2 (1+k) x^3-(1+c+k (4+k)) x^6+c \left (1+\frac {k (2+c+2 k)}{c}\right ) x^9+b \left (1-\frac {k (c+k)}{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}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (-1+2 (1+k) x^3-(1+c+k (4+k)) x^6+c \left (1+\frac {k (2+c+2 k)}{c}\right ) x^9+b \left (1-\frac {k (c+k)}{b}\right ) x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}+\frac {\left (9 (1+k) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (1-2 (1+k) x^3+(1+c+k (4+k)) x^6-c \left (1+\frac {k (2+c+2 k)}{c}\right ) x^9-b \left (1-\frac {k (c+k)}{b}\right ) x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}+\frac {\left (3 (1+k) (a+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+c+k (4+k)) x^6-c \left (1+\frac {k (2+c+2 k)}{c}\right ) x^9-b \left (1-\frac {k (c+k)}{b}\right ) x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}+\frac {\left (3 \left (-1-2 a-4 k-k^2\right ) \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+c+k (4+k)) x^6-c \left (1+\frac {k (2+c+2 k)}{c}\right ) x^9-b \left (1-\frac {k (c+k)}{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 = 5.33, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (1-2 (1+k) x+\left (1+c+4 k+k^2\right ) x^2-\left (c+2 k+c k+2 k^2\right ) x^3+\left (-b+c k+k^2\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.97, size = 109, normalized size = 1.00 \begin {gather*} -\text {RootSum}\left [b-c \text {$\#$1}^3-\text {$\#$1}^6\&,\frac {a \log (x)-a \log \left (\sqrt [3]{x+(-1-k) x^2+k x^3}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^3-\log \left (\sqrt [3]{x+(-1-k) x^2+k x^3}-x \text {$\#$1}\right ) \text {$\#$1}^3}{c \text {$\#$1}+2 \text {$\#$1}^4}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-RootSum[b - c*#1^3 - #1^6 & , (a*Log[x] - a*Log[(x + (-1 - k)*x^2 + k*x^3)^(1/3) - x*#1] + Log[x]*#1^3 - Log[
(x + (-1 - k)*x^2 + k*x^3)^(1/3) - x*#1]*#1^3)/(c*#1 + 2*#1^4) & ]

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

[Out]

Timed out

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(((a + k)*x^2 - (k + 1)*x + 1)*((k + 1)*x - 2)/(((c*k + k^2 - b)*x^4 - (c*k + 2*k^2 + c + 2*k)*x^3 +
(k^2 + c + 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 = -0.01 \begin {gather*} \int \frac {\left (x\,\left (k+1\right )-2\right )\,\left (\left (a+k\right )\,x^2+\left (-k-1\right )\,x+1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (x^4\,\left (k^2+c\,k-b\right )-x^3\,\left (c+2\,k+c\,k+2\,k^2\right )+x^2\,\left (k^2+4\,k+c+1\right )-2\,x\,\left (k+1\right )+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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