\(\int \frac {(-2+(1+k) x) (1-(1+k) x+(a+k) x^2)}{\sqrt [3]{(1-x) x (1-k x)} (1-2 (1+k) x+(1+c+4 k+k^2) x^2-(c+2 k+c k+2 k^2) x^3+(-b+c k+k^2) x^4)} \, dx\) [1593]

Optimal result

Integrand size = 95, antiderivative size = 109 \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (1-2 (1+k) x+\left (1+c+4 k+k^2\right ) x^2-\left (c+2 k+c k+2 k^2\right ) x^3+\left (-b+c k+k^2\right ) x^4\right )} \, dx=-\text {RootSum}\left [b-c \text {$\#$1}^3-\text {$\#$1}^6\&,\frac {a \log (x)-a \log \left (\sqrt [3]{x+(-1-k) x^2+k x^3}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^3-\log \left (\sqrt [3]{x+(-1-k) x^2+k x^3}-x \text {$\#$1}\right ) \text {$\#$1}^3}{c \text {$\#$1}+2 \text {$\#$1}^4}\&\right ] \]

[Out]

Unintegrable

Rubi [F]

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

[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*} \text {integral}& = \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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

Mathematica [A] (verified)

Time = 10.63 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.89 \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (1-2 (1+k) x+\left (1+c+4 k+k^2\right ) x^2-\left (c+2 k+c k+2 k^2\right ) x^3+\left (-b+c k+k^2\right ) x^4\right )} \, dx=-\text {RootSum}\left [b-c \text {$\#$1}^3-\text {$\#$1}^6\&,\frac {a \log (x)-a \log \left (\sqrt [3]{(-1+x) x (-1+k x)}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^3-\log \left (\sqrt [3]{(-1+x) x (-1+k x)}-x \text {$\#$1}\right ) \text {$\#$1}^3}{c \text {$\#$1}+2 \text {$\#$1}^4}\&\right ] \]

[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]

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

Maple [N/A] (verified)

Time = 1.42 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.52

method	result	size
pseudoelliptic	\(\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}+c \,\textit {\_Z}^{3}-b \right )}{\sum }\frac {\left (\textit {\_R}^{3}+a \right ) \ln \left (\frac {-\textit {\_R} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}}{x}\right )}{\textit {\_R} \left (2 \textit {\_R}^{3}+c \right )}\)	\(57\)

[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,method=_RETURNVERBOSE)

[Out]

sum(1/_R*(_R^3+a)*ln((-_R*x+((-1+x)*x*(k*x-1))^(1/3))/x)/(2*_R^3+c),_R=RootOf(_Z^6+_Z^3*c-b))

Fricas [F(-1)]

Timed out. \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (1-2 (1+k) x+\left (1+c+4 k+k^2\right ) x^2-\left (c+2 k+c k+2 k^2\right ) x^3+\left (-b+c k+k^2\right ) x^4\right )} \, dx=\text {Timed out} \]

[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

Sympy [F(-1)]

Timed out. \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (1-2 (1+k) x+\left (1+c+4 k+k^2\right ) x^2-\left (c+2 k+c k+2 k^2\right ) x^3+\left (-b+c k+k^2\right ) x^4\right )} \, dx=\text {Timed out} \]

[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

Maxima [N/A]

Not integrable

Time = 0.35 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.84 \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (1-2 (1+k) x+\left (1+c+4 k+k^2\right ) x^2-\left (c+2 k+c k+2 k^2\right ) x^3+\left (-b+c k+k^2\right ) x^4\right )} \, dx=\int { \frac {{\left ({\left (a + k\right )} x^{2} - {\left (k + 1\right )} x + 1\right )} {\left ({\left (k + 1\right )} x - 2\right )}}{{\left ({\left (c k + k^{2} - b\right )} x^{4} - {\left (c k + 2 \, k^{2} + c + 2 \, k\right )} x^{3} + {\left (k^{2} + c + 4 \, k + 1\right )} x^{2} - 2 \, {\left (k + 1\right )} x + 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}} \,d x } \]

[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)

Giac [N/A]

Not integrable

Time = 0.69 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.84 \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (1-2 (1+k) x+\left (1+c+4 k+k^2\right ) x^2-\left (c+2 k+c k+2 k^2\right ) x^3+\left (-b+c k+k^2\right ) x^4\right )} \, dx=\int { \frac {{\left ({\left (a + k\right )} x^{2} - {\left (k + 1\right )} x + 1\right )} {\left ({\left (k + 1\right )} x - 2\right )}}{{\left ({\left (c k + k^{2} - b\right )} x^{4} - {\left (c k + 2 \, k^{2} + c + 2 \, k\right )} x^{3} + {\left (k^{2} + c + 4 \, k + 1\right )} x^{2} - 2 \, {\left (k + 1\right )} x + 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}} \,d x } \]

[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)

Mupad [N/A]

Not integrable

Time = 8.11 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.84 \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (1-2 (1+k) x+\left (1+c+4 k+k^2\right ) x^2-\left (c+2 k+c k+2 k^2\right ) x^3+\left (-b+c k+k^2\right ) x^4\right )} \, dx=\int \frac {\left (x\,\left (k+1\right )-2\right )\,\left (\left (a+k\right )\,x^2+\left (-k-1\right )\,x+1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (x^4\,\left (k^2+c\,k-b\right )-x^3\,\left (c+2\,k+c\,k+2\,k^2\right )+x^2\,\left (k^2+4\,k+c+1\right )-2\,x\,\left (k+1\right )+1\right )} \,d x \]

[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)