Integrand size = 88, antiderivative size = 87 \[ \int \frac {x (-1+k x) (-1+(-1+2 k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(4-c) x+(-6+b+2 c+c k) x^2+(4-c-2 b k-2 c k) x^3+\left (-1+c k+b k^2\right ) x^4\right )} \, dx=\frac {1}{3} \text {RootSum}\left [b-c \text {$\#$1}^3-\text {$\#$1}^6\&,\frac {-\log \left (x+(-1-k) x^2+k x^3\right )+3 \log \left (-1+x+\sqrt [3]{x+(-1-k) x^2+k x^3} \text {$\#$1}\right )}{c \text {$\#$1}^2+2 \text {$\#$1}^5}\&\right ] \]
\[ \int \frac {x (-1+k x) (-1+(-1+2 k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(4-c) x+(-6+b+2 c+c k) x^2+(4-c-2 b k-2 c k) x^3+\left (-1+c k+b k^2\right ) x^4\right )} \, dx=\int \frac {x (-1+k x) (-1+(-1+2 k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(4-c) x+(-6+b+2 c+c k) x^2+(4-c-2 b k-2 c k) x^3+\left (-1+c k+b k^2\right ) x^4\right )} \, dx \]
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]
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]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {x (k x-1) ((2 k-1) x-1)}{\sqrt [3]{(1-x) x (1-k x)} \left (x^4 \left (b k^2+c k-1\right )+x^3 (-2 b k-2 c k-c+4)+x^2 (b+c k+2 c-6)+(4-c) x-1\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int -\frac {x^{2/3} ((1-2 k) x+1) (1-k x)}{\sqrt [3]{k x^2-(k+1) x+1} \left (\left (-b k^2-c k+1\right ) x^4-(-2 k c-c-2 b k+4) x^3+(-b-c (k+2)+6) x^2-(4-c) x+1\right )}dx}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {x^{2/3} ((1-2 k) x+1) (1-k x)}{\sqrt [3]{k x^2-(k+1) x+1} \left (\left (-b k^2-c k+1\right ) x^4-(-2 k c-c-2 b k+4) x^3+(-b-c (k+2)+6) x^2-(4-c) x+1\right )}dx}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {x^{4/3} ((1-2 k) x+1) (1-k x)}{\sqrt [3]{k x^2-(k+1) x+1} \left (\left (-b k^2-c k+1\right ) x^4-(-2 k c-c-2 b k+4) x^3+(-b-c (k+2)+6) x^2-(4-c) x+1\right )}d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 1395 |
| \(\displaystyle -\frac {3 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x} \int \frac {x^{4/3} ((1-2 k) x+1) (1-k x)^{2/3}}{\sqrt [3]{1-x} \left (\left (-b k^2-c k+1\right ) x^4-(-2 k c-c-2 b k+4) x^3+(-b-c (k+2)+6) x^2-(4-c) x+1\right )}d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {3 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x} \int \left (\frac {(1-2 k) (1-k x)^{2/3} x^{7/3}}{\sqrt [3]{1-x} \left ((1-k (c+b k)) x^4-4 \left (-\frac {c}{4}-\frac {1}{2} (b+c) k+1\right ) x^3+6 \left (\frac {1}{6} (-b-c (k+2))+1\right ) x^2-4 \left (1-\frac {c}{4}\right ) x+1\right )}+\frac {(1-k x)^{2/3} x^{4/3}}{\sqrt [3]{1-x} \left ((1-k (c+b k)) x^4-4 \left (-\frac {c}{4}-\frac {1}{2} (b+c) k+1\right ) x^3+6 \left (\frac {1}{6} (-b-c (k+2))+1\right ) x^2-4 \left (1-\frac {c}{4}\right ) x+1\right )}\right )d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x} \left (\int \frac {x^{4/3} (1-k x)^{2/3}}{\sqrt [3]{1-x} \left ((1-k (c+b k)) x^4-4 \left (-\frac {c}{4}-\frac {1}{2} (b+c) k+1\right ) x^3+6 \left (\frac {1}{6} (-b-c (k+2))+1\right ) x^2-4 \left (1-\frac {c}{4}\right ) x+1\right )}d\sqrt [3]{x}+(1-2 k) \int \frac {x^{7/3} (1-k x)^{2/3}}{\sqrt [3]{1-x} \left ((1-k (c+b k)) x^4-4 \left (-\frac {c}{4}-\frac {1}{2} (b+c) k+1\right ) x^3+6 \left (\frac {1}{6} (-b-c (k+2))+1\right ) x^2-4 \left (1-\frac {c}{4}\right ) x+1\right )}d\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\) |
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]
3.12.88.3.1 Defintions of rubi rules used
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e_.)*( x_)^(n_))^(q_.), x_Symbol] :> Simp[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + c*(x^n/e))^FracPart[p]) Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && E qQ[n2, 2*n] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && !IntegerQ[p] && !(EqQ[q, 1] && EqQ[n, 2])
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Not integrable
Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.99
\[\int \frac {x \left (k x -1\right ) \left (-1+\left (-1+2 k \right ) x \right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (-1+\left (4-c \right ) x +\left (c k +b +2 c -6\right ) x^{2}+\left (-2 b k -2 c k -c +4\right ) x^{3}+\left (b \,k^{2}+c k -1\right ) x^{4}\right )}d x\]
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)
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)
Exception generated. \[ \int \frac {x (-1+k x) (-1+(-1+2 k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(4-c) x+(-6+b+2 c+c k) x^2+(4-c-2 b k-2 c k) x^3+\left (-1+c k+b k^2\right ) x^4\right )} \, dx=\text {Exception raised: TypeError} \]
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="f ricas")
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (trace 0)
Timed out. \[ \int \frac {x (-1+k x) (-1+(-1+2 k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(4-c) x+(-6+b+2 c+c k) x^2+(4-c-2 b k-2 c k) x^3+\left (-1+c k+b k^2\right ) x^4\right )} \, dx=\text {Timed out} \]
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)
Not integrable
Time = 0.31 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.95 \[ \int \frac {x (-1+k x) (-1+(-1+2 k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(4-c) x+(-6+b+2 c+c k) x^2+(4-c-2 b k-2 c k) x^3+\left (-1+c k+b k^2\right ) x^4\right )} \, dx=\int { \frac {{\left ({\left (2 \, k - 1\right )} x - 1\right )} {\left (k x - 1\right )} x}{{\left ({\left (b k^{2} + c k - 1\right )} x^{4} - {\left (2 \, b k + 2 \, c k + c - 4\right )} x^{3} + {\left (c k + b + 2 \, c - 6\right )} x^{2} - {\left (c - 4\right )} x - 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}} \,d x } \]
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="m axima")
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)
Not integrable
Time = 0.68 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.95 \[ \int \frac {x (-1+k x) (-1+(-1+2 k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(4-c) x+(-6+b+2 c+c k) x^2+(4-c-2 b k-2 c k) x^3+\left (-1+c k+b k^2\right ) x^4\right )} \, dx=\int { \frac {{\left ({\left (2 \, k - 1\right )} x - 1\right )} {\left (k x - 1\right )} x}{{\left ({\left (b k^{2} + c k - 1\right )} x^{4} - {\left (2 \, b k + 2 \, c k + c - 4\right )} x^{3} + {\left (c k + b + 2 \, c - 6\right )} x^{2} - {\left (c - 4\right )} x - 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}} \,d x } \]
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="g iac")
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)
Not integrable
Time = 7.52 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.98 \[ \int \frac {x (-1+k x) (-1+(-1+2 k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(4-c) x+(-6+b+2 c+c k) x^2+(4-c-2 b k-2 c k) x^3+\left (-1+c k+b k^2\right ) x^4\right )} \, dx=-\int \frac {x\,\left (x\,\left (2\,k-1\right )-1\right )\,\left (k\,x-1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (\left (-b\,k^2-c\,k+1\right )\,x^4+\left (c+2\,b\,k+2\,c\,k-4\right )\,x^3+\left (6-2\,c-c\,k-b\right )\,x^2+\left (c-4\right )\,x+1\right )} \,d x \]
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)